Prasanna

You can Download Tangents Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 7 Tangents Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 7 Tangents Notes

Textbook Page No. 163

Tangents Class 10 Chapter 7 Kerala Syllabus Tangents Class 10 Kerala Syllabus Question 1.
In each of the two pictures below, a triangle is formed by a tangent to a circle, the radius through the point of contact and a line through the center:


Draw these in your notebook.
Answer:
Draw a circle with radius 2.5cm and center as O. Mark a point P at a distance 5cm from O. Taking OP as the diameter, draw a circle.

Mark A at the point where the two circles intersect each other. Join PA.
Radius of the circle = \(\sqrt{4^{2}-2^{2}}\)
= \(\sqrt{16-4}\) = √12 = 2√3 cm
= 2 × 1.73 = 3.46 cm ~ 3.5 cm
Draw a circle with radius 3.5 cm. OA is the radius. Draw AP = 2 cm, perpendicular to OA. Join O and P, such that PA is the tangent

Draw this picture in your notebook.

Answer:
Draw a circle of radius 4 cm and center O. Sdesof a rhombus are the tangents of the circa Let the angle between the tangents be 40°, then the center angle of arc between them = 180 – 40 = 140°.
Draw a rhombus by given measures.

Sslc Maths Chapter 7 Kerala Syllabus Question 3.
Prove that the tangents drawn to a circle at the two ends of a diameter are parallel.
Answer:

A tangent through P is perpendicular to PQ, PQ ⊥RP. A tangent through Q is perpendicular to PQ, PQ ⊥ SQ ∴∠SQP = ZRPQ = 90°. But they are co-interior angles. RP || SQ This means that RP is parallel to SQ. Hence proved.

Sslc Maths Chapter Tangents Kerala Syllabus Chapter 7 Question 4.
What sort of a quadrilateral is formed by the tangents at the ends of two perpendicular diameters of a circle?
Answer:
Let centre of the circle be O. Draw 2 mutually perpendicular diameters, PQ and SR.
Tangents drawn from a point outside to circle have equal length.

AP = AS, PD = PR, CQ = CR, QB = BS
AD + BC = AP + PD + BQ + CQ = AS + DR + SB + CR
= AS + SB + DR + CR
= AB + CD
Opposite sides are equal.
Hence ABCD is a square.

Textbook Page No. 166

Class 10 Maths Tangents Kerala Syllabus Chapter 7 Question 1.
Draw a circle of radius 2.5 centimeters. Draw a triangle of angles 40°, 60°, 80° with all its sides touching the circle.
Answer:
First, draw a circle with radius 2.5cm. PCOB is a quadrilateral, ∠COB = 360 – (90 + 90 + 40) = 140°. Divide the circle into three as 100°, 120°, 140°. Draw an arc equal distance from A and B find the point R. Similarly, find P and Q. Complete the figure,

Class 10 Tangents Kerala Syllabus Chapter 7 Question 2.
In the picture, the small (blue) triangle is equilateral. The sides of the large(red) triangle are tangents to the circumcircle of the small triangle at its vertices.
i. Prove that the large triangle is also equilateral and its sides are double those of the small triangle.
ii. Draw this picture, with sides of the smaller triangle 3 centimeters.

Answer:
i. Let O be the center of the radius
In ΔAOB
OA = OB (Radius)
∴ ∠OAB = ZOBA =30°
∴ ∠AOB = 120° ⇒ ∠APB = 60°
Similarly
∠AOC = 120° ⇒ ∠ARC = 60°
∠BOC = 120° ⇒ ∠CQB = 60°
Angle in the A PQR 60° each.
ΔPQRis an equilateral triangle.
In Δ APB, AP = PB
∠APB = 60°
∠PAB = ∠PBA = 60°
∴ Δ PAB is a equilateral triangle.
PA = PB=AB Similarly
AR = RC = AC and CQ = BQ =BC
Δ ABC is a equilateral triangle. Hence
PR = PA + AR = AB + AC = 2 AC
∴ The large triangle is also equilateral and its sides are double that of the small triangle.

Tangent Class 10 Solutions Kerala Syllabus Chapter 7 Question 3.
The picture shows the tangents at two points on a circle and the radii through the points of contact

i. Prove that the tangents have the same length.
ii. Prove that the line joining the center and the point where the tangents meet bisects the angle between the radii.

iii. Prove that this line is the perpendicular bisector of the chord joining the points of contact.

Answer:
i. Δ OAP, Δ OBP are similar triangles.
In.ΔOAP
OP²=OA²+AP²
AP²=OP^{2 –} OA²

Length of the tangents are equal,

ii. . Consider Δ AOP, ΔBOP
PA= PB
(tangents)
OA = OB (radii)
PO = PO (common side)
∴ ΔAO ~ ΔBOP (Since all the sides are same)
Hence triangles are similar.
∴ ∠OPA = ∠OPB
OP bisects ∠ APB

iii. As ΔOAP ~ ΔBOP
∠POA = ∠POB
Considering
ΔAOM, ΔBOM
OA =OB(radii)
OM = OM (common side)
∠AOM = ∠BOM
ΔAOM ~ ΔBOM (By SAS property).
∴ AM = BM, OP bisects AB

Sslc Maths Chapter 7 Notes Kerala Syllabus Question 4.
Prove that the quadrilateral with sides as the tangents at the ends of a pair of perpendicular chords of a circle is cyclic.
What sort of a quadrilateral do we get if one chord is a diameter? And if both chords are diameters?

Answer:
LetAB, AD be tangents
∠PON + ∠PCN = 180°
∠PON = ∠MOQ
∠MAQ + ∠PCN = 180°
(PQ ⊥ MN)
∠MOQ = 90°
∠MAQ = 180° – 90° = 90°,
Similarly
∠NCP = 90°, ∠AMQ || ∠NCP = 180°,
∠A+ ∠ C = 180°
ABCD is a cyclic quadrilateral. If one chord is the diameter then the quadrilateral is a trapezium. If two chords are diameters then the quadrilateral is a square.

Textbook Page No. 172

Tangents 10th Class Kerala Syllabus Chapter 7 Question 1.
In the picture, the sides of the large triangle are tangents to the circumcircle of the small triangle, through its vertices.

Calculate the angles of the large triangle
Answer:
In a circle, the angle which a chord makes with the tangents at its ends on any side are equal to the angle which it makes on the part of the circle on the other side.

In Δ ABC, ∠B = 80°, ∠C = 60°
∠R = 180 – (80 + 60) = 40
∠CAR = 80°= ∠ACR
∠QBC = ∠QCB = 40°
∠P = 60°, ∠Q = 100°
∠NCP = 20°
Then the angles of bigger triangle are 60°, 100°, 20°.

Sslc Tangent Questions And Answers Kerala Syllabus Chapter 7 Question 2.
In the picture, the sides of the large triangle are tangents of the circumcircle of the smaller triangle, through its vertices.

Calculate the angles of the smaller triangle
Answer:

∠ACB =180-(60+ 40) = 80°
In a circle, the angle which a chord makes with the tangents at its ends on any side are equal to the angle which it makes on the part of the circle on the other side.
AP = AR ∴ ∠APR = ∠ARP = 60°
CR = CQ ∴ ∠CRQ =∠CQR = 50°
PB = BQ ∴ ∠BPQ = ∠BQP = 70°
Angles of the smaller triangle are 50°, 60°, 70°

Hss Live Maths 10th Kerala Syllabus Chapter 7 Question 3.
In the picture, PQ, RS, TU are tangents to the circumcircle of ΔABC.
Sort out the equal angles in the picture

Answer:
∠QAB = ∠ACB = ∠ABS
∠RBC = ∠BAC = ∠BCU
∠PAC = ∠ABC = ∠TCA

10th Maths Tangents Kerala Syllabus Chapter 7 Question 4.
In the picture, the tangent the circumcle of a regular pentagon through a vertex is shown.
calculate the angle which the tangent makes with the two sides of the pentagon through the point of contact.

Answer:
Sides of the pentagon are equal.


Textbook Page No. 179

Tangent Class 10 Kerala Syllabus Chapter 7 Question 1.
In the picture, a triangle is formed by two mutually perpendicular tangents to a circle and a third tangent. Prove that the perimeter of the triangle is equal to the diameter of the circle.

Answer:

O is the center and OS perpendicular to PS
(Radius is perpendicular to tangent)
OQ perpendicular PQ
(Radius is perpendicular to tangent)
OS = PS = PQ = OQ
(PQRS is a square)
PS= PA + AS
AS = AM (Tangents from A is equal) PS= PB + BQ
Hence BM= BQ (Tangents from B is equal)
PQ = PB + BM
Therefore
PS+ PQ= PA+AM + PB + BM
= PA+PB+AM + BM
= PA+PB+AB = Perimeter of right angled triangle PS + PQ is diameter of the circle.
So, the perimeter of the triangle is equal to the diameter of the circle The picture shows a

Tangent Questions And Answers Kerala Syllabus Chapter 7  Question 2.
The picture shows a triangle formed by three tangents to a circle.

Calculate the length of each tangent from the corner of the triangle to the point of contact
Answer:

Sslc Maths Tangents Model Questions Kerala Syllabus Chapter 7 Question 3.
In the picture, two circles touch at a point and the common tangent at this point is drawn

i. Prove that this tangent bisects! another common tangent of these circles.

ii. Prove that the points of contact of these two tangents from the vertices of a right triangle.

iii. Draw the picture on the right in your notebook, using convenient lengths. What is special about the quadrilateral formed by joining the points of contact of the circles?

Answer:
i.

PB = PQ and AP = PQ (Since they are tangents from P). PQ is common side.
∴ PA = PB, Therefore, the first tangent bisects the second tangent

ii. AP = QP
=> ∠ PAQ = ∠ PQA = x
BP = QP
=> ∠BQP = ∠PBQ = y
In AAQB
∠QAP + ∠ABQ + ∠AQB = 180°
x + y + x + y = 180°
2(x + y) = 180°
x + y = 90°
∠AQB = 90°
Δ ABQ is a right angled triangle.

iii. It is a rectangle

Sslc Maths Tangents Notes Kerala Syllabus Chapter 7 Question 4.
In the picture below, AB is a diameter and p is a point on AB extended. A tangent from P touches the circle at Q.
What is the radius of the circle?

Answer:
AP = 8 cm PQ = 4 cm
QP² = AP × BP

AB = AP – BP
= 8 – 2 = 6
Diameter = 6
∴ Radius = \(\frac { 6 }{ 2 }\) = 3 cm

Kerala Syllabus 10th Standard Maths Guide Pdf Download Chapter 7 Question 5.
In the first picture below, the line joining two points on a circle is extended by 4 centimeters and a tangent is drawn from this point. Its length is 6 centimeters, as shown:

The second picture shows the same line extended by 1 centimeter more and a tangent drawn from this point. What is the length of this tangent?
Answer:
In the first circle
PC² = PB × PA
62 = 4 x PA
PA = \(\frac { 36 }{ 4 }\) = 9

AB = AP – PB = 9 – 4 = 5 cm In the second
PC² = PA × PB
PC²= 10 × 5 = 50
PC= √50 = 5 √2cm
= 7.05 cm

Textbook Rage No. 185

Kerala Syllabus 10th Standard Maths Guide Malayalam Medium Chapter 7 Question 1.
Draw a triangle of sides 4 centimetres, 5 centimetres, 6 centimetres and draw its incircle. Calculate its radius.
Answer:
Draw a line BC = 6 cm. Draw an arc with a radius of 4 cm with the B as center. Also, draw an arc with a radius of 5 cm taken C as center. Let the bisector of both arcs drawn and they meet at ‘A’. Complete Δ ABC.
Let the bisectors of ∠A and∠B be drawn and they meet at ‘O’. Draw an incircle to the triangle with center O.

10th Class Maths Malayalam Medium Kerala Syllabus Chapter 7 Question 2.
Draw a rhombus of sides 5 centimeters and one angle 50° and draw its incircle.
Answer:
Draw AB with length 5 cm. Draw a ray from A making angle 50° with AB. Mark D at 5 cm from A. Draw a perpendicular bisector to ∠A. That is meet BD atO.
Extend line AO up to C such that AO = OC. Join BC and DC. Draw a perpendicular OP to AB through O. Draw a circle with centre as O and OP as radius.

Question 3.
Draw an equilateral triangle and a semi-circle touching its two sides, as in the picture.

Answer:
Draw an equilateral triangle with side 4cm.
The bisector of ∠A is passing through the midpoint of BC. Take the midpoint of AB as P. Join OP. Let draw a circle with OP as radius, we get the required one.

Question 4.
What is the radius of the incircle of a right triangle having perpendicular sides of length 5 centimeters and 12 centimeters?
Answer:

Question 5.
Prove that if the hypotenuse of a right triangle is h and the radius of its incircle is r, then its area is r(h+r).
Answer:

Let the perpendicular sides of right-angled triangle be a, d and hypotenuse be h then

Question 6.
Prove the radius of the incircle of an equilateral triangle is half the radius of its circumcircle.
Answer:
The center of the circumcircle and the incircle are the same.

r = \(\frac { r }{ R }\), Hence the radius of the incircle of an equilateral triangle is half the radius of its circumcircle.

Tangents Orukkam Questions & Answers

Question 1.
ΔABC is an equilateral triangle. A circumcircle is drawn to it Prove that the triangle formed by the tangents to the circle at the vertices of ABC is another equilateral triangle.
Answer:

∴ ∠AOB = 120° ∠BOC = 120°
∠A = ∠B = ∠c = 60° (v Equilateral triangle)
∴ ∠AQC = 180 – 120° = 60°
(Sum of opposite sides of a quadrilateral is 180°)
∠BPC = 180 – ∠BOC = 60°
∠ARB = 180 –∠AOB = 60°
Three angles of the ΔPQR is 60° each.
So ΔPQR is an equilateral triangle.

Question 2.
Δ lf the perimeter of ΔABC is 10cm, calculate the perimeter of triangle PQR?
Answer:
Perimeter of ΔPQR = 2 × 10 = 20
One side =20/3
Area of ΔPQR =

Question 3.
What is the relation between the perimeters of ΔABC and ΔPQR?
Answer:
Area of Δ ABC =
Area of ΔPQR is four times of the area of ΔABC.

Question 4.
Draw the figure. Mark the circumcenter of AABC.
Answer:

Question 5.
Join OA, OB, OC. Note the cyclic quadrilaterals in it.
Answer:
OBPC, OAQC, OARB are the cyclic quadrilaterals by joining tangents and radius.

Question 6.
Since ∠B = 60°. ∠AOC will be 120° Write is ∠AOC?
Answer:
∠B = 60° hence ∠AOC = 120°
Angle made by an arc on the center which is twice the angle made by it on the alternate arc.

Question 7.
Find ∠P, ∠R . Write conclusions
Answer:
∠P = 60°, ∠R = 60°, both are equal

Question 8.
See three parallelograms like ABCQ. Find the perimeter of the outer triangle by the equality of opposite sides.
Answer:
ABCQ, CABP, CARB are three parallelograms.
In parallelogram. ABCQ , AB = CQ and BC = AQ . AB = BC
QA = QC (tangents from a outside point) = AB + BC = 2BC
PC = PB (tangents from a outside point) = 2 BC
RA = RB (tangents from a outside point) = 2 AC
∴ PQ + QR + PR =QC + PC + QA + RA + RB + PB
= 2 AB + 2 BC + 2 AC = 2(AB + BC + AC) = 2 × 10 (Perimeter of ΔABC is 10) = 20
Perimeter of ΔPQR =20

Question 9.
AC is the diagonal which divides the parallelogram by two equal triangles.
Answer:
The diagonal AC divides parallelogram ABCQ into two equal triangles.
Area of ΔACQ = Area of ΔABC
Similarly, area of ΔBPC
= Area of ΔABC
Area of ΔPQR
= Area (ΔACQ + ΔBPC + ΔABR + ΔABC)
= (4 × Area ΔABC)
∴ The perimeter of the outer triangle is twice of the perimeter of inner triangle.

Question 10.
Draw a circle and mark a point on it. Construct tangent to the circle at this point without using center.
Draw the circle and mark the point(P) DrawachordAB and joinAP and BP.
See the chord in the figure that you have drawn. This chord made the angle ∠PAB on one side. An equal angle will be formed on the other side of the chord at P with P B as one arm.(Use compass and scale method)
Answer:

All angles in the arc PB are equal.

Worksheet 6

Question 11.
In the figure, a circle touches the sides of?
ABC at P, Q, R.
If AB =AC then prove that BR = C R
Why is AP=AQ?

Answer:
AP = AQ
(Tangent drawn from a point A to circle have equal length)
Establish BP = C Q.
Establish BR = C R.
If AB = AC
AP + PB = AQ + QC
AP = AQ (Tangent drawn from a point A to circle have equal length)
∴ PB = QC
BP = BR (Tangent drawn from a point B to circle have equal length)
CR = CQ (Tangent drawn from a point C to circle have equal length)
∴ BR = CR

Question 12.
In the figure AP, BQ, P Qare tangents to the circle. The line AP is parallel to BQ. Find ∠POQ

Draw figure, mark OA, OB, OC.
Establish angle OAP, angle OCP equal.
Take ∠AOP, ∠COP as x.
Triangles BOQ, and COQ are equal.
∠BOQ, ∠COQ = Y
2x + 2y = 180. Write x + y and ∠POQ
Answer:
Δ OAP, Δ OCP are equal triangles.
So, ∠AOP = ∠COP = x
ΔBOQ, ΔCOQ are equal triangles.

So,
∠BOQ= ∠COQ = y
2x + 2y = 180°
Angle on semicircle.
So, x + y = 90°, ∠POQ = x + y = 90°

Question 13.
If a circle can be drawn by touching the sides of a parallelogram inside it will be a rhombus. Prove.
Answer:

Draw the figure AP = AS, BP = BQ, DR= DS, CR = CQ.
Using these equations prove the statement given as the 11^th point in the basic concepts.
2 × AB = 2 × AD.
Answer:
AP = AS, BP = BQ,
DR = DS, CR = CQ
(Tangent drawn from a point to circle have equal length)
Sum of opposite sides of a quadrilateral formed by joining the tangents on four points of a circle are equal.
So, 2 × AB = 3 × AD.
Therefore ABCD is a rhombus.

Question 14.
If r is the radius of the incircle of a right triangle prove that \(r=\frac{a+c-b}{2}\)

BP = BQ = r
AP = AR = c – r
CR = CQ = b – r
b = c – r + a – r
Answer:
In figure BP = BQ = r
∴ AP = AR = c – r
(Tangent drawn from a poijt A to circle have equal length)
CR = CQ = a – r
(Tangent drawn from a point C to circle have equal length)
AC = b = c – r + a – r
= c + a – 2r,
c + a – 2r = b
∴ \(r=\frac{a+c-b}{2}\)

Question 15.
Find the inradius of an equilateral triangle of side 10cm. User r = \(\frac { A }{ s }\)
Answer:

Tangents SCERT Questions and Answers

Question 16.
PQ is a tangent to the circle with center O.
a. Find ∠p?
b. If ∠O = 42° ,what is ∠Q? [Score: 3, Time: 5 minutes]

Answer:
∠P = 90°, ∠Q = 90 – 42 = 48°

Question 17.
Draw this figure using the given measurements. [Score: 3, Time: 4 minutes]
Answer:
Draw a circle of radius 4 centimeters. (1)
Draw a radius and a perpendicular to it. (1)
Complete the triangle after marking angle 60° at the center.

Question 18.
In the figure, AC and BC are tangents to the circle from C. Centre of the circle O.
i. Find ∠A
ii. If ∠C is 2 times ∠O, then what is ∠C ? [Score: 3, Time: 4 minutes]

Answer:
i. ∠A= 90° (1)
ii. ∠C + ∠O = 180° (0)
∠C= 60° (0)

Question 19.
The radius of a circle touching all sides of an equilateral triangle is 3centimetres. Draw this triangle. [Score: 3, Time: 4 minutes]
Answer:
Draw a circle of radius 3 cm. (1)
Mark 120° at the center of circle. (2)
Complete the equilateral triangle. (2)

Question 20.
Radius of an incircle to a triangle is 3 centimeters. Two angles of this triangle are 55° and 63°. Draw this triangle. [Score: 5, Time: 8 minutes]
Answer:
Draw a circle of radius 3 cm. (1)
Mark the angles 180 – 55 = 125°, 180 – 63 = 117°, at the centre (1)
Draw the tangents.
Complete the triangle. (3)

Question 21.
In the figure PA, PB are tangents through A and B of a circle with center O. If the radius of the circle is r, then prove that OP × OQ = r². [Score: 3, Time: 5 minutes]

Answer:
Δ OQA, ΔOPA are triangle with equal angles.
The ratio of sides opposite to the equal angles. (1)

Question 22.
In the figure, angles formed by the radius segment of the meeting points of the tan-gent to incircle are given. Find all angles of the triangle. [Score: 3, Time: 5 minutes]

Answer:
Angles 180 – 120 = 60° (1)
180-130 = 50° (1)
Third angle 180 – (60 + 50) = 70° (1)

Question 23.
The incircle triangle ABC touches the tri-angle sides at P, Q& R. as shown in the figure. Find all angles of PQR [Score: 3, Time: 5 minutes]

Answer:
The angles at the center of the circles are 180 – 70 = 110°,
180 – 80 = 100 and
360 + ( 110 + 100) = 150° (2)
The angles of triangle PQR are

Question 24.
Find all angles of triangle AOP and OPT. [Score: 4, Time: 7 minutes]

Answer:
The angles of ΔAOP are 32°, 32°, 116° (2)
The angles of ΔOPT are 64°, 26°, 90° (2)

Question 25.
QP is a tangent of the circle with center O. AR is a diameter. Find all angles of triangle PQR. [Score: 4, Time: 6 minutes]

Answer:
∠PQR = ∠QAR = 30° (1)
∠PRQ = 180 – 60 = 120° (1)
∠P = 180 – (120 + 30) = 30°

Question 26.
In the figure, PQ is a diameter and O is the center of the circle. ∠R = ∠T= 90°
1. Prove that ∠PSR – ∠OSQ.
2. Prove that ΔPSR and ΔSQT are similar. [Score: 3, Time: 5 minutes]
Answer:
1. ∠PSR = ∠PQS (1)
Q is a diameter ZPSQ = 90° (1)
∠PSR = 90 – ∠OSP = ∠OSQ

2. ∠PSR = ∠SQT (1)
∴ Triangle are similar (2 angles are same) (1)

Question 27.
Draw a circle of radiu$3 3 cm. Draw a chord AB = 4 cm of this circle. Draw tangents through A and B. [Score: 3, Time: 5 minutes]
Answer:

Question 28.
In the figure, O is the center. C is a point on the semicircle with diameter OA.
BC is a tangent through B, If OB = 1 cm AB = 3 cm then what is BC f Find all angles of triangle OBC ? [Score: 4, Time: 6 minutes]

Answer:
OB × AB = BC² (1)
1 × 3 = BC² = 3 (1)
BC = √3
The angles of ΔOBC 30°, 60°, 90°. (2)

Question 29.
In the figure, radius of the circle centered at O is 9 cm.
OA = 15 cm. Semicircle with diameter O A cuts the circle with center O at D and BC is a tangent through B.
1. What is the length of BC?
2. If the line PD is perpendicular to OA, then what is the length of PD? [Score: 4, Time: 7 minutes]

Answer:
1. BC² = OB × BA = 9 × 6 = 54
BC = √54cm (1)
2. OP × OA = r²

Question 30.
Hypotenuse of a right triangle is 18 cm and its inradius 3 cm. What is its perimeter? What is its area? . [Score: 3, Time: 5 minutes]
Answer:

Perimeter = 3 + x + x +y +y + 3
= 3 + 3 + 18 + 18 = 42 cm (1)
Area = \(\frac { 42 }{ 2 }\) × 3 = 21 × 3 = 63 sq.cm (1)

Question 31.
Area of a right triangle is 60 sq. centimetres and its inradius 3 cm. What is the length of its hypotenuse? [Score: 3, Time: 5 minutes]
Answer:
Area = 60 sq.cm, radius = 3

Tangents Exam Oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 32.
From a point P, the length of the tangent to a circle is 24 cm and distance of P from the center is 25cm. Find radius.
Answer:
OP² = OQ² + QP²

25² = OQ² + 24²
OQ² = 25² – 24²
OQ²= 625 – 576 = 49
OQ = 7 cm

Question 33.

Let PQ be a tangent to a circle at A and AB be a chord. Let C be a point on the circle such that ∠BAC = 54° and ∠BAQ= 62°. Find ∠ABC.
Answer:
∠ABC = l80° – ( ∠BAC + ∠ACB )
∠ABC = i8o° – (54° + 62°) = 64°

Question 34.
Draw a circle of radius 3.5cm and construct an equilateral triangle intersecting all sides with this circle. What is the radius of a circumcircle?
Answer:

Radius of circumcircle = 7 cm

Question 35.
In the figure, AB is the tangent at B of the circle centered at O. How much is ∠OBA? How much is ∠AOB?

Answer:
∠OBA = 90°
∠AOB = 55°

Question 36.
In the figure O is the center of the circle and P, Q,R are points on it. Find the angles of the triangle formed by the tangent at P, Q,R.

Answer:
Angle in a triangle and opposite angle in its center is supplementary. One angle is 40°, other angles are 70° each.

Question 37.
ABCD is a quadrilateral such that all of its sides touch a circle. If AB = 6cm, BC = 6.5cm and CD = 7cm, then find the length of AD.

Answer:
We have, AP = AS
BP = BQ
CR = CQ
DR = DS
Hence,
AP + BP+CR + DR = AS + BQ + CQ + DS
AB +CD = AD + BC
AD = AB + CD – BC = 6 + 7 – 6.5 = 6.5 cm

Short Answer Type Questions (Score 3)

Question 38.
Draw a circle of radius 2cm. Also, construct a regular hexagon intersecting all sides with this circle.
Answer:

ABCDEF is a regular hexagon.

Question 39.
AB 4cm and BC = 6cm. Find the length of AD? Draw a line segment of length √12 cm by using the same idea?

Answer:
AB × AC = AD²
AB = 4cm
AC = AB + BC = 4 + 6= 10cm
4 × 10 = AD2
∴ AD = √40
Inorder to draw √12 , draw AC = 6cm by taking BC = 4 cm and AB = 2 cm.
When a tangent is drawn from A, its length is √12 cm.

Question 40.
The radius of a circle with center O is 8cm. P is a point outside the circle. PQ and PR are the tangents drawn from P to the circle. If ∠QOR = 60° then find the lengths of PQ, PR, and OP.
Answer:
In Δ OPQ the angles are 30°, 60°, and 90°.
∴ The sides are in the ratio 1 : √3: 2

Question 41.
In the figure PA, PB and RS are the three tangents to the circle. Prove that the perimeter of ΔPRS is the sum of the lengths PAandPB.

Answer:
PA = PB (Tangents from an outside point are equal)
RA = RT
ST = SB
Perimeter of ΔPRS = PR + RS + PS
= PR + (RT + ST) + PS
= (PR + RA) + (SB + PS)
(Since RT = RA, ST= SB)
= PA + PB

Question 42.
In the figure, tangents PA and PB are drawn to a circle with center O from an external point P. If CD is a tangent to the circle at E and AP=25cm, find the perimeter of ∠PCD,
Answer:
We know that the lengths of the two tangents from an exterior point to a circle are equal.

CA = CE, DB = DE and PA = PB Now,
the perimeter of = PC + CD + DP
= PC + CE + ED + DP
= PC + CA + DB + DP
= PA + PB = 2PA(PB = PA)
Thus, the perimeter of ∠PCD = 2 × 25 = 50cm

Question 43.
In the figure, PQ is the diameter and RS is a tangent to the circle.

If ∠SPQ = 34°,fmd
a) ∠SQP
b) ∠RSQ
c) ∠SRP
Answer:
∠PSQ = 90° (Angle in a semicircle)
∴ ∠ SQP
= 180 – (90 + 34)
= 180 – 124 = 56°
b. ∠RSQ = ∠SPQ = 34° (Angle between chord and tangent = Angle in the segment on the other side)
c. ∠SRP= 180 – (124 + 34) s = 180 – 158 = 22°

Question 44.
Prove that the angles formed by the tangents from the endpoints of a chord are equal.
Answer:
Each angle between a chord and the tangent at one of its ends is equal to the angle in the segment on the other side of the chord. [Angles in the alternate segments are equal].
∴ ∠P = x ie ∠RBA = x.
ie ∠QAB = ∠RBA

Long Answer Type Questions (Score 4)

Question 45.
In the following figure, PQ is the tangent and PB is the scent. Prove that PQ2 = PA × PB

Answer:
Considering the triangles PAQ and PQB.
∠APQ = ∠BPQ (Common angle)
∠PQA = ∠QBP (Angle between tangent and chord = the angle made by the chord on its . complimentary arc)
ΔPAQ ~ ΔPQB
[A.A Similarly]

[Ratio of the similar sides are equal]

PQ² = PA × PB

Question 46.
AB, BC, and AC are the tangents of the circle. If AR = 5cm, BP = 4cm and BC = 10cm, find the perimeter of the triangle.
Answer:
AR = AQ = 5 cm
BP = BR = 4 cm
BC – BP = PC
∴ PC = 10 – 4 = 6cm,
PC = PQ = 6cm
∴Perimeter = 2 × 5 + 2 × 4 + 2 × 6
= 2 (5 + 4 + 6 ) = 30 cm

Long Answer Type Questions (Score 5)

Question 47.
AB and AC are the tangents of the circle with center O.

Show that the quadrilateral ABOC is a cyclic quadrilateral.
Answer:
AB and AC are tangents
∠B = ∠C = 90°
∠B + ∠C = 180°
The sum of the four angles of a quadrilateral is 360°. ie, ∠A + ∠O = 180°
The opposite angles of quadrilateral ABOC are complementary.
∴ ABOC is a cyclic quadrilateral.

Question 48.
AB, BC, CD and AD are the tangents of the circle. If AP = x, BP = y, CR = z, SD =w then show that the perimeter of the quadrilateral ABCD is 2(x + y + z + w).

Answer:
AP = AS = x;
BP = BQ = y
CR = CQ = z;
SD = DR = w
ie, AB = x + y
BC = y + z;
CD = z + w
AD = x + z
∴ Perimeter = 2x + 2y + 2z + 2w = 2(x + y + z + w)

Question 49.
In the picture PQ, RS, and TU are the tan-gents of the circle with center O. Find the pairs of angles with same measures

Answer:
∠QAB = ∠RBA ; ∠QAB = ∠ACB
∠RBA =∠ACB ; ∠SBC = ∠UCB
∠SBC = ∠BAC ; ∠UCB = ∠BAC
∠ACT = ∠PAC ; ∠ACT = ∠ABC
∠PAC = ∠ABC

Tangents Memory Map

Tangent to the circle is perpendicular to the radius through the point of tangency.

The tangents from an exterior point to a circle and radii to the points of tangency form a cyclic quadrilateral. In figure PAOB is a cyclic quadrilateral.
Tangents from an exterior point to a circle are equal. If PA, PB are the tangents then PA=PB.

The angle between a chord of a circle and the tangent at one end of the chord is equal to angle formed by the chord in the other side of the circle.

If a circle touches the sides of a quadrilateral, that circle will be the incircle of that quadrilateral. Sum of the opposite sides of such quadrilateral are equal. In the figure, ABCD is a quadrilateral having incircle AB + CD = AD + BD.

If P is an exterior point to a circle, a line from P touches the circle at T on the circle and a line intersects the circle at A and B then PA × PB = PT2.

The center of the circle which touches two lines will be a point on the bisector of the angle between the lines. The bisectors of the angles of a triangle passes through a point. That point will be the incenter of the triangle.

The circle drawn inside a triangle which touches the sides of the triangle is called incircle. The circle drawn outside the triangle which touches the sides of the triangle are excircles.

The radius of the incircle of a triangle is obtained by dividing area of the circle by its semi perimeter.

If a, b, c are the sides of a triangle then the area of the triangle
A= \(\sqrt{\mathrm{S}(\mathrm{S}-\mathrm{a})(\mathrm{S}-\mathrm{b})(\mathrm{S}-\mathrm{c})}, s=\frac{a+b+c}{2}\)
This is popularly known as Hero’s formula.

Students can Download Physics Chapter 5 Refraction of Light Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Physics Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Physics Chapter 5 Refraction of Light Questions and Answers Malayalam Medium

SCERT 10th Standard Physics Textbook Chapter 5 Solutions Malayalam Medium

You can Download Energy Management Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Physics Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Physics Chapter 7 Energy Management Textbook Questions and Answers

SCERT Class 10th Standard Physics Chapter 7 Energy Management Solutions

Textbook Page No. 147

Sslc Physics Chapter 7 Kerala Syllabus Question 1.
It is said that energy can neither be created nor destroyed; then how does energy crisis occur?
Answer:
When energy is transformed from one form to another, some part of it gets lost in other forms. Such a loss is the main cause for energy crisis.

Textbook Page No. 148

Energy Management Class 10 Kerala Syllabus Question 2.
List down the different forms of energy you use for various purposes from the time you wake up till you reach your school.
Answer:

• Muscular energy – for different physical activities
• Chemical energy – for cooking
• Mechanical energy – for moving
• Sound energy – to call friends.

Sslc Physics Chapter 7 Notes Kerala Syllabus Question 3.
From which sources are you getting these forms of energy?
Answer:
From different sources like sun, fuels and power stations.

Textbook Page No. 149

Sslc Physics Chapter 7 Energy Management Kerala Syllabus Question 4.
Complete the table 7.1

Answer:

Physics Chapter 7 Class 10 Kerala Syllabus Question 5.
Take three papers of the same size. Keep one stretched. Crumble the next. Make the third paper wet using water. Burn each of them over a candle flame using pincers. Compare the burning of each.
Complete the table 7.2
.
Physics Chapter 7 Answer:

Textbook Page No. 150

Hss Live Guru 10th Physics Kerala Syllabus Question 6.
Write down the situations/specialties for partial combustion.
Answer:

• Partial dryness
• Insufficient availability of O₂
• Lack of facilities for the removal of oxygen.

Sslc Physics Chapter Wise Questions And Answers Kerala Syllabus Question 7.
What are the drawbacks of partial combustion?
Answer:

• Fuel loss
• Wastage of time
• Economic loss
• Atmospheric pollution
• More smoke is produced
• CO₂, CO (carbon monoxide) are produced as byproducts.

Sslc Physics Textbook Solutions Kerala Syllabus Question 8.
What are the advantages of using smokeless choolahs at home? Note them down in the science diary.
Answer:
Makes home neat, lung disease can be reduced, does not affect the oxygen-carrying capacity of blood. Reduces wastage of time and fuel loss.

Class 10 Physics Kerala Syllabus  Question 9.
Visit a nearby pollution testing center, interact with the staff there and prepare a note on the permissible pollution rate.
Answer:
The carbon monoxide produced as a result of combustion of fuels causes environmental pollution. More carbon monoxide may be produced if the vehicles are not working properly. Smoke testing is conducted to know what quantity of carbon monoxide is present in the smoke coming out of vehicle.

Kerala Syllabus 10th Standard Physics Question 10.
Which are the fuels that are used in vehicles and industries?
Answer:
Petrol, Diesel, LPG, CNG etc

Hss Live 10th Physics Kerala Syllabus Question 11.
Tabulate the category to which these fuels belongs to.

Answer:

Textbook Page No. 151

Hsslive Physics 10th Kerala Syllabus Question 12.
Which are the products obtained from fractional distillation of petroleum?
Answer:
Petrol (or gasoline), naptha, kerosene, diesel oil, lubricating oil, fuel oil, grease wax, and some residue.

Physics Class 10 Chapter 7 Kerala Syllabus Question 13.
Which is the cooking gas that we get in cylinders for domestic use?
Answer:
LPG

10th Scert Physics Solutions Kerala Syllabus Question 14.
How will you know if there is leakage in a LPG cylinder?
Answer:
The smell of LPG is felt

Physics Kerala Syllabus 10th Standard Question 15.
Never switch on or switch off electricity when there is a leakage of LPG Why?
Answer:
It is usually advised not to switch on or off any of the electric switches if you detect a gas leakage. It is because the fumes of gas are highly flammable and even smallest of sparks can ignite a huge fire.

Class 10 Kerala Syllabus Physics Solutions Question 16.
If there is a leakage of LPG does it rise up or come down in the atmosphere? Why?
Answer:
Another reason why care should be taken during storage is that LPG vapor is heavier than air, so any leakage will sink to the ground and accumulate in low lying areas and may be difficult to disperse. LPG expands rapidly when its temperature rises

Class 10 Physics Chapter 7 Kerala Syllabus Question 17.
If there is leakage of LPG it is mandatory to open the doors and windows. Why?
Answer:
Immediately open all the doors and windows to your house so that the gas can escape. Never open electrical fans or even an exhaust fan. Let the gas escape naturally. Once you do that, go outside the house and isolate the main electric supply. Be sure to evacuate yourself and others from the area.

Kerala Syllabus 10 Physics  Question 18.
What precautions are to be taken to avoid accidents due to LPG leakage? Discuss and record in the science diary.
Answer:

• Examine the rubber tube at regular intervals and ensure that it does not have a leakage.
• Turn on the knob of stove only after the regulator is turned on.
• Always buy LPG cylinders from authorized franchisees only
• Check that the cylinder has been delivered with the company seal and safety cap intact, do not accept the cylinder if the seal is broken.
• Please look for the due date of test, which is marked on the inner side of the cylinder stay plate and if this date is over, do not accept the cylinder
• Disconnect LPG regulator and affix safety cap on the cylinder when your gas stove is not in use for prolonged period.
• Always store the LPG cylinder in an upright position and away from other combustible and flammable materials.
• Check for gas leaks regularly by applying soap solution on cylinder joints and Surakshapipes.

Textbook Page No. 153

Physics Class 10 Kerala Syllabus Question 19.
If a gas leak is suspected or if the fire spreads on a cylinder, what else could be done? Think it over.
Answer:
If you are convinced that there is a gas leak, disconnect electricity from outside the home (switch off the main switches). Switch off the regulator and shift the cylinder to an empty space. Keep the windows and doors open. Request help from the Fire Force by calling in the toll-free number 108.

Well trained rescue operators can put out the fire by covering the top end of the cylinder with wet sack to prevent the contact with oxygen. If the fire is in flat or the top story, then one should not try to escape using lifts. Only staircase should be used. Cover the nose and the mouth with soft cloth to avoid the intake of smoke or gases.

Kerala Syllabus 10th Standard Physics Notes Question 20.
Haven’t you seen bio-wastes dumped in public places? Don’t you experience a putrid smell, when you pass them by? Which are the gases responsible for this smell?
Answer:
Methane gas is responsible for the putrid smell.

Physics 10th Class Notes Kerala Syllabus Question 21.
Besides air pollution, what are the problems that may arise when garbage is heaped? Discuss and record.
Answer:

• Soil contamination
• Air contamination
• Water contamination
• Bad impact on human health
• Impact on animals and marine life.
• Disease-carrying pests
• Adversely affect the local economy
• Missed recycling opportunities.

Textbook Page No. 154

Question 22.
Some of you may be using LPG in your houses. What is the weight of the LPG filled in the cylinders supplied to your homes?
Answer:
14.2 kg

Question 23.
Using this quantity of LPG for how many days can you cook?
Answer:
This must be sufficient for a month.

Question 24.
How many days can you cook using firewood of the same weight?
Answer:
14.2 kg firewood would be sufficient for a few a days only.

Question 25.
What difference do you feel in the efficiency of these two fuels?
Answer:
The LPG with more calorific value has more fuel efficiency.

Textbook Page No. 155

Question 26.
Based on its calorific value, which fuel can be considered as the most efficient?
Answer:
Hydrogen

Question 27.
Which are the instances when hydrogen is used as fuel?
Answer:
Hydrogen used as fuel in rockets.

Question 28.
Why is hydrogen not used as a domestic fuel?
Answer:
The combustion rate is very high for hydrogen, the possibility of explosion is also more. And it is also difficult to store hydrogen.

Textbook Page No. 156

Question 29.
What is the energy conversion in a generator?
Answer:
Mechanical energy → Electrical energy

Question 30.
From where do we get energy required for the working of a generator?
Answer:
The mechanical energy required can be provided by engines operating on fuels such as diesel, petrol, natural gas, etc. or via renewable energy sources such as a wind turbine, water turbine, solar-powered turbine, etc

Question 31.
Power stations can be classified based on the nature of the source providing the energy required to operate the generator.
Answer:
Flowing water -Hydroelectric power station.
Nuclear energy – Nuclear power station
Coal – Thermal power station

Textbook Page No. 157

Question 32.
On the basis of notes and discussions, complete the table :

Answer:

Textbook Page No. 158

Question 33.
We get different forms of energy from the Sun. Attempts are underway now to utilize solar energy of its maximum. What are the devices used for this?
Answer:

• Solar panel
• Solar water heater
• Solar cooker
• Solar water heater
• Solar thermal power plant

Textbook Page No. 159

Question 34.
What is the energy transformation that takes place in a solar cell?
Answer:
In a solar panel light energy is converted into electrical energy.

Question 35.
There are certain situations in which a solar panel cannot be put to use. Which are they?
Answer:
When the sky is cloudy, during the rain and night time we can not use solar panels.

Question 36.
What are the situations where solar panel alone is depended on?
Answer:
In space stations, satellite, and in remote islands where there is no electricity, etc solar panels are used.

Question 37.
Take two conical flasks. Paint the outer surface of one with black paint and the other, with white paint. Fill both of them with water and expose them to direct sunlight for the same period of time. Which one gets heated first? What may be the reason?
Answer:
The conical flask which is painted black seems to be more hot when kept under the sun for an hour because the black body absorbs more heat and do not let it out from it. It is the property of black body to absorb heat and there is no chance of heat to escape.

Textbook Page .159

Question 38.
List down the specialties of a solar cooker by examing Fig. 7.7.

1. A box with blackened interior
2. A glass cover for the box
3. A mirror outside the each? What is the function of each?
Answer:
1. A box with blackened interior:
The black color of the vessel lets it to absorb more heat as dark colors absorb more heat.
2. A glass cover for the box:
The glass lid will allow the light and heat energy of the sun to come inside but the light energy will go out but the heat will be trapped inside the cooker to cook food.
3. A mirror outside the each?:
The plane mirror will reflect the maximum light of the Sun inside the cooker.

Question 39.
Find out the working of solar cookers other than the box type and record them in the science diary.
Answer:
Panel solar cookers are inexpensive solar cookers that use reflective panels to direct sunlight to a cooking pot that is enclosed in a clear plastic bag.

Textbook Page .160

Question 40.

Observe the figure. Discuss and record how hot water is formed in the tank of solar heater.
Answer:
Solar water heating (SWH) is the conversion of sunlight into heat for water heating using a solar thermal collector. A sun-facing collector heats a working fluid that passes into a storage system for later use. SWH are active (pumped) and passive (convection-driven). They use water only, or both water and a working fluid. They are heated directly or via light-concentrating mirrors. They operate independently or as hybrids with electric or gas heaters. In large- scale installations, mirrors may concentrate sunlight into a smaller collector.

Textbook Page .162

Question 41.
‘Ocean as a source of energy: its possibilities and limitations’ Prepare a seminar paper on this.
Answer:
Hints: Oceans cover 70 percent of the earth’s surface and represent an enormous amount of energy. Although currently under-utilized, Ocean energy is mostly exploited by just a few technologies: Wave, Tidal, Current Energy and Ocean Thermal Energy.
Limitations:

• Few limited sites/places where this wave, tidal or ocean thermal energy can be obtained.
• The cost of construction of plants is very; high
• The efficiency of producing energy is also low.

Question 42.
Why is it said that geothermal power plants are not possible in Kerala? Discuss and record.
Answer:
There are some minor environmental is-sues associated with geothermal power. Geothermal power plants can in extreme cases cause earthquakes. There are heavy upfront costs associated with both geothermal power plants and geothermal heating/cooling systems. Very location-specific (most resources are simply not cost-competitive). Some countries have been blessed with great resources – Iceland and Philippines meet nearly one-third of their electricity demand with geothermal energy. If geothermal energy is transported long distances by the means of hot water (not electricity), significant energy losses has to be taken into account. Geothermal power is only sustainable (renewable) if the reservoirs are properly managed.

Textbook Page .164

Question 43.
What are the different methods by which energy is produced from the nucleus?
Answer:
Nuclear fusion, nuclear fission.

Question 44.
Even if the matter converted is very small, the energy produced is very large. What is the reason?
Answer:
According to Einstein’s equation E = mc². The amount of energy produced is high, even though the transformed mass is very less. Here’ represents the converted mass, c is the speed of light (3 x 10⁸ m/s) and E is the amount of energy obtained.

Question 45.
What is the reason for an uncontrolled fission reation ending in an explosion?
Answer:
Nuclear fission is the process in which the nuclei of greater atomic mass are split into lighter nuclei using neutrons. The mass of nucleus produced in such a process is less than of its parent nuclei. Thus there will be loss of matter in fission process. The mass lost during fission converts into energy. The two or three neutrons produced during the process, bombards with other nucleus and fission process continue rapidly and ends in big explosion.

Textbook Page . 165

Question 46.
Complete the Table 7.5

Answer:

Textbook Page .166

Question 47.
Classify the energy from the follow-ing sources as green energy and bown energy. Solar cells, atomic reactors, tidal energy, hydroelectric power, diesel engines, windmills, thermal power stations.
Answer:

Question 48.
What must be done to ensure maximum utilization of green energy while constructing a house?
Answer:

• Sufficient sunlight should be available in the rooms during day time.
• Comfortable warmth, coolness and air circulation must be available without the help of electricity.
• Using the sun for heating through south facing windows during the winter lowers heating costs. Shading those same windows in summer lowers cooling costs.
• Grid-tied solar photovoltaic (PV) panels currently provide the most cost- effective form of renewable energy for a zero energy home.
• Fresh filtered air and moisture control are critical to its success. This need for ventilation has a silver lining:

Let Us Assess

Question 1.
In a way most of the important energy sources of today can be said to be solar. Which of the following does not belong to the solar energy?
a. fossil fuel
b. energy from the mind
c. nuclear energy
d. biomass
Answer:
Nuclear energy

Question 2.
Which of the following is a green energy?
a. coal
b. naptha
c. biogas
d.petroleum gas
Answer:
Biogas

Question 3.
Write down the advantages and limitations of solar cooker.
Answer:
Advantages:

• Eco-friendliness
• Maintain better air quality indoors, reduce carbon monoxide emissions
• Less expensive
• Don’t cause any environmental pollution

Limitation:

• Cooking with solar cookers obviously requires sunlight, which makes it difficult to use during winter months and on rainy days.
• Cooking also takes a significantly longer time compared to conventional methods.
• Solar cookers are not as efficient at retaining heat as conventional cooking devices. Factors such as wind,  rain, and snow can seriously hinder operation, and in such weather conditions, even after the food is cooked, it  will lose its warmth very quickly.

Question 4.
Kerala has a long coastal land. Still, ocean is not considered as a major source of energy. Why?
Answer:
In Kerala usually, waves are of low. So they have less energy. Also during tides Kerala sea coasts, the sea attitude does not increasingly rise.

Question 5.
The graph regarding the calorific value of certain fuels is given below. Analyze the graph and answer the following Questions.

a. Which fuel has the highest calorific value? Which has the lowest?
b. How many kilogram of dried cow- dung is to be burnt to obtain the same amount of heat produced when 1 kg of LPG burns?
c. From the graph find out the most suitable fuel for household pur-pose. Justify your answer.
Answer:
a. Highest – Hydrogen
Lowest – Cowdung (Dried)
b. Energy released when 1kg LPG bums = 54000 kJ.
Dried cow-dung is to be burnt to obtain the same amount of heat \(\frac { 54000 }{ 6000 }\) = 9 kg
c. Biogas, because it is a renewable energy. Comparatively high calorific value.

Extended Activities

Question 1.
Find out the scope of hydrogen as a fuel with a high calorific value and prepare an essay.
Answer:
The amount of heat liberated by the complete combustion of 1kg of fuel is its calorific value. Hydrogen has high calorific value and also possess high combustion rate too. Hydrogen does not create much atmospheric pollution like other gases. Therefore seek method to combust it in moderate rates and use fuel cells.

Question 2.
Visit a hydroelectric power station and
try to understand different stages of the production of electricity. Make use of this principle and find out the scope of mini hydroelectric power project.
Answer:
There are only some primary expenses to build a hydrogen-electric power station. Hydroelectric power stations must be constructed only in areas of heavy rainfall and good streamflow. If these required conditions are satisfied, the production of electricity from such power station is very profitable one.

Question 3.
Visit a biogas plant and explore the possibility of establishing a community biogas plant in your region.
Answer:
Generally, the biomass required to construct a biogas plant may not be available from a single house and it is not profitable to construct separate biogas for each house. We know that the social cleanliness is equally important as personal cleanliness, due to several reasons, like.economic advantages, it is necessary to run a social biogas plant. And it will be useful to more people.

Question 4.
Write a short play to make the public aware of the need for making use of solar energy.
Answer:
Sun is the major source of energy available in earth, during photosynthesis the light energy is converted into chemical energy and is stored in plant cells. Since this plant reaches in animals as food, the stored energy reaches in animals too. The remainings of the animals hurried in earth transforms under high pressure and temperature and becomes fuels. Write drama using these hints.

Question 5.
Solar energy has an incredible future in the field of transportation. We are in its infant stage. Write an essay on the topic “Prospects of solar energy”.
Answer:
Solar energy is the ultimate energy source. All energy sources are related and connected with solar energy in one or the other way. Expand these points and complete the essay.

Question 6.
Find out the advantages and disadvantages of main energy sources and tabulate them.
Answer:

Question 7.
A nuclear reactor is about to be established in Kerala. What is your reaction to this proposal? Justify.
Answer:
Kerala is a highly populated state in India. Therefore there are not much vacant places available for the construction a nuclear reactors here. Since the radiations given out from such power plant are very harmful to human life, it may cause unpredictable effects. I strongly disagree with this opinion.

Question 8.
A man pointing at a car running on petrol says. “This car is running on solar energy”. Write down your responses about this matter.
Answer:
We use fossil fuels like petrol or diesel in cars. The plants which lived millions of year ago have also used solar energy for photosynthesis to make food. Humans and animals grew up by taking these plants and fruits. Then this humans and animals are burned in earth after their death and became fossil fuels. Fuels are formed out of them. This fuel is being used in cars so the ultimate source of energy here is the solar energy itself. The vehicles connected with solar panels uses solar energy in the direct form.

Energy Management Orukkam Questions and Answers

Question 1.
Observe the fast and complete combustion of a flat paper. Understand that combustion of a crumbled paper make more smoke.
a. List the conditions for the complete combustion.
b. Name the gases produced during complete combustion.
c. Name the gases produced during partial combustion.
d. How does the partial combustion causes environmental pollution?
e. How are the fossil fuels formed?
f. Why are they considered as non-renewable?
g. Write the full form of LNG and CNG
h. Which is the main constituent of LNG? What are the uses of CNG?
i. Name the main constituent of CNG.
Answer:
a. Conditions for complete combustion. Dryness, Fast evaporation, Materials should reach at a certain temperature that is required for combustion.
b. Carbon dioxide, Steam, Heat, and light.
c. Carbon monoxide, soot and a little of car-bondioxide.
d. Besides the fuel loss and time loss, the carbon monoxide produced during partial combustion
e. Fossil fuels are formed by the transformation of plants and animals that went under the earths crust millions of years ago. Eg: Coal, petroleum and natural gases.
f. They are not replenished or renewed in proportion to their consumption.
g. LNG – Liquefied Natural Gas CNG – Compressed Natural Gas
h. Methane. CNG is used as fuel in vehicles, thermal power stations.
i. Methane

Question 2.
1. Write the full form of LPG, what is its main constituent?
2. Why is ethyl mercaptan added to LPG?
3. Which element is the main component of coal?
4. Name the different types of coal?
5. Name the process by which the components of coal are separated?
6. Write the products obtained by the distillation of coal.
Answer:
1. Liquefied Petroleum Gas, Butane
2. To identify the leak of LPG.
3. Carbon
4. Peat, Lignite, Anthracite and Bituminous coal.
5. Distillation
6. Ammonia, Coal gas, Coal tar, and Coke.

Work Sheet:

Question 1.
Identify the relation and fill in the blanks
a. LPG: Butane
LNG ………….
b. LNG: Fuel in Vehicles
LPG …………..
Answer:
a. Methane
b. For domestic use.

Question 2.
Name the chemical used to identify the leakage of cooking gas.
Answer:
Ethyl mercaptan

Question 3.
Find the odd one out and explain the reason
(Peat, Anthracite, Bauxite, Lignite)
Answer:
Bauxite, Others are types of coal.

Question 3.
1. Is the heat produced by the combustion of different fuels the same? Discuss
2. What do you mean by fuel efficiency?
3. What is its unit?
4. List the qualities of a good fuel.
5. Compare biomass and biogas
6. Name the device that converts solar energy into electrical energy.
7. What is the reason for electric current in this device?
8. In a Solar Panel Solar energy is converted into electrical energy. What do you mean by a Solar panel?
9. Write down the situation where solar panels are used.
10. What is the energy change in a solar heater?
11. What is the difference between a solar voltaic power plant and solar thermal power plant?
Answer:
1. No, the heat produced by combustion of different fuels is not the same because their calorific value is not the same.
2. We use firewoods, kerosene, LPG etc in our homes as fuels. Each of them liberates different amounts of heat. This is indicated as fuel efficiency.
3. Kilo Joule/ Kilogram
4. Should be easily available Should be of low cost
Should cause minimum atmospheric pollution on combustion.
5. Biogas: Biogas refers to a mixture of different gases produced by the breakdown of organic mater in the absence of oxygen. Biomass: The body parts of plants and animals are known as Biomass.
6. Solar panel
7. The flow of electrons from p – region to n – region.
8. A large number of solar cells are suitably assembled to form a solar panel.
9. In lighting street lamps, In artificial satellites
10. Solar energy to heat energy
11. Both photovoltaic and solar thermal are the two established solar power technologies. Photovoltaics use semi conductor technology to directly convert sunlight into electricity. Solar thermal works by using mirrors to concentrate sunlight.

Question 4.
Watch the animation video of nuclear fission and fusion
a. Which nuclear reaction happens in an atom bomb?
b. E=mc²
E = energy, then m = ………., c = ………….
c. What will you call the nuclear reaction in which small atoms combine?
d. Which nuclear reaction is carried out in stars?
e. What is energy change in a nuclear reactor?
f. Classify the given energy sources into conventional and non-conventional energy.
(Fossil fuels, Solar energy, Nuclear energy, Biomass, Hydro-Electric Power)
g. What will call the energy sources that lead to global warming?
h. What do you mean by green energy?
i. Why is it instructed to control the use of brown energy sources?
j. List the names of renewable energy sources.
k. Write down the causes an remedies of energy crisis.
Answer:
a. Nuclear Fission
b. m = mass, c = velocity of light
c. Nuclear fusion
d. Nuclear fusion
e. Nuclear energy to electical energy
f. Conventional energy sources – Fossil fuels, Biomass, Hydroelectric power Nonconventional energy sources – Solar energy, Nuclear energy
g. Brown energy
h. Green energy is the energy produced from natural sources which does not cause environmental pollution.
i. Brown energy are the sources which cause environmental problems including global warming, so it is essential to control the use of brown energy sources.
j. Solar energy, Wind energy, energy from biomass.
k. Judicious utilization of energy, Maximum utilization of energy Timely repairing of machines

Work Sheet

Question 1.
Classify the given energy sources into renewable and nonrenewable.
(Solar energy, Petroleum, Nuclear energy, Coal, Geothermal energy)
Answer:
Renewable energy resources – Solar energy, Geothermal energy, Coal

Question 2.
Identify the relation and fill in the blanks.
a. Hydrogen bomb: Nuclear fusion
Atom bomb: …………
b. Solar energy: Green energy
Nuclear energy: ……….
Answer:
a. Nuclear fission
b. Brown energy

Question 3.
Why green energy is called clean energy?
Answer:
Green energy is the energy produced from natural sources which does not cause environmental pollution. Hence it is known as clean energy.

Question 4.
What is the fuel used in a nuclear reactor?.
Answer:
Enriched Uranium

Energy Management SCERT Question Pool Questions and Answers

Question 5.
Though energy is available in many forms we depend mostly on electrical energy.
a. Write down two forms of energy, other than electrical energy, used in daily life.
b. Electrical energy is used much in daily life. Why?
c. Does increase in population and mechanization lead to energy crisis? How?
Answer:
a. Heat energy, Light energy.
b. Electrical energy can be easily converted into many other forms.
c. Small increase in population cause large increase in the use of energy consumption, mechanization will lead to excessive use of sources of energy and leads to energy crisis

Question 6.
The heat energy obtained on burning 2 kg of hydrogen, coal and petrol are given below:
Petrol – 9 × 107 J
Hydrogen- 3 × 10⁸ J
Coal – 6 × 10⁷ J
a. Which of these is the most efficient fuel?
b. How much is the calorific value of hydrogen?
c. Arrange these fuels in the ascending order of their calorific values.
d. Which of the above will you select as a good fuel? What is the basis of your selection?
Answer:
a. Hydrogen
b. 150000 kj/kg
c. Coal, petrol, hydrogen.
d. Petrol.
Easy to handle: less atmospheric pollution, Calorific value is high.

Question 7.
It is advisable to keep stirring heap of wastes while burning them.
a. Write down two essential situations for the complete burning of fuels.
b. How does the stirring help the combustion? Explain.
c. Write down two disadvantages, of partial combustion.
Answer:
a. To increase the availability of oxygen.
i. Increase the surface area exposed to air
ii. Make available the temperature needed for burning.
b. Sufficient oxygen is made available which increases the rate of combustion.
c. Partial combustion pollutes the air by giving excess of smoke, soot, carbon monoxide an carbon dioxide.

Question 8.
a. What is meant by nonrenewable sources of energy?
b. How did such fuels originate in nature?
c. Write down any two examples for it.
Answer:
a. sources that cannot be renewed.
b. The biowastes (bio residue) that went into the earth change into petroleum by undergoing chemical changes due to high temperature, in the absence of air.
c. Petrol, coal.

Question 9.
a. What are fuels? Write down the names of two fuels each from the various category of fuels.
b. How does the excessive use of fuels in-fluence global warming? Explain.
c. English the need for prohibiting diesel vehicles in our country in the context of environmental pollution.
Answer:
a. Fuels are substances that release large quantities of heat on burning; solid: logs: liquid: kerosene.
b. Excessive of fuels produce greenhouse gases which increase the temperature of atmosphere and cause global warming.
c. The amount of CO² and SO² released by diesel vehicles is very large. This accelerates global warming.

Question 10.
a. How does the biomass change into biogas in a biogas plant?
b. Of these which is the fuel that is advantageous?
c. What are the advantages of having community biogas plants?
Answer:
a. Biogas is formed in the biogas plants by the action of bacteria on biomass in the absence of oxygen.
b. Biogas has higher calorific value; environmental pollution is less.
c. Community biogas plants help in con-trolling environment pollution, centralised energy source minimizes expense.

Question 11.
a. On what basis are the sources of energy classified as green energy and brown energy?
b. Classify the following into green energy and brown energy Solar cell, nuclear reactor, tidal energy, hydroelectric, diesel engine, windmill, thermal power station
c. Give any two suggestions to use the green energy to the maximum level.
Answer:
a. Green energy – green energy is the energy produced from natural resources that do not cause environmental pollution Brown energy – Energy from petroleum, coal, etc., and nuclear energy.
b. Green energy: solar cell, tidal energy, electricity from water, windmill.
Brown energy: nuclear reactor, diesel engine, thermal power station.
c. i. Install biogas plants
ii. Use solar panels

Question 12.

The figure indicates the fission reaction of a heavy nucleus of Uranium.
a. Write down an activity to make energy available by using lighter nucleii.
b. Calculate the energy obtained if Ig matter changes into energy during nuclear fission reaction (c = 3 x 10⁸ m/s).
c. What will be the result if the reaction shown in the above figure continues?
d. Why are we establishing nuclear reactors despite of the realization that nuclear reactors are controlled nuclear bombs?
Answer:
a. Nuclear fusion
b. E = mc² m = lg

c. Continuous fission reaction ends in an explosion (atom bomb).
d. Energy crisis and industrial growth promote production of energy.

Question 13.
Coal is the most abundant fossil fuel on the earth.
a. How did the fossil fuels originate?
b. Which is the main constituent of coal?
c. What are the substances obtained when coal is allowed to undergo distillation in the absence of air?
d. How does the excessive use of fossil fuels cause global warming?
Answer:
a. Plants and animals that went under the earth many years ago changed into fossil fuels by undergoing changes under high pressure, high temperature, and absence of oxygen.
b. Carbon
c. Ammonia, coal gas, coaltar, coke
d. The greenhouse gases given out by the burning of fossil fuels increase the temperature of atmosphere and causes global warming

Question 14.
LNG and CNG are made from natural gases.
a. What is the advantage of LNG over CNG?
b. Compare LNG and CNG from the point of view of fuel efficiency.
Answer:
a. LNG – liquefied natural gases can be conveniently transported by liquefaction. Can be changed into gaseous state at the same temperature and can be distributed through pipes.

Question 15.
Hydrogen is a fuel with a high calorific value.
a. What is the limitation of hydrogen as a fuel?
b. Which is the substance added to hydro-gen to make a hydrogen fuel cell? ‘
Answer:
a. Hydrogen easily catches fire and has a high chance of explosion. It is difficult to store and transport.
b. Oxygen.

Energy Management Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 16.
Which of the following is a renewable energy source?
(Petroleum, Solar energy, Coal)
Answer:
Solar energy

Question 17.
What is the fuel used in a nuclear reactor?
Answer:
Enriched Uranium

Question 18.
Using the relation from the first pair, complete the other.
LNG – Methane
LPG –
Answer:
Butane

Question 19.
Write the full expansion of LPG.
Answer:
Liquified Petroleum Gas

Question 20.
Which nuclear reaction is carried out in stars?
Answer:
Nuclear Fusion

Question 21.
Which material is used along with hydro-gen to make Hydrogen fuel cells?
Answer:
Oxygen

Question 22.
Using the relation from the first pair, complete the other. Solar cell – Green energy Atomic reactor –
Answer:
Brown energy

Question 23.
Find the odd one in the group and write the reason.
[Coal gas, Coaltar, Butane, Coke]
Answer:
Butane. Others are those materials got from distillation of coal.

Question 24.
Which nuclear reaction takes place in an atom bomb?
a.Nuclear fission
b. Nuclear fusion
c. Radioactivity
d. Combustion
Answer:
a.Nuclear fission

Question 25.
Using the relation from the first pair, complete the other.
Atomic bomb – Nuclear fission
Hydrogen bomb – …………….
Answer:
Nuclear fusion

Short Answer Type Questions (Score 2)

Question 26.
What is meant by Biomass? Give examples for biomass.
Answer:
The body parts of plants and animals are known as biomass. Examples for biomass are Wood, Excreta, etc.

Question 27.
L.P.G is one of the common fuel for domestic use.
a. Which is the main constituent of LPG?
b. Write full form of LPG.
Answer:
a. Butane b. Liquified Petroleum Gas

Question 28.
Calorific value of Hydrogen is 1,50,000 kj / kg. What does it mean?
Answer:
The amount of heat liberated by the complete combustion of 1 kg of hydrogen is 1,50,000 kj/kg.

Question 29.
a. Write any two qualities of hydrogen as a fuel.
b. What is the limitation of hydrogen as a domestic fuel?
Answer:
a. High calorific value, High availability b. Hydrogen easily catches fire and has a high chance of explosion. It is difficult to store and transport.

Question 30.
Complete the flowchart.

Answer:
a. Methane
b. LNG

Short Answer Type Questions (Score 3)

Question 31.
Write down the remedies of energy crisis.
Answer:

• Judicious utilization of energy.
• Maximum utilization of solar energy
• Minimizing the wastage of water
• Making use of public transportation as far as possible
• Construction and beautifying of houses and roads in a scientific manner.
• Controlling of the street lamps with LDR.

Question 32.
Classify the following suitably.
Solar cell, Atomic reactor, Tidal energy, Hydroelectric power, Diesel engine, Thermal power station, Windmill, Biogas

Question 33.

Various situations which uses solar energy are given in the diagram. What are the advantages & disadvantages of solar energy list them
Answer:

Question 34.
a. What are fossil fuels?
b. Why it is called as fossil fuels?
Answer:
a. Coal, Petroleum, Natural gas
b. Fossil fuels are formed by the transformation of animals and plants that went under the earth millions of years ago.

Question 35.
High calorific value is one among the properties that a substance must possess to be considered as a good fuel.
a. What do you mean by calorific value?
b. What are the other properties a substance must have in order to be considered as a good fuel?
Answer:
a. The amount of heat liberated by the complete combustion of 1kg of fuel is its calorific value.
b. Low price, more availability, less atmospheric pollution, portability, and easiness to store, etc. are the properties of a good fuel.

Long Answer Type Questions (Score 4)

Question 36.
Firewood, Cowdung cake, Kerosene, Petrol, LPG, etc are fuels.
a. Which among these has high calorific value?
b. In which state it exists?
c. What are the qualities of a good fuel?
d. Complete the table.

Answer:
a. LPG
b. Gaseous state
c. Fuel with high calorific value is considered as good fuel
d. 1. Butane
2. Methane
3. Methane

Question 37.
A few news related to energy crisis are given below. State two reasons of energy crisis and write how we can overcome them.

Answer:
Reasons: Increase in population, Overexploitation of nonrenewable energy resources, Industrialisation, Urbanisation. Remedial measures:- Population control, Use of renewable energy sources, Industrialisation after finding required energy sources.

Question 38.
Choose the correct answer from the bracket
Fossil fuel, Hydrogen, electrical energy
methane, Butane, Petroleum, Solar panal
a. The most abundant element in sun is ………
b. A solar cell converts solar energy into ………..
c. ……….. is the main constituent of biogas.
d. ……….. provides the power required for artificial satellite.
Answer:
a. Hydrogen
b. Electrical energy
c. Methane
d. Solar panel

Question 39.
a. Among the energy sources identify which will give green energy.
Energy from fossil fuel. Nuclear energy, Solar energy, Energy from wind
b. What are the precautions to be taken for green energy in household buildings.
c. When fuels v are partially burned it can cause certain hazards. List out some.
Answer:
a. Solar energy, Energy from wind
b. Use solar panels, Use biogas plants.
c. Energy loss, Pollution

Question 40.
a. What are the properties that a good fuel must-have?
b. What is the full form of LNG? List out its characteristics.
Answer:
a. 1. Should be of low cost
2. Should be easily available
3. Should cause minimum atmospheric pollution on combustion.
b. Liquefied Natural gas. LNG is a natural gas that can be liquefied and transported to long distances conveniently. It can be again converted into gaseous format atmospheric temperature and distributed through pipelines.

Question 41.
a. What is the full form of LNG? List out its characteristics.
b. Compare Nuclear fission and Nuclear Fusion.
Answer:
a. Liquefied Natural gas. LNG is a natural gas that can be liquefied and transported to long distances conveniently. It can be again converted into gaseous form atmospheric temperature and distributed through pipelines.
b. Nuclear fusion: The process in which lighter nuclei are combined to form heavier ones. Nuclear Fission: The process by which the nuclei of greater mass are split into lighter nuclei, using neutrons. The mass of small nuclei is less than that of parent nucleus.

You can Download Arithmetic Sequences Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 1 Arithmetic Sequence Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 1 Arithmetic Sequences Notes

Textbook Page No. 10

Arithmetic Sequence Questions And Answers 10th Standard Question 1.
Make the following number sequences, from the sequence of equilateral triangles, squares, regular pentagons and so on, of regular polygons:
Number of sides 3, 4, 5, ………
Sum of inner angles
Sum of outer angles
One inner angle
One outer angle
Arithmetic Sequence worksheet with Answer:
No. of triangles of a regular polygon having no. of sides are 3, 4, 5 ………… n respectively is given as 1, 2, 3, 4. Sum of interior angles of a triangle is 180°.

Then the sequence of Sum of interior angles
= 180, 180 × 2, 180 × 3 …. = 180, 360, 540, ……………….
Sum of exterior angles of any geometric structure having any number of sides is always 360°.
Sum of exterior angles = 360, 360, …………..
One interior angle
= \(\frac { 360 }{ 3 }\), \(\frac { 360 }{ 4 }\), \(\frac { 360 }{ 5 }\), ……………. = 120, 90, 72, ………….

Find the sequence calculator can determine the terms (as well as the sum of all terms) of the arithmetic, geometric, or Fibonacci sequence.

Arithmetic Sequence Class 10 Kerala Syllabus Question 2.
Look at these triangles made with dots. How many dots are there in each ?

Compute the number of dots needed to make the next three triangles.
Arithmetic Sequence worksheet with Answer:
3, 6, 10
The number of dots needed to make the next three triangles will be:
10 + 5 = 15
15 + 6 = 21
21 + 7 = 28
15, 21, 28 Dots

Arithmetic Sequence Question 3.
Write down the sequence of natural numbers leaving remainder 1 on division by 3 and the sequence of natural numbers leaving remainder 2 on division by 3.
Answer:
The numbers that leave 1 as remainder when divided by 3 are 1, 4, 7, 10, 13, ………..
(Sequence each with difference of 3 and starting from 1)
The numbers that leave 2 as remainder when divided by 3 are 2, 5, 8, 11, 14, ………….
( Sequence each with difference of 3 and starting from 2)

Kerala Syllabus 10th Standard Maths Chapter 1 Question 4.
Write down the sequence of natural numbers ending in 1 or 6 and describe it in two other ways.
Answer:
1, 6, 11, 16, 21, ……………..
Numbers, each with difference of 5 and starting from 1.
Numbers, when divided by 5, leaves 1 as remainder.

SSLC Maths Arithmetic Sequence Question 5.
A tank contains 1000 litres of water and it flows out at the rate of 5 litres per second. How much water is there in the tank after each second? Write their numbers as a sequence.
Answer:
Water in the tank initially = 1000 litre
Water in the tank after first second
= 1000 – 5 = 995 litre
Water in the tank after next second
= 995 – 5 = 990 litre
Water in the tank after third second
= 990 – 5 = 9.85 litre
Sequence 1000, 995, 990, 985, 980, ………………

Textbook Page No. 15

Sslc Maths Chapter 1 Question Answer Question 1.
Write the algebraic expression for each of the sequences below:
i. Sequence of odd numbers
ii. Sequence of natural numbers which leave remainder 1 on division by 3.
iii. The sequence of natural numbers ending in 1.
iv. The sequence of natural numbers ending in 1 or 6.
Answer:
i. Sequence of odd numbers = 1, 3, 5, 7 …………
That is 2 × 1 – 1,2 × 2 – 1,
Common difference a = 2,
First term, a + b=1, b = 1
Algebraic expression xn = an + b = 2n – 1
n=1, 2, 3 ………..

ii. Sequence of natural numbers which leave remainder 1 on division by 3.
1, 4, 7, 10, 13, 16
x₁ = 1
x₂ = 1 + 3(1) = 4
x₃ = 1 + 3(2) = 7
………………………
……………………..
x_n= 1 + 3(n – l) = 3n – 2
Algebraic expression xn = 3n – 2
n = 1, 2, 3 …………..

iii. The sequence of natural numbers ending in 1 is 1, 11, 21, 31……….
That is 10 × 1 – 9, 10 × 2 – 9, ……………….
Algebraic expression x_n = 10n – 9
n = 1,2,3 ………….

iv. The sequence of natural numbers ending in 1 or 6 is 1, 6, 11, 16, 21 ………….
That is 5 × 1 – 4, 5 × 2 – 4,
Algebraic expression xn = 5n – 4
n=1,2,3 ……………….

An arithmetic common difference calculator is an online free tool to find the arithmetic sequence with the given common difference.

Arithmetic Sequence Pictures Question 2.
For the sequence of regular polygons starting with an equilateral triangle, write the algebraic expressions for the sequence of the sums of inner angles, the sums of the outer angles, the measures of an inner angle, and the measure of an outer angle.
Answer:
Let n be the number of sides
Sum of interior angles: 180°, 360°, 540°, 720°, ……………
Algebraic expression x_n = 180n
Sum of exterior angles 360°, 360°, 360°,………..
Algebraic expression xn = 360°
Sequence of one interior angle:
\(\frac { 180° }{ 3 }\), \(\frac { 360° }{ 4 }\), \(\frac { 540° }{ 5 }\) ………………
Algebraic expression xn = \(\frac{180^{0} n}{n+2}\)
Sequence of one exterior angle:
\(\frac { 360° }{ 3 }\), \(\frac { 360° }{ 4 }\), \(\frac { 360° }{ 5 }\), …………..
Algebraic expression xn = \(=\frac{360^{0}}{n+2}\)

Class 10 Maths Chapter 1 Kerala Syllabus Question 3.
Look at these pictures:

The first picture is got by removing the small triangle formed by joining the midpoints of an equilateral triangle. The second picture is got by removing such a middle triangle from each of the red triangles of the first picture. The third picture shows the same thing done on the second.
i. How many red triangles are there in each picture?
ii. Taking the area of the original uncut triangle as 1, compute the area of a small triangle in each picture.
iii. What is the total area of all the red triangles in each picture?
iv. Write the algebraic expressions for these three sequences obtained by continuing this process.
Answer:
i. The red triangles in the first picture = 3
Red triangles in the second picture = 9
Red triangles in the third picture = 27

ii. Area of the original uncut triangle = 1
Area of the first triangle = \(\frac { 1 }{ 4 }\)
Area of red triangles in the second picture = \(\frac { 1 }{ 16 }\)
Area of red triangles in the third picture = \(\frac { 1 }{ 64 }\)

iii. Total area of red triangle in the first picture = = \(3 \times \frac{1}{4}=\frac{3}{4}\)
Total area of red triangles in the second picture = \(9 \times \frac{1}{16}=\frac{9}{16}\)
Total area of red triangles in the third picture = \(27 \times \frac{1}{64}=\frac{27}{64}\)

This free Recursive sequence formula calculator can determine the terms (as well as the sum of all terms) of the arithmetic, geometric, or Fibonacci sequence.

Textbook Page No. 18

Arithmetic Sequence Class 10 SCERT Question 1.
Check whether each of the sequences given below is an arithmetic sequence. Give reasons. For the arithmetic sequences, write the common difference also.
i. Sequence of odd numbers.
ii. Sequence of even numbers.
iii. Sequence of fractions got as half the odd numbers.
iv. Sequence of powers of 2.
V. Sequence of reciprocals of natural numbers.
Answer:
i. Sequence of odd numbers 1, 3, 5, 7,…….
Difference of adjacent numbers = 2
Common difference = 2

ii. Sequence of even numbers 2, 4, 6, 8,…….
Difference of adjacent numbers = 2 Common difference = 2
∴ This is an arithmetic sequence.

iii. Sequence of half of the odd numbers

Difference of adjacent numbers = 1
Common difference = \(\frac{3}{2}-\frac{1}{2}=1=\frac{5}{2}-\frac{3}{2}\)
∴ This is an arithmetic sequence.

iv. Sequence of powers of 2 21, 22, 23, 24,
2, 4, 8,16, ………………
Difference of adjacent numbers = 4 – 2 ≠ 8 – 4
It does not have a common difference
∴ This is not an arithmetic sequence

v. Sequence of reciprocals of natural numbers

Difference of adjacent numbers

It does not have a common difference
∴ This is not an arithmetic sequence

SSLC Maths Chapter 1 Question 2.
Look at these pictures:

If the pattern is continued, do the numbers of coloured squares form an arithmetic sequence? Give reasons.
Arithmetic Sequence Questions and Answer:
Numbers of coloured squares in each picture 8, 12, 16, ……………
Common difference =12 – 8 = 4 = 16 – 12
∴ This is an arithmetic sequence

10th Class Maths Chapter 1 Arithmetic Sequence Question 3.
See the pictures below:

i. How many small squares are there in each rectangle?
ii. How many large squares?
iii. How many squares in all?
Continuing this pattern, is each such sequence of numbers, an arithmetic sequence?
Answer:
i. No., of small squares in the first picture = 2
No. of small squares in the second picture = 4
No. of small squares in the third picture = 6
No. of small squares in the fourth picture = 8

ii. No. of big squares in first picture = 0
No. of big squares in second picture = 1
No. of big squares in third picture = 2
No. of big squares in fourth picture = 3

iii. Total squares in first picture = 2
Total squares in second picture = 5
Total squares in third picture = 8
Total squares in fourth picture = 11
Sequence of no. of small squares 2, 4, 6, 8
This is an arithmetic sequence with common difference = 2
Sequence of no. of big squares, 0, 1, 2, 3
This is an arithmetic sequence with common difference = 1
Sequence of Total no. of squares 2, 5, 8, 11
This is an arithmetic sequence with common difference = 3

Question 4.
In the staircase shown here the height of the first step is 10 centimetres and the height of each step after it is 17.5 centimetres.
i. How high from the ground would be some-one climbing up, after each step?
ii. Write these numbers as a sequence

Answer:
i. High from the ground climbing up first step = 10 cm
High from the ground climbing up sec-ond step = 10 + 17.5 = 27.5cm
High from the ground climbing up third step = 27. 5+ 17.5 = 45 cm
High from the ground climbing up fourth step = 45 + 17.5 = 62.5 cm
High from the ground climbing up fourth step = 62.5 + 17.5 = 80 cm
High from the ground climbing up fourth ‘ step = 80 + 17.5 = 97.5cm

ii. Sequence of height
10, 27.5, 45, 62.5, 80, 97.5 ……………

Question 5.
In this picture, the perpendiculars to the bottom line are equally spaced. Prove that, continuing like this, the lengths of perpendiculars form an arithmetic sequence.

Answer:

In figure ΔOAB, ΔOCD are similar right-angled triangles. That is if one side and its included angles of a triangle are equal to the one side and included angles of another triangle, then the triangles are similar.
∴ sides are proportional.
\(\frac{O A}{O C}=\frac{x}{2 x}=\frac{A B}{C D} \quad \Rightarrow C D=2 A B\) similarly
EF = 3AB. Sequence of length of perpendicular AB, 2AB, 3AB,………..
That is AB is the common difference, so perpendicular lengths are in arithmetic sequence

Question 6.
The algebraic expression of a sequence is x_n = n³ – 6n² + 13n – 7
Is it an arithmetic sequence?
Answer:
Algebraic expression = n³ – 6n² + 13n – 7
n = 1 ⇒ =1 – 6 + 13 – 7 = 1
n = 2 ⇒ = 8 – 24 + 26 – 7 = 3
n = 3 ⇒ =27 – 6 × 9 + 39 – 7
= 27 – 54 + 32 = 5
n = 4 ⇒ = 64 – 96 + 52 – 7 = 13
1, 3, 5, 13,………. This is not an arithmetic sequence.lt does not have a common difference

Textbook Page No. 21

Question 1.
In each of the arithmetic sequences below, some terms are missing and their positions are marked with . Find them.

Answer:

Common difference = 42 – 24 = 18
Numbers to be found out = 42 + 18 = 60
60 + 18 = 78
Arithmetic sequence 24, 42, 60, 78, ………….

Common difference = 42 – 24 = 18
Numbers to be found out = 24 – 18 = 6
42 + 18 = 60
24 – 18, 24, 42, 42 + 18
Arithmetic sequence 6, 24, 42, 60, ………….

Common difference = 18
Numbers to be found out = 24 – 18 = 6
6 – 18 = – 12 6 – 18, 24 – 18, 24, 42
Arithmetic sequence – 12, 6, 24, 42, ………

Common difference = 9
[(42 – 24)/2 = 18/2 = 9)
Numbers to be found out =
24 + 9 = 33, 33 + 9 = 42, 42 + 9 = 51
Arithmetic sequence 24, 33, 42, 51 ……..

Common difference = (42 + 24)/2 = 18/2 = 9
Numbers to be found out = 24 – 9, 24, 24 +9, 42
Arithmetic sequence 15, 24, 33, 42

Let numbers to be found out be x and y If common differences are equal = x – 24 = 42 – y

Question 2.
The terms in two positions of some arithmetic sequences are given below.
Write the first five terms of each:
i. 3^rd term 34
6^th term 67
ii. 3^rd term 43
6^th term 76
iii. 3^rd term 2
5^th term 3
iv. 4^th term 2
7^th term 3
v. 2^nd term 5
5^th term 2
Answer:
i. Third term =34
Sixth term =67
We get the 6th term from the 3 rd term, we must add the common difference (6 – 3 = 3)3 time.
Three times of common difference 67 – 34 = 33
Common difference = \(\frac { 33 }{ 3 }\) = 11
Second term = 34 – 11 = 23
First term = 23 – 11 = 12
First five terms 12, 23, 34, 45, 56

ii. Third term =43
Sixth term = 76
Thrice the common difference = 76 – 43 = 33
Common difference = \(\frac { 33 }{ 3 }\) = 11
Second term = 43 – 11 = 32
First term =32 – 11 =21
First five terms 21, 32, 43, 54, 65.

iii. Third term = 2
Fifth term = 3
Twice the common difference= 3 – 2 = 1
Common difference = 1/2
Second term =2 – 1/2 = 1 1/2
First term = 1 1/2 – 1/2 = 1

iv. Fourth term =2
Seventh term = 3
Thrice the common difference = 3 – 2 = 1
Common difference = 1/3

v. Second term = 5
Fifth term = 2
Thrice the common difference = 2 – 5 = -3
Common difference = –3/3 = –1
First term = 5 – –1 = 6
First five terms 6, 5, 4, 3, 2

Question 3.
The 5th term of an arithmetic sequence is 38 and the 9th term is 66. What is its 25th term?
Fifth term = 38
Ninth term = 66
Term difference = 66 – 38 = 28
Position difference = 9 – 5 = 4 – 2 8
Common difference = 28/4 = 7
25th term = Fifth term = 20 × Common difference
25th term = Fifth term + 20 × Common difference = 38 + 20 × 7 = 38 + 140 = 178

Question 4.
Is 101 a term of the arithmetic sequence 13, 24, 35, …? What about 1001?
Answer:
13, 24, 35,………..
Common difference = 24 – 13 = 11
Difference between 101 and 13
= 101 – 13 = 88 = 8 × 11
That is we can obtained 101 by adding 8 times of common difference with 13.
So, ‘101’ is the 9th term of this arithmetic sequence
101 is the 9th term of this sequence
Difference between 1001 and 13
= 1001 – 13 = 988
This is not a multiple of 11.
Therefore ‘1oo1’ is not a term of this arithmetic sequence.

Question 5.
How many three-digit numbers are there, which leave a remainder 3 on division by 7?
Answer:
101, 108, …………………………. 997
First-term = 101
Common difference = 7
Last term = 997
Term difference = 997 – 101 = 896 = 128 × 7
Last term = First term + 128 × common difference
i. e., 997 is the 129th term.
129 three-digit numbers are there which leave remainder 3 on division by 7

Question 6.
Fill up the empty cells of the given square such that the numbers in each row and column form arithmetic sequences:

What if we use some other numbers instead of 1, 4, 28 and 7?
Answer:
1. In the first row differences between 1st and 4th terrm =3
Position difference = 4 – 1 = 3
Common difference = 3/3 = 1
Arithmetic sequence 1, 2, 3,4

2. In the first coloumn Term difference = 7 – 1 = 6
Position difference = 4 – 1 = 3
Common difference = 6/3 = 2
Arithmetic sequence 1, 3, 5, 7……….

3. In the fourth row
Termdifference = 28 – 7 = 21
Position difference = 4 – 1=3
Common difference = 21/3 = 7
Arithmetic sequence 7, 14, 21, 28

4. In the second coloumn 2, –,– 14
Term difference = 14 – 2 = 12
Position difference = 4 – 1=3
Common difference = \(\frac { 14 – 2 }{ 3 }\) = 4
Arithmetic sequence 2, 6, 10, 14

5. In the third coloumn 3, –, –, 21
Term difference = 21 – 3 = 18
Position difference = 4 – 1=3
Common difference = \(\frac { 21 – 3 }{ 3 }\) = 6
Arithmetic sequence 3, 9,15, 21

6. In the fourth coloumn 4, –, –, 28
Term difference = 28 – 4 = 24
Position difference = 4 – 1=3
Common difference = \(\frac { 28 – 4 }{ 3 }\) = 8
Arithmetic sequence 4,12, 20, 28

Question 7.
In the table below, some arithmetic sequences are given with two numbers against each. Check whether each belongs to the sequence or not.

Answer:
i. Sequence 11, 22, 33, …………..
Difference = 22 – 11 = 11
Difference between 123 and 11.
=123 – 11 = 112
112 is not a multiple of 11, so 123 is not a term of this arithmetic series.

Difference between 132 and 11
= 132 – 11 = 121
121 is a multiple of 11, so 132 is a term of this arithmetic series.

ii. Sequence 12, 23, 34,
Common difference = 23 – 12 = 11 Difference between 100 and 12
= 100 – 12 = 88
88 is a multiple of 11, so 100 is a term of this arithmetic series.

Difference between 1000 and 12.
= 1000 – 12 = 988
988 is not a multiple of 11, so 1000 is not a term of this arithmetic series.

iii. Sequence 21, 32, 43, ……………
Common difference = 32 – 21 = 11
Difference between 100 and 21.
= 100 – 21 = 79
79 is not a multiple of 11, so 1000 is not a term of this arithmetic series.

Difference between 1000 and 21
=1000 – 21 = 979
979 is a multiple of 11, so 1000 is a term of this arithmetic series.

Textbook Page No. 24

Question 1.
The 8th term of an arithmetic sequence is 12 and its 12th term is 8. What is the algebraic expression for this sequence?
Answer:
Difference between terms = 8 – 12 = -4
To get the 12th term from the 8td term, we must add the common difference (9 – 5 = 4) 4 times.
Common difference = – 4 / 4 = -1
Algebraic expression for arithmetic series
x_n = an + b
12 = – 1 × 8 + b ⇒ 12 = – 8 + b
b = 12 + 8 = 20
Algebraic expression for arithmetic series =
xn = –1 × n + 20 = 20 – n

Question 2.
The Bird problem in Class 8 (The lesson, Equations) can be slightly changed as follows.
One bird said:
“We and we again, together with half of us and half of that, and one more is a natural number”
Write all the possible number of birds starting from the least. For each of these, write the sum told by the bird also.
Find the algebraic expression for these two sequences.
Answer:
If we take x as the number of birds

∴ 11x + 4 is a multiple of 4.
⇒ 11x is a multiple of 4.
x is a multiple of 4.
No. of birds 4, 8, 12, 16 ……………
(a= 4, a+b = 4, b = 0)
Algebraic expression of the series = an + b = 4n No. of sums
\(\left(\frac{11 x}{4}+1, x=4,8,12 \ldots .\right)=12,23,34,45, \ldots \ldots\)
Algebraic expression for the series = 12 + (n -1)11 = 11 n + 1

Question 3.
Prove that the arithmetic sequence with first term 1/3 and common difference 1/6 contains all natural numbers.
Answer:
f = \(\frac { 1 }{ 3 }\)
d = \(\frac { 1 }{ 6 }\)
\(x_{n}=d n+(f-d)=\frac{1}{6} n+\frac{1}{6}=\frac{1}{6}(n+1)\)
As n = 5, 11, 17,………….
∴ All natural numbers will occur in this arithmetic sequence

Question 4.
Prove that the arithmetic sequence with first term 1/3 and common difference 2/3 contains all odd numbers, but no even number.
Answer:

Value of (2n -1) is always an odd number, Therefore there is no even number.

Question 5.
Prove that the squares of all the terms of the arithmetic sequence 4,7,10,…. belong to the sequence.
Answer:

This is in the form 3k + 1, so squares of all terms are in this series. (When we divide 3k + 1 by 3 get 1 as remainder)

Question 6.
Prove that the arithmetic sequence 5,8,11,……. contains no perfect squares.
Answer:
f = 5
d = 3
\(x_{n}=d n+(f-d)=3 n+(5-3)=3 n+2\)
That is when we divide a perfect square with 3 we get remainder as 1 or 0. Here we can divide 3k + 2 by 3 to get 2 as remainder and it doesn’t contain any perfect square.

Question 7.
Write the whole numbers in the arithmetic sequence \(\frac { 11 }{ 8 }\), \(\frac { 14 }{ 8 }\), \(\frac { 17 }{ 8 }\), ………..
they form an arithmetic sequence?
Answer:

If ‘n’ is a multiple of 8 then 4, 7, 10, … is a sequence having whole numbers
Common difference = 3,This is an arithmetic sequence.

Textbook Page No. 28

Question 1.
Write three arithmetic sequences with 30 as the sum of the first five terms.
Answer:
If the five terms are
x, x + 1, x + 2, x +3, x + 4
x + x+ 1 + x + 2 + x + 3 + x + 4 = 30
5x + 1o = 30
5x = 20
x = 4
The arithmetic sequence is 4, 5, 6, 7, 8

If the five terms are
x, x + 2, x + 4, x +6, x + 8
x + x + 2 + x + 4 + x + 6 + x + 8 = 30
5x + 20 = 30
5x = 30 – 20 = 10
x = 2
The arithmetic sequence is 2, 4, 6, 8, 10

If the five terms are
x, x + 3, x + 6, x + 9, x + 12
x + x + 3+ x+ 6 + x + 9+ x + 12 = 30
5X + 30 = 30
5x = 0
x = 0
The arithmetic sequence is 0, 3, 6, 9,12
The arithmetic sequences whose sum of first five terms give 30 are:
4, 5, 6, 7, 8, …………..
2, 4, 6, 8, 10, …………..
0, 3, 6, 9, 12, ………….

Arithmetic Sequence Question 2.
The first term of an arithmetic sequence is 1 and the sum of the first four terms is 100.
Find the first four terms.
Answer:
First term f = 1
If first four terms are
f, f+d, f+2d, f+3d
f + f + d + f + 2d + f + 3d = 100
4f + 6d = 100
= 2f + 3d = 50
2+3d = 50
= 3d = 50 – 2 = 48
d = 16
The arithmetic sequence is 1, 17, 33, 49, ………….

Question 3.
Prove that for any four consecutive terms of an arithmetic sequence, the sum of the two terms on the two ends and the sum of the two terms in the middle are the same.
Answer:
First four terms of an arithmetic sequence is x,
x+ d, x+ 2d, x+ 3d
Sum of the two terms on the two ends = 2x + 3d
Sum of the two terms in the middle
= (x + d) +(x + 2d) = 2x + 3d
They both are equal

Question 4.
Write four arithmetic sequences with 100 as the sum of the first four terms.
Answer:
If the first four terms are
x – 3d, x – d, x + d, x + 3d
x – 3d + x – d + x + d + x + 3d = 100
4x = 100
x = 25
i. If d= 1
The arithmetic sequence is 22, 24, 26, 28

ii. If d= 2
The arithmetic sequence is 19, 23, 27, 31

iii. If d = 3
The arithmetic sequence is 16, 22, 28, 34

iv. If d = 4
The arithmetic sequence is 13, 21, 29, 37

Question 5.
Write the first three terms of each of the arithmetic sequences described below:
i. First-term 30; the sum of the first three terms is 300.
ii. First-term 30; the sum of the first four terms is 300.
iii. First-term 30; the sum of the first five terms is 300.
iv. First-term 30; the sum of the first six terms is 300.
Answer:
i. First-term = 30
Sum of first three terms = 300
In the arithmetic sequence sum of any three consecutive natural numbers is thrice the middle number
∴ Second term = \(\frac { 300 }{ 3 }\) = 100
∴ Common difference = 300 – 30 = 270
∴ Third term =100 + 70 = 170
Sequence 30, 100, 170, ……….

ii. First-term = 30
Sum of first four terms = 300
Four consecutive terms of an arithmetic sequence, the sums of the first and the last is equal to the sum of the second and the third.
First term + Fourth term = Second term + Third term = \(\frac { 300 }{ 2 }\) = 150
∴ Fourth term = 150 – 30 = 120
30, ………, ………., 120
Term difference = 120 – 30 = 90
Position difference = 4 – 1 = 3
Common difference = 90/3 = 30
∴ Sequence = 30, 60, 90, 120,………….

iii. First term = 30
Sum of first five terms = 300
Sum of the five consecutive terms of arithmetic sequence is five times of its middle term.
Third term = 300/5 = 60
Common difference = 15
∴ Sequence = 30, 45, 60, 75, 90, ……….

iv. First term =30
Sum of first six terms =300
Fist term + Sixth term = Second term + Fifth term = Third term + Fourth term = 300/3 = 100
Sixth term = 100 – First term = 100 – 30 = 70
Term difference = 70 – 30 = 40
Position difference =6 – 1 = 5
Common difference = \(\frac { 40 }{ 5 }\) = 8
Sequence = 30, 38, 46, 54, 62, 70,

Question 6.
The sum of the first five terms of an arithmetic sequence is 150 and the sum of the first ten terms is 550.
i. What is the third term of the sequence?
ii. What is the eighth term?
iii. What are the first three terms of the sequence?
Answer:
i. Sum of first five terms = 150
Sum of the five consecutive terms of arithmetic sequence is five times of its middle term.
Third term = \(\frac { 150 }{ 5 }\) = 30

ii. First term + Tenth term = Second term + Nineth term = Third term + Eighth term = Fourth term + Seventh term = Fifth term + Sixth term = \(\frac { 550 }{ 5 }\) = 110
Third term + Eighth term = 110
Eighth term =110 – Third term
= 110 – 30 = 80

iii. Third term = 30
Eighth term = 80
Term difference = 80 – 30 = 50
Position difference = 8 – 3 = 5
Common difference = \(\frac { 50 }{ 5 }\) = 10
∴ Sequence = 10, 20, 30,…………….

Question 7.
The angles of a pentagon are in arithmetic sequence. Prove that its smallest angle is greater than 36°.
Answer:
Let the smallest angle in the pentagon be x
x + x + d + x + 2d + x + 3d + x + 4d = 540
5x + 10d = 540,
x + 2d = 108
x = 36°, then
d = 36,
angles are 36, 72, 108, 144, 180.
180° will not an angle of a pentagon (Sum of exterior and interior angle is 180°).
Therefore the smallest angle in a pentagon will always be greater than 36°

Textbook Page No. 35

Question 1.
Find the sum of the first 25 terms of each of the arithmetic sequences below.
i. 11, 22, 33,….
ii. 12, 23, 34,…
iii. 21, 32, 43,….
iv. 19, 28, 37,…
vi. 1, 6, 11,….
Answer:

Question 2.
What is the difference between the sum of the first 20 terms and the next 20 terms of the arithmetic sequence 6, 10, 14,…?
Answer:

Differences = 2480 – 880 = 1600

Question 3.
Calculate the difference between the sums of the first 20 terms of the arithmetic sequences 6, 10, 14,… and 15, 19, 23,……
Answer:
6, 10, 14,
Algebraic expression of the above arithmetic sequence = 4n + 2

15, 19, 23, …… Algebraic expression of the arithmetic sequence = 4n +11

Differences = 1060 – 880 = 180

Question 4.
Find the sum of all three-digit numbers, which are multiples of 9.
Answer:
First 3 digit number divisible by 9 = 108
Last 3 digit number divisible by 9 = 999 108, 117,126, …………… 999

Question 5.
The expressions for the sum to n terms of some arithmetic sequences are given below. Find the expression for the nth term of each:
i. n² + 2n
ii. 2n² + n
iii.n² – 2n
iv.2n² – n
v. n² – n
Answer:
i. n² + 2n, First term =12+ 2 × 1 = 3
Sum of first two terms = 22 + 2 × 2 = 8
Second term = 8 – 3 = 5
The arithmetic sequence is 3, 5,…………..
nth term = dn + (f – d) = 2n + (3 – 2)
= 2n + 1

ii. 2n² + n
First term =2 + 1 = 3
Sum of first two terms =2 × 22 + 2
= 8 + 2 = 10
Second term = 10 – 3 =7
The arithmetic sequence is 3, 7, …………

iii. n² – 2n
First term = 1 – 2 = – 1
Sum of first two terms =22 – 4 = o
Second term = 0-(-1) = 0 + 1 = 1
The arithmetic sequence is –1, 1, ……………

iv. 2n² – n
First term = 2 – 1 = 1 Sum of first two terms = 2 – 22 – 2
=8 – 2=6
Second term = 6 – 1 = 5
The arithmetic sequence is 1, 5,……….

v. n² – n
First term = 1 – 1 = 0
Sum of first two terms = 22 – 2 = 2
Second term = 2 – 0 = 2
The arithmetic sequence is 0, 2,……….

Question 6.
Calculate in head, the sums of the following arithmetic sequences.

Question 7.
Prove that the sum of any number of terms of the arithmetic sequence 16, 24, 32,….. starting from the first, added to 9 gives a perfect square
Answer:
16, 24, 32,…………. Algebraic expression of the arithmetic sequence = 8n + 8

Hence it is a perfect square

Question 8.

i. Write the next two lines of the pattern above.
ii. Write the first and the last numbers of the 10th line.
iii. Find the sum of all the numbers in the first ten lines.
Answer:
i. First row contains one element, second row contains two elements, therefore j fifth row will contain five elements and j sixth row will contain six elements.
Consider the first row
1, 2, 4, …………..
1, 1+1, 2 + 2, 4 + 3, …………
Therefore
1, 1+1, 2 + 2, 4 + 3, 7 + 4, 11+5 …………..
1, 2, 4, 7, 11, 16 …………
Generally it written as
1 + 1 (1+2 + 3 + ……….)
Fist term in the fifth row = 1 + 1(1 + 2 + 3 + 4) = 1 + 10 = 11
Fist term in the sixth row= 1 + 1(1 + 2 + 3 + 4+ 5) = 1 + 15 = 16
Common differences in each row = 1
Fifth row 11, 12, 13, 14, 15.
Sixthrow 16, 17, 18, 19, 20, 21.

ii. Fist term in the tenth row = 1 + 1(1 + 2 + 3+ 4 + 5 + 6+ 7 + 8 + 9) = \(=1+\frac{1 \times 9 \times 10}{2}=46\)
Last term in the tenth row = 46 + 1 × 9 = 46 + 9 = 55

iii. Numbers in the first ten lines 1, 2, 3, 4, 5, ………….. 55
Sum of all the numbers in the first ten lines = \(\frac{n(n+1)}{2}=\frac{55 \times 56}{2}=1540\)

Question 9.

Write the next two lines of the pattern above. Calculate the first and last terms of the 20th line.
Answer:
4, 7, 13, 22, 34, 49,…….
(4, 4+3, 7+6, 13+9, 22 + 12, 34+15),
Generally it written as 4 + 3(1 + 2 + 3 + ………..)
First term in the fifth row = 4 + 3(1 +2 + 3+ 4) = 4 + 30 = 34
First term in the sixth row =
4 + 3(1 + 2+ 3 + 4 + 5) = 4 + 45 = 49
Common differences in each row = 3
Fifth row 34, 37, 40, 43, 46,..
Sixth row 49, 52, 55, 58, 61, 64…
4
7 10
13 16 19
22 25 28 31
34 37 40 43 46
49 52 55 58 61 64
………………………….
First term in the 20th row = 4 + 3(1 + 2 + 3 + 4 +……………. +19)

Arithmetic Sequences Orukkam Questions & Answers

Worksheet 1
Answer the following questions

Question 1.
Write the sequence of natural numbers
Answer:
1,2,3,4,

Question 2.
Write the sequence of odd numbers
Answer:
1,3,5,7,

Question 3.
Write the sequence of even numbers
Answer:
2,4,6,8,10,

Question 4.
Write the sequence of multiples of 3.
Answer:
3,6,9,12,15,

Question 5.
Write the sequence of numbers which leaves the remainder 1 on dividing by 4.
Answer:
0 × 4 + 1, 1 × 4 + 1, 2 × 4 + 1, 3 × 4 + 1, 4 × 4 + 1, 5 × 4 + 1, …………
=1, 5, 9, 13, 17, 21,

Question 6.
Write the sequence of prime numbers.
Answer:
2,3,5,7,11,13,17,

Question 7.
Write the sequence of perfect squares.
Answer:
1, 4, 9, 16, 25, 36,

Question 8.
Write the sequence of numbers which leaves the remainder 0 on dividing by 6
Answer:
6, 12, 18, 24, 30, 36, ………….

Question 9.
Write the sequence starting from 1 and is added subsequently
Answer:

Question 10.
Write the sequence starting from 1/2 and 3/4 is added subsequently
Answer:

Question 11.
Write the sequence starting from 60 and 0 is added subsequently
Answer:
60, 60, 60,…………..

Worksheet 2

Question 12.
Write the sequence of the perimeters of the equilateral triangles having sides 1cm, 2cm, 3cm.
a. Write the sequence of area
b. Write the sequence of sum of angles.
Answer:
’If we Increase 1 cm of sides of an equilateral triangle having sides 1cm, 2cm, 3cm.
a. Area =1 cm2, Area after increasing
sides by 1 cm \(\frac{\sqrt{3} \times 2^{2}}{4}=\sqrt{3}\) cm2
Area after increasing thelength of side 2cm by
V3x32 9V3 ,
1cm = \(=\frac{\sqrt{3} \times 3^{2}}{4}=\frac{9 \sqrt{3}}{4}\) cm2
Area after increasing thelength of side 3cm by

c. The sum of angles of a triangle will be 180° always for any measures of sides. Sequence of sum of angles 180, 180, 180,……..

Question 13.
Write the sequence of numbers which leaves the remainder 3 on dividing by 5 and 10.
Answer:
The number which can be divided by 5 and 10 must be divisible by 10.
∴ The sequence of numbers which leaves the remainder 3 on dividing by 5 and 10 is 3, 13, 23, 43,………

Question 14.
Look at the sequence 1 + (1 + 5), 2 + (2 + 5), 3 + (3 + 5) …….
a. Write next two terms
b. Write its algbraic form.
Answer:
a. Next two terms
4 + (4 + 5), 5 + (5 + 5),

b. Algebraic expression
x_n = n + (n + 5) = n + n + 5
x_n = 2n + 5

Question 15.
Write the terms of the sequence 5 × (1+6), 10 × (2+6), 15 × (3+6), 20 × (4+6) in the form :
first term 5 × 1(1 + 6),
second term 5 × 2(2 + 6).
Write its algebraic expression
Answer:
5 x (1+6), 10 x (2 + 6), 15 x (3 + 6), 20 x (4 + 6), … this sequence can be written as 5 x 1(1 + 6), 5 x 2 (2 + 6), 5 x 3 (3 + 6), 5 x 4 (4 + 6). Algebraic expression = 5 x n (n + 6)
i.e. x_n = 5 n² + 30n.

Worksheet 3

Question 16.
Write eighth terms of an arithmetic sequence using the numbers given below. (22, 15, 18, 4, 10, 14, 6, 12)
Answer:
6, 10, 14, 18, 22, ……………

Question 17.
Write the missing terms in the arithmetic sequence given below

Question 18.
The difference between 12th and 8th term of an arithmetic sequence is 20. Find the common difference.
Answer:

Question 19.
The tenth term of an arithmetic sequence is 65 and its 15^th term is 80. Is 200 a term of this sequence?
Answer:
’To get the 15^th term from the 10th term, we must add the common difference 5 times.
5 times of common difference = 80 – 65 = 15
Common difference (d) = 15/5 = 3
Dividing 65 by 3 we get the remainder as 2.
Dividing 200 by 3 we get the remainder as 2.
Therefore 200 is a term of this sequence.

Question 20.
The 20^th term of an arithmetic sequence is 64 and its 21^th term is 70. Can the difference between two terms 46? Why?
Answer:
20^th term = 64
21^th term = 70
Common difference = 70 – 64 = 6 Difference between any two terms of this sequence will be a multiple of 6.
Dividing 46 by 6 we get the remainder as 4, So difference between any two terms of the sequence will not be 46.

Question 21.
The angles of a quadrilateral are in an arithmetic sequence. The largest angle is 150°. Find other angles.
Answer:
Sum of four angles of a quadrilateral = 360° Angles are in arithmetic sequence, so first angle + 3 x common difference (d) = 150°
i.e., f + 3d= 150

The four angles are f, 70, f + 2d, 150 .
\(\mathrm{f}+2 \mathrm{d}=\frac{150+70}{2}=110\)
Common difference = 110 – 70 = 40
First angle = 70 – 40 = 30
So, the four angles are 30, 70, 110, 150

Question 22.
What will be the remainder on dividing a term of the sequence 3n + 7 by its common difference?
Answer:
Dividing 3n + 7 by 3 we get the remainder as 1.

Worksheet 4

Question 23.
Write the algebra of the following sequences and its sum of n terms
1.5, 10, 15, 20, ………..
2. 6, 11, 16, 21, ………
3. 4, 9,14,19, ……….
4. 3, 8, 13, 18, …………..
Answer:
1. 5, 10, 15, 20,
First term f = 5
Common difference = d = 10 – 5 = 5
General form xn = dn + (f – d)
= 5n + (5 – 5) = 5n

2. 6, 11, 16,21, …………..
First term f = 6

3. 4, 9, 14, 19, ………….
First term f = 4

4. 3, 8, 13, 18, …………
First term f = 3

Worksheet 6

Question 24.
Write the sequence of the squares of all odd numbers. What is its algebra?
Answer:
Odd numbers = 1, 3, 5, 7, 9, 11, ………
f= l, d = 3 – 1 =2, f – d= 1 – 2 = – 1
Algebraic form of odd numbers
x_n= dn+ (f – d) = 2n – 1
The sequence of the squares of all odd numbers = 1, 9, 2 5, 49,……….
Algebraic form = (2n -1 )²

Question 25.
Write the sequence formed by the number of diagonals from a vertex of a triangle, a quadrilateral, a pentagon etc. What is its algebra?
Answer:
Diagonal drawn from one vertex of a triangle = 0 = 3 – 3 = 0
Diagonal drawn from one vertex of a quadrilateral = 4 – 3 = 1
Diagonal drawn from one vertex of a pentagon = 5 – 3 = 2
Diagonal drawn from one vertex of a n sided polygon = n – 3
Sequence of number of diagonals 0, 1, 2, 3, 4, ……….
Algebraic expression x_n = n – 3

Question 26.
Write the sequence of the number of diagonals in a quadrilateral, pentagon, hexagon etc. What is its algebra?
Answer:

Question 27.
Can the difference between any two terms of an arithemetic sequence having common difference 6 be 2016? Justify your answer.
Answer:
\(\frac{\text { Differences of terms }}{\text { Common difference }}=\frac{2016}{6}=336\)
It is a natural number, so the difference between any two terms of an arithemetic sequence having common difference 6 be 2016.

Question 28.
Write algebra of the sum of the sequence 6n + 5. Can the sum 2000 ? Why?
Answer:
nth term = 6n + 5
1st term = 6 + 5 = 11
Sum of n terms = n/2 (11 + 6n + 5)
= n/2 (6n + 16) = n(3n + 8) = 3n² + 8n
Algebraic form of the sum = 3n² + 8n
To check wheater the sum is 2ooo
3n² + 8n = 2000
i.e., 3n² + 8n – 2000 = 0

It is not a natural number. So 2000 will not be a sum.

Worksheet 7

Question 29.
Prove that the squares of the sequence 1, 3, 5,……..belongs to that sequence itself.
Answer:
Odd-numbered arithmetic sequence will be 1, 3, 5, …
Here the arithmetic sequence have common difference 2. Their squares are also odd numbers. Therefore the squares of the sequence 1, 3, 5, …belongs to that sequence itself. nth term = 2n – 1.

Question 30.
In an arithmetic sequence having terms natural numbers, prove that if one of the terms is a perfect square, it will have more that this as the perfect square term.
As we know when a definite number of common difference of an arithmetic sequence is added to a term we get another term of the same sequence. If n2 is a perfect square term, add (2n + d) times d to n². n² + (2n + d) x d = (n + d)². This is nothing but a perfect square term.

Question 31.
If the angles of a right triangle are in an arithmetic sequence, find them by making suitable equations.
Answer:
Let angles be f – d, f, f + d f – d + f + f+d = 180
3f = 180, f = 60, f + d = 90, d = 30 Angles are 30, 60 and 90

Question 32.
If ten times tenth term of an arithmetic sequence is equal to fifteen times fifteenth term, find 25^th term. Calculate the product of first 25^th terms.
Answer:
10 × (f + 9d) = 15 × ( f + 14 d)
⇒ 10 f + 90 d = 15 f + 210 d
⇒ 5 f+ 120 d = 0
⇒ f + 24 d=0
25^th term = f + 24 d = 0
25^th term is 0. Product of 25 terms is 0.

Question 33.
Write the algebraic form of 1,4,7,10,… is 100 a term of this sequence. Why? Prove that the square of any term of this sequence belongs to that sequence.
Answer:
1,4, 7, 10,…
f = 1, d = 3
x_n= dn + (f – d) = 3n + (1 – 3) = 3n – 2
When 4 is divided by 3 we get remainder as 1 When 100 is divided by 3 we get remainder as 1 So, 100 is a term of the arithmetic sequence.
Square = (3n – 2)² = 9n² – 12n + 4 (9n² – 12n + 4) + 3
When 9n² – 12n + 4 is divided by 3the remainder is 1.
So square of the term also in the sequence.
∴ The square of any term of this sequence belongs to that sequence.

Question 34.
Write the sequence obtained by adding two adjacent consecutive terms in counting numbers starting from 1.
Write the algebraic expression of this sequence.
[Score: 3, Time: 4 minutes]
Answer:
Counting numbers:
1, 2, 3, 4, 5, … (1)
Sequence obtained by adding:
: 1+2, 2 + 3, 3 + 4, 4 + 5, … (1) Two adjacent consecutive terms
3, 5, 7, 9, …
Algebraic expression of above sequence :
n + (n+l) = 2n + 1 (1)

Question 35.
Consider circles, points on its circumference and chords as shown in the figure. Mark two points on the circle and draw a chord. Mark one more point and draw three chords. Continue this process by adding one more point each time. [Score: 4, Time: 7 minutes]
a. Write the number of chords in each figure as a sequence.
b. Write the algebraic expression of this sequence.
c. Find the number of chords in the 10^th figure.

Answer:
a. No. of chords in figure 1 = 1
No. of chords in figure 2 = 1 + 2 = 3
No. of chords in figure 3 = 1+ 2 + 3 = 6 (1)
Sequence of number of chords
= 1, 3, 6, 10, … (1)

b. No. of chords in figure n
\(=1+2+3+\ldots+n=\frac{n(n+1)}{2}\) (1)

c. No. of chords in the 10^th figure
\(=\frac{10 \times 11}{2}=55\) (1)

Question 36.
A pattern is formed using sticks of equal length as shown below:
[Score: 4, Time: 9 minutes]

a. Write the sequence of number of sticks used in each figure.
b. Write the sequence of number of squares and rectangles in each figure,
c. Write the algebraic expression in the above two sequences.
d. Find the number of sticks and squares in the 10^th figure.
Answer:
a. No of sticks in figure 1 = 1 + 3 = 4
No of sticks in figure 2
= 1 + 3 + 3 = 1+2 × 3 =7 No of sticks in the figure 3
= 1 + 3 × 3 = 10 No of sticks in the figure 4
= 1+4 × 3 = 13 Sequence of number of sticks
= 4, 7, 10, 13, … (i)

b. No of squares and rectangles in the figure 1 = 1
No of squares and rectangles in the figure 2 = 2 + 1 = 3
No of squares and rectangles in the figure 3 =3 + 2+ 1=6
No ofsquares and rectangles in the figure 4 = 4 + 3 + 2 + 1 = 10
Sequence of squares and rectangles = 1, 3, 6, 10, …. (1)

c. No of sticks in the nth figure 1
= 1+ n × 3 = 3n + 1
No of squares and rectangles in the nth figure = 1 + 2+ 3….. +n = \(=\frac{n(n+1)}{2}\) (1)

d. No of sticks in the 10th figure = 3 × 10 + 1 = 31
No of squares and rectangles in the 10th figure = \(\frac{10 \times 11}{2}=55\) = 55 (1)

37. Consider an arithmetic sequence with common difference 6 and 7^th term 52. Find the 15^th term of the arithmetic sequence. Is it possible, to get a difference of 100 between any two terms of this sequence?
[Score: 3, Time: 5 minutes]
Answer:
15^th term can be obtained by adding 8 times the common difference to the 7^th term.
x₁₅ = x₇ +8d (1)
= 52 + 8 × 6 = 100 (1)
The difference between any two terms of an Arithmetic sequence will be a multiple of com-mon difference. 100 can’t be the difference be-tween any two terms of this sequence, since it is not a multiple of 6. (1)

Question 38.
Consider an arithmetic sequence whose 7^th term is 34 and 15^th term is 66.
[Score: 3, Time: 5 minutes]
a. Find the common difference,
b. Find the 20th term.
Answer:
a. 15^th term can be obtained by adding 7^th term ‘ and 8 times the common difference.
x₁₅ = x₇ + 8d (1)
66 = 34 + 8d
8d = 66 – 34 = 32
d = \(\frac { 32 }{ 8 }\) = 4 (1)

d. 20^th term can be obtained by adding 15^th term and 5 times the common difference.
x₂₀ = x₁₅ + 8d (1)
= 66 + 5 × 4 = 86

Question 39.
Consider an arithmetic sequence \(\frac { 17 }{ 7 }\), \(\frac { 20 }{ 7 }\), \(\frac { 23 }{ 7 }\), ……….
a. Write the algebraic expression of the sequence.
b. Write the sequence of counting numbers in the above given sequence. Is the newly obtained sequence an arithmetic sequence. [Score:. 4, Time: 6 minutes]
Answer:
a. Common Difference = \(\frac { 20 }{ 7 }\) – \(\frac { 17 }{ 7 }\) = \(\frac { 3 }{ 7 }\) (1)
Algebraic expression of the sequence

Question 40.
Let the algebraic expression of an arithmetic sequence be 5n + b. If there is no perfect square in this sequence, find the counting number less than 5 that can be the value of ‘b’. [Score:4, Time: 5minutes]
Answer:

Any perfect square when divided by 5 leaves remainder 0, 1,4 So if remainder is 2 and 3 then it will not be a perfect square.
If there is no perfect square in the sequence, the only possibility for value of ‘b’ less than 5 is 2 and 3. (1)

Arithmetic Sequences Exam Oriented Questions & Answers

Short Answer Type Questions (Score 2)

Question 41.
—, 18, —, 28 are four consecutive terms of an arithmetic sequence. Fill in the blanks.
Answer:
18 + 2d = 28
2d = 28 – 18 = 10
d = 5
Sequence 13, 18, 23, 28.

Question 42.
98 is a term of the arithmetic sequence having common difference 7. is 2016 a term of this sequence. Why?
Answer:
If (2016 – 98) is a multiple of common difference 7, then 2016 is a term of the arithmetic sequence.
2016 – 98 = 1918
1918 is a multiple of 7.
(Quotient = 274, Remainder = 0)
∴ 2016 is a term of this sequence.

Question 43.
For the arithmetic sequence 6, 12, 18,………
a. What is the common difference?
b. Find the 10th term?
Answer:
a. Common difference = 6
b. 10th term = f + 9d = 6 + (9 × 6)
= 6 + 54 = 60

Question 44.
The algebraic form of an arithmetic sequence is 3 + 2n.
a What is the first form of the sequence?
b. What will be the remainder if the terms of the sequences are divided by 2?
Answer:
a. Firstterm = 3 + 2 x 1 = 5
b. d = 2 (coefficient of n be the common difference)
The remainder divided by 2 = 1

Short Answer Type Questions (Score 3)

Question 45.
The nth term of an arithmetic sequence is an = 5 – 6n. Finditssumofnterms?
Answer:

Question 46.
Consider the multiples of 7 in between 100 and 500.
a. What are the first and last numbers?
b. How many terms are there in the sequence?
Answer:
a. Firstteim=100 – 2 + 7 = 105
Last term = 500 – 3 = 497

Question 47.
For an arithmetic sequence 22, 26, 30, …………..
a. What is the common difference?
b. Will 50 be a term of this sequence? Why?
c. Can the difference between any two terms of this sequence be 50? Justify your answer?
Answer:
a. Common differenced = 26 – 22 = 4
b. \(\frac{50-22}{4}=7\)
So, 50 is a term of this sequence,
c. 50 is not a multiple of 4. So, 50 is cannot be a difference of two terms.

Question 48.
Find the smallest 3 digit number which is the multiple of 6. Find the sum of all the three-digit numbers which are the multiple of six.
Answer:
Smallest 3 digit number which is the multiple of 6 = 102,
Highest =996
Common difference = 6
Arithmetic series : 102, 108, …………….. 996
Number of three digit numbers

Question 49.

Answer:

Question 50.
Find the sum of first 24 terms of the list of numbers whose nth term is given by a_n = 3 + 2n.
Answer:

Long Answer Type Questions (Score 4)

Question 51.
a. Write the arithmetic sequence with first term 2 and common difference 3.
b. Check whether 100 is a term in this sequence.
c. Check whether the difference of any two terms of this sequence will be 2015.
d. Find the position of the term 125 in this sequence.
Answer:
a. f = 2
d. = 3
Sequence is 2, 5, 8, 11, ……….

b. If (100 – 2) is not a multiple of common difference 3, then 100 is not a term of the arithmetic sequence.
c. 2015 is not a multiple of common difference 3, so 2015 will not be the difference of any two terms of this sequence.

d. x_n= 3n – 1
3n-1 = 125
3n= 126
n = 42

Question 52.
When 60 added to the first term of an A.P, we get its 11^th term. Which number should be added to its first term to get the 19^th term? Can 75 be the difference between any of the two terms of this sequence?
Answer:
Differences between first term and 11th term is = 60
(11 – 1 = 10)
10 times of common difference = 60
Common difference = 60/10 = 6
Differences between first term and 19^th term is = 18 x common difference = 18 x 6 = 108 When 108 is added to the first term, we get 19^th
term. The difference between two terms in an A.P is the multiple of common difference. 75 is not the multiple of common difference 6. So 75 cannot be the difference between two terms in the series.

Question 53.
i. What is the sum of first 20 natural numbers?
ii. The algebraic form of an arithmetic sequence is 6n + 5. Find the sum of first 20 terms of this sequence?
Answer:

Question 54.
23rd term of an arithmetic sequence is 32. 35^th term is 104. Then
a. What is the common difference?
b. Which is the middle term of first 35 terms of this sequence?
c. Find the sum of first 35 terms of this sequence.
Answer:

b. Middle term offirst 35 terms = \(\sqrt { 4 } \)
c. 18 th term = X₁₉ = X₂₃ – 5d = 32 -5 × 6 = 32 – 30 = 2
Sum of 35 terms = 18^th term × 35 = 70

Question 55.
The difference between the 15^th term and the 5^th term of an A.P is 40. Which number is to be added to its 12 Answer:
The difference between the 15^th term and the 5^th term = 40
i.e., ten times an of a common difference = 40
Common difference = 40/10 = 4
When we add 8 x common difference we will get 20^th term.
The difference between the 12^th term and the 20^th term = (20 – 12) × common difference
= 8 x 4 = 32
The difference between the first term and the 21^th term = (21 – 1) × Common difference
= 20 × = 80

Long Answer Type Questions (Score 5)

Question 56.
Consider the arithmetic sequence 171, 167, 163,………..
i) Is ‘0’ is a term of this sequence? Why?
ii) How many positive terms are in this sequence?
Answer:
Arithmetic senes : 171, 167, 163 …………….

∴ 0 will not be a term of the sequence

Question 57.

a. How many numbers are there in the 30th row of this number pyramid?
b. Which is the last number in the 30th row?
c. Which is the first number in the 30th row?
d. What is the sum of all terms in the first 30 rows?
Answer:
a. Total numbers in each sequence can be written as 1, 3, 5, …. xn = 2_n – 1
Numbers in the 30th row = 2 x 30 – 1 = 59

b. Last number in the first row = 12 = 1
Last number in the second row = 22 = 4
Last number in the third row = 32 = 9
Last number in thew 30th row = 302 = 900

c. Number of terms in the 30th row = 59
Last number in the 30th row = 900
First number in the 30th row + 58d = Last term in the 30th row
First number in the 30th row + 58 x 1 = 900 First number in the 30th row = 900 – 58 = 842

d. The sum of all terms in the first 30 rows = \(\begin{array}{l}{\frac{900 \times 901}{2}} \\ {=450 \times 901=405540}\end{array}\)

Arithmetic Sequences Memory Map

Students can Download Maths Chapter 10 Polynomials Questions and Answers, Notes PDF, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Maths Chapter 10 Polynomials Questions and Answers Malayalam Medium

SCERT 10th Standard Maths Textbook Chapter 10 Solutions Malayalam Medium

You can Download Periodic Table and Electronic Configuration Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Chemistry Chapter 1 Periodic Table and Electronic Configuration Textbook Questions and Answers

SCERT Class 10th Standard Chemistry Chapter 1 Periodic Table and Electronic Configuration Solutions

Kerala Syllabus 10th Standard Chemistry Chapter 1 Text Book Page No: 7

→ What is the basis of classification of elements in the periodic table?
Answer:
Atomic Number

Sslc Chemistry Chapter 1 Questions And Answers Text Book Page No: 8

→ Atomic number of sodium is 11 Electronic configuration – 2,8,1
GroupNumber — …………..
Period number — …………
Answer:
Group Number — 1
Period number — 3

→ Is the group 1 element a metal or a nonmetal?
Answer:
Metal

→ Write the electronic configuration of sodium and argon and complete the Table.

Answer:

This article mainly deals with the chemical formula of lithium oxide, the structural lithium oxide formula with its properties and uses.

Periodic Table And Electronic Configuration Class 10 Notes Text Book Page No: 9

→ How many electrons are present in the M shell, the outermost shell of argon?
Answer:
8

→ What is the maximum number of electrons that can be accommodated in the M Shell?
Answer:
18

→ The ‘K’ shell, which is the first shell, has 1 subshell. The next ‘L’ shell has 2, and so on. What will be the number of subshells in the ‘M’ shell and ‘N’
M = ……………… , N = ……………….
Answer:
M = 3, N = 4

→ Which subshell is common to all shells?
Answer:
S

Sslc Chemistry Chapter 1 Notes Kerala Syllabus Text Book Page No: 10

→ Complete the Table 1.3

Answer:

→ Complete the Table 1.4

Answer:

→ What is the maximum number of electronics that can be accommodated in the ‘s’?
Answer:
2

→ What may be the maximum number of electrons to be filled in the ‘p’ subshell?
Answer:
6

Periodic Table And Electronic Configuration Class 10 Text Book Page No: 11.

The atomic number of hydrogen is 1(₁H)

→ How many electrons are present?
Answer:
1

→ In which shell is the electron filled?
Answer:
‘K’ shell

→ In which subshell?
Answer:
S

→ How many electrons are present in helium (₂He)?
Answer:
2

Follow along with the alkene reactions cheat sheet.

Chemistry Class 10 Chapter 1 Kerala Syllabus Text Book Page No: 12

→ Complete the subshell electronic configuration?
Answer:
1s²

→ Write the electronic configuration of Lithium (3Li)
Answer:
1s² 2s¹

→ Complete the electronic configuration of beryllium?
Answer:
Be[Z=4] -1s² 2s²

→ Write the electronic configuration of Boron
Answer:
B[Z=5] -1s¹ 2s² 2p¹

→ Write the electronic configuration of Carbon
Answer:
C[Z=6] – 1s² 2s² 2p²

→ Complete the Table 1.6

Answer:

Class 10 Chemistry Chapter 1 Periodic Table And Electronic Configuration Text Book Page No: 13

→ How was the shell wise electronic configuration of potassium written?
Answer:
2, 8, 8, 1

→ Compare the energies of Is and 2s subshells. Which one has lower energy?
Answer:
1s < 2s

Sslc Chemistry Notes Chapter 1 Kerala Syllabus Question 14.
Among the 3s & 3p subshells which has higher energy?
Answer:
3s < 3p

→Among the 3d & 4s subshells which has higher energy?
Answer:
4s < 3d

→ Write down the subshells in the increasing order of their energies.
Answer:
1s <2s <2p <3s <3p <4s <3d <4p

→ Write the subshell wise electronic configu-ration of potassium.
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

→The electronic configuration of scandium (2lSc) is
Answer:
s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s²

Periodic Table And Electronic Configuration Kerala Syllabus Text Book Page No: 14

→ Write the electronic configu ration of 22Ti, 23V, the two elements after Sc.
Answer:
₂₂Ti — 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
₂₃V — 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²

→Which is the noble gas preceding sodium (11Na)?
Answer:
Neon(Ne)

→ Write its subshell electronic configuration.
Answer:
₁₀Ne – 1s² 2s² 2p⁶

→ Subshell electronic configuration of sodium?
Answer:
₁₁Na – 1s² 2s² 2p⁶ 3s¹

Periodic Table And Electronic Configuration Class 10 Important Questions Text Book Page No: 15

→ Using the symbol of neon, write the subshell electronic configuration of sodium?
Answer:
[Ne] 3s¹

→ Complete the Table 1.7

Answer:

→ Write the subshell electronic configuration of 24Cr
Answer:
₂₄Cr – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

→ On the basis of this, identify the correct electronic configuration of 29Cu from those given below:
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s² – False
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ – True

Periodic Table And Electronic Configuration Class 10 Question Paper Text Book Page No: 16

If the subshell wise electronic configuration of an atom is 1s² 2s² 2p⁶ 3s², find answers to the following:

→ How many shells are present in this atom?
Answer:
3

→Which are the subshells of each shell?
Answer:
K — Is, L — 2s, 2p, M — 3s

→Which is the subshell to which the last electron was added?
Answer:
3s

→ What is the total number of electrons in the atom?
Answer:
12

→ What is its atomic number?
Answer:
12

→ How can the subshell electronic configuration be written in a short form?
Answer:
[Ne]3s²

Periodic Table And Electronic Configuration Class 10 Questions And Answers Text Book Page No: 17

→ Complete the Table 1.8

Answer:

→ Which is the subshell of lithium to which the last electron was added?
Answer:
S

→ What about the subshell to which the last electron of nitrogen was added
Answer:
p

→ What is the relation between the subshell to which the last electron was added and the block to which the element belongs?
Answer:
The subshell in which the last electron enters represent the block in which the element belongs.

→ Write the subshell electronic configuration of the following elements and find the blocks to which they belong.
a. 4Be: ………………..
b. 26Fe……………..
c. 18Ar: ……………
Answer:
a. 4Be : 1s² 2s² — s block
b. 26Fe : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² — d block
c. 18Ar : 1s² 2s² 2p⁶ 3s² 3p⁶ — p block

Text Book Page No: 18

→ Complete the Table 1.9

Answer:

→ Complete the Table 1.10

Answer:

→ What is the relation between number of electrons present in the last ‘s’ subshell and their group number?
Answer:
The number of electrons in the outermost ‘s’
subshell = The group number

Text Book Page No: 20

→ When the s block elements react, do they donate or accept electrons?
Answer:
They donate electrons.

→ Which type of chemical bond is usually formed?
Answer:
Ionic bonds

→ How many electrons are donated by the first group elements in chemical reactions ?-
Answer:
One

→ How many electrons are donated by the second group elements in chemical reaction?
Answer:
Two

→ Complete the table 1.11

Answer:

→ ‘s’ block elements are present at the extreme left side of the periodic table. Relating to their position, what other characteristics can be listed out?
Answer:

• More metallic character s
• Less ionization energy
• Less electronegativity
• Lose of electrons in chemical reaction
• Compounds are mostly ionic
• Oxides and hydroxides are basic in nature

Text Book Page No: 21

→ Which are the group included in the p block
Answer:
13, 14, 15, 16, 17, 18

→ In which subshell did the filling of the last electron take place?
Answer:
p subshell

→ Complete the table 1.13.

Answer:

Text Book Page No: 22

The outermost subshell wise electronic configuration of an element Y (Symbol is hot real) is 3s2 3p4.

→ To which period and group does this element belongs to?
Answer:
Period = 3, Group = 16

→ Write down the outermost subshell electronic configuration of the element coming just below it in the same group?
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁴

→ Find out examples of elements in such different states with the help of the periodic tables?
Answer:
Solid – Li, Be, B, C, Na, Mg, Al, Si
Liquid – Br
Gas – H, He, N, O, F, Ne

→ Which element has the highest ionization energy in each period?
Answer:
Group 18 elements.

Text Book Page No: 23

→ The elements having the highest electronegativity is in the p block. Find its name and position?
Answer:
Fluorine F, Period – 2, p block, Group 17

→ Analyze the general characteristics of the p block elements and prepare a note on this?
Answer:

• The outermost p subshell of the p block elements contains 1 to 6 electrons.
• Elements showing positive oxidation state and negative oxidation state are members of this block.
• There are metals and nonmetals in these blocks.
• Elements in the solid, liquid and gaseous states are present in p block.

→ Complete the table 1.14

Answer:

→ Which element has a valency 1?
Answer:
Y

→Which element shows metallic character?
Answer:
X

→ Which element has the highest ionization energy?
Answer:
Y

→ Write the chemical formula of the compound formed by the combination of X and Y and label the oxidation states?
Answer:
Compound: X Y₂
Oxidation state: X²⁺, Y^1-

→ Where is the position of d block elements in the periodic table?
Answer:
3rd Group to 12th Group

→ From which period onwards does the d block begin?
Answer:
4

Text Book Page No: 24

→ Put a tick mark ✓’ against the statements below, which are applicable to d block elements.
Answer:
1. ‘✓’ These are metals.
2. ‘✓’ The last electron is filled in the penultimate shell.
3. ‘✗’ In the case of these elements in the 4th period, the last electron is filled in 4s.
4. ‘✓’ These are found in groups 3 to 12 of the periodic table.

Text Book Page No: 25

→ Complete the table 1.16

Answer:

→ How does Fe change to Fe²⁺?
Answer:
By losing 2 electrons from 4s valence subshell.

Text Book Page No: 26

→ Write down the subshell electronic configuration of Fe21.
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶

There is only a small difference of energy between the outermost s subshell and the penultimate d subshell of transition elements.

→ If so, which will be the subshell from which iron loses the third electron?
Answer:
From 3d sub-shell

→ Write the electronic configuration of Fe3+ on the basis of this.
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

→ Write the subshell electronic configuration of Manganese (Mn).
Answer:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²

→ Complete the table 1.17

Answer:

Text Book Page No: 27

→ Examine these compounds available. Find more colored compounds and extend the list.
Answer:

• Copper sulfate CuSO₄.5H₂O – blue,
• Copper nitrate Cu(NO₃)₂.6H₂O – pink.
• Potassium permanganate KMnO₄ – violet.
• Ferrous sulfate FeSO₄.7H₂O – Green,
• Ferrous nitrate (Fe(NO₃)₂.6H₂O) – light green

Text Book Page No: 28

→ List out the dements of the s block?
Answer:
A, B

→ Which elements shows+2 oxidation state?
Answer:
B, C, D

→ Which elements contains 5 electrons in the outermost shell?
Answer:
E

→ Which is the element that has 5 p electrons in the outermost shell?
Answer:
G

→ Which are the elements in which the last electron enters the d subshell?
Answer:
C, D

→ Which element has the highest ionization energy?
Answer:
H

→ Which is the highly reactive nonmetal?
Answer:
G

→ Which elements show -2 oxidation state?
Answer:
F

Text Book Page No: 29

The outermost electron configuration of an element in this is 2s² 2p⁶

→ Which is the element?
Answer:
H

→ Write down the complete subshell electronic configuration?
Answer:
Is² 2s² 2p⁶

→ Write any two characteristics of this element?
Answer:

• Noble element / gases.
• The outermost shell is completely filled

→ Write the chemical number of questions, the answer of which is an element in the table
Answer:
A G

Periodic Table and Electronic Configuration Let Us Assess

Question 1.
Based on the hints given, find out the atomic number and write down the subshell electronic configuration of elements (Symbols used are not real).
i. A – period 3 group 17
ii. B – period 4 group 6
Answer:
A₁₇ — 1s² 2s² 2p⁶ 3s² 3p⁵
B₂₄ — 1S² 2s² 2p⁶ 3S² 3p⁶ 3d⁵ 4S¹

Question 2.
When the last electron of an atom was filled in the 3d subshell, the subshell electronic configuration was recorded as 3d8 Answer the questions related to this atom.
1. Complete subshell electronic configuration
2. Atomic number
3. Block
4. Period number
5. Group number
Answer:
1. Complete subshell electronic configuration:
1 s², 2s², 2p⁶, 3s², 3p⁶, 3d⁸, 4s²
2. Atomic number: 28
3. Block : d
4. Period number: 4
5. Groupnumber : 8 + 2 = 10

Question 3.
Pick out the wrong ones from the subshell electronic configuration given below.
a. 1s² 2s² 2p⁷
b. 1s² 2s² 2p²
c. 1s² 2s² 2p⁵ 3s²
d. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
e. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
Answer:
Wrong electronic configuration
a. 1s², 2s², 2p⁷
(2p maximum 6 electrons only)

c. 1s², 2s², 2p⁵, 3s¹ (electrons are filled in 3s only after filling 6 electrons in 2p)

d. 1s², 2s², 2p⁶, 3s², 3p⁶, 3d², 4s¹ (electrons are filled in 3d only after filling 2 electrons in 4s)

Question 4.
The element X in group 17 has 3 shells. If so,
a. Write the subshell electronic configuration of the element.
b. Write the period number,
c. What will be the chemical formula of the compound formed if the element X reacts with element Y of the third period which contains one electron in the p subshell?
Answer:
a. Three shells are K, L, M. The subshells are 1s, 2s, 2p, 3s, 3p, 3d
Group number: 17
Electrons in last shell: 7
Shell electronic configuration: 2,8,7
Sub-shell electronic configuration 1s², 2s², 2p⁶, 3s², 3p⁵

b. Period-3

c. Y – Third period
∴ shells – 3
1 electron in p – subshell
Total electrons in valence shell 2+1=3 (2 electrons in s + 1 electron in p)
Valency of x – 1(1 electron is recieved – electro negative atom)
Valency of y – 3 (3 electrons are lost – electro positive atom)
Therefore they combine to form compounds with chemical formula YX3
(Symbol of electropositive element first followed by electro negative element).

Question 5.
The element Cu with atomic number 29 undergoes chemical reaction to formation with oxidation number +2.
a. Write down the subshell electronic configuration of this ion.
b. Can this element show variable valency? Why?
c. Write down the chemical formula of one compound formed when this element reacts with chlorine (17CI).
Answer:
a. ₂₉Cu — 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s¹
Cu²⁺ — 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁹

b. Yes. One electron can be lost from 4s subshell and can’exist as Cu⁺ ion, It is a d-block element.

c. Copper react with chlorine to form two compounds Cu⁺, Cu²⁺ ions react with chlorine to form CuCl and CuCl2 respectively.

Question 6.
Certain subshells of an atom are given below. 2s, 2d, 3f, 3d, 5s, 3p
a. Which are the subshells that are not possible?
b. Give the reason.
Answer:
a. Not possible sub-shells are 2d, 3f

b. d – subshell is not possible in 2^nd shell
f – subshell is not possible in 3^rd shell

Periodic Table and Electronic Configuration Extended Activities

Question 1.
Prepare the comprehensive table which indicates the name, symbol, electron configuration, subshell configuration of elements having atomic number 1 to 36?
Answer:

Question 2.
Some information related to the elements of the p bllock in the 17th group of the periodic table are given in the table below. Complete the table and analyze the following questions?
Answer:

a. What is the family names of elements belonging to the 17^th group?
Answer:
Halogen

b. What is their common valency?
Answer:
1

c. Which element has the highest electro negativity ?
Answer:
F

d. Which element has the highest ionization energy?
Answer:
F

e. List out the name and chemical formula of the compounds formed by these elements with block elements?
Answer:

• sodium chloride – NaCl
• potassium chloride – KCl
• magnesium chloride- MgCl₂
• calcium chloride- CaCl₂
• magnesium fluoride- MgF₂
• calcium fluoride – Ca F₂
• sodium iodide – Nal
• potassium iodide – KI
• potassium bromide – KBr
• potassium fluoride – KF

Periodic Table and Electronic Configuration Orukkam Questions and Answers

Question 1.
Complete the table of details about shells and subshells.

a. No of electrons in KLMN shell.
b. No of electrons in each shell.

c. Which subshell is common to all sub-shells?
d. Write names of subshells in accordance with increasing energy level,
e. Identify the incorrect subshell electronic configuration.
– 1s²
– ls² 2p⁶
– ls² 2s² 2p⁶
– 1s² 2s² 2p⁶ 3s² 3p²
Answer:

a, K – 2 ; L – 8 ; M – 18 ; N – 32

c. s- Subshell

d. 1s <2s <2p <3s <3p <4s <3d <4p <5s <4d <5p <6s <4f <5s

e. 1s² 2p⁶

Question 2.
Atomic number of iron is 26. It exhibits Fe²⁺, Fe³⁺ oxidation state. Write the subshell electronic configuration.

Answer:

Question 3.
Manganese, a d-block element exhibits I different oxidation state. Why?
a. Include chemical formulae of more compounds of manganese in the table, write their ; oxidation state and subshell electronic configuration.

b. Write the oxidation number and subshell electronic configuration K, Cl and O.
Answer:
Manganese shows different oxidation states because in manganese 4s and 3d subshell electrons take part in chemical reactions.

Question 4.
Find out atomic number, group, block period using subshell electronic configuration and then complete the table.

Answer:

Question 5.
Write down the characteristics of s,d,p, f block elements
Answer:
s-block elements:
Elements in which last electron enters into s-subshell are called s-block elements. It contains group I elements (Alkali metals) and group II elements (Alkaline earth metals).

1st group elements lose one electron during chemical combination. Therefore its oxidation state is +1.

2nd group elements lose two electrons from valence shell during chemical combination and their oxidation state is +2.

The highest shell number in a sub-shell electronic configuration is the period number of that element.

1. Group number characteristics = no.of electrons in valence sub-shell.
2. s block ionization energy & electro negativity decreases downwards.
3. Metallic character & reactivity increases downwards.
4. Lose electrons during chemical combination j and they form ionic compounds.
Their oxides and hydroxides are basic.
Their atomic radii are high in a period.

p-block elements:

• Last electron enters into p-subshell.
• Group 13 -18 elements.
• Highly reactive elements are non-metals – group 17,
• These are elements with positive and negative oxidation state.

Group number of p-block elements = electrons in last p-subshell + 12

d-block elements:

• Last electron enters into penultimate d-subshell
• Known as transition elements.
• Metals
• Shows similarity in group and period.
• Variable oxidation states.
• Form coloured compounds.

Group = electrons in ‘d’-subshdl + electrons in s-subshell.

f-block elements:

• Last electron enters into antepenultimate f sub-shell.
• Contains Lanthanoids and Actinoids.
• Variable oxidation state.
• Most of the Actinoids are radioactive.
• Most of the elements are artificial.
• U, Th, Pu are used in nuclear reactors.
• Some elements are used as catalyst in pet-roleum industry.

Periodic Table and Electronic Configuration Evaluation Questions

Question 1.
Write down subshell electronic configuration of Cu¹⁺ and Cu²⁺
Answer:
Cu¹⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰
Cu²⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹

Question 2.
How many ‘s’ subshell electrons are in 1s², 2s², 2p⁶, 3s², 3p²
Answer:
6 Electrons

Question 3.
11,17,10 are the atomic number of elements X, Y, and Z.
a. Write down their subshell electronic configuration, group, block, period,
b. Write the molecular formulae of the compound formed when any two of the above elements are combined.
c. Write down the oxidation numbers of the elements in those compounds. Write the subshell electronic configuration of both ions.
Answer:

b. X Y

Question 4.
Element ‘X’ is having atomic number 28, it gives two electrons to element ‘Y’.
a Write down the electronic configuration of ‘X’ and its ion
b. In which block ‘X’ belongs?
c. Write down the characteristics of that block
Answer:
a. X₂₈ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²
X²⁺– 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸

b. d block Compound

c. 1. It exhibits variable oxidation states
2. Forms colored compounds
3. Last electron enters d subshell

Question 5.

a. Write down the group and period of each element.
b. What are the use of writing electronic configuration this fashion?
Answer:

b. Group and period of the element can be identified easily. In the Same way long electron configuration can be avoided.

Question 6.
₂₄Cr – [Ar] 3d⁵ 4s¹
₂₉Cu – [Ar] 3d¹⁰ 4s¹
Why chromium and copper exhibits such electronic configuration ?
Answer:
Half filled and completely filled subshells are most stable. Change in the electronic configuration of ₂₄Cr &, ₂₉Cu is due to .this. The electrons in these elements are arranged in such away to give these elements stability.

Periodic Table and Electronic Configuration SCERT Questions and Answers

Question 1.
The electronic configuration of the elements A, B, C, Dare given below.
A – 1s² 2s² 2p⁶ 3s² 3p⁴
B – 1s² 2s² 2p⁶ 3s²
C – 1s² 2s² 2p⁶ 3s² 3p⁵
D – 1s² 2s² 2p⁶ 3s¹
a. Which of these elements show +2 oxidation state?
b. Which metal belongs to 17^th group?
c. Which is the period number of the element A ? What is the basis of your findings?
d. Which of these elements can form basic Oxides?
Answer:
a. B

b. C

c. Period number: 3, Period number = No.of shells

d. B, D

Question 2.
Two compounds of iron are jpven below.
FeSO₄ Fe₂(SO₄)₃
(The oxidation state of sulfate radical is-2)
a. Which ofthese compounds show +2 oxidation state for Fe?
b. Which compounds has Fe³⁺ ion?
c. Write the subshell electronic configuration of Fe³⁺ ion.
d. Why do transition elements show variable oxidation states?
Answer:
a., FeSO₄
b. Fe₂(SO₄)₃
c. Fe³⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

d. The energy difference between the outer most ‘s’ subshell and penultinate ‘d’ subshell is very small. Hence under suitable conditions, the electrons in ‘d’ subshell also take part in chemical reaction.

Question 3.
Identify the incorrect electronic configurations and correct them.
i) 1s² 2s² 2p³
ii) 1s² 2s² 2p⁶ 3s¹
iii) 1s² 2s² 2p⁶ 2d⁷
iv) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴
Answer:
iii). 1s² 2s² 2p⁶ 3s² 3p⁵ .
iv). 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²

Question 4.
Complete the table.

Answer:
a. – 2
b. 1
c. 17,
d. – 1
e. 12
f. +12

Question 5.
a. Two compounds XY₂, XZ₄ are given. The oxidation state of Z is 1. What will be the oxidation state of Y ?
b. Write the molecular formula of the compound formed by Y when it combines with aluminum (Al) having oxidation state +3.
Answer:
a. Y= – 2 (oxidation state of X is +4)
b. Al₂ Y₃

Question 6.
Pick out the statements which suit to f-block elements.
a. All of them are naturally occurring elements.
b. Uranium and Thorium are f block elements.
c. Last electrons is filled in the shell pre-ceding the outermost shell.
d. last electrons are filled up in the antepenultimate shell.
e. Includes some radioactive elements.
f. Some of them are used as catalyst in petroleum industry.
Answer:
b, d, e, f

Question 7.
The atomic number of four elements are given below. (The symbols ore not real)
A – 8
B – 10
C – 12
D – 18
a. Write the sub-shell electronic configuration of the elements,
b. Which of them are inert gases?
c. Write the chemical formula of the compound formed by two elements other than inert gases.
Answer:
a. A – 1s² 2s² 2p⁴
B – 1s² 2s² 2p⁶
C – 1s² 2s² 2p⁶ 3s²
D – 1s² 2s² 2p⁶ 3s² 3p⁶

b. B, D

c. CA, (C₂ A₂ is simplified and written as CA)

Question 8.
The subshell electronic configuration of two elements ends as follows. (Symbols are not real)
P – 3s² Q – 3p⁴
a. Write the complete subshell electronic configuration.
b. Find out the oxidation state of each element.
c. The chemical formula of the compound formed by these elements is PQ. Is this statement correct? Justify your answer.
Answer:
a. P – 1s² 2s² 2p⁶ 3s²
Q – 1s² 2s² 2p⁶ 3s² 3p⁴

b. P = +2, Q = – 2 :

c. Right, valency of both P and Q is ‘2’

Question 9.
Match the following.

Answer:

Question 10.
The atomic number of two elements are given below.
Si – 14 Ni – 28
a. Write the subshell electronic configu-ration of these elements.
b. Find out the group and period of each element.
Answer:
a. Si – 1s² 2s² 2p⁶ 3s² 3p²
Ni – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²
b. Si – Period Number – 3, Group number – 14 Ni – Period Number – 4, Group number – 10

Question 11.
The element ‘X’ has 4 shells and its 3d subshell has 6 electrons. (Symbol is not real)
a. Write the complete electronic configu-ration of the element.
b. What is its group number? Which is the block?
c. Write any two characteristics of the block to which element X belongs to.
d. From which subshell the electrons are lost when the element X shows +2 oxidation state.
Answer:
a. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²
b. Group number – 8, Block – d
c. All of them are metals
d – block elements are placed in group 3 to group 12
d. s – Sub shell

Question 12.
The outermost electronic configuration of the element A is 2s² 2p². (Symbol is not real)
a. Find out the group number and block of the element.
b. Write the chemical formula of the compound formed by A when it combines with chlorine.
c. Write the complete electronic configuretion of the element just below ‘A’ in the j periodic table.
Answer:
a. Group number – 14, Block – P
b. ACl₄
c. 1s² 2s² 2p⁶ 3s² 3p²

Question 13.
The figure of an incomplete periodic table is given below.

a. Which one of these elements shows -2 oxidation state?
b. Which of these elements have 3 electrons in their outermost p subshell?
c. Which element has the highest atomic radius? Which one has the least?
d. Which of these elements show variable oxidation state?
e. Which of these elements has the highest ionization energy?
Answer:
a. G

b.F

c. The element having highest atomic radius – A
The element having lowest Atomic radius – H

d. D, C

e. H

Question 14.
Examine the given electronic configurations.
A – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s²
B – 1s² 2s² 2p⁶ 3s¹
C – 1s² 2s² 2p¹ 3s² 3p⁶
D – 1s² 2s² 2p⁶ 3s²
E – 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
a. Which of these elements belongs to 4th period?
b. Which elements belongs to the same group ?
c. Which element doesn’t participate in chemical reactions generally ?
d. Which element has highest metallic character ?
Answer:
a. A, E
b. B, E
c. C
d. E

Question 15.
The atomic number of the elements X and Y are 20, 26 respectively. When these elements combine with chlorine, three compounds XCl₂, YCl₂, YCl₃ are formed.
a. What is the specialty of the oxidation number of Y, compared to that of X?
b. Explain the reason for this, on the basis of the subshell based electronic configuration.
Answer:
a. Element X has constant oxidation state. Y shows variable oxidation states.
b. X₂₀ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s²
X₂₆ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁶
Y is a transitional element. In chemical reactions only two elections in ‘ s’ subshell or besides ‘s’ subshell electrons ‘d’ sub shell electrons also take part.

Periodic Table and Electronic Configuration Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
Arrange the following sub-shells in the in-creasing order of energy 5p, 2s, 4f, 3s, 4s, 3d, 6s
Answer:
2s < 3s < 4s < 3d < 5p < 6s < 4f

Question 2.
Last electron in f-block elements goes to
a. Which shell? Outer shell/Penultimate shell /Antepenultimate shell
b. Which sub-shell? Outer f-subshell Penultimate f-subshell/Antepenulti mate f-subshell.
Answer:
a. Antepenultimate shell
b. Antepenultimate f-sub-shell

Question 3.
Sub-shell electronic configuration of X is given below.
1s², 2s², 2p⁵
a. The element Y is coming just below the element in same group. Then write the sub-shell electronic configuration of Y.
b. Write the sub-shell electronic configuration of the element next to X in same period.
Answer:
a. 1s², 2s², 2p⁶, 3s², 3p²
b. Is², 2s², 2p⁶

Question 4.
A compound of vanadium pentoxide (V20;) is used as catalyst.
a. What is the oxidation state of vanadium in this compound?
b. How vanadium ion is represented?
c. Write the sub-shell electronic configuration of this ion (V – 23)
Answer:
a. +5
b. V⁵⁺
c. 1s², 2s², 2p⁶, 3s², 3p⁶

Short Answer Type Questions (Score 2)

Question 5.
Find the wrong electronic configurations from the following. What is wrong in these?
a. 1s²,2s²,2p⁶,3s²,3p⁶,3d⁹,4s²
b. 1s²
c. 1s², 2s¹, 2p⁶
d. 1s², 2s², 2p⁶, 3s², 3p²
Answer:
(a) and (c) are wrong electronic con figurations.

In (a) one electron from 4s is to be trans-ferred to 3d since completely filled configu-rations are more stable. So the correct electronic configuration is 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s¹

In (c) electrons are filled in 2p only after filling electrons in 2s.

Question 6.
Group and period number of two elements are given.
P – group 17, period – 3
Q – group 2, period – 3
a. Write the sub-shell electronic configuration of each.
b Write the chemical formula of the compound formed by their combination.
Answer:
Answer:
a. P – 1s², 2s², 2p⁶, 3s², 3p⁵
Q – 1s², 2s², 2p⁶, 3s²

b. Q is electropositive. P is electro negative;
∴Chemical formula QP2

Question 7.
Write the reason for the statement given
below.
a. d-block elements in the same period show similarity.
b. Transition elements show variable oxidation state.
Answer:
a. Valence shell electrons of d-block elements in same periods are almost same. Valence shell electrons are entering in chemical reaction. Therefore they shows similarity.

b. Energy of electrons in s-subshell and inner d- subshells are almost same. Therefore s- electrons or s and d electrons take part in chemical reaction and show variable oxidation state.

Short Answer Type Questions (Score 3)

Question 8.
Write the sub-shell electronic configuration of following elements. Predict the block, group and period. (Symbols are not real)
a. M – 27 b. N – 19 c. P – 15
Answer:
a. 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁷, 4s²
block – d; group – 9; period – 4.
b. 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹
block – s; group – 1; period – 4
c. 1s², 2s², 2p⁶, 3s², 3p³
block – p; group – 15; period – 3

Question 9.
Observe the model of periodic table.

a. Which element is having S electrons in valence shell?
b. Which elements are having 2 electrons in valence sub-shell?
c. Which element is having last electron in3p?
d. Which element ends with electronic configuration 4d⁵, 5s¹ ?
Answer:
a. B;
b.A, C;
c. C, D;
d. E

Question 10.
Calculate oxidation state of transition elements in the following compounds.
Answer:
a. KMnO4 – Mn – 7⁺
b. Cr₂ O₃ – Cr – 3⁺
c. K₂Cr₂O₇ – Cr – 6⁺

Question 11.
Atomic number of some elements are given. A – 15, B – 8, C – 11, D – 18, E – 20, F – 34, G – 10
a. Which are the elements in same period?
b. Which are the elements in same group?
Answer:
A -1s², 2s², 2p⁶, 3s², 3p³
(group -15 period – 3)
B – 1s², 2s², 2p⁴ (group –16 period – 2)
C – 1s², 2s², 2p⁶, 31 (group – 1 period – 3)
D – 1s², 2s², 2p⁶, 3s², 3p⁶ (group – 18 period – 3)
E – 1s², 2s², 2p⁶, 3s², 3p⁶, 4s².(group – 2 period – 4)
F – 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁴ (group – 16 period – 4)
G – 1s², 2s², 2p⁶ (group – 18 period – 2)
a. A, D same period ;
B, G same period 1
b. B, F same group ;
D, G same group

Long Answer Type Questions (Score 4)

Question 12.
Electronic configuration of some elements are given. Write answers to the following questions.
i. [Ne] 3s²
ii. [Ar ] 3d²,4s²
iii. [Xe] 6s²
iv. [Ne]3s²
v [Ne] 3s²,3p⁵
a. Which metal is having high reactivity?
b. Which is having possibility of formation of colored compounds?
c. Which is the non-metal?
d Which element shows the possibility of +2 oxidation state?
Answer:
a. [Xe] 6s¹
b. [Ar ] 3d², 4s²
c. [Ne] 3s², 3p⁵
d. [Ne] 3s²,[Ar] 3d², 4s²

Question 13.
Pick the wrong statement from the following.
a. Elements with atomic number 5 belong to group 15.
b. Electronic configuration of scandium (Atomic number 21) is 2,8,8,3.
c. d-block elements are known as transition elements.
d. All s-block elements are metals.
Answer:
a. Wrong. It belongs to group 13.
b. Wrong. Electronic configuration
2,8,9,2 (1s², 2s², 2p⁶, 3s², 3p⁶,3d¹, 4s²)
c. Correct .
d. Correct

Question 14.
Look at the Bohr model of X-atom.

a. Write the sub-shell electronic configuration of this atom.
b. Mention the compounds in which d-subshell electrons are taking part in chemical reaction during their formation.
XCl₂, XO₂, X₂O₇
c. Write the electronic configuration of X ions in the above three compounds.
Answer:
a. 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵, 4s²

b. XCl₂ – ion X²⁺ (electrons in 4s only)
XO₂ – ion X+⁴ (2 electrons in 4s and 2 electrons in 3d)
X₂O₇ – ion X⁷⁺ (2 electrons in 4s and 5 electrons in 3d)
XO₂, X₂O₇ d – subshell electrons are taking part in chemical reaction during the formation of X207

c. X²⁺ – 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵
X⁴⁺ – 1s², 2s², 2p⁶, 3s², 3p⁶, 3d³
X⁷⁺ – 1s², 2s², 2p⁶, 3s², 3p⁶

Question 15.
Select the suitable one from the following columns A, B, C.

Answer:

Students can Download Maths Chapter 3 Mathematics of Chance Questions and Answers, Notes PDF, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Maths Chapter 3 Mathematics of Chance Questions and Answers Malayalam Medium

SCERT 10th Standard Maths Textbook Chapter 3 Solutions Malayalam Medium

Students can Download Maths Chapter 2 Circles Questions and Answers, Notes PDF, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Maths Chapter 2 Circles Questions and Answers Malayalam Medium

SCERT 10th Standard Maths Textbook Chapter 2 Solutions Malayalam Medium

Students can Download Maths Chapter 11 Statistics Questions and Answers, Notes PDF, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Maths Chapter 11 Statistics Questions and Answers Malayalam Medium

SCERT 10th Standard Maths Textbook Chapter 11 Solutions Malayalam Medium

Students can Download Physics Chapter 3 Electromagnetic Induction Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Physics Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Physics Chapter 3 Electromagnetic Induction Questions and Answers Malayalam Medium

SCERT 10th Standard Physics Textbook Chapter 3 Solutions Malayalam Medium

You can Download Production of Metals Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Chemistry Chapter 4 Production of Metals Textbook Questions and Answers

SCERT Class 10th Standard Chemistry Chapter 4 Production of Metals Solutions

→ All ores are minerals, but are all minerals ores?
Answer:
NO

→ Which metal’s ore is calamine?
Answer:
Zinc (Zn)

→ Which is the ore of aluminum?
Answer:
Bauxite (Al₂O₂ 2H₂O)

→ Which metals have sulfide ores?
Answer:
Copper, Zinc

Production Of Metals Class 10 Kerala Syllabus Text Book Page No: 65

→ Complete the table 4.2.

Answer:

→ Write suitable method of concentration for the ores given in the Table (4.3)

Answer:

Sslc Chemistry Chapter 4 Kerala Syllabus Text Book Page No: 67

→ Complete The table 4.4

Answer:

→ Haematite, magnetite, iron pyrites, etc. are the minerals of iron. Which are the ores of iron among these minerals?
Answer:
Haematite, magnetite.

Chemistry Class 10 Chapter 4 Kerala Syllabus Text Book Page No: 69

→ Complete the table 4.5.

Answer:

→ Which alloy steel is used for the production of heating coils? Explain the reason.
Answer:
Nichrome, because of high resistance, easily it becomes hot.

→ Even though nichrome and stainless steel contain the same components they possess different properties. Find out the reason.
Answer:
The ratio of the component elements are different.

→ Which alloy steel is used for making permanent magnets?
Answer:
Alnico

Sslc Chemistry Chapter 4 Notes Kerala Syllabus Text Book Page No: 70

→ Complete the table 4.7

Answer:

→ How ¡s alumina obtained from this aluminum hydroxide?
Answer:
The precipitate is separated, washed and then heated strongly to get alumina.

Production Of Metals Class 10 Notes Kerala Syllabus Text Book Page No: 71

→ Complete the flow diagram, related to concentration of bauxite, which is given below.
Answer:

→ Complete the chemical equation for the reaction taking place when Aluminium hydroxide is heated.
Ans. 2Al(OH)₃ → Al₂O₃ + 3H₂O

→ Which method can be used for separating aluminum from alumina?
Answer:
Electrolysis

→ Can we use carbon as the reducing agent? Why?
Answer:
Can’t use carbon as the reducing agent, because the reactivity of aluminum is very high hence they require very strong reducing agent.

Sslc Chemistry Chapter 4 Notes Pdf Kerala Syllabus Text Book Page No: 72

→ To which electrode does Al3+ move?
Answer:
Towards negative electrode (Cathode)

→ What about oxide ion?
Answer:
Towards positive electrode (Anode)

→ Complete the table related to the electrolysis of Alumina.
Answer:

Production of Metals Let Us Assess

Sslc Chemistry Chapter 4 Notes English Medium Question 1.
Which of the properties of metals is utilized in the following instances?
a. Aluminum utensils are used for cooking.
b. Copper is used for making vessels.
c. Gold wires are used in ornaments.
Answer:
a. Heat conductivity, lightweight, can be molded in any shape, low price, etc.
b. Malleability
c. Ductility

Sslc Chemistry 4th Chapter Kerala Syllabus Question 2.
What are the factors to be considered while selecting minerals for the extraction of metals?
Answer:
High availability
Extraction of metal should be easy
The percentage of metal content in the mineral should be comparatively high.
Cost of production should below.

Kerala Syllabus 10th Standard Chemistry Chapter 4 Question 3.
Write the different stages involved in metallurgy.
Answer:
a. Concentration of the ore methods are:

• Levigation/hydraulic washing
• Froth floatation
• Magnetic separation
• Leaching

b. Extraction of metal from the concentrated ore has 2 stages.

• Conversion of ore into its oxide. The different methods are Calcination, Roasting etc.
• Reduction of oxidised ore using suit-‘ able reducing agents.

c. Refining of metals. Different methods are:

• Liquation
• Distillation
• Electrolytic refining

Production Of Metals Class 10 Notes Pdf Kerala Syllabus Question 4.
What are the different methods for the refining of metals?
Answer:

• Liquation
• Distillation
• Electrolytic refining

Chemistry Chapter 4 Class 10 Kerala Syllabus Question 5.
How is iron extracted industrially?
Answer:
The ore is crushed into small lumps and then wash out the soluble impurities in running water. Then it undergoes roasting. Thus impurities like sulfur, arsenic, moisture, etc. get removed. The roasted ore is mixed with coke and limestone and the mixture is changed into a blast furnace.
Reaction in blast furnace:

At high temperature, CaCO₃ (limestone) decomposes to form CaO.
CaCO₂ (s) → CaO_(s) + CO₂ (g)

This CaO combines with acidic gangue, SiO₂ forming slag.
CaO_(s)+ SiO₂(g) → CaSiO₃ (s)

Coke reacts with oxygen and form CO₂
C_(s) + O₂(g) → CO₂(g)

CO₂ combines with more carbon and produces CO.
CO₂(g) + C_2(s) + Heat → 2 CO (g)
This CO acts as reducing agent.

Fe₂O₃ is reduced to form Fe.
Fe₂O_3(s)+ 3 CO(g) → 2Fe_(s)+ 3CO₂(g)

Hss Live Guru 10th Chemistry Kerala Syllabus Question 6.
Write the uses of the following:
a. Nichrome
b. Stainless steel
c. Alnico
Answer:
a. Nichrome – For making heating coils
b. Stainless steel – For the manufacture of utensils, parts of vehicles
c. Alnico-To make permanent magnets

Kerala Syllabus 10th Standard Chemistry Guide Question 7.
Explain the process of producing alumina from bauxite.
Answer:
Bauxite is treated with hot concentrated NaOH. Aluminum oxide reacts with NaOH solution forming sodium aluminate solution (NaAlO₂). The unreacted impurities are removed from the solution. To the remaining solution of sodium aluminate, add Al(OH)₃ in small quantity and dilute with water. Then the whole aluminum in the solution gets precipitated as Al(OH)₃. The Al(OH)3 precipitated is separated from the solution. Then it is washed and heated strongly. Then Al(OH)₃ decomposes to give pure AL₂O₃ or alumina.

2Al(OH)_3(s)→ Al₂O_3(s) + 3H₂O(1).

Hsslive Chemistry Class 10 Kerala Syllabus Question 8.
Explain the method of obtaining pure aluminum from alumina by electrolysis. In this process, the carbon rods are replaced from time to time. Why?
Answer:
Pure Al₂O₃ mixed with cryolite undergo electrolysis.

At cathode :
Al^₂₊ + 3e^– → Al
At anode :
2O₂ → O₂ + 4e^–
Carbon at anode reacts with O₂and form CO₂. Thus anode gets used up. So anode is replaced frequently.

Production of Metals Extended Activities

Hss Live Guru Class 10 Chemistry Kerala Syllabus Question 1.
You know that metals can be separated from molten compounds of metals by electrolysis. Find out how metals like Na, Ca and Mg are extracted.
Answer:
Na
Produced by the electrolysis of molten NaCl. KCl is added to reduce the melting point of NaCl.

Chemical equation:
NaCl → Na++Cl-
At cathode : Na⁺+ le^– → Na
At anode : 2Cl^– → Cl₂ + 2e^–

Ca:
Mixture of CaCl₂ and 16% CaF₂ are melted and undergo electrolysis.

Mg:
By the electrolysis of molten mixture of camalite (KCl.MgCl₂.6H₂O) and NaCl in equal amounts.

Production of Metals Orukkam Questions and Answers

Hss Live Guru 10th Chemistry Malayalam Medium Question 1.
Complete the table

Answer:

Chemistry Solutions Class 10 Kerala Syllabus Question2.
Features of ore and impurity are given in the table. Write down the method used for the separation of the ore.

a. How can we convert ore into its oxide form? Explain with proper examples
b. ZnCO₃/Cu₂S in these two calcination is used for ………. and Roasting is used for ………..
c. Give Examples for reducing agents for reducing oxide ores.
d. Strongest reduction agent
e. Which reducing agent used fur reducing ZnO, Fe₂O₃, Al₂O₃ ?
Answer:
i. Levigation/Hydraulic washing
ii. Magnetic Separation
iii. Froth Floatation
iv. Leaching

a. Calcination
Calcination is the process of heating the concentrated ore at a temperature below its melting point to remove the volatile impurities. When subjected to calcination, impurities like water, organic matter, and other volatile impurities are expelled from the ore. Metal Carbonate and hydroxides decompose to form oxides,
eg: ZnCO is converted to ZnO by calcination

Roasting:
Roasting is the process of heating the concentrated ore at a temperature below its melting point in a current of air. During roasting the ore gets converted into its oxide. When the concentrated ore is subjected to roasting, the water present in it is removed as vapor. Other impurities like sulfur, phosphorus and organic matter are oxidized and expelled. The sulfide combines with oxygen to form oxide eg: Cu₂S ore is converted to Cu₂O by roasting;
b. Calcination – ZnCO₃, Roasting – Cu₂S
c. Carbon monoxide, Carbon, Electricity
d. Electricity
e. Carbon Monoxide is used as the reducing agent to extract iron from haematite. (Fe₂O₃)
ZnO :- Carbon is used as the reducing agent to extract Zinc from Zinc Oxide. Electricity is used to extract aluminum from Al₂O₃.

Kerala Syllabus Class 10 Chemistry Solutions Question 3.
Complete the table:

Answer:
a. Low Melting point.
b. Low Boiling points
c. Metals having high electropositivity

Question 4.
a. Complete the table.

b. Stainless steel and Nichrome are having same content (Fe, Ni, Cr, C). But nature of both alloys are different Why?

c. Bauxite and clay are minerals of aluminum. But bauxite is the only ore of Aluminium. Why?.
Answer:
i. Obtained from Blast Furnace Contains 4% Carbon and other impurities like manganese silicon, phosphorus, etc.

ii. Pig iron mixed with scrap iron and coke melted in a special furnace contains 3% carbon.

iii. Made by purifying cast Iron.

iv. Prepared by varying the amount of carbon from 0.1 to 1.5%.

b. Stainless steel and Nichrome are having the same content but the nature of both alloys are different because ratio of constituent elements are different.

c. A mineral from which a metal is economically, easily and quickly extracted is called the core of the metal. Among the minerals of aluminum, bauxite possesses these properties. Hence bauxite is the ore of Aluminium. Since clay does not possess these properties it is not an ore of Aluminium.

Question 5.
Given below are the equations for the reaction taking place inside the blast furnace.
C + O₂→ CO₂
CO₂+ C → 2CO
CaCO₃ + SiO₂ → CaSiO₃
Fe₂O₃+3CO → 2Fe + 3CO₂
a. Name the ore of iron.
b. Which is the gangue in iron ore?
c. Name the flux used in blast furnace,
d. Gangue + flux → ………… Which product is formed in blast furnace?
e. Reducing agent used in blast furnace.
f. Subjects dropped in blast furnace are …………., …………….
Answer:
a. Haematite
b. SiO₂
c. CaO
d. slag(CaSiO₃ )
e. Carbon monoxide
f. A mixture of roasted haematite, coke and limestone.

Question 6.
a. Write down the names of Anode, Cathode, Electrolyte used in the Electrolyte cell for the manufacturing of copper,
b. Write down the equations for the reaction in anode and cathode.
Answer:
Anode – Copper to be refined
Cathode – Pure Copper
Electrolyte – Aqueous Copper Sulphate Solution mixed with H2SO4

b. Anode – Cu → Cu₂ + 2e^–
Cathode – Cu²⁺ +2e^– → Cu

Question 7.
Manufacturing of iron.
a. Name the furnace used for producing iron.
b. Name the materials using for producing iron.
c. Write down the reaction occurring on coke when hot is blasted on it?
d. Why CaCO₃ is dropping inside the furnace?
e. Write down the nature of gangue with iron ore.
f. Gangue + flux → ……….. Write down the uses of the product formed in blast furnace.
g. Reducing agent in blast furnace
h. Write down the reactions taking place inside the blast furnace.
i. Iron formed from the blast furnace is called ……………
j. How can we change iron into steel?
k. What are the different types of steel?
Answer:
a. Blast Furnace
b. Mixture of roasted haematite, coke and limestone.
c. At the bottom of the blast furnace, coke combines with oxygen in the hot current of air. The CO₂ which rises up along with the hot air current is reduced by coke.
d. CaCO₃ in the furnace decomposes to form CaO and CO₂. CaO which is basic combines with SiO₂ (acidic) to form slag. Molten slag which is lighter floats over the heavier molten iron. CaCO₃ is added to the furnace for the production of slag.
e. Acidic
f. Slag. (CaSiO₃) used for the production of cement and in the Construction of Road.
g. Carbon monoxide
h. CaCO₃(s) → CaO + CO₂
CaO+SiO₂ → CaSiO₃
C+O₂ CO₂ + Heat
CO₂(g) + C_(s) + Heat → 2CO (g)
Fe₂O₃+3CO → 2Fe + 3CO₂
i. Pig iron
j. Different types of steel can be prepared by varying the amount of carbon from 0.1 to 1.5
k. Mild steel, Medium steel, High carbon steel
Mild steel:- If the carbon content in steel is from 0.05% to 0.2 % then it in called mild steel. I

Medium steel :- Medium steel contains carbon from 0.2 % to 0.6 % medium steel

High carbon steel: – If the content of carbon is from 0.61 % to 15 % then it is known as high carbon steel.

Question 8.
Flow chart of Manufacturing Aluminium.

a. Draw the Electrolyte cell and then write answers for the following questions.
b. Anode, Cathode in this cell are …………….
c. Write down the reactions taking place in Anode and Cathode.
d Why Carbon power dropped above the electrolyte?
e. Which gas is evolved out from graphite.
f. Uses of Cryolite.
Answer:
a.

b. Anode – Carbon Rod, Cathode – Carbon lining
c. Anode -2O₂– → O₂ + 4e-
Cathode – AI³⁺ +3e^– → AI
d. To prevent the chemical reaction of carbon rods with the oxygen in atmosphere.
e. Oxygen
f Cryolite is added to alumina to reduce its melting point and increase its electrical conductivity.

Production of Metals SCERT Questions and Answers

Question 1.
Nature of some ores are given. Pick ore concentration from the bracket.
(Magnetic Separation, Froth Floatation, Levigation, Leaching)
i. Ores are lighter and impurities are heavier.
ii. Ore is magnetic. But impurities are non-magnetic.
iii. Uses a solution which dissolves the ore.
iv. Ore is heavier and impurities are lighter.
Answer:
i. Froth floatation
ii. Magnetic separation
iii. Leaching
iv. Levigation

Question 2.
Some metals and ores are given. Match them suitably

Answer:

Question 3.
Calcination is used to convert zinc carbonate into zinc oxide. But cuprous sulfate is converted into cuprous oxide by roasting.
a. What is the difference between calcination and roasting?
b. What happens to the ore when it is subjected to calcination?
Answer:
a. Calcination: It is the process in which the ore is heated in the absence of air at a temperature below its melting point so that the moisture content of ore, volatile impurities, bio substances, etc. can be removed. Along with this, metal carbonates and hydroxides are converted into its oxide.

Roasting: It is the process in which the ore is heated in the presence of air at a temperature below its melting point so that the moisture content of ore, sulfur, phosphorous, etc, are converted into its oxide and can be removed. Sulfide ores also get converted into oxides.

b. Zinc carbonate ore is converted to Zinc oxide.

Question 4.
a. Some metals and their methods of concentration are given. Match them suitably.
Mercury, Zinc, Tin, Copper, Lead
Liquation, Electrolytic refining, Distillation
b. Write the reason for selecting the methods for the concentration of mercury and tin.
Answer:
a. Liquation – Tin, Lead
Electrolytic refining – Copper Distillation – Mercury, Zinc
b. Boiling point of mercury is low. Melting point of Tin is low

Question 5.
The order the reactivity of some metals are given. Answer the following questions by analyzing it.
Al >Zn >Cu >Au
a Which metal is produced bytfie electrolysis of its molten salt ?
b. Metal occur in free state in nature,
c. Metal produced by the self oxidation reduction reaction.
d. Metal ore which is reduced by carbon.
Answer:
a. Al
b. Au
c. Cu
d. Zn

Question 6.
A reducing agent is required to extract the metal from its ore. Why ? Explain with ex ample.
Answer:
In ores metals are in positive oxidation state, reducing agent is needed (electron giving substance) to get the metal.
Eg: Carbon is used to extract zinc from zincoxide. Zno_(s) + C_(s) → Zn_(s) + CO(g)

Question 7.
The equations of the production of iron in the blast furnace are given. Answer the following questions.
C + O₂ → CO₂
CO₂ + C → 2CO
CaCO₃ → CaO + CO₂
CaO + SiO₂ → CaSiO₃
Fe₂O₃+3CO → 2Fe + 3CO₂
a. Which substance reduces haematite in the metallurgy of iron? How this reducing agent is produced in the furnace?
b. Which is the main impurity found in haematite? Which substance is used to remove the gangue?
c. Write the chemical equation of the formation of slag in blast furnace.
Answer:
a. CO, Oxygen in the blast of hot air reacts with coke, to form CO₂. This CO₂ again reacts with coke to produce CO.
b. SiO₂ – Impurity; CaO – remove gangue
c. CaO + SiO₂ → CaSiO3

Question 8.
a. How pig iron is converted into cast iron?
b. Molten cast iron is poured into moulds to make different shapes. Which speciality if cast iron is based for it?
Answer:
a. Pig Iron mixed with scrap iron and coke, is melted in a special furnace to make cast iron.
b. Molten cast iron expands a little on solidifiation.

Question 9.
Alloys containing iron are given. Find out a, b, c and

Answer:
a. Fe, Ni, Al, Co
b. For the manufacture of permanent magnets
c. Stainless steel
d. For making heating coils

Question 10.
Aluminium is prepared industrially by Hall-Heroult process. Various steps in the concentration of ore are given below. Write them
in the correct order.
i. The precipitate formed is separated, washed and strongly heated to get alumina,
ii. Crushed bauxite is leached with hot sodium hydroxide solution.
iii. Impurities are removed from the sodium aluminate solution by filtration.
iv. Solution is diluted after adding a little aluminium hydroxide, to precipitate aluminum hydroxide.
Answer:
Order:
(ii),
(iii),
(iv),
(i)

Question 11.
a. Carbon monoxide cannot be used as reducing agent to extract aluminium from alumina. Why?
b. The electrolytic cell for alumina is given below.

i. Al₂O₃ dissolved in molten cryolite is used as the electrolyte. What is the purpose of adding cryolite to alumina?
ii. Anode is replaced from time to time while producing aluminium. Why?
iii. Write the chemical equation of the reaction at the cathode.
Answer:
a. Aluminium compounds are very stable,
b. i. The melting point of alumina is very high. Cryolite is added to alumina to reduce its melting point and increase its electrical conductivity, .
ii. Oxygen liberated at the anode reacts with carbon, forming CO
iii. Al³⁺ + 3e^– → Al

Question 12.
a. Illustrate the arrangement of refining copper and label the anode, cathode and electrolyte.
b. Write the chemical equations at the anode and cathode and sustain it as a redox reaction.
Answer:

As oxidation and reduction takes place it is a redox reaction

Question 13.
Clay, cryolite and bauxite are the minerals of aluminium.
a. Which among them is the ore of aluminium? What is its chemical formula?
b. What are the features of an ore?
Answer:
a. Bauxite, Al₂O₃, 2H_2O
b. 1. Easily available
2. Metal can be separated easily
3. High content of metal

Question 14.
The chemical reaction of calcium carbonate while heating is given. .
\(\mathrm{CaCO}_{2} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}\)
How the reaction is made use in the metallurgy of iron?
Answer:
CaO is framed by the decomposition of Ca- CO₃. This CaO acts as a flux and combines with SiO₂ (gangue) to form CaSiO₃ (Slag).

Question 15.
Find the relation and answer the following.
a. Zinc Sulphate : Roasting,
Calcium Carbonate:…………
b. Haematite: Magnetic Separation;
Bauxite:…………..
Answer:
a. Calcination
b. Leaching

Question 16.
The flowchart of the process of concentrate of aluminium ore is given. Complete the flowchart.

Answer:
a. NaAlO₂(Sodium aluminate)
b. Al(OH)₃ /Aluminum hydroxide)
c. The precipitate is separated, washed well and strongly heated;
d. Al₂O₃

Production of Metals Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
What is the importance of adding cryolite in the electrolysis of alumina? ‘
Answer:
Alumina has a very high melting point. Cryolite lowers the melting point of alumina and improves the conductivity of alumina.

Question 2.
What are gangue, flux and slag?
Answer:
Gangue: Impurities in the metal ore 1
Flux: Chemical substances used to convert impurities which are not easily removable (gangue) into slag.
Slag: substances which are easily removable when gangue and flux are combined.

Short Answer Type Questions (Score 2)

Question 3.

a. Choose the ore from those given below which can be used in the above figure. Bauxite, Tin stone, Copper pyrites, Calamine
b. Give reason for your answer:
Answer:
a. Tin stone (SnO₂)
b. This figure depicts magnetic separation. SnO₂ is a magnetic ore. But the gangue, iron tungstate present in this also has magnetic property.

Question 4.
Haematite, the ore of iron undergo roasting.
a. Which impurity is not removed by this method?
b. How is it removed then? Explain.
Answer:
a. Silicon dioxide (SiO₂)
b. It is removed during metallurgy. At high temperature, CaC03 (limestone) decomposes to form CaO, which is the flux. CaO combines with SiO₂ and forms CaSiO₃ (slag).
CaCO_3(s) → CaO_(s) + CO₂(g)
CaO_(s) + SiO_2(s) → CaSiO_3(s)

Question 5.
During the concentration process of bauxite,
a. Why is hot concentrated NaOH used?
b. Why isAl(OH)3 added in small quantity and diluted with water to sodium aluminate solution?
Answer:
a. Only bauxite is soluble in hot concentrated-. ted NaOH. As the impurities are insoluble, they can be easily filtered. This process is known is leaching.
b. When Al(OH)₃ is added in small quantity and diluted with water to sodium aluminate solution, the whole aluminium in the solution gets precipitated as Al(OH)₃. The Al(OH)₃ precipitate is separated from the solution. Then it is washed and heated strongly to form pure Al₂O₃ or alumina.

Question 6.
Define:
a. Cast iron
b. Wrought iron
Answer:
a. Cast iron is formed by heating pig iron, scrap iron and coke. Cast iron contains about 3% of carbon. Cast iron expands on solidification. So these are used to make molds. Though it is strong, it is brittle.
b. When cast iron is purified, it becomes wrought iron which is comparatively a pure form of iron. Wrought iron contains about 0.2% to 0.5% of carbon. Small amounts of phosphorus, silicon, etc. are also present.

Short Answer Type Questions (Score 3)

Question 7.
Match columns A, B and C suitably.

Answer:

Question 8.
The concentration methods of certain ores are given below. Why these methods are used?
a. Bauxite – Leaching
b. Magnetite – Magnetic separation
c. Copper pyrites – Froth floatation
Answer:
a. Bauxite gets dissolved in hot concentrate NaOH. But the impurities does not dis-solve in this.
b. Magnetite, the ore of iron has magnetic properties but the impurities does not have.
c. Copper pyrites have a lesser density and impurities have higher density. Also, only the ore particles float in pine oil.

Question 9.
Explain the relationship between reactivity series of metals and metallurgy.
Answer:
Metals such as K,”Na, Ca, Mg and Al are placed above in the reactivity series. Strong reducing agent like electricity is used in the production of these metals by the electrolysis of their molten metallic compounds. Weak reducing agent such as carbon/carbon monoxide is used for the production of Zn, Fe, Ni, Sn and Pb. Copper is produced by the oxidation-reduction reactions of metal sulfide. Ag, Au, etc

Question 10.

Purification of copper is depicted here.
a Identify the anode, cathode and electrolyte.
b. Write the chemical equation during electrolysis.
c. What is seen below the positive electrode?
Answer:
a. Anode : Impure copper
Cathode : Pure copper
electrolyte : aqeous solution of CuSO4 to which H2SO4 is added.

b. At anode : Cu → Cu²⁺ + 2e^–
At cathode : Cu²⁺ + 2e^– → Cu

c. Anode mud – impurities in impure copper

Question 11.
Certain alloys are given below.
i. Nichrome
ii. Stainless steel
a. What are the constituent elements in them?
b. What is the reason for the difference in their properties?
c. Write one use of each.
Answer:
i. Nichrome: Fe, Ni, Cr, C
ii. Stainless steel: Fe, Cr, Ni, C
b. The constituent elements are same in both but their ratios are different. So their properties also differ.
c. Nichrome : for making heating coils Stainless steel: to make utensils

Long Answer Type Questions (Score 4)

Question 12.
Describe the following:
a. Calcination
b. Roasting v
Answer:
Extraction of metals from the concentrated ore have 2 stages:-
a. Calcination: It is the process in which the ore is heated in the absence of air at a temperature below its melting point so that the moisture content of ore, volatile impurities, bio substances, etc. can be removed. Along with this, metal carbonates and hydroxides are converted into its oxide.

b. Roasting: It is the process in which the ore is heated in the presence of air at a temperature below its melting point so that the moisture content of ore, sulfur, phosphorous, etc, are converted into its oxide and can be removed. Sulfide ores also get converted into oxides.

Question 13.
Complete the table.

Answer:
a. Haematite
b. Levigation
c. Roasting
d. Impurities with less density are removed.
e. Moisture content is removed. Sulfur, Arsenic, phosphorous, etc. are converted into its oxides. As they are in gaseous state, these are also removed.

Question 14.
Minerals of certain metals are given below. Write down the refining method of each.
a. Tin
b. Copper
c. Zinc
d. Lead
e. Cadmium
f. Silver
g. Mercury
Answer:
a. Tin – Liquation
b. Copper – Electrolysis
c. Zinc – Distillation
d. Lead – Liquation
e. Cadmium – Distillation
f. Silver – Electrolysis
g. Mercury – Distillation

Question 15.
The figure depicts the electrolysis of mixture of alumina and cryolite. Label anode, cathode and electrolyte in the figure.

Answer:

You can Download Mathematics of Chance Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 3 Mathematics of Chance Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 3 Mathematics of Chance Notes

Textbook Page No. 71

Mathematics Of Chance Questions And Answers Kerala Syllabus 10th Standard Question 1.
A box contains 6 black and 4 white balls. If a ball is taken from it. What is the probability of it being black? And the probability of it being white?
Answer:
Total no. of balls = 10
No. of black balls = 6
No. of white balls = 4
Probability of it to be black = \(\frac { 6 }{ 10 }\) = \(\frac { 3 }{ 5 }\)
Probablitiy of to be white = \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)

Mathematics Of Chance Class 10 Kerala Syllabus Kerala Syllabus Question 2.
There are 3 red balls and 7 green balls in a bag, 8 red and 7 green balls in another.
i. What is the probability of getting a red ball from the first bag?
ii. From the second bag?
iii. If all the balls are put in a single bag, What is the probability of getting a red ball from it?
Answer:
i. Total no. of balls in the first bag = 10
No. of red balls = 3
P(red ball) = 3/10

ii. Total no. of balls in the second bag= 15 No. of red balls = 8
P(red ball) = 8/15

iii. Total no. of balls in both bags = 25
Total no. of red balls = 11
P(red ball) = 11/25

Mathematics Of Chance Questions And Answers Pdf Kerala Syllabus 10th Standard Question 3.
One is asked to say a two-digit number. What is the probability of it being a perfect square?
Answer:
Total no. of two-digit numbers = 90
Perfect squares from 10 to 99 are 16, 25, 36, 49, 64, 81.
There are 6 favourable numbers Probability = \(\frac { 6 }{ 90 }\) = \(\frac { 1 }{ 15 }\)

Mathematics Of Chance Extra Questions And Answers Kerala Syllabus 10th Standard Question 4.
Numbers from 1 to 50 are written on slips of paper and put in a box. A slip is to be drawn from it; but before doing so, one must make a guess about the number, either prime number or a multiple of five. Which is the better guess? Why?
Answer:
Probablity of getting prime numbers from 1 to 50 = \(\frac { 15 }{ 50 }\) = \(\frac { 3 }{ 10 }\)
Probability of getting a multiple of 5 from 1 to 50 = \(\frac { 10 }{ 50 }\) = \(\frac { 1 }{ 5 }\)
∴ Better guess would be prime numbers as probability of that is more.

Sslc Maths Mathematics Of Chance Kerala Syllabus 10th Standard Question 5.
A bag contains 3 red beads and 7 green beads. Another contains one red and one green more. The probability of getting a red from which bag is more?
Answer:
Total no. of beads in first bag = 10
No. of red beads in first bag = 3
P(red ball) in first bag = \(\frac { 3 }{ 10 }\)
Total no. of beads in second bag = 12
No. of red beads in second bag = 4
P(red ball) in second bag = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)
The probability of getting a red bead is more from the second bag since, \(\frac{1}{3}>\frac{3}{10}\left(\frac{10}{30}>\frac{9}{30}\right)\)

Textbook Page No. 72

In each picture below, the explanation of the green part is given. Calculate in each, the probability of a dot put without looking to be within the green part.

Mathematics Of Chance Pdf Kerala Syllabus 10th Standard Question 1.
A square got by joining the mid points of a bigger square.

Answer:
∴ Probability = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)

Class 10 Maths Chapter 3 Mathematics Of Chance Kerala Syllabus Question 2.
A square with all vertices on a circle

Answer:
Side of square = a = 2cm Radius

ie., The area of the square is 2/π the area of the circle.

Probability = \(\frac{a^{2}}{\frac{\pi}{2} a^{2}}=\frac{4}{2 \pi}=\frac{2}{\pi}\)
Probablity ofthe dot falling on the square is \(\frac { 2 }{ π }\) = 0.64

Sslc Maths Chapter 3 Questions And Answers Kerala Syllabus Question 3.
Circle exactly fitting inside a square.

Answer:
If radius of circle is r
Side of the square = 2r
Area of the square = 4r²
Area of the circle= πr²

Mathematics Of Chance Extra Questions Kerala Syllabus 10th Standard Question 4.
A triangle got by joining alternate vertices of a regular hexagon.

Answer:
The area of regular hexagon having side a = \(\frac{3 \sqrt{3}}{2} a^{2}\)
Side of the triangle got by joining alternate vertices = √3a
Area of triangle = \(\frac{\sqrt{3}}{4}(\sqrt{3} a)^{2}=\frac{3 \sqrt{3}}{4} a^{2}\)
∴ Probability of the dot falling on the tri-angle = \(\frac{3 \sqrt{3}}{4} a^{2} \times \frac{2}{3 \sqrt{3} a^{2}}=\frac{2}{4}=\frac{1}{2}\)

Sslc Maths Chapter 3 Solutions Kerala Syllabus 10th Standard Question 5.
A regular hexagon formed by two overlapping equilateral triangles.

Answer:
Area of two equilateral triangles

Area of regular hexagon = \(\frac{3 \sqrt{3}}{2} a^{2}\)
Area of regular hexagon is half of the area of equilateral triangle.
∴ Probability of dot falling on the hexagon

Textbook Page No. 75

Class 10 Maths Chapter 3 Kerala Syllabus Question 1.
Raj ani has three necklaces and free pairs of earrings, of green, blue and red stones. In what all different ways can she wear them? What is the probability of her wearing the necklace and earrings of the same color? Of different colors?
Answer:
Possible ways of wearing:

Raj ani can wear the ornaments in 9 different ways.
Probability of wearing necklace and earring of same colour = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)
Probability of wearing necklace and earring of different colour = \(\frac { 6 }{ 9 }\) = \(\frac { 2 }{ 3 }\)

Sslc Maths Chapter 3 Notes Question 2.
A box contains four slips numbered 1, 2, 3, 4 and another box contains two slips numbered 1,2. If one slip is taken from each, what is the probability of the sum of numbers being odd? What is the probability of the sum being even?
Answer:

Probability of sum being odd = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)
Probability of sum being even = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)

Kerala Syllabus 10th Standard Maths Chapter 3 Question 3.
A box contains four slips numbers 1, 2, 3, 4 and another contains three slips numbered 1, 2, 3. If one slip is taken from each, what is the probability of the product being odd? The probability of the product being even?
Answer:

Probability of sum being odd = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)
Probability of sum being even = \(\frac { 8 }{ 12 }\) = \(\frac { 2 }{ 3 }\)

10 Th Maths Text Book Questions And Answers Question 4.
From all two-digit numbers with either digit 1, 2 or 3 one number is chosen.
i. What is the probability of both digits being the same?
ii. What is the probability of the sum of the digits being 4?
Answer:
Two digit numbers
11, 12, 13, 21, 22, 23, 31, 32, 33.
i. P (both digits being same) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)
ii. P(sum of digits being 4) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)

10th Class Maths Textbook Answers Question 5.
A game for two players. First, each has to decide whether he wants odd number or even number. Then both raise some fingers of one hand. If the sum is odd, the one who chose odd at the beginning wins; if it is even, the one who chose even wins. In this game, which is the better choice at the beginning, odd or even?
Answer:
The results in the order when the first player raises one finger and second player raises one finger and so on.

Textbook Page No. 78

Maths Chapter 3 Class 10 Kerala Syllabus 10th Standard Question 1.
In class 10A, there are 30 boys and 20 girls. In 10B, there are 15 boys and 25 girls. One student is to be selected from each class.
i. What is the probability of both being girls?
ii. What is the probability of both being boys?
iii. What is the probability of one boy and one girl?
iv. What is the probability of at least one boy?
Answer:
i. Total no. of possible pairs = 50 × 40 = 2000
No. of pairs in which both are girls = 20 × 25 = 500
Probability of both being girls = \(\frac { 500 }{ 2000 }\) = \(\frac { 1 }{ 4 }\)

ii No. of pairs in which both are boys = 30 × 15 = 450
Probability of both being boys = \(\frac { 450 }{ 2000 }\) = \(\frac { 9 }{ 40 }\)

iii. No. of pairs in which one is boy and one is girl = 2000 – (500 + 450) = 2000 – 950 = 1050
Probability of one being boy and one girl = \(\frac { 1050 }{ 2000 }\) = \(\frac { 21 }{ 40 }\)

iv. No. of pairs in which atleast one is boy = 1050 + 450 = 1500
Probability in which atleast one is boy = \(\frac { 1050 }{ 2000 }\) = \(\frac { 3 }{ 4 }\)

Question 2.
One is asked to say a two-digit number.
i. What is the probability of both digits being the same?
ii. What is the probability of the first digit being larger?
iii. What is the probability of the first digit being smaller?
Answer:
i. Probability of two digits being same = \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)
(11, 22, 33, 44, 55, 66, 77, 88, 99)

ii. Numbers in which first digit is greater than second digit are 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70, 71, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 97, 98,
45 outcomes, Required Probability = \(\frac { 45 }{ 90 }\) = \(\frac { 1 }{ 2 }\)

iii. Numbers in which first digit is smaller than second digit are
12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89,
36 total no.of favourable outcomes = 36
Probability =

Question 3.
Each two-digit number is written on a paper slip and these are all put in a box. What is the probability that the product of the digits of a number drawn is a prime number? What if three-digit numbers are used instead?
Answer:
Total two-digit numbers = 90
Product of the digits of a number drawn is a prime number 12, 13, 15, 17, 21, 31, 51, 71.
(1 is not a prime number)
Total number whose product of the digits drawn is a prime number = 8
Probability of product of the digits drawn is a prime number = \(\frac { 8 }{ 90 }\) = \(\frac { 4 }{ 45 }\)
Total three-digit numbers = 900
Product of the digits of a number drawn is a prime number 112, 113, 115, 117, 121, 131, 151, 171.
Total number whose product of the digits drawn is a prime number = 8
Probability of product of the digits drawn is a prime number = \(\frac { 8 }{ 900 }\) = \(\frac { 2 }{ 225 }\)

Question 4.
Two dice with faces numbered from 1 to 6 are rolled together. What are the possible sums? Which of these sums has the maximum probability?
Answer:
Each dice has the following numbers:

Mathematics of Chance Orukkam Questions & Answers

Worksheet 1

Question 1.
How many odd numbers are there below 25
Answer:
12 odd numbers

Question 2.
How many prime numbers are there below 30?
Answer:
10 prime numbers

Question 3.
Find the number of two-digit even numbers?
Answer:
90 two-digit even numbers

Question 4.
How many two digits perfect squares are there?
Answer:
6

Question 5.
Write all three-digit numbers that can be written using the digits 3, 6, 8 without repeating the digits.
Answer:
368, 386, 683, 638, 836, 863

Question 6.
How many multiples of 7 are there in between 100 and 300?
Answer:
First multiple of 7 = 105
Last multiple of 7 = 294
Number of multiples
=

Question 7.
There are 50 children in a class. Thirty of them are girls. There are 40 children in another class. 25 of them are boys. One student is taken from each class at random. What is the number of outcomes? How many outcomes contain both boys. How many outcomes contain both girls. How many outcomes have one boy and one girl?
Answer:
Total pairs = 50 × 40 = 2000
No. of pairs in which both are boys = 20 × 25 = 500
No. of pairs in which both are girls = 30 × 15 = 450
No. of pairs in which one is a boy and the other a girl = 30 × 25 + 20 × 15 = 750 + 300 = 1050

Worksheet 2

Question 8.
A fine dot is placed into the picture with-out looking into it. What is the probability of falling the dot in the small semicircle? What is the probability of falling the dot outside the small semicircle but inside the big semicircle?
Answer:
Let the radius of circle be r, then the radius of

Question 9.
P, Q, R are the midpoints of the sides of triangle ABC. Another triangle is drawn by joining these points. A fine dot is placed into the figure without looking into the picture. What is the probability of falling the dot in triangle PQR?
What is the probability of falling the dot outside the triangle?
Answer:
Area of each small triangle is 1/4^th of area of large triangle ABC.

Probability of the dot falling on triangle PQR is = 1/4
Probability of the dot falling on small triangle is = 1/4
Probability of the dot falling inside the triangle PQR is \(\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\) less than that of outside the triangle.

Question 10.
What is the probability of occurring 53 Sundays in a leap year
Answer:
Leap year have 366 days.
That is 52 weeks and 2 days.
There two days are
Sunday – Monday, Monday – Tuesday, Tuesday – Wednesday, Wednesday – Thursday, Thursday – Friday, Friday – Saturday, Saturday – Sunday.
∴ Probability for to occur 53 Sunday is 2/7.

Question 11.
You can see a triangle inside a square. ABCD is a square. P, Q are the midpoints of C D and C B. A fine dot is placed into the figure without looking into the figure. What is the probability of falling the dot in triangle APQ?
Answer:

Question 12.
The value of 21, 22, 23… 250 are written in small papers and put it in the box. A paper is taken at random. What is the probability of getting a number having 4 in ones place? What is the probability of falling 8 in the one’s place?
Answer:
The 13 numbers having 4 in one’s place are 22, 26, 210 …, 250.
∴ Probability = \(\frac { 13 }{ 50 }\)
The 12 numbers having 8 in one’s place are 23, 27, 211 …, 247.
∴ Probability = \(\frac { 12 }{ 50 }\)

Worksheet 3

Question 13.
Numbers from 1 to 10 are written in small papers and placed in a box. One number is taken from the box at random. What is the probability of getting a prime number?
Answer:
Probability of getting a prime number = \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)

Question 14.
Two boxes contain tokens on which numbers 1, 2, 3, 4 are written One token is taken from each box. What is the probability of getting sum of the face numbers a prime number
Answer:
Pairs of numbers in tokens are
(1.1) , (1,2), (1,3), (1,4)
(2.1), (2, 2), (2, 3), (2, 4)
(3, 1), (3,2), (3; 3), (3,4)
(4, 1), (4, 2), (4, 3), (4,4).
Pairs getting sum as prime numbers
(1.1) , (1,2), (1,4), (2, 1), (2, 3), (3, 2), (3, 4), (4, 1), (4, 3)
Probability of getting sum of the face numbers a prime number = \(\frac { 9 }{ 16 }\)

Question 15.
One box contains 8 black balls and 12 white balls. Another box contains 9 black and 6 white balls. One ball is taken from each box at random. What is probability of getting both black? What is the probability of getting both white? What is the probability of getting one black and one white?
Answer:
Total pairs = 20 x 15 = 300
Number of pairs getting both black= 8 x 9 = 72
Probability of getting both black = \(\frac { 72 }{ 300 }\) = \(\frac { 6 }{ 25 }\)
Probability of getting one black and one white

Question 16.
In the figure, a triangle is drawn by joining the alternate vertices of a regular hexagon. A fine dot is placed into the figure at random. What is the probability of falling the dot in the triangle?

Answer:

Probability of the dot falling on triangle

Question 17.
What is the probability of occurring four Wednesdays in 23 consecutive days in a month?
Answer:
23 days = 3 weeks + 2 days
Wednesday comes on Tuesday + Wednesday, Wednesday + Thursday when two days are taken.
Total probabilities (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
∴ Total probabilities = 7
∴ Probabilities = 2/7

Mathematics of Chance SCERT Question Pool Questions & Answers

Question 18.
One is asked to say a two digit number. What is the probability of being the number not a perfect square? [Score : 3, Time : 3 Minutes]
Answer:
Total number of two digit number : 90 (1)
Total number of two digit perfect squares: 6 (1)
Number of two digit numbers which are not perfeet squares : 90 – 6 = 84,
Probability = \(\frac { 84 }{ 90 }\) = \(\frac { 42 }{ 45 }\) (1)

Question 19.
A bag contains 10 blue balls and 12 yellow balls. Another contains 15 blue balls and 7 yellow balls.
a. What is the probability of getting a yellow ball from the first bag?
b. What is the probability of getting a yellow ball from the second bag?
C. If all the balls are put in a single bag, what is the probability of getting a yellow ball from it? [Score : 4, Time : 4 Minutes]
Answer:
a. Total number of balls in the first bag = 10 + 12 =22, Number of yellow balls = 12
Probability of getting a yellow ball = \(\frac { 12 }{ 22 }\) = \(\frac { 6 }{ 11 }\) (1)

b. Total number of balls in the second bag = 15 + 7 = 22, Number of a yellow ball = 7
Probability of getting a yellow ball = \(\frac { 7 }{ 22 }\) (1)

c. Total number of balls = 22 + 22 = 44
Number of yellow balls = 12 + 7 = 19 (1)
Probability of getting a yellow ball = \(\frac { 19 }{ 44 }\) (1)

Question 20.
A regular hexagon is drawn with its vertices on a circle. Without looking into the I picture, if one put dot in that picture, what is the probability of being the dot not in the regular hexagon? [Score: 4, Time: 3 Minutes] .

Answer:
Area of circle = πr²

Question 21.
In the figure, all the four shaded semicircles have same area. If we put a dot in the figure without looking into it, what is the probability of being the dot in the shaded semicircles? [Score : 3, Time: 5 Minutes]

Answer:
If the radius of the shaded semicircle is r,

Question 22.
What is the probability of getting 5 Sundays in December in a calendar year? [Score : 3, Time : 5 Minutes]
Answer:
There are 31 days in December. That means 4 full weeks and 3 days.
The probable three days are as shown below.

Question 23.

Two semicircles are drawn in a square as shown. If we put a dot in the figure, without looking into it, what is the probability of being the dot in the shaded region? [Score: 3, Time: 5 Minutes]
Answer:
If a is the side of a square,

Mathematics of Chance Exam Oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 24.
A bag contains 6 red balls, 8 green balls, and 8 white balls. One ball is drawn at random from the bag, find the probability of getting
i. A white or green ball
ii. Neither green ball nor a red ball.
Answer:
Red balls = 6
Green balls = 8
White balls = 8
Total number of balls = 6 + 8 + 8 = 22
a. The probability of getting a white or green ball = 16/20
b. The probability of getting neither green balls nor a red ball = 8/20

Question 25.
20 cards numbered 1, 2, 3, 4, ….19, 20 are put in a box. One boy draws a card from the box. Find the probability that the number on the card is:
i. Prime
ii. Divisible by 3
Answer:
Total number of outcomes = 20
a. Prime numbers from 1 to 17 are 2, 3, 5, 7, 11,13, 17, 19.
Number of outcomes = 8
The probability that the card drawn is prime number = 8/20

b. Numbers are divisible by 3 are 3, 6, 9, 12, 15, 18.
Number of outcomes = 6
The probability that the card drawn is divisible by 3 = 6/20

Question 26.
In a bag, there were 3 white balls and 5 black balls. From this one ball is taken, then
a. What is the probability of being blackball?
b. What is the probability of being white ball?
Answer:
Total no. of balls is 8 and in that 5 of then are black balls.
a. Probability of getting black balls = 5/8
b. 3 of the balls were white, so the probability of getting white balls = 3/8

Question 27.
a. How many two-digit natural numbers are there in all?
b. If we choose one number from the two-digit numbers, what is the probability that the sum of digits of that number will be 10?
Answer:
a. Number of two-digit natural numbers = 90

b. Numbers whose sum of digits will be 10 is 19, 28, 37, 46, 55, 64, 73, 82, 91
Probability that the sum of digits of that number will be 10 = \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)

Short Answer Type Questions (Score 3)

Question 28.
In selecting a two-digit number up to 50.
a. What is the probability of the digit in the ten’s place to be larger than the digit in the one’s place?
b. What is the probability of the digit in the tens place to be smaller than the digit in the one’s place?
Answer:
Total numbers of two-digit numbers up to 50 = 41
a. Number of numbers with the digit in the tens place to be larger than the digit in the units place = 11
Probability that the number with tens place digit is larger than the digit in the unit place = 11/41

b. Probability that the number with the tenth place digit is smaller than the digit in the unit place = 26/41

Question 29.
A black dice and a white dice are thrown at the same time,
a. Write all the possible outcomes.
b. What is the probability that the sum of the two numbers is to be 8.
c. What is the probability that being the same number to be on both dice?
Answer:
a. Total outcomes
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6) ‘
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6) .
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4.1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5.1) , (5,2), (5,3), (5,4), (5,5), (5,6)
(6.1), (6,2), (6,3), (6,4), (6,5), (6,6)
Total = 36

b. Sum of two numbers is to be 8 = 5
(2,6), (3,5), (4,4), (5,3), (6,2),
Probability = 5/36

c. Probability of being same number =(1,1),(2,2), (3,3), (4,4), (5,5), (6,6),
Total = 6
Probability = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

Question 30.
A box contains 400 electronic toy cars. Among them, 12 are defective. One toy is taken out at random. What is the probability that
a. It is a defective toy.
b. It is a non-defective toy.
Answer:
Total number of cars = 400
Number of defective toys = 12
Number of non-defective toys = 400 – 12 = 388
a. Probability of getting a defective toy = \(\frac { 12 }{ 400 }\) = \(\frac { 3 }{ 100 }\)

b. Probability of getting a non-defective toy = \(\frac { 388 }{ 400 }\) = \(\frac { 97 }{ 100 }\)

Long Answer Type Questions (Score 4)

Question 31.
Natural numbers from 1 to 30 are written on paper slips and kept in a box. If one slip is taken from the box,
a. What is the probability of this number to be even?
b What is the probability of this number to be a multiple of 3?
c. What is the probability of this number to be a multiple of 3 and 5?
d. What is the probability that this number be a natural number?
Answer:

Question 32.
There is one spot at one side of the cube, two on another side, three on the third side and so on. There are spots on all the six faces; in this order. Another cube which is marked in the same way is taken.
a. If both the cubes are thrown, what is the probability that the total number of spots on the upper faces is 6?
b. What is the probability that the sum of the spots on the upper faces is 9?
c. What is the probability that the sum of the spots be one?
d. What is the probability that the sum of the spots be a prime number?
Answer:

Question 33.
In a pack of 52 cards, half are red and the rest are black. There are 4 Suits of 13 cards each and having the signs ‘hearts’, ‘spade’, ‘clubs’ and ‘diamond’. If a card is picked from this pack.
a. What is the probability of it being black?
b. Whatistheprobabilityofitbeingaspade?
c. What is the probability of it being a spade or a diamond?
Answer:
a. Total number of cards = 52
Number of black cards = 26
Probability of the picked card being black = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

b. No. of spade cards = 13
Probability of a picked card being spade = \(\frac { 12 }{ 52 }\) = \(\frac { 1 }{ 4 }\)

c. No. of cards spade or diamond =13 + 3 = 26
Probability of a picked card being spade or diamond = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

Long Answer Type Questions (Score 5)

Question 34.
A man is asked to say a 3 digit number,
a. What is the probability that the first and last digits be equal?
b. What is the probability that the last two digits be ‘O’?
c. What is the probability that the last digits being greater than the first?
Answer:
Total 3 digit numbers = 900
a. Numbers with first and last digits are equal
101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393
There are 90 such numbers.
∴ Probability = \(\frac { 90 }{ 900 }\) = \(\frac { 1 }{ 10 }\)

b. The numbers with last two digits zero are 100, 200, 300,……… 900 total numbers 9.
∴ Probability = \(\frac { 9 }{ 900 }\) = \(\frac { 1 }{ 100 }\)

c. The numbers with the last digit greater than the first digit is 36
∴ probability = \(\frac { 36 }{ 90 }\) = \(\frac { 2 }{ 5 }\)

Question 35.
There are 10 black pearls and 5 white pearls in the box A. There are 8 black pearls and 7 white pearls in box B.
a. Which box has more probability of be ing the pearls black, when a pearl from each of the boxes is taken?
b. What is the probability to get a white pearl from the box A?
c. What is the probability to get a black pearl from box B?
d. If all the pearls in the box B is dropped in the box A, then what is the probability to get a black pearl from it?
Answer:
a. Box A because it has more black pearl

Question 36.
Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number of the card is
a. an even number
b. a number less than 16
c. a number which is a perfect square
d. a prime number less than 25
Answer:

Mathematics of Chance Memory Map

When probabilities are explained in terms of numbers it is the ratio of number of favorable outcomes to the total number of outcomes.

The least probability will be 0 and the highest will be 1. The probability will be a number between 0 and 1.

If an event can be completed in ‘m’ ways and another event can be completed in ‘n’ ways then both the events can be completed one after the other in m x n ways.

Students can Download Chemistry Chapter 5 Compounds of Non-Metals Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Chemistry Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Chemistry Chapter 5 Compounds of Non-Metals Questions and Answers Malayalam Medium

SCERT 10th Standard Chemistry Textbook Chapter 5 Solutions Malayalam Medium

You can Download Nomenclature of Organic Compounds and Isomerism Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemistry Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Chemistry Chapter 6 Nomenclature of Organic Compounds and Isomerism Textbook Questions and Answers

SCERT Class 10th Standard Chemistry Chapter 6 Nomenclature of Organic Compounds and Isomerism Solutions

Text Book Page No: 97

Nomenclature Of Organic Compounds And Isomerism Class 10 Question 1.
The given structures indicate the valency of carbon. Imagine that hydrogen atoms are added to these structures. Complete the given structures.

Answer:

Text Book Page No: 98

Sslc Chemistry Chapter 6 Kerala Syllabus Question 2.
Complete the table (6.2) given below.

Answer:

Sslc Chemistry Chapter 6 Solutions Kerala Syllabus Question 3.
What is the relationship between the number of atoms of carbon and hydrogen in alkane?
Answer:
Number of hydrogen atoms are equal to adding 2 with twice of number of carbon atoms.

Chemistry Chapter 6 Class 10 Kerala Syllabus Question 4.
If an alkane contains ‘n’ carbon at-oms, how many hydrogen atoms will be there?
Answer:
2n + 2

Chemistry Class 10 Chapter 6 Kerala Syllabus Question 5.
If so, can you deduce a general formula for alkanes?
Answer:
C_nH_2n+2

Text Book Page No: 99

Nomenclature Of Organic Compounds And Isomerism Kerala Syllabus Class 10 Question 6.
What is the difference between the number of carbon atoms and hydrogen atoms in CH₄ and C₂H₆ ?
Answer:
One carbon and two hydrogen atoms (CH₂)

Sslc Chemistry Chapter 6 Notes Kerala Syllabus Question 7.
Is the difference same in the case of C₂H₆ and C₃H₈.
Answer:
Yes

Nomenclature Of Organic Compounds Class 10 State Syllabus Question 8.
What is the difference between the molecular formulae of any two successive alkanes?
Answer:
CH₂

Class 10 Chemistry Chapter 6 Question 9.
Complete the table given below (Table 6.3).

Answer:

Text Book Page No: 100

Class 10 Chemistry Chapter 6 Kerala Syllabus Question 10.
Analyze Table 6.3 and find the number of hydrogen atoms in an alkene with ‘n’ carbon atoms.
Answer:
2n

Sslc Chemistry Chapter 6 Questions Kerala Syllabus Question 11.
If so, can a general formula of alkenes be deduced? Try to write it.
Answer:
C_nH₂n

Sslc Chemistry Chapter 6 Notes Pdf Kerala Syllabus Question 12.
Complete the table 6.4.

Answer:

Chemistry Class 10 Chapter 6 Kerala Syllabus Question 13.
Analyse Table 6.4 and find out how many hydrogen atoms would be present in an alkyne within carbon atoms.
Answer:
2n – 2

Sslc Organic Chemistry Kerala Syllabus Question 14.
If so, can a general formula of alkynes be deduced? Try to write the general formula of alkynes?
Answer:
C_nH_2n-2

Sslc Chemistry Nomenclature Of Organic Compounds Kerala Syllabus Question 15.
Check whether the alkynes given in the above table are members of a homologous series.
Answer:
Yes

Text Book Page No: 102

Chemistry Chapter 6 Class 10 Kerala Syllabus Question 16.
Write the IUPAC names of all the alkanes in Table 6.2.
Answer:
CH₄ — Methane
C₂H₆ — Ethane
C₃H₈ — Propane
C₄H₁₀ — Butane
C₅H₁₂ — Pentane

Text Book Page No: 103

Class 10 Chemistry Chapter 6 Notes Kerala Syllabus Question 17.

number for the carbon atom carrying the branch?
b. Number of carbon atoms in the main chain.
c. Word root
d. Suffix
e. Name of the alkyl radical coming as branch.
f. Position of the branch.
Answer:
a. 2
b. 4
c. But
d. -ane
e. methyl
f. 2

Kerala Syllabus 10th Standard Chemistry Chapter 6  Questions 18.
Complete the table 6.3.

Answer:

Text Book Page No: 104

Hsslive Guru 10th Chemistry Kerala Syllabus Question 19.
Some structural formula are given below. Name them.

Answer:

• Number of carbon atoms in the main chain: 5
• Number of branch/branches: 2
• Position of the first branch while numbering from left to right: 2
• Position of the first branch while numbering from right to left: 2
• Is there any change in the position number: Nil
• IUPAC name: 2, 4 dimethyl pentane

Text Book Page No: 105

10th Class Chemistry Chapter 6 Malayalam Kerala Syllabus Question 20.

Number the carbon atoms in the main chain of the above compound given above. Put an ✓ against the correct position numbers of the branches.

Answer:

a. What is the IUPAC name?
Answer:
2,4 – Dimethyl hexane.

b. Which is the second branch?
Answer:
CH₃

c. When does this branch get the lowest number? Put a ✓ mark against the correct one.
d. While numbering from left to right
e. While numbering from right to left
Answer:
While numbering from right to left ✓

Class 10 Organic Chemistry Questions Kerala Syllabus Question 21.
Write the IUPAC name of the compound given below.

Answer:
2, 3, 6 – trimethylheptane

Hsslive Guru Chemistry Kerala Syllabus Question 22.

See the given compound
Answer:
No. of branches in this compound: 2
Names of the branches: methyl
Position numbers of branches: 2, 2
IUPAC name: 2, 2-dimethylbutane

Text Book Page No: 106

Hss Live Guru 10th Chemistry Kerala Syllabus Question 23.
How can the structure of 2, 3- dimethylbutane be written?
Answer:

a. How many carbon atoms are present in its main chain?
Answer:
4

b. Let us represent the main chain.
Answer:
C – C – C- C

c. Which are the branches?
Answer:
Methyl groups – 2 in number

d. What are the positions?
Answer:
2, 3

Question 24.
Complete the table given below. (Table 6.4).

Answer:

Text Book Page No: 107

Question 25.
Complete the table 6.5.

Answer:

Question 26.
Can you write the structural formula of the compound C₂H₄ ?
Answer:
CH₂ = CH₂

Question 27.
If so, what will be structural formula of But – 2- ene?
Answer:
CH₃ – CH = CH – CH₃

Text Book Page No: 108

Question 28.
Which is IUPAC name of the com-pound CH₃ – CH₂ – CH = CH – CH₃? Tick (✓) the right one.
Pent – 3 – ene
Pent – 2 – ene
Answer:
Pent – 2 – ene ✓

Question 29.
CH₃ – C = C – CH₃ But – 2 – yne
How many hydrocarbons can be written by changing the position of triple bond in this compound? Try to write their IUPAC names also.
Answer:
CH₃ = C – C – CH₃ But – 1 – yne
CH₃ – C = C – CH₃ But – 2 – yne
CH₃ – C – C = CH₃ But – 1 – yne