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Students can Download Chapter 3 Motion in a Straight Line Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 3 Motion in a Straight Line

Plus One Physics Chapter 3 Notes Pdf Summary
Motion In A Straight Line
In this chapter, we shall learn how to describe motion. For this, we develop the concepts of velocity, acceleration and relative velocity. We also develop a set of simple equations called Kinematic equations.

• Motion: Motion is change in position of an object with time.
• Rectilinear motion: The motion along a straight line is called rectilinear motion.
• Point object: If the distance travelled by the body is very large compared with its size, the size of the body may be neglected. The body under such a condition may be taken as a point object. The point object can be represented by a point.

Example:

• The length of bus may be neglected compared with the length of the road it is running.
• The size of planet is ignored compared with the size of the orbit in which it is moving.

Position, Path Length And Displacement
1. Reference point, Frame of reference:
In order to specify position of object, we take reference point and a set of axes. Consider a rectangular coordinate system consisting of three mutually perpendicular axes, labelled x, y, and z axes. The point of intersection of these three axes is called origin (O) and serves as the reference point.

The coordinates (x, y, z) of an object describe the position of the object. To measure time, we place a clock in this coordinate system. This coordinate system along with a clock is called a frame of reference.

Straight-line motion in coordinate system

To describe the motion along a straight line we can choose x-axis. The position of a carat different time are given in the figure 3.1. The position to the right of 0 is taken as positive and to the left of 0 as negative. The position coordinates of point P and Q are +360m +240m. The position coordinate of R is-120m.

2. Path Length (Distance):
The total length of the path travelled by an object is called path length.
Explanation:
Consider a car moving along straight line. The positions of car at different time are given in the x-axis. (See figure 3.1)
Case-1:
The car moves from 0 to P. In this case the distance moved by car is OP = +360.
Case-2:
The car moves from 0 to P and then moves back from P to Q.
In this case, the distance travelled is OP + PQ = +360 + (+120) = +480m.

3. Displacement:
The distance between initial point and final point is called displacement.

OR

The change of position of the particle in a particular direction is called displacement.
Explanation:
Consider a car moving along a straight line. The positions of car at different time is given in the x-axis.
See figure (3.1)
Let us take two cases
Case-1:
The car moves from 0 to P, in this case displacement = (360 – 0) = 360
Case-2:
The car moves from 0 to P and moves back from P to Q.
In this case,
Displacement = 240m
Let x₁ and x₂ be the positions of an object at time t₁ and t₂. Then displacement in time Dt = (t₂ – t₁) can be written as Dx = x₂ – x₁
If x₁ < x₂, Dx is positive and if x₂ < x₁, Dx is negative.
Note: The magnitude of displacement may or may not be equal to the path length traversed by an object.

4. Position Time Graph:
Motion of an object can be represented by a position-time graph.
Position time graph for a stationary object:
For a stationary object, the position does not change with time. Hence the position time graph will be a straight line parallel to time axis.

Position time graph in a uniform motion:
Uniform motion:
A body is said to be uniform motion, if it undergoes equal displacements in equal intervals of time. In uniform motion velocity is constant The figure below shows the positiontime graph of such a motion.

Plus One Physics Chapter 3 Notes Question 1.
The position-time of a car is given below. Analyze the graph and explain the motion of car.

Answer:
The car starts from rest a time t=0s from the origin 0 and picks up speed till t=10s. After 10 sec, the car moves with uniform speed till t=18 sec. Then the brakes are applied and the car stops at t = 20s and x = 296m.

Class 11 Physics Chapter 3 Notes Question 2.
Draw the position-time for an object

1. moving with positive velocity
2. moving with negative velocity.

Answer:
1.

2.

Average Velocity And Average Speed
1. Average Velocity:
The average velocity of a particle is the ratio of the total displacement to the time interval.

Explanation:
To explain average velocity, consider a position time graph of a body given below.

Let x₁ be the position of body at a time t₁ and x₂ be the position at t₂.
The average velocity during the time interval Dt = (t₂ – t₁)

where Dx = x₂ – x₁, and Dt = t₂ – t₁,
\(\overline{\mathbf{v}}\) is the average velocity.

Motion In A Straight Line Class 11 Notes Pdf Question 3.
Find the slope of position time graph given below of uniform motion and explain the result.

Answer:

Slope of displacement time graph gives average velocity.

Class 11th Physics Chapter 3 Notes Question 4.
Displacement time graph of a car is given below.

1. Find the average velocity during the time interval 5 to 7 sec.
2. Find the average velocity by taking slope in the interval 5 to 7 sec.

Answer:
1.

2. Slope, tan q

In this case, slope and average velocity are equal in the same interval.

2. Average Speed:
Average speed of a particle is the ratio of the total distance to total time taken.

Physics Chapter 3 Class 11 Notes Question 5.
A car is moving along a straight line. Say OP in figure. It moves from 0 to P in 18s and returns from P to Q in 6s. What are the average velocity and average speed of the car in going?

1. From 0 to P? and
2. from 0 to P and back to Q. (See Figure 3.1)

Answer:
1. Average velocity

Average speed

In this case the average speed is equal to the magnitude of the average velocity.

2. In this case
Average velocity

Average speed

In this case the average speed is not equal to the magnitude of the average velocity. This happens because the motion here involves change in direction. So that the distance is greater than displacement.
Note: In general, the velocity is always less than or equal to speed.

Instantaneous Velocity And Speed
Nonuniform Motion:
A body is said to be nonuniform motion, if it undergoes unequal displacements in equal intervals of time.

OR

A body moving with varying velocity is called nonuniform motion.

1. Instantaneous Velocity:
Chapter 3 Physics Class 11 Notes Question 6.
Why the concept of instantaneous velocity is introduced?
Answer:
In nonuniform motion the average velocity tells us how fast the object has been moving over a given interval. But it does not tell us how it moves at different instants during that interval. For this we define instantaneous velocity. The velocity at an instant is called instantaneous velocity.
Explanation:
Position-time of a body moving along a straight line is given below.

Let us find average velocity in the interval 2 sec (3s to 5s), centered at t = 4 sec. In this case, the slope of line P₁P₂ give the value of average velocity, ie. Slope of P₁P₂,

Decrease the value of Dt from 2.to 1 sec. (ie. 3.5 to 4.5 sec). Then line P₁P₂ becomes Q₁Q₂. Then the slope of gives average velocity overthe interval 3.5 sec to 4.5sec.
ie. slope of Q₁Q₂

In the limit Dt ® 0, gives the instantaneous velocity at t = 4sec and its value is nearly 3.84m/s.

Motion In A Straight Line Class 11 Notes Question 7.
When average velocity of a body becomes instantaneous velocity?
Answer:
In the limit, Dt goes to zero, the average velocity becomes instantaneous velocity.

But lim \(\lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}=\frac{d x}{d t}\)
\Instantaneous velocity,

Here \(\frac{d x}{d t}\) is the differential coefficient of x with respect to time. It is the rate of change of position with respect to time at an instant.

Class 11 Physics Notes Chapter 3 Question 8.
The table given below gives the value of \(\frac{\Delta x}{\Delta t}\) for Dt equal to 2s, 1s, 0.55, 0.1s and 0.01s centered at t = 4 sec. (See figure given above). What conclusions can be made from this table?
Answer:
The value of average velocity \(\left(\frac{\Delta x}{\Delta t}\right)\) becomes instantaneous velocity (3.8 m/s), in the limit of Dt goes to zero, (ie Dt is infinitesimally small).

Class 11 Physics Chapter 3 Notes Pdf Download Question 9.
The position of an object moving along x-axis is given by x = a + bt² where a = 8.5m, b = 2.5 m/s² and t is measured in seconds

1. What is the velocity at t = 0s and t = 2s.
2. What is the average velocity between t = 2s and t = 4s?

Answer:
1.

when t = 0
we get v = 2 × 2.5 × 0
v = 0
when t = 2sec
v = 2 × 2.5 × 2 v = 10m/s.

2. The average velocity

Note: If a body is moving with constant velocity, the average velocity is the same as instantaneous velocity at all instants.

2. Instantaneous Speed:
The speed at an instant is called instantaneous speed.
Note:

• The average speed over a finite interval of time is greater or equal to the magnitude of the average velocity.
• Instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant.

Motion In Straight Line Notes Pdf Acceleration
1. Average Acceleration:
Average acceleration of a particle is ratio of the change in velocity to the time interval.

Explanation
Consider a body moving along a straight line. Let v₁ and v₂ be the instantaneous velocities at time t₁ and t₂ respectively.


where Dv = change in velocity, Dt = Time interval

2. Instantaneous Acceleration:
Acceleration at any instant is called instantaneous acceleration.
Explanation
In the limit Dt ® 0, (Dt goes to zero) the average acceleration becomes instantaneous acceleration.
ie. Instantaneous acceleration

Instantaneous acceleration is the rate of change of velocity with respect to time.

3. Uniform Acceleration:
A body is said to be in uniform acceleration, if velocity changes equally in equal intervals of time.

Motion In A Straight Line Class 11 Pdf Question 10.
The velocities of two bodies A and B are given in the tables. From this table, find which body is moving with uniform acceleration. Explain.
Body A

Body B

Answer:
The body A is moving with uniform acceleration be-cause the velocity of body increases at the rate of 2 m/s².
The body B is moving with constant velocity. Hence this motion is called uniform motion.

4. Velocity-Time Graph For Uniformly Accelerated Motion:

An example for velocity-time of a uniformly accelerated motion is given in the above figure.
Let v_t1 and v_t2 be the velocities at instants t₁ and t₂respectively.
The slope of graph in the interval (t₂ – t₁) can be written as,

∴ tan q = acceleration
Thus the slope of the velocity-time gives the acceleration of the particle.

Motion In A Straight Line Notes Pdf Question 11.
Velocity-time of a body is given below. From this graph draw corresponding acceleration time graph.

Answer:
The slope of velocity-time graph increases in the interval (0 – 10) sec which means that acceleration of the body increases in this interval.

Velocity is constant in the interval (10 – 18) sec. Hence ’ the slope is zero which means that acceleration is zero in this range.

The slope in the interval (18 – 20) sec is constant and negative. Hence acceleration in this is a negative value. The acceleration – time graph for the above motion is given below.

Motion In Straight Line Class 11 Notes Question 12.
The position-time graph of a car is given below.

1. Draw corresponding velocity-time graph. Explain the reason for your answer.
2. From velocity-time graph draw acceleration-time graph and identify the regions of

• positive acceleration
• Negative acceleration
• zero acceleration.

Answer:
1. In the time interval (0 – t₁) sec, the slope of x – t graph increases which means that velocity is increasing in this time interval.

In the time interval (t₁ – t₂) sec, slope is constant. Hence velocity remains constant in this time interval.

In the time interval (t₂ – t₃) sec, the slope is decreasing and finally becomes zero. Which means that velocity decreases to zero.

2. Slope is constant throughout the interval (0 – t₁) sec which means that acceleration constant.

In the interval (t₁ – t₂) sec, slope is zero. Which means that acceleration is zero in this region.

Slope is constant (but negative) in the interval (t₂ – t₃)sec. Hence acceleration is constant and negative in this time interval.

Motion In A Straight Line Class 11 Notes Pdf Download Question 13.
Find the region of

1. positive acceleration
2. zero acceleration
3. negative acceleration from the above x-t graph

Answer:

1. Region OA – Positive acceleration
2. Region AB – zero acceleration
3. Region BC – Negative acceleration

Straight Line Class 11 Notes Question 14.
Match the following.

Answer:
1) – d, 2) – c, 3) – b, 4) – a.

5. Area Under Velocity-Time Graph:
Area under velocity-time graph represents the displacement over a given time interval.
Explanation
Consider a body moving with constant velocity v. Its velocity-time graph is given below.

The area of the rectangle has height v and bast t. Therefore,

Note: The acceleration and velocity of a body cannot change values abruptly at an instant. Changes are always continuous.

Kinematic Equations For Uniformly Accelerated Motion
For uniformly accelerated motion, we can derive some simple equations.

1. Velocity-time relation
2. Position-time relation
3. Position-velocity relation

These equations are called kinematic equations for uniformly accelerated motion.
1. Velocity-Time Relation:

Consider a body moving along a straight line with uniform acceleration ‘a’. Let ‘u’ be initial velocity and ‘v ‘ be the final velocity at time t.
We know acceleration a = \(\frac{\text { Change in velocity }}{\text { Time interval }}\)
a = \(\frac{v-u}{t}\)
at = v – u

2. Position-Time Relation:

Consider a body moving along a straight line with uniform acceleration a. Let ‘u’ be initial velocity and ‘v’ be the final velocity. ‘S’ is the displacement travelled by the body during the time interval ‘t‘.
Displacement of the body during the time interval t,
S = average velocity × time
\(S=\left(\frac{v+u}{2}\right) t\) _____(1)
But v = u + at ____(2)
Substitute eq.(2) in eq.(1), we get

3. Position-Velocity Relation:
\(S=\left(\frac{v+u}{2}\right) t\) _____(1)
But v = u + at
\(\frac{v-u}{a}\) = t _____(2)
Substitute eq.(2) in eq.(1)

2as = v² – u²
v² – u² = 2as

Free-fall:
An object released (near the surface of earth) is accelerated towards the earth. If air resistance is neglected, the object is said to be in free fall. The acceleration due to gravity near the surface of earth is 9.8 m/s².
Note: Free-fall is a case of motion with uniform acceleration.

Question 15.
A body is allowed to fall freely. Draw the following graph.

1. Acceleration-time
2. Velocity-time
3. Position-time

Answer:
1.

2.

3.

Stopping distance of vehicles:
When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance.

Question 16.
Derive an expression for stopping distance of a vehicle in terms of initial velocity (u) and retardation (a).
Answer:
Let the distance travelled by the vehicle before it stops be ‘s’.
Then we can find ‘s’ using the formula
v² = u² + 2as
0 = u² + -2as

3.7 Relative Velocity
Suppose the distance between two bodies changes with time in magnitude, or in direction or in both. Then each body is said to have a velocity relative to the other.

For example, consider two cars A and B moving in the same direction with equal velocities. To a person in A, the car B would appear to be rest.

Hence the velocity of B relative to A is zero.
ie. V_BA = 0
Similarly, the velocity of A with respect to B is zero.
or V_AB = 0
Let A be moving with a velocity V_A and B be moving with a greater velocity V_B in the same direction. Then the person in A feels that the car B is moving away from him with a velocity V_BA. The velocity of B relative to A

For an observer in B, the car A is going back with a velocity. The velocity of A relative to B
V_AB = -(V_B – V_B).

Question 17.
The position-time graph of two bodies A and B (at different situations) are given in the following graphs. Find the relative velocities of the following graph.

Answer:
a) The slope of Aand B are equal. Hence velocity of A and B are equal. So velocity of A with respect to B, V_AB = 0

b) The body A and B meet at t = 3sec

Velocity of A w.r. to B, V_AB = V_A – V_B
= 20-10 = 10 m/s Velocity of B w.r. to A, V_BA = V_B – V_A
= 10 – 20 = -10 m/s

c) The body A and B meet at t = 1 sec.
The velocity of body in the interval t = 1 sec,

Velocity of A w. r. to B,
V_AB = V_A – V_B
= 20 – 10 = 30 m/s
Similarly velocity of B w.r. to A,
V_BA = V_B – V_A
= 10 – +20 = -30 m/s
The magnitude of V_BA or V_AB (=30 m/s) is greater than the magnitude of velocity A or that of B.

Students can Download Chapter 1 Sets Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 1 Sets

Plus One Maths Sets Three Mark Questions and Answers

Plus One Maths Chapter Wise Previous Questions And Answers Question 1.
There are 200 individuals with a skin disorder, 120 had been exposed to the chemical A, 50 to chemical B and 30 to both chemical A and B, Find the number of individuals exposed to

1. Chemical A but not chemical B. (1)
2. Chemical B but not.chemical C. (1)
3. Chemical A or chemical B. (1)

Answer:
1. Given; n(U) = 200; n(A) = 120;
n(B) = 50; n(A∩B) = 30
n (Chemical A but not chemical B)
= n(A ∩ B’) = n(A) – n(A ∩ B) = 120 – 30 = 90

2. n (Chemical B but not chemical A)
= n(A’ ∩ B) = n(B) – n(A ∩ B) = 50 – 30 = 20

3. n (Chemical A or chemical B)
= n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 120 + 50 – 30 = 140.

Plus One Maths Sets Previous Questions And Answers Question 2.
In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as taking both apple as well as orange juice. Find how many students were taking neither apple juice nor orange juice.
Answer:
Let A – Apple juice; O – Orange juice be the sets.
Given; n(U) = 400; n(A) = 100;
n(O) = 150; n(A ∩ O) = 75
n (neither apple juice nor orange juice)
= n(A’ ∩ O’) = n((A ∪ O)’)
= n(U) – n(A ∪ O)
= 400 – [n(A) + n(O) – n(A ∩ O)]
= 400 – [100 + 150 – 75] = 400 – 175 = 225.

Plus One Maths Sets Questions And Answers Pdf Download Question 3.
In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speaks at least one of these two languages?
Answer:
Let F – French; S – Spanish be the sets.
Given; n(F) = 50; n(S) = 20;w(F ∩ S) = 10
n (speaks at least one of these two languages)
= n(F ∪ S) = n(F) + n(S) – n(F ∩ S)
= 50 + 20 – 10 = 60.

Plus One Maths Textbook Questions And Answers Question 4.
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis, how many like tennis only and not cricket? How many like tennis?
Answer:
Let C – Cricket; T – Tennis be the sets.
Given;
n(C’ ∪ T) = 65; n(C) = 40; n(C ∩ T) = 10
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 65 = 40 + n(T) – 10 ⇒ n(T) = 35
n (tennis only and not cricket)
= n(T ∪ C’) = n(T) – n(T ∩ C) = 35 – 10 = 25.

Plus One Maths Chapter Wise Questions And Answers Pdf Question 5.
Let A and B be two sets such that n( A) = 20, n(A ∪ B) = 42, n(A ∩ B) = 4. Find

1. n(B) (1)
2. n(B – A) (1)
3. n(A – B) (1)

Answer:

1. n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
   ⇒ 42 = 20 + n(B) – 4 ⇒ n(B) = 26
2. n(B – A) = n(B) – n(A ∩ B) = 26 – 4 = 22
3. n(A – B) = n(A) – n(A ∩ B) = 20 – 4 = 16.

Plus One Maths Chapter Wise Questions And Answers Question 6.
A = {x: x is a natural number less than 8}

1. Write in roster form. (1)
2. Write a subset of A containing all even numbers in A. (1)
3. Which of the following could not be the number of elements of power set of a set [2, 8, 10, 16]? (1)

Answer:

1. A = {1, 2, 3, 4, 5, 6, 7}
2. {2, 4, 6} or {2, 4, 6, 7}
3. 10. (since other are powers of 2.)

Plus One Maths Sets Four Mark Questions and Answers

Plus One Maths First Chapter Questions And Answers Question 1.
Observe the Venn diagram.

1. Write in roster form. (1)
2. Verify that (A – B) ∪ (A ∩ B) = A (2)
3. Find (A ∩ B)’ (1)

Answer:

1. A = {1, 3, 4, 8} ; B = {2, 3, 5}
2. A – B = {1, 4, 8}; A ∩ B = {3}
   ⇒ (A – B) ∪ (A ∩ B) = {1, 3, 4, 8}
   Hence; (A – B) ∪ (A ∩ B) = A
3. (A ∩ B)’ = {1, 2, 4, 5, 6, 7, 8, 9}

Plus One Maths Sets Practice Problems Questions and Answers

Plus One Maths Chapter Wise Questions Question 1.
Write the following sets in roster form.

1. A = {x: x xis an integer and -3 < x < 7}
2. B = {x: x ∈ N; x ≤ 6}
3. C = {x : x is a vowel in English alphabet}
4. D = {x : x is a two-digit natural number such that the sum of its digits is 8}
5. E = {x: x ∈ Z; \(-\frac{1}{2}<x<\frac{9}{2}\)}

Answer:

1. A = {-2, -1, 0, 1, 2, 3, 4, 5, 6}
2. B = {1, 2, 3, 4, 5, 6}
3. C = {a, e, i, o, u}
4. D = {17, 71, 26, 62, 35, 53}
5. E = {1, 2, 3, 4}

Plus One Maths Text Book Questions And Answers Question 2.
Write the following sets in Set builder form.

1. A = {3, 6, 9, 12}
2. B = {2, 4, 8, 16, 32}
3. C = \(\left\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}\right\}\)
4. D = {5, 25, 125, 625}
5. E = {2, 4, 6……..} (1 score each)

Answer:

1. A = {x: x = 3n, n ∈ N, n< 4 }
2. B = {x: x = 2ⁿ; n ∈ N; n < 5 }
3. C = {x: x = \(\frac{n}{n+1}\); n ∈ N, n ≤ 6}
4. D = {x: x = 5ⁿ; n ∈ N, n ≤ 4 }
5. E = {x: x is an even number}

Plus One Maths Chapter Wise Questions And Answers Pdf Hsslive Question 3.
Write the following in interval form.

1. {x: x∈ R, -4 < x ≤ 6}
2. {x: x∈ R, 0 ≤ x < 7 }
3. {x: x∈ R, 3 ≤ x ≤ 4 } (1 score each)

Answer:

1. (-4, 6]
2. [0, 7)
3. [3, 4]

Plus One Maths Sets Questions And Answers Pdf Question 4.
Write the following in set builder form.

1. [0, 10]
2. [-2, 7)
3. (3, 4)

Answer:

1. {x: x∈ R, 0 ≤ x ≤ 10 }
2. { x: x ∈ R, -2 ≤ x < 7 }
3. {x: x ∈ R, 3 < x < 4 }

Plus One Maths Questions And Answers Question 5.
Find Set A, B and Universal set U (1 score each)

Answer:
A = {e, f, d}; B = {a, b, c, d} and U = {a, b, c, d, e, f, g, h}

Hsslive Maths Textbook Answers Plus One Question 6.
Write all subset of the following

1. {1, 2}
2. {a, b, c}
3. Φ (1 score each)

Answer:

1. Φ, {1}, {2}, {1, 2}
2. Φ, {a}, {b}, {c} ,{a, b}, {a, c}, {b, c}, {a, b, c}
3. Φ

Plus One Mathematics Questions And Answers Question 7.
Let A = {1, 2, {3, 4}, s, d, θ} , Which of the following statements are true/false and why?

1. 3 ∈ A
2. {1, {3, 4}} ∈ A
3. {1, 2, 3} ⊂ A
4. Φ ∈ A
5. 1 ⊂ A (1 score each)

Answer:

1. False
2. True
3. False
4. False
5. False

Important Questions For Class 11 Maths Sets Question 8.
If A = {1, 2, 4, 6, 7, 8}; B = {2, 5, 7, 9, 10} and C = {4 , 5, 9, 10}. Find

1. A ∪ B
2. B ∪ C
3. A ∪ C
4. A ∩ B
5. B ∩ C
6. A∪ B ∪ C
7. A ∩ B ∩ C
8. (A ∩ B) ∪ (C ∩ A) (1 score each)

Answer:

1. A ∪ B = {1, 2, 4, 5, 6, 7, 8, 9, 10}
2. B ∪ C = {2, 4, 5, 7, 9, 10}
3. A ∪ C = {1, 2, 4, 5, 6, 7, 8, 9, 10}
4. A ∩ B = {2, 7}
5. B ∩ C = {5, 9, 10}
6. A ∪ B ∪ C = {1, 2, 4, 5, 6, 7, 8, 9, 10}
7. A ∩ B ∩ C = Φ
8. (A ∩ B) ∪ (C ∩ A) = {2, 7} u {4} = {2, 4, 7}

Plus One Maths Important Questions And Answers Question 9.
If A = {2, 4, 6, 7, 8, 12}; B = {2, 7, 9, 10} and C = {5, 9, 10, 12}. Find

1. A – B
2. B – C
3. A – C
4. B – A
5. C – A
6. (A ∪ B) – C
7. A – {B ∩ C)
8. (A ∩ B) – (C ∩ A) (1 score each)

Answer:

1. A – B = {4, 6, 8, 12}
2. B – C = {2, 7}
3. A – C = {2, 4, 6, 7, 8}
4. B – A = {9, 10}
5. C – A = {5, 9, 10}
6. (A ∪ B) – C = {2, 4, 6, 7, 8, 9, 10, 12} – {5, 9, 10, 12} = {2, 4, 6, 7, 8}.
7. A – (B ∩ C) = {2, 4, 6, 7, 8, 12} – {9, 10} = {2, 4, 6, 7, 8, 12}
8. (A ∩ B) – (C ∩ A) = {2, 7} – {12} = {2, 7}

Plus One Maths Previous Questions And Answers Question 10.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}; A = {2, 7, 9, 10} B = {5, 9, 10, 12} and C = {1, 4, 5, 7, 11}. Find

1. A’
2. B’
3. A’ – C
4. (B – A)’
5. B’ ∩ C’
6. (A ∪ B)’
7. A’ ∩ B’ (1 score each)

Answer:

1. A’ = { 1, 3, 4, 5, 6, 8, 11, 12}
2. B’ = { 1, 2, 3, 4, 6, 7, 8, 11}
3. A’ – C = {1, 3, 4, 5, 6, 8, 11, 12} – {2, 3, 6, 8, 9, 10, 12} = {1, 4, 5, 11}
4. (B – A)’ = {5, 12}’ = {1, 2, 3, 4, 6, 7, 8, 9, 10, 11}
5. B’ ∩ C’ = {1, 2, 3, 4, 6, 7, 8, 11} – {2, 3, 6, 8, 9, 10, 12} = {2, 3, 6, 8}
6. (A ∪ B)’ = {2, 5, 7, 9, 10, 12}’ = {1, 3, 4, 6, 8, 11}
7. A’ ∩ B’ = {1, 3, 4, 5, 6, 8, 11, 12} ∩ {1, 2, 3, 4, 6, 7, 8, 11} = {1, 3, 4, 6, 8, 11}

Plus One Maths Text Book Answers Question 11.
If X and Y are two sets such that X ∪ Y has 50 elements, X has 28 elements and Y has 32 elements, how many elements does X ∩ Y have?
Answer:
Given; n(X ∪ Y) = 50; n(X) = 28; n(Y) = 32
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 50 = 28 + 32 – n(X ∩ Y)
⇒ n(X ∩ Y) = 60 – 50 = 10

Students can Download Chapter 4 Motion in a Plane Notes, Plus One Physics Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Physics Notes Chapter 4 Motion in a Plane

Summary
Motion In A Plane Class 11 Notes Pdf Introduction
In this chapter, we will study, about vector, its ’ addition, substraction and multiplication We then discuss motion of an object in a plane. We shall also discuss uniform circular motion in detail.

Plus One Physics Chapter 4 Notes Scalars And Vectors

a. Scalars:
A quantity which has only magnitude and no direction is called a scalar quantity.
Eg: length; volume, mass, time, work etc.

b. Vectors:
(i) The need for vectors:
In one dimensional motion, there are only two possible directions. But in two or three dimensional motion, infinite number of directions are possible. Hence quantities like displacement, velocity, force etc. cannot be represented by magnitude alone: Therefore in order to describe such quantities, not only magnitude but direction also is essential.

(ii) Vector:
A physical quantity which has both magnitude and direction is called a vector quantity.
Eg: Displacement, Velocity, Acceleration, Force, momentum.

1. Position and Displacement Vectors:
Position vector:
Consider the motion of an object in a plane. Let P be the position of object at time tw.r.t.origin given O.

A vector representing the position of an object P with respect to an origin O is called position vector \(\overrightarrow{\mathrm{OP}}\) of the object. This position vector may be represented
by an arrow with tail at O and head at P.

The length of the line gives the magnitude of the vector and arrow head (tip) indicates its direction in space. The magnitude of OP is represented by |\(\overrightarrow{\mathrm{OP}}\)|.

Displacement vector:

Consider the motion of an object in a plane. Let P be the position of a moving object at a time t and p¹ that at a later time t¹. \(\overrightarrow{\mathrm{OP}}\) and \(\overrightarrow{\mathrm{OP}^{1}}\) are the position vectors at time t and t¹ respectively. So the vector \(\overrightarrow{\mathrm{PP}^{1}}\) is called displacement vector corresponding to the motion in the time interval (t – t¹).

2. Equality of vectors:
Two vectors are said to be equal if they have the same magnitude and direction.

The above figure shows two vectors \(\vec{A}\) and \(\vec{B}\) having the same magnitude and direction.
∴ \(\vec{A}\) = \(\vec{B}\).

Motion In A Plane Class 11 Notes Hsslive Question 1.
Observe the following figures (a) and (b) and find which pair does represents equal vectors?

Answer:
Figure a represent that A and B are equal vectors. Two vectors A¹ and B¹ are unequal, because they were in different directions.

Multiplication Of Vectors By Real Numbers
Multiplying a vector \(\vec{A}\) with a positive number I gives a vector whose magnitude is changed by the factor λ.

The direction λ\(\vec{A}\) is the same as that of \(\vec{A}\).
Examples:

A vector \(\vec{A}\) and the resultant vector after multiplying \(\vec{A}\) by a positive number 2.

A vector A and resultant vector after multiplying it by a negative number-1 and -1.5.

Addition And Subtraction Of Vectors – Graphical Method
Vectors representing physical quantities of the same dimensions can be added or subtracted. The sum of two or more vectors is known as their resultant.

1. When two vectors are acting in the same direction:

2. When two vectors act in opposite direction:
In this case, the angle between the vectors is 180°.

The resultant of the two vectors is a new vector whose magnitude is the difference between the magnitudes of the two vectors and whose direction is the same as the direction of the bigger vector.

3. When two vectors are inclined to each other:
The sum of two vectors inclined at an angle q can be obtained either by

• the law of triangle of vectors
• the parallelogram law of vectors

(i) Triangle method:
This law states that if two vectors can be represented in magnitude and direction by the two sides of a triangle taken in the same order, then the resultant is represented in magnitude and direction by the third side of the triangle taken in the reverse order.

Explanation
Consider two vectors \(\vec{A}\) and \(\vec{B}\) as shown in figure.

(ii) Parallelogram law of vector addition:
This law states that if two vectors acting at a point can be represented in magnitude and direction by the two adjacent sides of a parallelogram, then the diagonal of the parallelogram through that point represents the resultant vector.
Explanation
Consider two vectors \(\vec{A}\) and \(\vec{B}\) as shown in figure.

To find \(\vec{A}\) + \(\vec{B}\), we bring theirtails to a common origin Q as shown below.


The diagonal of parallelogram OQSP, gives the resultantof (\(\vec{R}\) = \(\vec{A}\) + \(\vec{B}\)) of two vectors \(\vec{A}\) and \(\vec{B}\).
Note: Triangle and parallelogram law of vector addition gives the same result, ie. the two methods are equivalent.

4. Substraction of vectors:

To substract \(\vec{B}\) from \(\vec{A}\), reverse the direction of \(\vec{B}\).

Then add –\(\vec{B}\) with \(\vec{A}\) using parallelogram law or tri¬angle law.

The resultant of \(\vec{A}\) and \(\vec{B}\) is given by \(\vec{R}\).
Null vector or zero vector:
A vector having zero magnitude is called a zero vector or null vector. Null vector is represented by \(\vec{O}\). Since the magnitude is zero, we don’t have to specify its direction.
Properties of null vector:

Class 11 Physics Motion In A Plane Notes Pdf Question 2.
Explain a zero vector using an example.
Answer:
Suppose that an object which is at P at time t, moves to p¹ and then comes back to P. In this case displacement is a null vector.

Resolution Of Vectors Unit Vectors
A vector divided by its magnitude is called unit vector along the direction of that vector. A unit vector in the direction of \(\vec{A}\) is written as \(\hat{A}\).

Orthogonal unit vectors:

In the Cartesian coordinate system, the unit vectors along the X, Y and Z directions are represented by \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) respectively and are known as orthogonal unit vectors.
For unit vectors

Resolution of vector into rectangular components:
The components of a vector in two mutually perpendicular directions are called its rectangular components.
Explanation

Consider a vector \(\overrightarrow{\mathrm{A}}\) that lies in x-y plane as shown in figure. To resolve \(\overrightarrow{\mathrm{A}}\), draw lines from the head of \(\overrightarrow{\mathrm{A}}\) perpendicularto the coordinate axes as shown below.

The quantities A_x and A_y are called x and y components of the vector \(\overrightarrow{\mathrm{A}}\). Hence the vector \(\overrightarrow{\mathrm{A}}\) can be written in terms of rectangular components as

Magnitude of \(\overrightarrow{\mathrm{A}}\):

From the figure, the magnitude of \(\overrightarrow{\mathrm{A}}\) can be written as,

Vectors Physics Class 11 Notes Pdf Question 3.
A vector \(\overrightarrow{\mathrm{A}}\) in xyz plane is given below. A_x, A_y and A_z are the perpendicular components in x,y and z directions respectively.

1. Write \(\overrightarrow{\mathrm{A}}\) in terms of rectangular components.
2. Write the magnitude of \(\overrightarrow{\mathrm{A}}\).

Answer:

The magnitude of vector \(\overrightarrow{\mathrm{A}}\) is

Vector Addition – Analytical Method
The graphical method of adding vectors helps us in visualizing the vectors and the resultant vector. But this method has limited accuracy and sometimes tedious. Hence we use analytical method to add vectors.
Explanation

The vectors obey commutative and associative laws. Hence

Projectile Motion Class 11 Notes Question 4.
Find the magnitude and direction of the resultant of two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{A}}\) in terms of their magnitudes and angle between them.
Answer:

Consider two vectors \(\vec{A}(=\overrightarrow{O P}) \text { and } \vec{B}(=\overrightarrow{O Q})\) making an angle q. Using the parallelogram method of
vectors, the resultant vector \(\overrightarrow{\mathrm{R}}\) can be written as,

SN is normal to OP and PM is normal to OS. From the geometry of the figure
OS² = ON² + SN²
but ON = OP + PN
ie. OS² = (OP+PN)² + SN² ______(1)
From the triangle SPN, we get
PN = Bcosq and SN = Bsinq
Substituting these values in eq.(1), we get
OS² = (OP + Bcosq)² + (Bsinq)²
But OS = R and OP = A
R² = (A + Bcosq)² + B²sin²q
= A² + 2ABcosq + B²cos²q + B²sin²q
R² = A² + 2 ABcosq + B²

The resultant vector \(\overrightarrow{\mathrm{R}}\) make an angle a with \overrightarrow{\mathrm{A}}. From the right angled triangle OSN,

But SN = Bsinq PN = Bcosq

Motion In A Plane

1. Position vector and displacement vector Position vector:

Consider a small body located at P with reference to the origin O. The position vector of the point ‘P’

Displacement vector


where Dx = x¹ – x¹, Dy = y¹ – y
Velocity:
If Dt is the time taken to reach from P to P¹
The average velocity, \(\overrightarrow{\mathrm{v}}_{\mathrm{av}}=\frac{\overrightarrow{\Delta r}}{\Delta \mathrm{t}}\) ____(3)
Substitute eq.(2) in eq.(3), we get

The direction of average velocity is the same as that of \(\overrightarrow{\Delta r}\).
The instantaneous velocity can be written as

Acceleration:
If the velocity of an object changes from \(\overrightarrow{\mathrm{v}} \text { to } \overrightarrow{\mathrm{v}^{1}}\) in time Dt, then its average acceleration is given by

Instantaneous acceleration:
The acceleration at any instant is called instantaneous acceleration. When Dt goes to zero, the average acceleration becomes instantaneous acceleration.
ie. Instantaneous acceleration

Motion In A Plane With Constant Acceleration
Consider an object moving in xy plane with constant acceleration ‘a’. Let \(\vec{u}\) be the initial velocity at t=0 and \(\vec{v}\) be the final velocity at time t.
Then by definition acceleration

In terms of components
v_x = u_x + a_xt
v_y = u_y + a_yt
Displacement in a plane
If \(\overrightarrow{\mathrm{r}_{0}}\) and \(\vec{r}\) be position vectors of particle at t = 0 and time t respectively, then
displacement = \(\vec{r}-\vec{r}_{0}\) _______(1)
For uniformly accelerated motion, displacement,

In terms of components
x = x₀ + u_xt + 1/2 a_xt²
y = y₀ + u_yt + 1/2 a_yt²
The eq.(2) shows that, the above motion in xy plane can be treated as two separate one dimensional motions along two perpendicular directions.

Relative Velocity In Two Dimensions
Consider two bodies A and B moving along a plane with velocities \(\overrightarrow{\mathrm{V}}_{\mathrm{A}}\) and \(\overrightarrow{\mathrm{V}}_{\mathrm{B}}\). Then velocity of A relative to that of B is,

Similarly velocity of B relative to that of A

Projectile Motion
Projectile:
A body is projected into air and is allowed to move under the influence of gravity is called projectile.

Consider a body which is projected into air with a velocity u at an angle q. The initial velocity ‘u’ can be divided into two components ucosq along horizontal direction and using along vertical direction.

1. Time of flight:
The time taken by the projectile to cover the horizontal range is called the time of flight. Time of flight of projectile is decided by usinq. The time of flight can be found using the formula
s = ut + 1/2 at²
Taking vertical displacement s = 0, a = -g and initial vertical velocity = usinq, we get
0 = usinqt – 1/2gt²
1/2 gt² = usinqt

2. Vertical height:
Vertical height of body is decided by vertical component of velocity (usinq). The vertical displacement of projectile can be found using the formula v² = u² + 2as
When we substitute v=0, a = -g, s = H and u = usinq, we get
0 = (usinq)² + 2 – g × H
2gH = u²sin²q

3. Horizontal Range:
If we neglect the air resistance, the horizontal velocity (ucosq) of projectile will be a constant.
Hence the horizontal distance (R) can be found as
R = horizontal velocity × time of flight

The eq.(3) shows that, R is maximum when sin2q is maximum, ie. When q₀ = 45°.
The maximum horizontal range

Equation for path of projectile

Motion In A Plane Class 11 Pdf Question 5.
What is the shape of path followed by the projectile? Show that the path of projectile is parabola. The vertical displacement of projectile at any time t, can be found using the formula.
Answer:
S = ut+ 1/2at²
y = usinqt – 1/2gt²
But we know horizontal displacement, x = ucosq × t

In this equation g, q and u are constants. Hence eq.(4) can be written in the form
y = ax + bx²
where a and b are constants. This is the equation of parabola, ie. the path of the projectile is a parabola.

Uniform Circular Motion
The motion of an object along the circumference of a circle is called circular motion.
Uniform circular motion:
When an object follows a circular path at a constant speed, the motion is called uniform circular motion.
Period:
The time taken by the object to complete one full revolution is called the period.
Frequency:
The number of revolutions completed per second is called the frequency u of the circular motion.
If the period of a circular motion isT, its frequency

Angular Displacement (Dq):
The angle Dq in radians swept out by the radius vector in a given interval of time is called the angular displacement of the object.
Angular velocity:
The rate of change of angular displacement is called the angular velocity.

If T is the period of an object, then its radius vector sweeps out an angle of 2p radian.
Therefore in one second it sweeps out an angle \(\frac{2 \pi}{T}\).
∴ Angular velocity of the object

Expression for velocity and acceleration in uniform circular motion:

The direction of velocity is in the direction of tangent at that point. The change in velocity vectors \((\overrightarrow{\Delta v})\) is obtained by triangle law of vector as shown in figure (b).

a. Speed and angular speed in uniform circular motion:
Let the Dq be the angle constructed by the body during the time interval ∆t. The angular velocity can be written as

If the distance travelled by the object during the time Dt is Dr (ie. PP¹ = Ds) then speed

But Ds = RDq
where R = \(|\vec{r}|=|\overrightarrow{r^{\prime}}|\)
Substituting Dr = RDq in eq.(1)
we get

b. Acceleration in uniform circular motion:

The direction of this acceleration should be in the direction of \(\overrightarrow{\Delta V}\). The fig(b) shows that \(\overrightarrow{\Delta V}\) is towards the centre of the circular path. Hence the acceleration is directed towards the centre of the circle and is called centripetal acceleration.

The force which produces this centripetal acceleration is called centripetal force.
Centripetal force can be written as
F = ma_c

But ω = \(\frac{V}{R}\). Hence we get K

Students can Download Chapter 1 Introduction to Accounting Questions and Answers, Plus One Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Accountancy Chapter Wise Questions and Answers Chapter 1 Introduction to Accounting

Plus One Accountancy Introduction to Accounting One Mark Questions and Answers

Plus One Business Studies Chapter 1 Notes Pdf Question 1.
Who among the following are not a user of accounting information?
(a) Management
(b) Investors
(c) Advertisers
(d) Lenders
Answer:
(c) Advertisers

Plus One Business Studies Chapter 1 Notes Question 2.
Spot the odd one out and state reason
(a) Loose Tools
(b) Copy Write
(c) Patent
(d) Goodwill
Answer:
(a) Loose Tools, is a Fixed assed, all others are intangible assets.

Plus One Business Studies Chapter 1 Question 3.
A Person who owes money to the business is a
(a) Debtor
(b) Investor
(c) Creditor
(d) Borrower
Answer:
(a) Debtor

Plus One Business Studies Notes Question 4.
Book-keeping is concerned with
(a) Analysis of transaction
(b) Recording of transaction
(c) Classification of transaction
Answer:
(b) Recording of transaction

Plus One Business Studies Notes In English Question 5.
Amount spent for purchasing fixed asset is a
(a) Revenue Expenditure
(b) Capital Expenditure
(c) Deferred Revenue Expenditure
Answer:
(b) Capital Expenditure.

Plus One Business Studies Chapter 1 In Malayalam Question 6.
Which quantitative characteristics of accounting in-formation is reflected when accounting information is clearly presented?
(a) Understandability
(b) Relevance
(c) Comparability
(d) Reliability
Answer:
(a) Understandability.

Plus One Business Studies Notes Pdf Question 7.
Which of the following is an example of a business transaction?
(a) Appointed Mr.Ram as the Manager of the business with a salary of Rs. 15,000.
(b) Obtain a loan of Rs. 1,00,000 to the business from Bank of India.
(c) Sent a quotation to Matha Traders worth Rs. 20,000.
Answer:
(b) Obtained a loan of Rs. 1,00,000 from Bank of India. Loan is taken meant for business. So it is a business transaction.

Hsslive Plus One Business Studies Notes Question 8.
Find the odd one out and state the reasons,
(a) share capital
(b) Debentures
(c) Sundry creditors
(d) Long-term loans.
Answer:
(c) Sundry creditors, all others are long term liabilities.

Business Studies Plus One Notes Question 9.
Value of goods remaining unsold at the end of an accounting period is termed as …………
Answer:
Closing Stocks

Plus One Business Studies Chapter Wise Notes Question 10.
Arun, a sole trader, draw Rs. 500 from the business for paying tuition fees to his child. This amount is termed as ……….
Answer:
Drawings.

Nature And Purpose Of Business Notes Pdf Question 11.
Assets minus liabilities are called ……….
Answer:
Capital.

Nature And Purpose Of Business Class 11 Notes Pdf Download Question 12.
………….. assets are those assets, which do not have any real value.
Answer:
Fictitious Assets.

Hss Live Business Studies Plus One Notes Question 13.
The amount earned by a business concern through sale of its products or providing services to customers is called ……
Answer:
Revenues.

Plus One Business Studies Notes In English Pdf Question 14.
The assets bought for long-term use in the business are termed as …………….. assets
Answer:
Fixed.

Hsslive Business Studies Plus One Notes Question 15.
Analysis of recorded data to bring entries of similar nature to one plane is called …………..
Answer:
Classifying.

Hsslive Plus One Business Studies Malayalam Notes Question 16.
A Person who is entitled to get money from the business is termed as …………
Answer:
Creditor

Question 17.
Information in financial reports is based on …………….. transaction.
Answer:
Economic.

Question 18.
All claims against the business are called ………………..
Answer:
Equity

Question 19.
The transaction is one wherein payment or receipt of money is postponed for a future date.
Answer:
Credit transaction.

Question 20.
Mr. Ismail, who is the owner of a Provision shop, took 50 kg. of rice worth Rs. 600 for his house-hold use. He should record this as
Answer:
Drawings.

Question 21.
Ravi, a trader purchased 100 notebooks from ‘Shyni stores’ on credit. How is Shyni stores related to Ravi?
Answer:
Shyni stores is the creditor of Ravi.

Question 22.
Who was the inventor of double-entry bookkeeping?
Answer:
Luca Pacioli

Question 23.
Expand AICPA
Answer:
The American Institute of Certified Public Accounts.

Question 24.
Identify the events not used to the accounting treatment.
(a) Commenced business with cash
(b) Bought Machinery for cash
(c) Cash Purchase of goods.
(d) The firm appointed an efficient Manger.
Answer:
(d) The firm appointed an efficient Manger.

Plus One Accountancy Introduction to Accounting Two Mark Questions and Answers

Question 1.
Define Accounting.
Answer:
According to American Institute of certified Public Accountants, “Accounting is the art of recording, classifying and summarizing in a significant manner and in terms of money, transactions and events which are in part, at least, of a financial character and interpreting the results thereof.”

Question 2.
‘Accounting is the language of the business’. Why?
Answer:
Accounting is the language of the business:
The performance of business in terms of profit or losses is conveyed to users of accounting information in a systematic manner. The financial position of the business concerned is revealed through accounting information.

Question 3.
“Raju sold goods to Rahim on credit”. What relation exists between them? What are the accounting terms involved in it?
Answer:

• Rahim – Debtor
• Raju – Creditor

Question 4.
You are the accountant of a firm. What are the functions to be performed by you?
Answer:
Accounting provides information regarding the financial status of a business and results of its operations. The following are the important functions of accountant of a firm.

1. Recording transactions by referring source documents.
2. Preparing journal, subdivision of journal.
3. Preparing ledger accounts.
4. Summarising.
5. Making statements of interpretation 0 Reporting to Management.

Question5.
Users of accounting are classified as under:

1. Internal users – Management, Investor, Creditor, Bank, Employees, Stock exchange.
2. External Users – Customers, Government, Researchers, Lenders.

Do you agree with this classification, if not correct it?
Answer:
No,
1. Internal Users:

• Management
• Employees

2. External Users:

• Investors
• Creditors
• Bank
• Stock Exchange
• Government
• Customers
• Lenders
• Researchers

Question 6.
All business transactions are events. But all events are not business transactions. Comment.
Answer:
Events can be anything, some events can be expressed in monetary terms while others are not. Only those events which can be expressed in money, terms are business transactions. Transaction is an event or economic activity of a businessman in his business having exchange of money or money’s worth. While events is part of the business transaction.

Question 7.
Classify the following expenses into capital expenditure and revenue expenditure.

• Machinery purchased
• Rent paid
• Interest paid
• Purchased building
• The amount for repair of the building

Answer:
1. Capital Expenditure:

• Machinery Purchase
• Purchased building

2. Revenue Expenditure:

• Rent paid
• Interest
• paid Amount for repair of building

Question 8.
How will you define Revenue and Expenses?
Answer:
1. Revenue:
These are the amounts earned by a business concern through sale of its products or providing services to customers. The common items of revenues are sales, commission received, rent received, interest received, etc.

2. Expenses:
The amount spent in the process of earning revenue is termed as expenses. Examples are Wages, Salaries, rent, Interest paid, electricity charges, etc.

Question 9.
What is a capital expenditure? Give some examples?
Answer:
Capital expenditure represents the amount spent for the acquisition of assets, the benefit from which is derived over a period that extends beyond the accounting year. It is long term in nature.
Examples: Furniture purchased, Land Purchased, Building purchased, etc.

Question 10.
Explain the meaning of Gain and Profit.
Answer:
1. Profit: The excess of revenues of a period over its related expenses during an accounting year is profit. Profit increases the investment of the owners.

2. Gain: A profit that arises from events or transactions which are incidental to business such as sale of fixed assets, winning a court case, receipt of interest and dividend, etc. Gain is irregular in nature. Gains are part of capital receipt. Gains are also known as “non-operating income.”

Question 11.
Match the following.

Answer:

Question 12.
Name the branches of accounting.
Answer:

1. Financial Accounting
2. Cost Accounting
3. Management Accounting

Plus One Accountancy Introduction to Accounting Three Mark Questions and Answers

Question 1.
Classify the following assets into suitable head Goodwill, Building, Land, Patent, Cash, Oilwell, Copy-write, Debtors, Stock, mines, Bill receivable, Preliminary expenses.
Answer:

Question 2.
Define assets, Liabilities, and capital.
Answer:
1. Assets:
Assets are properties and things of value owned by the business which can be expressed in monetary terms. Examples of Machinery, Buildings, Stock, Debtors, Furniture, etc.

2. Liabilities:
Liabilities are the obligations that an enterprise owes. These represent the amount payable by the business in the future. They represent the claim against the asset of business. Examples Loans, Creditors, Bills payable, etc.

3. Capital:
Capital is the investment made by the owners for use in the business. It is owner’s claim on the total assets of the business and is also called “owners equity”.

Question 3.
Distinguish between:

1. Goods and Assets
2. Expense and Loss

Answer:
1. Goods and Assets:

• Goods refer to things in which the trader deals. But assets refers to things with which the trader deals.
• Goods are meant for resale, while assets are kept in the business permanently with the help of which the business is carried on.

2. Expense and Loss:

• Costs incurred by a business in the process of earning revenue are known as expense.
• The excess of expenses of a period over its related revenues is termed as loss. It decreases in owner’s equity.

Plus One Accountancy Introduction to Accounting Four Mark Questions and Answers

Question 1.
Accounting has certain objectives to business Explain.
Answer:
The following are the important objectives of accounting.

1. Keeping of records of business transactions.
2. Ascertainment of Profit or Loss.
3. Ascertainment of financial position of business enterprises.
4. Providing meaningful information to different groups of people having interest in the business.

1. Keeping of records of business transaction:
The main purpose of accounting is to identify business transactions of financial nature and enter into appropriate books of accounts. The accounting records should be made properly and systematically, so that requisite information may be obtained at a glance.

2. Ascertainment of Profit or Loss:
The result of business (Profit or Loss) is available from the statement prepared for ascertaining it, called the Profit and Loss Account.

3. Ascertainment of financial position:
At the end of an‘ accounting year, a position statement known as the ‘Balance Sheet’ is prepared. The value of assets and liabilities are depicted in the Balance Sheet. The Balance sheet gives a true and fair view of the state of affairs of the concern.

4. Providing meaningful information to different groups of people having interest in the business:
Accounting records provide meaningful information to different groups of people having interest in the business.

Question 2.
Accounting information must possess certain qualitative characteristics. What are they?
Answer:
The following are the qualitative characteristic of accounting information.

1. Reliability: Accounting information will be reliable if it is free from error and faithfully represents what it seeks to represent.
2. Relevance: Information should be relevant and must be available in.time.
3. Understandability: Accounting information that is relevant must be capable of being understood by all its users.
4. Timeliness: Information must be available timely. If not, it loses its ability to influence decision.
5. Comparability: Accounting information should facilitate inter-firm comparison as well as interfirm comparison.
   Maximum Cputtishers

Question3.
Accounting and Book-keeping are viewed as distinct functions. Mention any four differences between Accounting and Book Keeping.
Answer:

Question 4.
“Accounting gives number of advantages to the business”. What are the important advantages of accounting?
Answer:
The following are the advantages of accounting
1. Provide Quantitative information:
Accounting helps in gathering quantitative information on profits earned by the business or loss sustained by them.

2. Helps in ascertaining financial position of the business that is, total assets owned and total liabilities owed.

3. Helps in making a systematic record of transactions, which can be used for future reference and appropriate retrieval.

4. Acts as an information system:
It provides adequate information to the interested users in a processed form.

5. Beneficial to different interested users of accounting information.

Question 5.
Accounting has certain ‘Limitations’. Explain.
The following are the limitations of accounting.
Answer:
1. It records only transactions which can be recorded in monetary terms:
Qualitative aspects like managerial skill, Services of experts, etc. are not recorded.

2. Accounting is a post mortem survey:
It records events as they have taken place. For example, expenses are recorded as incurred, assets are recorded at their cost of purchase. There is no scope for ascertaining what the appropriate expenditure or cost of acquisition should have been.

3. Effect of price level changes are not considered:
Transactions are always recorded in the books at cost price and not at market price.

4. Inexactness:
Accounting transactions are not exact. Different firms have their own different methods, so the results of the business will change in the practice.

Plus One Accountancy Introduction to Accounting Six Mark Questions and Answers

Question 1.
What are the different types of assets? Explain briefly.
Assets are things of value owned. They may be subdivided into the following.
Answer:
1. Fixed Assets:
Fixed Assets are assets held on long term basis, such as land, buildings, machinery, plant, furniture, etc. These assets are used for the normal operations of the business.

2. Current Assets:
These are assets held on a short-term basis such as debtors, bills receivables, stock, cash in hand, cash at bank, etc. It is also known as “Floating asset.”

3. Fictitious Assets:
These are those assets, which do not have a physical form. They do not have any real value. Actually, they are not the real assets but they are called assets on legal and technical grounds. Examples are preliminary expenses, discount on issue of shares or debentures, etc.

4. Tangible Assets:
Assets having physical existence which can be seen, touched are known as tangible assets. These assets are land, building, plant, equipment, etc.

5. Intangible Assets:
These assets have no physical existence which cannot be touched, seen or felt. Examples are Goodwill, trademark, patent, copyright.

6. Wasting Assets:
Assets, whose value goes on declining with the passage of time, are known as wasting assets. Mines, oilwells, quarries are its examples.

7. Liquid Assets:
Liquid assets are those assets, which can be converted into cash at short notice. The examples of liquid assets are cash in hand, cash at bank, debtors, bills receivable, etc. Liquid assets = Current Assets – (Stock + Prepaid Expenses)

Question 2.
“Accounting provides information to various users.” Discuss accounting as an information system.
Answer:
Accounting plays a significant role in society by providing information to management at all levels (internal users) and to those having a direct financial interest in the enterprise (external users), such as present and potential investors, creditors. Accounting information is also important to those having an indirect financial interest, such as regulatory agencies, tax authorities, customers, labour unions, stock exchange, and others.

Internal users, mainly management, need timely information on cost of sales, profitability, etc. for planning, controlling and decision making. External users who have limited authority, ability and resources to obtain the necessary information have to rely on financial statements. The external users are interested in the following.

1. Investors and Potential investors:
Information on the risks and returns on investments.

2. Suppliers and Creditors:
Information on whether amounts owed will be repaid when due and on the continued existence of the business.

3. Customers:
Information on the continued existence of the business and thus the profitability of a continued supply of products, parts, and after-sales services.

4. Employees:
They are interested in getting their salary, welfare measures, bonus, working conditions, etc. which are all related to financial performance of the business.

5. Lenders:
Information on the creditworthiness of the business and its ability to repay loans and pay interest.

6. Government and other regulators:
Information on the allocation of resources and the compliance to regulators.

Students can Download Chapter 1 Nature and Purpose of Business Notes, Plus One Business Studies Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Business Studies Notes Chapter 1 Nature and Purpose of Business

1st Standard Malayalam Textbooks Chapter 1 Contents

• Economic and non-economic activities
• Business and its characteristics
• Business, Profession and Employment
• Classification of business activities
• Industry and commerce
• Objectives of business
• Business risk
• Factors to be considered before starting a business.

Activities which human beings undertake are known as human activities. We can divide these activities into two categories.

1. Economic activities
2. Non-economic activities

1. Economic Activities:
The human activities that are undertaken with an objective to earn money or livelihood is known as economic activities. eg: A worker working in a factory, a doctor operating in his hospital, a manager working in the office, a teacher teaching in a school etc.

2. Non-economic activities:
Activities are undertaken to derive psychological satisfaction are known as non economic activities. eg: Mother preparing food for her children, Praying, listening to radio or watching television, playing football with friends, etc.

Types of economic activities:

• Business
• Employment
• Profession

1st Standard Malayalam Chapter 1 Business:
Business may be defined as an economic activity involving the production or purchase and sale of goods and services with the main object of earning profit by satisfying human needs in the society.
Characteristics of business

1. Business is an economic activity with the object of earning profit.
2. Business includes all the activities concerned with the production or procurement of goods and services.
3. There should be sale or exchange of goods and services for the satisfaction of human needs.
4. Business involves dealings in goods or services on a regular basis. Normally, one single transaction of sale or purchase is not treated as business.
5. One of the main objectives of business is to earn maximum profit.
6. Business involves risk and uncertainty of income. Risk means the possibility of loss due to change in consumer taste and fashion, strike, lockout competition, fire, theft etc.

Malayalam Text Book Std 1 Pdf Download Employment:
Employment refers to that type of economic activity in which people engage in some work for others regularly and get salary or wages in return of their services.
Characteristics of Employment

1. There must exist employer-employee relationship.
2. There must be a service contract between the employer and employee.
3. Employees get salary or wages for their services
4. Regularity in service.

First Standard Malayalam Textbooks Chapter 1 Profession:
Profession refers to an occupation which requires specialized knowledge, skip and training. Its objective is to provide service to the society.
Characteristics of Profession

1. A profession requires specialized knowledge, training and skill
2. The membership of a professional body is a must
3. Professionals have a code of conduct
4. They charge fee in return of their service.

Comparison of Business, Profession and Employment:

Classification of Business Activities: Business activities may be classified into two categories

• Industry
• Commerce.

Chart showing business activities

1 Standard Malayalam Textbooks Chapter 1 Industry:
Industry refers to economic activities, which are connected with conversion of resources into useful goods. Industries may be divided into 3 categories.
They are
1. Primary industries:
Primary industries are connected with the extraction and production of natural resources and reproduction and development of living organisms, plants, etc. Such industries are further divided into two.
(i) Extractive industries:
These industries extract products from natural resources. eg: mining, farrqing, hunting, fishing etc.

(ii) Genetic industries:
These industries are engaged in activities like rearing and breeding of animals, birds and plants. eg: diary faming, paultry farming, floriculture, pisciculture etc.

2. Secondary industries:
Secondary industries deal with materials extracted at the primary stage. Such goods may be used for consumption or for further production. Secondary industries are classified into two.
They are:
(i) Manufacturing industries:
Manufacturing industries engage in converting raw materials into finished goods. eg: Conversion of rubber into cotton, timber into furniture rubber into tyres etc. Manufacturing industries may be further divided into four categories. They are,

• Analytical industry which analyses and separates different elements from the same materials. eg: Oil refinery
• Synthetical industry which combines various ingredients into a new product. eg: cement
• Processing industry which involves successive stages for manufacturing finished products. eg: Sugar and paper industry.
• Assembling industry which assembles different component parts to make a new product. eg: television, car, computer, etc.

(ii) Construction industries:
These industries are involved in the construction of buildings, dams, bridges, roads etc.

3. Tertiary industries:
These are concerned with providing support services to primary and secondary industries. eg: Transport, banking, insurance, warehousing, communication, advertising etc.

1st Standard English Book Chapter 1 Commerce:
Commerce is defined as all activities involving the removal of hindrances in the process of exchange of goods. It includes all those activities, which are necessary for the free flow of goods and services from the producer to the consumer. Commerce includes trade and auxiliaries to trade.
Commerce = Trade + auxiliaries to trade
Functions of commerce:

Malayalam 1st Standard Chapter 1 Trade:
Trade refers to sale, transfer or exchange of goods. Trade may be classified into two broad categories
They are:

1. Internal trade
2. External trade

1. Internal, domestic ot dome trade:
is concerned with the buying and selling of goods and services within the geographical boundaries of a country. This may further be divided into two. They are:-
(a) Wholesale trade:
Under wholesale trade, the trader purchases goods in large quantities from the producers, and sells them in smaller quantities to the retailers.

(b) Retail trade:
Under the retail trade, the trader buys in comparatively smaller quantities from the wholesalers or producers and sells them to ultimate consumers.

2. External or Foreign trade:
Foreign trade consists of exchange of goods and services between two or more countries. Foreign trade may be divided in to three.

• Import trade: If goods are purchased from a foreign country, it is called import trade.
• Export trade: When goods are sold to a foreign country, it is known as export trade.
• Entrepot trade: When goods are imported for export to other countries, it is known as entrepot trade.

Auxiliaries to Trade (Aids to trade):
Activities which assist trade are called aids to trade or auxiliaries to trade.

1. Transport & Communication:
Transport facilitates the movement of raw material to the place of production and the finished products from factories to the place of consumption. Communication helps the producers, traders and consumers to exchange information with one another.

2. Banking & Finance:
Banking helps business activities to overcome the problem of finance. Commercial banks lend money in the form of overdraft, cash credit, loans and advances etc… and they also provide many services required for the business activity.

3. Insurance:
The goods may be destroyed while in production process or in transit due to accidents, or in storage due to fire or theft, etc. Insurance provides protection in all such cases.

4. Warehousing:
The goods should be stored carefully from the time they are produced till the time they are sold. This function is performed by warehouses.

5. Advertising:
Advertising helps in providing information about available goods and services and create in them a strong desire to buy the product.

Multiple Objectives of Business:
The main objectives of a business are:
1. Market standing:
A business firm can succeed only when it has a good market standing. Market standing refers to the position of an enterprise in relation to its competitors.

2. Innovation:
Innovation means developing new product or services orfinding new ideas and new methods of production and distribution. Innovation accelerates the growth of an enterprise.

3. Productivity:
Productivity is ascertained by comparing the value of output with the value of input. Every enterprise must aim at greater productivity through the best use of available resources.

4. Physical and financial resources:
The business must aim at maximum utilization of available physical and financial resources, i.e. men, material, money and machine in the best possible manner.

5. Earning Profit:
Earning maximum profit is the primary objective of every business. Profit is required for survival and growth of a business.

6. Manager performance and development:
Efficient managers are needed to conduct and co-ordinate business activities. So it is the objective of an enterprise to implement various programs for motivating the managers.

7. Worker performance and attitude:
Every enterprise must aim at improving its workers performance by providing fair salary, incentives, good working conditions, medical and housing facilities.

8. Social responsibility:
It refers to the obligation of business firm to contribute resources for solving social problems and work in a socially desirable manner.

Malayalam Text Book Std 1 Chapter 1 Business Risks:
The term ‘business risks’ refers to the possibility of inadequate profits or even losses due to uncertainties or unexpected events. Business enterprises may face two types of risk, i.e. speculative risk and pure risk.

Speculative risks involve both the possibility of gain as well as the possibility of loss. It arise due to change in demand, change in price etc. Pure risks involve only the possibility of loss or no loss. The chance of fire, theft or strike is examples of pure risks.
Nature of Business Risks:

1. Business risks arise due to uncertainties.
2. Risk can be minimized, but cannot be eliminated. It is an essential part of business.
3. Degree of risk depends mainly upon the nature and size of business.
4. Profit is the reward for risk taking.

Causes of Business Risks: Business risks arise due to a variety of causes.
They are:

1. Natural Causes: it includes natural calamities like flood, earthquakes, lightning, heavy rains, famine, etc.
2. Human Causes: Human causes include dishonesty, carelessness or negligence of employees, strikes, riots, management inefficiency, etc.
3. Economic causes: These include change in demand, change in price, competition, technological changes etc.
4. Political Causes: Change in Govt, policies, taxation, licensing policy etc.

Starting a Business – Basic Factors:
Factors to be considered for starting a business:
1. Selection of line of business:
The first thing to be decided by any entrepreneur of a new business is the nature and type of business to be undertaken.

2. Size of the firm:
If the market conditions are favorable, the entrepreneur can start the business at a large scale. If the market conditions are uncertain and risks are high, a small size business would be better choice.

3. Choice of form of ownership:
The selection of a suitable form of business enterprise i.e. Sole proprietorship, Partnership or a Joint stock company is an important management decision. It depends on factors like nature of business, capital requirements, liability of owners, legal formalities, continuity of business etc.

4. Location of business enterprise:
Availability of raw materials and labour, power supply and services like banking, transportation, communication, warehousing, etc., are important factors while making a choice of location.

5. Financing:
Proper financial planning must be done to determine (a) the requirement of capital, (b) source from which capital will be raised and (c) the best ways of utilizing the capital in the firm.

6. Physical facilities:
Availability of physical facilities including machines and equipment, building and supportive services is a very important factor to be considered at the start of the business.

7. Plant layout:
Layout means the physical arrangement of machines and equipment needed to manufacture a product.

8. Competent worked force:
Every enterprise needs competent and committed employees to perform various activities so that physical and financial resources are converted into desired outputs.

9. Tax planning:
The promoter must consider in advance the tax liability under various tax laws and its impact on business decision.

Students can Download Chapter 1 Biological Classification Notes, Plus One Botany Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Botany Notes Chapter 1 Biological Classification

Two kingdom classification:

• It was proposed by Linnaeus, include two Kingdoms-Plantae & Animalia.
• In two kingdom classification following things are not considered

Cell structure, nature of wall, mode of nutrition, habitat methods of reproduction and evolutionary relationship .

Five kingdom classification:

• It was first proposed by R H.Whittaker (1969).
• It includes Monera, Protista, Fungi, Plantae and Animalia
• Blue green algae are placed in kingdom Monera.
• Chlamydomonas, Chlorella with Paramoecium and Amoeba are placed in kingdom protista
• Chlorophyll less and non cellulosic (chitin)type plants are placed in kingdom Fungi
• All photosynthetic plants are placed in-kingdom plantae
• All animals with mode of nutrition(ingestion) placed in kingdom animalia.
• The characteristics of classification are

Kite Victers Plus One Botany Notes Pdf Kingdom monera:
Types of bacteria:
Based on shape, bacteria are of 4 types

1. Spherical – Coccus
2. Rod – shaped – Bacillus
3. Comma – shaped – Vibrium
4. Spiral – Spirillum.

Based on nutrition, bacteria are of 3 types:

1. Photosynthetic autotrophic: Bacteria can synthesise their own food by using chlorophyll in the presence of light.
2. Chemosynthetic autotrophic: Bacteria can synthesise their own food from inorganic substrates.
3. Heterotrophs: They are depend on other organisms for food.

Plus One Botany Notes Archaebacteria:

• These bacteria can live in extreme conditions.
• Its cell wall structure is different from other bacteria

Types of archaebacteria:

1. Halophiles: They are found in salty areas
2. Thermoacidophiles: They are found in hot springs.
3. Methanogens: They are found in marshy areas and guts of ruminant animals eg-cows, buffaloes etc.

Importance in industry:
They are responsible for the production of methane (biogas) from the dung.

Eubacteria Cyanobacteria (blue-green algae):

• They have chlorophyll a similar to green plants called as photosynthetic autotrophs.
• Some of them found in polluted water bodies.

Significance of cyanobacteria:
They can fix atmospheric nitrogen in specialised cells called heterocysts and increases fertility of soil eg: Nostoc and Anabaena

Chemosynthetic autotrophs:
They oxidise various inorganic substances such as nitrates, nitrites and ammonia and use the released energy for their ATP production.

Significance chemosynthetic autotrophs:
They play a great role in the recycling of nutrients like nitrogen, phosphorous, iron and sulphur.

Heterotrophic bacteria:

• They depend upon others for getting energy.
• Most of them are decomposers.
• They are helpful in making curd from milk, production of antibiotics, fixing nitrogen in legume roots, etc.

Disease caused by bacteria:
Some are pathogens causing disease to plants and animals, eg: Cholera, typhoid, tetanus, and citrus canker in plants.

Cell division in bacteria:
Reproduction:

• Bacteria reproduce mainly by fission.
• Some bacteria produce spores during unfavourable conditions.
• In sexual reproduction, transfer of DNA from one bacterium to the other takes place

Mycoplasmas:
Salient features:

• They are the smallest living cells can survive without oxygen
• They are pathogenic in animals and plants
• They lack a cell wall.

Plus One Botany Chapter Wise Notes Kingdom Protista:
Salient features:

• They are single-celled eukaryotes.
• Their cell body contains a well defined nucleus and other membrane-bound organelles.
• They are mainly aquatic.

Types of protist:
The kingdom include chrysophytes, Dianoflagellates, Euglenoids, Slime moulds and Protozoans.

Plus One Botany Chapters Chrysophytes:

• They are diatoms(chief ‘producers’ in the oceans) and golden algae (desmids).
• Most of them are photosynthetic.

Salient features:

• In diatoms the cell walls form two thin overlapping shells,which fit together as in a soap box.
• Their cell wall contain silica

Economic value:
Their cell wall deposited in ocean floor over billions of years in large amount called as ‘Diatomaceous earth’. It is used in polishing, filtration of oils and syrups.

Plus One Botany Notes Hsslive Dinoflagellates:
Salient features:

• They are marine and photosynthetic
• Most of them have two flagella one lies longitudinally and the other transversely.

Harmful effects:

• Red dianoflagellates -Gonyaulax undergoes rapid multiplication and sea appear red (red tides).
• Toxins released by them kill other marine animals such fishes.

Plus One Botany Victers Notes Euglenoids:
Salient features:

• They are fresh water organisms found in stagnant water.
• Pigments of Euglenoids are identical to those of higher plants.
• They have a protein rich layer called pellicle which makes their body flexible.
• They have two flagella, a short and a long one.

Mode of nutrition:
They are photosynthetic in the presence of sunlight (autotrophic) and predating on other smaller organisms in the absence of sunlight (heterotrophs). Hence nutrition is mixotrophic. Example: Euglena.

Plus One Botany Notes Victers Slime Moulds:
Salient features:

1. They are saprophytic protists.
2. They form an aggregation during favourable conditions called plasmodium which may grow and spread over several feet.
3. The plasmodium differentiates and forms fruiting bodies bearing spores at their tips during unfavourable conditions.
4. They are resistant and survive for many years under adverse conditions.

Plus One Botany Notes Pdf Protozoans:
They are heterotrophs and live as predators or parasites.
Type of protozoans:
1. Amoeboid protozoans:

• They live in freshwater or sea water
• They capture their prey by pseudopodia (false feet) as in Amoeba.
• Some are parasites as in Entamoeba

2. Flagellated protozoans:

• They possesss flagella.
• The parasitic forms cause diseases such as sleeping sickness. Example: Trypanosoma.

3. Ciliated protozoans:

• They are aquatic and have a cavity (gullet) that collect food from outside.
• They move with the help of cilia. Example: Paramoecium.

4 Sporozoans:
They are infectious due to the spore-like stage in their life cycle. eg: Plasmodium (malarial parasite) which causes malaria.

Hsslive Plus One Botany Notes Kingdom Fungi:
They are heteterotrophs, mainly 2 types

1. Saprophytes: They can absorb soluble organic matter from dead substrates
2. Parasites: They are depend on living plants and animals

Symbiotic associations:

1. Lichens-Fungi forms an association with algae
2. Mycorrhiza-Fungi forms association with roots of higher plants.

Disease caused bv fungi:
They cause diseases in plants and animals. eg: wheat rust disease by Puccinia.

Salient features:

• Yeast is unicellular fungus but others are multicellular.
• Mycelium: lt is body of fungi contains many hyphae
• Hyphae It is the long, slender thread-like structures
• Some hyphae with multinucleated cytoplasm are called coenocytic hyphae.
• Others have septae or cross walls in their hyphae. eg: Penicillium.
• The cell walls of fungi are composed of chitin and polysaccharides

Types of Reproduction:

1. Vegetative method: It takesplace by fragmentation, fission and budding.
2. Asexual method: It takesplace by spores called conidia orsporangiospores or zoospores.
3. Sexual reproduction: It takesplace by oospores, ascospores and basidiospores.
4. The spores are produced in special structures called fruiting bodies.

Steps of sexual cycle:

In this, two haploid hyphae come together and fuse results in diploid cells (2n).

Dikarvotic stage in funai:

1. In ascomycetes and basidiomycetes, after plasmogamy dikaryotic stage (n + n ) occurs for some time. Later these nuclei fuse and the cells become diploid.
2. The fungi form fruiting bodies in which reduction division occurs and forms haploid spores.

Hss Live Plus One Botany Notes Phvcomycetes:
Salient features:

• They are found in aquatic habitats and on decaying wood.
• Their mycelium is aseptate and coenocytic.

Types of reproduction:
1 Asexual reproduction:
It takes place by zoospores (motile) or by aplanospores (non-motile). These spores are produced in sporangium.

2 Sexual reproduction:
It is the fusion of gametes have similar structure (isogamous) or dissimilar structure (anisogamous or oogamous) and after fusion zygospore is formed. Examples are Mucor, Rhizopus (the bread mould) and Albugo (the parasitic fungi on mustard).

Hss Live Plus One Biology Notes Ascomycetes:
Salient features:

• They are commonly known as sac-fungi, eg-unicellular- yeast (Sacharomyces) or multicellular Penicillium.
• Some are coprophilous (growing on dung).
• Mycelium is branched and septate.
• The asexual spores are conidia produced on conidiophores.
• Sexual spores are called ascospores which are produced in sac like ascus.
• Spores are arranged in fruiting bodies called ascocarps.
• Examples are Aspergillus, Claviceps and Neurospora.

Economic value:

• Neurospora is used in genetic studies.
• Edible members are morels and buffles.

Plus One Botany Notes Chapter 1 Basidiomycetes:
Salient features:

• Their mycelium is branched and septate.
• Some members grow as parasites and disease causing organisms e.g. rusts and smuts
• Common basidiomycetes are mushrooms, bracket fungi or puffballs

Reproduction:

1. The asexual spores and sex organs are not found but vegetative reproduction by fragmentation.
2. In sexual reproduction, plasmogamy occur by fusion of two vegetative cells of different strains. It results dikaryotic mycelia which gives rise to basidium. Later, karyogamy and meiosis takeplace in the basidium and producing four basidiospores.
3. The basidia are arranged in fruiting bodies called basidiocarps.
4. Examples -Agaricus (mushroom) Ustilago (smut) and Puccinia (rust fungus).

Hsslive Botany Plus One Notes Deuteromvcetes:
Salient features:

• They are called as imperfect fungi because perfect stage or sexual reproduction is absent.
• Their asexual reproduction takes place with the help of conidia. ,
• The mycelium is septate and branched.

Economic value:

• Some members are decomposers play an important role in the mineral cycling.
• Examples are Alternaria, Colletotrichum and Trichoderma.

Hsslive Botany Notes Plus One kingdom Plantae

1. Majority members are eukaryotic, chlorophyll-containing organisms.
2. Few members are partially heterotrophic- insectivorous plants or parasite. eg: Bladder wort and Venus fly trap are examples of insectivorous plants and Cuscuta is a parasite.

Different types of plant group:
The kingdom Plantae includes algae, bryophytes, pteridophytes, gymnosperms and angiosperms.

Life cycle:
It has two distinct phases – the diploid sporophytic and the haploid gametophytic – that alternate with each other.

Plus One Botany Biological Classification Notes Kingdom Animalia:
Salient features:

• It includes heterotrophic eukaryotic organisms.
• They are multicellular and their cells lack cell wall.
• Their mode of nutrition is holozoic (ingestion of food).
• Most of them are capable of locomotion.

Plus One Botany Kite Victers Notes Pdf Viruses, Viroids And Lichens:
R.H. Whittaker not placed acellular organisms such as viruses, viroids and lichens in five kingdom classification.

VIRUSES:
Historical aspects and Discovery

1. Name virus that means venom or poisonous fluid was given by Pasteur D.J. Ivanowsky.
2. Extract of the infected plants of tobacco could cause infection in healthy plants and called the fluid as Contagium vivum fluidum (infectious living fluid).It was identified by M.W. Beijerinek (1898)
3. Viruses could be crystallised and crystals consist of proteins outside. It was identified by W.M. Stanley. (1935)

Salient features:

• Viruses are non living particle outside the living cell.
• It has an inert crystalline structure .
• They have living state inside the host and multiply by using host cell machinery-Ribosome. So they are called as obligate parasites.
• They are smaller than bacteria because they passed through bacteria-proof filters.

Structure of viruses:
Viruses contain proteins coat outside, either RNA or DNA inside, (i.e either single or double stranded RNA or double stranded DNA).

Symptoms and disease caused by viruses:
Disease:
Mumps, smallpox, herpes, influenza and AIDS
Symptoms in plants:

Viroids:
Strucure:
It has free RNA without protein coat.

Discovery:
T .O .Diener found that this infectious agent was smaller than viruses.

Disease:
It causes potato spindle tuber disease

Lichens:
They are symbiotic associations between algae and fungi.

Types of component:

1. The algal component is called phycobiont: (autotrophic) prepare food for fungi
2. The fungal component is called mycobiont: (heterotrophic) provide shelter and absorb mineral nutrients and water for algae.

Significance:
Lichens are very good pollution indicators i.e they do not grow in polluted areas.

Ncert Supplementary Syllabus
Six kingdom classification:
It was proposed by Carl Woese. It includes kingdoms like Archaebacteria, Eubacteria Protista, Mycota, Plantae and Animalia.

Students can Download Chapter 1 The Living World Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examination

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 1 The Living World

Plus One The Living World One Mark Questions and Answers

Plus One Zoology Exercise Answers Chapter 1 Question 1.
As we go from species to kingdom in a taxonomic hierarchy, the number of common characteristics
(a) Will decrease
(b) Will increase
(c) Remain same
(d) May increase or decrease
Answer:
(a) Will decrease

Plus One Zoology Chapter Wise Questions And Answers Pdf Question 2.
Which of the following ‘suffixes’ used for units of classification in plants indicates a taxonomic category of ‘family’.
(a) -Ales
(b) -Onae
(c) -Aceae
(d) none of the above
Answer:
(c) -Aceae

Zoology Chapter Wise Questions And Answers Question 3.
The term ‘systematics’ refers to:
(a) Identification and classification of plants and animals
(b) Nomenclature and identification of plants and animals
(c) Diversity of kinds of organisms and their relationship
(d) Different kinds of organisms and their classification
answer:
(c) Diversity of kinds of organisms and their relationship

Plus One Zoology Chapter Wise Questions And Answers Pdf Download Question 4.
Genus represents
(a) An individual plant or animal
(b) A collection of plants or animals
(c) Group of closely related species of plants or animals
(d) None of these
Answer:
(c) Group of closely related species of plants or animals

Plus One Zoology Previous Question Papers Chapter Wise  Question 5.
The taxonomic unit ‘Phylum’ in the classification of animals is equivalent to which hierarchical level in classfication of plants.
(a) Class
(b) Order
(c) Division
(d) Family
Answer:
(c) Division

Plus One Zoology Chapter Wise Questions And Answers Question 6.
Correct and rewrite the following if there is any mistake.
Sativa Oriza, Tigris Panthera
Answer:
Oriza sativa, Panthera tigris

Plus One Zoology Chapter Wise Previous Questions And Answers Question 7.
Kingdom – Carnivora
Phylum – Mammalia
Class – Felidsffe
Order – Chordata
Family – Tigris
Genus – Animalia
Sps – Panthera

Correct the mistakes of the flow chart to get the hierarchical arrangement of tiger in the ascending order.
Answer:
Kingdom – Animalia
Phylum – Chordata
Class – Mammalia
Order – Carnivora
Family – Felidae
Genus – Panthera
Sps – Tigris

Hsslive Plus One Zoology Chapter Wise Questions And Answers Question 8.
Match the following:

1. A Unit of classification – kingdom
2. Lowest taxonomic category – Genus
3. Panthera – Species
4. Highest Taxonomic category – Taxon

Answer:

1. A Unit of classification – Taxon
2. Lowest taxonomic category – Species
3. Panthera – Genus
4. Highest Taxonomic category – kingdom

Plus One Zoology Previous Question Papers Chapter Wise Hsslive Question 9.
Expand the term ICBN and ICZN.
Answer:
1. ICBN: International Code for Botanical Nomenclature.
2. ICZN: International Code for Zoological Nomenclature.

Plus One Zoology Chapter Wise Questions And Answers Pdf Hsslive Question 10.
Given below is the scientific name of Frog. Identify the correctly written name.
(a) Rana Tigrina
(b) Rana tigrina
Answer:
(b) Rana tigrina

Plus One Zoology Animal Kingdom Questions Question 11.
Rearrange the order of classification.
Genus, Family, Phylum, Species, Class, Kingdom, Order
Answer:
Species, Genus, Family, Order, Class, Phylum, Kingdom

The Living World Class 11 Questions And Answers Question 12.
Arrange the given terms in their taxonomic hierarchy.
Primata, Homosapien, Chordata, Mammalia, Hominidae.
Answer:
Homosapiens, Hominidae, Primata, Mammalia, Chordata.

Plus One The Living World Two Mark Questions and Answers

The Living World Class 11 Important Questions With Answers Question 1.
You are provided with a stuffed rabbit and a dried leaf. As a student of biology which taxonomic aid will you choose to store them.
Answer:
1. Stuffed rabbit – Museum
2. Dried leaf – Herbarium.

Question 2.
Find the odd one out and give reason.
Herbarium, botanical garden, museum, aquarium
Answer:
Aquarium – Aquarium is not a taxonomical aid.
All other are taxonomical aids.

Question 3.
Once you visited a museum. There are different kinds of animals and plants are preserved. How these plants and animals all preserved in a museum.
Answer:
Plants and animals are preserved in the containers or jars in preservative solutions. These may also be preserved as dry specimen. Insects are preserved in insect box.

Larger animals like birds and mammals are usually stuffed and preserved. Museums often have collections of skeleton of animals too.

Question 4.
Zoological parks are different from museums. Give reason?
Answer:
Zoological parks are the places where wild animals are kept in protected environments, that the conditions similar to their natural habitats. Museums have collection of preserved plants and animal specimens for study and reference.

Question 5.
Raju collected a skull of an animal and a living rare animal during a study tour. Select the suitable location for each from the list given in the brackets. (Botanical garden, Zoological park, Herbarium, Museum)
Answer:
Skull of an animal – museum
Living rare animal – Zoological park

Question 6.
Distinguish between taxonomic category and taxonomic hierarchy.
Answer:
Classification involves hierarchy of steps in which each step represents a rank or category. Since category is a part of overall taxonomic arrangement, it is called taxonomic category. All the categories together constitute taxonomic hierarchy.

Question 7.
Write the scientific name of following animals.

1. Lion
2. Frog
3. Housefly
4. Tiger

Answer:

1. Pantheraleo
2. Ranatigrina
3. Musca domestica
4. Panthera tigris

Question 8.
Fill in the blank spaces in the table given below.

Answer:

(a) Live
(b) Museum

Question 9.
Panthera jeo is the scientific name of Lion. List the rules you follow to write this scientific name.
Answer:

1. Biological names are generally in Latin and written in italics.
2. When handwritten, the words in a biological name are separately underlined.
3. The first word in a biological name represents the genus and second word represents the species name.
4. The first name (Genus) starts with capital letter and the second name (species) starts with small letter.

Question 10.
List the advantages of

1. Taxonomical key
2. Herbarium

Answer:
1. axonomical key is a taxonomical aid used for identification of plants and animals based on the similarities and dissimilarities. Key are generally analytical in nature.

2. Herbarium is a store house of collected plant specimens that are dried, pressed and preserved on sheets. These specimens, along with their descriptions on herbarium sheets, became a store house or repository for future use. Herbarium serves as quick reference system in taxonomical studies.

Question 11.
Give the terms.

1. The actual account of habital and distribution of plants of a given area.
2. Providing information for identification of names of species found in an area.
3. Contain information on any one taxon.
4. Identification of plants and animals based on the similarities and dissimilarities.

Answer:

1. Flora
2. Manuals
3. Monographs
4. Taxonomical key

Question 12.
Define a taxon. Give some examples of taxon at different hierarchical levels.
Answer:
Each category in the taxonomical hierarchy is considered as a taxonomic unit and is known as a taxon. The taxon used in the classification of animals are kingdom, phylum, class, order, family, genus and species.

Plus One The Living World NCERT Questions and Answers

Question 1.
What do we learn from identification of individuals and populations?
Answer:
In a diverse country like India can learn following things from identification of individuals and population:

1. Native place
2. MotherTongue
3. Costumes
4. Cuisine
5. Religion
6. Caste
7. Socio-economic Background

Question 2.
Can you identify the correct sequence of taxonomical categories?
(a) Species → Order → Phylum → Kingdom
(b) Genus → Species → Order → Kingdom
(c) Species → Genus → Order → Phylum
Answer:
As clear from the table in previous answer, (a) and (c) are showing the correct order.

Question 3.
Why are the classification system changing every now and then? ,
Answer:
In any branch of science nothing is written in concrete. Theories keep on changing as more relevant and correct theories are being discovered. In case of living beings certain species become extinct and some new species is being formed in every era.

This process of addition and deletion of species necessitates the continuous change of the classification system.

Question 4.
Define a taxon. Give some examples of taxa at different hierarchical levels.
Answer:
A taxon is a particular level of hierarchy in the system of classification of living beings. The following figure gives taxa at different hierarchical levels:

Plus One The Living World Multiple Choice Questions and Answers

Question 1.
Binomial nomenclature is described in the book
(a) Genera Plantarum
(b) Historia Plantarum
(c) Systema Naturae
(d) Flora Japonica
Answer:
(c) Systema Naturae

Question 2.
Which of the following is not the main criteria for five kingdom system of classification?
(a) Cell structure and thallus organization
(b) Mode of nutrition and reproduction
(c) Phylogenetic relationship.
(d) Gram staining
Answer:
(d) Gram staining

Question 3.
Two plants can be conclusively said to belong to the same species if they
(a) Can reproduce freely with each other and form seeds
(b) have more than 90 percent similar genes
(c) look similar and possess identical secondary metabolites
(d) have same number of chromosomes
Answer:
(a) Can reproduce freely with each other and form seeds

Question 4.
ICZN stands for
(a) International code of Botanical Nomenclature
(b) International code of Zoological Nomenclature
(c) international code of Viral Nomenclature
(d) International code of Zoo Nomenclature
Answer:
(b) International code of Zoological Nomenclature

Question 5.
In Mangifera indica L. Generic epithet is
(a) Indica
(b) Mangifera
(c) Linnaeus
(d) None of these
Answer:
(b) Mangifera

Question 6.
ICBN stands for
(a) Indian Congress of Biological Names
(b) International Code of Botanical Nomenclature
(c) International Congress of Biological Names
(d) Indian Code of Botanical Nomenclature
Answer:
(b) International Code of Botanical Nomenclature

Question 7.
Taxon is the
(a) taxonomic group of any rank
(b) procedure to assign a scientific name
(c) process of classification
(d) process by which anything is grouped into convenient categories based on characters
Answer:
(a) taxonomic group of any rank

Question 8.
Which of the following is famous for stating that “Population increases much faster than its food supply”?
(a) Fredrick Losch
(b) R Vircow
(c) T R Malthus
(d) Karl Von Baer
Answer:
(c) T R Malthus

Question 9.
Reproduction is the characteristic feature of living organisms. Which of the following can not reproduce?
(a) Amoeba and Paramecium
(b) Fungi and filamentous algae
(c) Humans and Ayes
(d) Mules and worker bees
Answer:
(d) Mules and worker bees

Question 10.
Nicotiana sylvestris flowers only during long days and N.tobacurh flowers only during short days. If raised in the laboratory under different photoperiods, they can be induced to flower at the same time and can be cross fertilized to produce self-fertile offspring. What is the best reason for considering N. sylvestris and N. tobocum to be separate species?
(a) They are physiologically distinct
(b) They are morphologically distinct
(c) They cannot interbreed in nature
(d) They are reproductively distinct
Answer:
(c) They cannot interbreed in nature

Question 11.
Which of the following book is not associated with Carolus Linnaeus, the father of Taxonomy and Nomenclature?
(a) Systema Naturae
(b) Genera Plantarum
(c) Species Plantarurn
(d) Historia Generalis Plantarum
Answer:
(d) Historia Generalis Plantarum

Question 12.
Philosophie Zoologique, a book written by Jean Baptiste de Lamarck is based on
(a) Survival of the fittest
(b) Natural Selection
(c) Inheritance of acquired characters
(d) Biogenetic law
Answer:
(c) Inheritance of acquired characters

Question 13.
The book “Philosophic Zoologique” was written by
(a) Lamarck
(b) Mendel
(c) Haeckel
(d) Hugo deVries
Answer:
(a) Lamarck

Question 14.
Binomial nomenclature was introduced by
(a) Linnaeus
(b) Darwin
(c) Aristotle
(d) deCandoile
Answer:
(a) Linnaeus

Students can Download Chapter 2 Animal Kingdom Questions and Answers, Plus One zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examination

Kerala Plus One Zoology Chapter Wise Questions and Answers Chapter 2 Animal Kingdom

Plus One Animal Kingdom One Mark Questions and Answers

Plus One Zoology Animal Kingdom Questions Chapter 2 Question 1.
Fill in the blanks.

Answer:
i) B – Nereis
ii) B – Ctenoplanna
C – Ctenophora
iii) A – radula

Plus One Zoology Chapter Wise Questions And Answers Pdf Chapter 2 Question 2.
Write any two members of the Phylum Aschelminthes which are found parasitic on Human beings.
Answer:

1. Ascaris (Roundworm)
2. Wuchereria (Filaria worm)

Plus One Zoology Previous Question Papers Chapter Wise Chapter 2 Question 3.
In some animal groups, the body is found divided into compartments with at least some organs/ organ repeated. This characteristic feature is named
(a) Segmentation
(b) Metamerism
(c) Metagenesis
(d) Metamorphosis
Answer:
(b) Metamerism

Presence Of Column And Metamerism Are The Most Important Characters In Chapter 2 Question 4.
Given below are types of cells present in some animals. Each one is specialized to perform a single specific function except
(a) Choanocytes
(b) Interstitial cells
(c) Gastrodermal cells
(d) Nematocytes
Answer:
(b) Interstitial cells

Plus One Zoology Chapter Wise Questions And Answers Chapter 2 Question 5.
Which one of the following sets of animals share a four chambered heart?
(a) Amphibian, Reptiles, Birds
(b) Crocodiles, Birds, Mammals
(c) Crocodiles, Lizards, Turtles
(d) Lizards, Mammals, Birds
Answer:
(b) Crocodiles, Birds, Mammals

Hsslive Plus One Zoology Chapter Wise Questions And Answers Chapter 2 Question 6.
Which of the following pairs of animals has non glandular skin.
(a) Snake and Frog
(b) Chameleon and Turtle
(c) Frog and Pigeon
(d) Crocodile and Tiger.
Answer:
(c) Frog and Pigeon

Plus One Zoology Previous Question Papers Chapter Wise Hsslive Chapter 2  Question 7.
Birds and mammals share one of the following characteristics as a common feature.
(a) Pigmented skin
(b) Alimentary canal with some modification
(c) Viviparity
(d) Warm blooded nature
Answer:
(d) Warm blooded nature

Plus One Zoology Chapter Wise Questions And Answers Pdf Download Chapter 2 Question 8.
Note the relationship between the first two words and find a suitable word for the fourth place,

1. Coelenterata: radial symmetry, platyhelminthes, _______
2. Lizard: Poikilothermous, crow, _________

Answer:

1. bilaterally symmetrical
2. Homoiothermous

Plus One Zoology Chapter Wise Previous Questions And Answers Chapter 2 Question 9.

1. Annelida: Parapodia :: __________ : Comb plates
2. _________: Water vascular system :: Coelenterata : cnidoblast

Answer:

1. Ctenophora
2. Echinodermata

Plus One Zoology Chapter Wise Questions And Answers Pdf Hsslive Chapter 2 Question 10.
Malpighian tubule is the excretory organ of which phylum?
(a) Phylum porifera
(b) Phylum arthropoda
(c) Phylum Coelenterata
(d) Phylum mollusca
Answer:
(b) Phylum Arthropoda

Pick Out The Acoelomate Organisms From The Following Chapter 2 Question 11.
A chordate animal having flame cells as the excretory organ.
Answer:
Amphioxus

Animal Kingdom Class 11 Important Questions And Answers Chapter 2 Question 12.
From the pictures given below, find out the poikilothermic animals.

Answer:
B and D are poikilothermic animals.

Question 13.
Segmentation in the body is first observed in which of the following:
(a) Platyhelminthes
(b) Aschelminthes
(c) Annelida
(d) Arthropoda
Answer:
(c) Annelida

Question 14.
Shark has to swim continuously, otherwise, it will sink down. Give reason.
Answer:
Due to absence of air bladder.

Plus One Animal Kingdom Two Mark Questions and Answers

Question 1.
Arrange the phylum in order.
Arthropoda → Platyhelminthes → Porifera → Ctenophora → Cnidaria → Mollusca → Annelida → Echinodermata → Aschelminthesip
Answer:

• Presence of milk producing mammary gland.
• Presence of hair on skin.
• External ear or pinnae is present.
• Different types of teeth are present in the jaw.

Question 2.
Complete the blanks.

Answer:

Question 3.
Triploblastic animals are more complex than diploblastic animals. Do you agree with this statement? Justify.
Answer:
Triploblastic animals have more cell layers so they have the possibility of greater degree of cellular specialisation.

Question 4.
All vertebrates are chordates but all chordates are not vertebrates. Justify.
Answer:
Notochord is present in all vertebrates but vertebral column is present only in vertebrates and not in all chordates.

Question 5.
Copy and complete the table.

Answer:

Question 6.
Arrange the phylum in order.
Arthropoda → Platyhelminthes → Porifera → Ctenophora → Cnidaria → Mollusca → Annelida → Echinodermata → Aschelminthes
Answer:
Porifera → Cnidaria → Ctenophora→ Platyhelminthes → Aschelminthes → Annelida → Arthropoda → Mollusca → Echinodermata

Question 7.
Categorise and classify the following organisms and arrange them in a table with separate columns and provide appropriate headings.
Exocoetus, Physalia, Ascaris, Apis, Locusta, Corvus, Pita, Hydra, Sepia, Ancylostoma.
Answer:

Question 8.

1. Identify the phylum which exhibit metagenesis or alternation of generation.
2. What is meant by alternation of generation?

Answer:

1. Cnidaria
2. Alternation of sexual and asexual forms of organism: ie., Polyp asexually produce medusa, medusa sexually product polyp.

Question 9.
Name the following.

1. Phylum in which flatworms are included
2. Excretory organs of Annelids.
3. Largest phylum.
4. An oviparous mammal.

Answer:

1. Platyhelminthes
2. Nephridia
3. Arthropoda
4. Platypus

Question 10.
Apis, Prawn, Locust, Spider
Following animals have different habit and habitat. But they have many common characters.

1. Mention the common characters.
2. Identify their phylum.

Answer:

1. Joint footed animals
   • Metameric segmentation
   • Chitinous exoskeleton
2. Arthropoda

Question 11.
A list of animals are given below. Arrange them according to increase in complexity of organization. Scorpion, Earthworm, Liver fluke, Pigeon, Seaanemon, Sycon, Elephant, Anabas.
Answer:
Sycon, Seaanemon, Liver fluke, Earthworm, Scor¬pion, Anabas, Pegeon, Elephant.

Question 12.
Nithin is Studying in Std. XI. He collected some specimens during the field trip conducted by the Science Club of his School. Help Nithin to Classify the Animal in respective Phylum.
Prawn, Slug worm, Butterfly, Pila, Grass Hopper, Crab
Answer:

Question 13.
During a field trip Raju has collected some organisms with the following characters. Help him to identify the phyla of those organisms.

1. Metamerically segmented body.
2. Body covered with calcareous shell.
3. Dorso-ventrally flattened leaf like body.
4. Body divided into head, thorax and abdomen.

Answer:

1. Annelida
2. Mollusca
3. Platyhelminthes
4. Arthropoda

Question 14.
Categorise the following fishes into Osteichthyes and Chondrichthyes?

1. Exocoetus
2. Trygon

Answer:

1. Exocoetus – Oesteichthyes
2. Trygon – Chondrichthyes

Question 15.
Arrange the following terms in two columns correctly. Malpighian tubules, radula, metamerism, Bioluminescence, choanocytes, nematocytes, Phylum-coelenterate, phylum-Arthropoda, phylum- ctnophora, phylum-Mollusca, Phylum-Porifera.
Answer:

Question 16.
Match the following

Answer:

Question 17.
Observe the given organisms

1. Place these animals in proper phylum.
2. Segmentation in the body is first observed in which of the above phylum?

Answer:

1.  i) Arthropoda,
   ii) Porifera,
   ii) Annelida
2. Annelida

Question 18.
Observe the table given below and fill the blank columns A, B, C and from the animals given in brackets. (Ascaris, Starfish, Fasciola, Earthworm)

Answer:

Question 19.
1. Identify the animal given below.

2. Write one well marked property of the above animal.
Answer:

1. Pleurobrachia
2. Bioluminescence is well marked property of pleurobrachia

Question 20.
Representatives of some vertebrate classes are introducing themselves. Write down the name of the class in which they belong.

1. My gills are covered by operculum. I have bony endoskeleton.
2. I give birth to young ones. My body is covered by hair.
3. My skin is glandular? Have trilocular heart.
4. I live only in marine water. My endoskeleton is made up of cartilage.

Answer:

1. Osteichthyes
2. Mammalia
3. Amphibia
4. Chondrichthyes

Question 21.
During classroom discussion a student said that sponges are more complex than cnidarians. Do you agree with him. Justify.
Answer:
NO. Sponges are asymmetrical and body is formed of loose aggregate of cells. Cells are not organised to from tissues and organs. Cnidarians are radially symmetrical and tissue grade of organisation. So cnidarians are more complex than sponges.

Question 22.
Due to the absence of air bladder, fishes belonging to the class Chondrichthyes have to swim constantly. How important is the presence of air bladder in these fishes?
Answer:
Due to the absence of air bladder in chondrichthyes, they have to swim constantly to avoid sinking. If air bladder is present which regulates buoyancy.

Question 23.
Match the column A, B &C in the table given below:

Answer:

Question 24.
Identify the phylum whose larvae are bilaterally symmetrical, but adults are radially symmetrical.

1. Annelida
2. Arthropoda
3. Mollusca
4. Echinodermata

Mention two salient features of the phylum.
Answer:
4. Echinodermata
Salient features Presence of Echinodermata:

• Water vascular system
• Spiny bodies, Endoskeleton of Calcareous ossicles

Question 25.
Write the name of phylum.

1. Diploblastic, tissue grade of organisation, radially symmetrical, polymorphic animals.
2. Soft bodied, Unsegmented, Bilaterally symmetrical animals with open type circulation.
3. Triploblastic, bilaterally symmetrical, true coelomic animals with metamerism, closed circulation.
4. Triploblastic, chitinous exoskeleton and open circulation.

Answer:

1. Cnidaria
2. Mollusca
3. Annelida
4. Arthropoda

Question 26.
Observe the table given below and fill the blank columns a, b, c, and d from the animals given in brackets.

(Hydra, Shark, Spongilla, Obelia)
Answer:
(a) Hydra/Obelia
(b) Shark
(c) Spongilla
(d) Hydra/Obelia

Question 27.
Copy and complete the table.

Answer:

Question 28.
You are provided with two fishes Catla (Bony fish) and Shark (Cartilagenous fish). Prepare a table showing difference in:
(a) Position of mouth
(b) Air bladder
(c) Scales
(d) Fertilization
Answer:

Question 29.
Prepare a list of some animals that are found parasitic on human beings.
Answer:
Tapeworm (Taenia), Ascaris (Roundworm), Wuchereria (Filaria worm), Ancylostoma(Hookworm)

Question 30.

1. Identify the 2 forms of Cnidarians.
2. Mention any 2 difference between them.

Answer:

1. A – polyp
   B – medusa
2. Polyp: Asexual, sessile, mouth upwards Medusa: Sexual, Free swimming, Mouth downwards

Plus One Animal Kingdom Three Mark Questions and Answers

Question 1.
Classify the given organisms and arrange them in the order of their phylum.
Limulus, Corvus, Spongilla, Ascaris, Physalia, Nereis, Catla, Sepia, Echinus, Taenia, Pleurobrachia, Tiger, Viper, Toad.
Answer:

Question 2.
Match the following.

Answer:

Question 3.
Observe the diagram

1. Identify the phylum of this hypothetical organism.
2. List out the features that helps in identifying it.
3. Write about the fate of notochord in Urochordata, Cephalochordata and Chordata.

Answer:

1. Chordata
2. Notochord, Dorsal nerve cord, Pharyngeal gill slits, Post anal tail
3. In Urochordatarfiotochond is present only in larval tail. In Cephalochondata, notochord extends from head to tail region is persistent throughout their life. In vertebrata, the notochord is replaced by a cartilagenous or bony vertebral column in the adults.

Question 4.
Select the following items into their appropriate phylum.
Radula, Parapodia, Comb plate, Nephridia, Choanocytes, Flame cells.
Answer:

• Radula – Mollusca
• Parapodia – Annelida
• Comb plate – Ctenophora
• Nephridia – Annelida
• Choanocytes – Porifera
• Flame cells – Platyhelminthes

Question 5.

(a) Fill and complete the chart given below.

(b) Write any two fundamental characters of the phylum chordata.
(c) Classify Tetrapoda into classes:
Answer:
Male accessory ducts store and transport the sperms from testis to the outside through urethra. Male accessory glands secrete seminal Plasma, Which is rich in fructose, citrate, prostaglandins and certain enzymes. The secretion of cowper’s glands lubricate the penis.

Question 6.
Write down the functions of the following (any two) structures and assign their phyla.

1. Radula
2. Flame cells
3. Parapodia

Answer:

1. Radula: File like rasping organ for feeding.
   Phylum: Mollusca
2. Flame cells: Osmoregulation and excretion
   Phylum: Platyhelminthes
3. Parapodia: help in swimming
   Phylum: Annelida

Question 7.
1. Identify the organism A and B.

2. Which features of these organism enable you to identify them?
Answer:

1. A-Tapeworm
   B – Earthworm
2. Features of these organism:
   • Tapeworm: Scolex is present, suckers and hooks are present. No true segmentation.
   • Earthworm: True segmentation, Absence of Scolex, hook and suckers, Clitellum is present.

Question 8.

1. Which of the following animals exhibit metagenesis? (Ascaris Obelia Earthworm Crab)
2. To which phylum does it belong?
3. Write any two features of the phylum.

Answer:

1. Obelia
2. Cnidaria
3. Two features of the phylum:
   • Presence of Cnidoblasts
   • Cnidarians exhibit two basic forms called polyp and medusa.

Question 9.
Pick out the appropriate one from the term given within bracket and put against the corresponding phylum.

1. Porifera
2. Coelenterata
3. Platyhelminthes
4. Annelida
5. Arthropoda
6. Mollusca
7. Echinodermata
8. Chordata

(Hirudin, Flame cell, Choanocytes, Cnidoblast, Jointed legs, Radula, Notochord and Dermal Ossicles)
Answer:

1. Porifera – Choanocytes
2. Coelenterata – Cnidoblast
3. Platyhelminthes – Flame cells
4. Annelida – Hirudin
5. Arthropoda – Jointed legs
6. Mollusca – Radula
7. Echinodermata -Dermal ossicles
8. Chordata – Notochord

Question 10.

1. Identify the organism (A) and (B) and which class do they belong?
2. On which basis do you classify these animals?

(Hint: Write any 2 Identifying characters)
Answer:

1. Organism in (A) and (B)
   • A – Bony fish – Osteichthyes
   • B – Cartilagenous fish – Chondrichthyes

2.

Question 11.
From a fish market, you got a fish, on a close watching you friend says it is a cartilaginous fish.

1. Which characters helped him to identify it as a cartilaginous fish, (any four characters.)
2. Name the class it belongs.

Answer:

1. Characters:
   • a – Gillslits are separate and without operculum
   • b – Placoidscale
   • c – Mouth is located ventrally
   • d – heterocercal caudal fin
2. Chondrichthyes

Question 12.
Presence or absence of a cavity between the body wall and the gut wall is very important in classification.

1. Identify the different types of body cavities.
2. Give examples to each

Answer:

1. Different types of Body Cavities:
   • a – Coelomate
   • b – Pseudocoelomate
   • c – Acoelomate
2. Examples:
   • Coelomate – Chordates
   • Pseudocoelomate – Aschelminthes
   • Acoelomate – Platyhelminthes

Question 13.
From the following general characters find out corresponding/Class with an Example

1. Exclusively marine, triploblastic, spines on the skin, radially symmetrical in the adult and bilaterally symmetrical in the larval stage.
2. Marine, they migrate towards freshwater for spawning, then their larvae return to ocean after metamorphosis.
3. Triploblastic, bilaterally symmetrical, coelomate and metamerically segmented animals.

Answer:

1. Echinodermata
   • eg: Starfish
2. Class – Cyclostomata
   • eg: Petromyzon
3. Annelida
   • eg: Earthworm

Question 14.
Arrange the following points in a two-column table and give suitable heading for each column.

1. Notochord present
2. Post anal tail absent
3. Pharynx Perforated by gill slits
4. Notochord absent
5. Posts anal tail present
6. Gill slits are absent

Answer:

Question 15.
Observe the figure and answer the questions.

1. Identify the structure.
2. Name the phylum which possess this structure.
3. How this structure help the organism?

Answer:

1. Cnidoblastornematocyst
2. Cnidaria
3. Capture of Prey
   • Defense organ.

Question 16.
Note the relationship between first two words and suggest suitable words for the 4th place.

1. Planaria: Flartfe cells: Earthworm: _________
2. Jaw present: Gnathostomata:: Jaw absent: ___________

Answer:

1. Nephridia
2. Agnatha

Question 17.
Give reasons for the following.

1. Respiratory and circulatory system are absent in parasitic platyhelminthes and Aschelminthes.
2. Arthropods are the most successful invertebrate.
3. Body of endoparasites are covered with cuticle.

Answer:

1. Parasitic Plalyhelminthes and Aschelminthes lives in anaerobic condition. So respiratory and circulatory systems are absent in these parasitic forms.
2. Arthropods are most successful, because of the presence of unique chitinous cuticle.
3. In these parasites, the cuticle helps in escaping from the action of digestive enzymes.

Question 18.
Arthropodes are organisms with chitinous exoskeleton. Suppose exoskeleton is absent in arthropods. List the difficulties arthropodes has to face.
Answer:

1. Their body will be dried up due to evaporation
2. They couldn’t escape from predators.
3. They couldn’t live in all environments.

Question 19.
Suppose during your field visit for collection from a rocky seashore you have got some live specimens such as sea urchin, sea cucumber, sea anemone. Is it possible to keep them on an aquarium in your school. Give reasons for your answer.
Answer:
No. It is not possible.
Marine animals cannot live on freshwater because it leads to endosmosis and death occurs.

Question 20.

1. Why nematocysts are more concentrated on the oral end and tentacles of cnidarians?
2. What are the difficulties that coelenterate have to face if nematocysts were absent in body.

Answer:

1. Nematocysts are concerned with defence and offence. Tentacles are usually used for defence, offence and food collection. Hence nematocysts are more concentrated in the oral end.
2. The major difficulties cnidarians has to face in the absence of nematocysts are for food collection and escaping from enemies.

Question 21.

1. Which of the following show the body cavity of earthworm?
2. Identify the names of germlayers ‘a’ and ’b’.

Answer:

1. Figure C is the body cavity of earthworm. Because it is a true coelom
2. The names of germlayers
   • a-ectoderm
   • b-endQderm

Question 22.
Identify the characters listed below and put (✓) mark on appropriate places

Answer:

Question 23.

1. Identify the organisms.
2. Which feature of these organisms enable you to identify them?

Answer:

1. The organisms:
   • A-Tape worm
   • B – Ascaris
   • C – Earthworm
2. Features:
   • Tapeworm – Body is dorso-ventrally flattened and segmented.
   • Ascaris – Body is cylindrical and vermiform
   • Earthworm – Body is divided into similar segments and clitellum is present.

Question 24.
During a classroom discussion a student said that sponges are more complex than cnidarians. Do you agree with him. Jusftfy.
Answer:
1. Sponges: Cellular grade of organisation and cell aggregate body plan.

2. Cnidarians: Tissue level of organisation and blind sac body plan. So Cnidarians are more complex than sponges.

Question 25.
While comparing the digestive system of a roundworm and flatworm, a boy noted some differences. List out the differences.
Answer:
1. Flatworms: Digestive system incomplete, has only a single opening, ingestion and egestion occurs through the same opening.

2. Round worm: Complete digestive system, it has both mouth and anus. A muscular pharynx is also present in their digestive system.

Question 26.

1. Identify the invertebrate whose mouth is on ventral or lower side and anus is on dorsal or upper side.
2. Mention the phylum.
3. Comment on its General characters.

Answer:

1. Starfish
2. Echinodermata
3. Its General characters are:
   • Water vascular system
   • Tube feet
   • Coelomate
   • Spiny body

Question 27.
Match the following.

Answer:

• Arthropoda – Bombyx – Joint footed animals
• Annelida – Earthworm – Little rings
• Echinodermata – Antedon – Spiny bodies animals
• Mollusca – Pila – Soft bodies animals

Question 28.
Animals showing metameric segmentation are included under annelida. Body of tapeworm has numerous segments, but the animal is not included under annelida. How will you account for it?
Answer:
Tapeworm show false segmentation. In true segmentation number of segments is fixed and age of the segments are all same.

Question 29.
Complete the given flow chart showing the flow of water through canal system in sponges.

Answer:

Question 30.
Mention a single word for the following.

1. Sexes are not separate.
2. Body wall with three layers.
3. File like rasping organs in Mollusca.
4. Segmentation of animal body.
5. The property of a living organism to emit light.
6. Alternation of generation in Cnidarians.

Answer:

1. Hermaphrodite
2. Triploblastic
3. Radula
4. Metamerism
5. Bioluminescence
6. Metagenesis

Question 31.
The birds are well adapted for flying.

1. Write the general characters of Aves.
2. Give some of the flight adaptations seen in birds.

Answer:

1. General characters of Aves:
   • Presence of feathers
   • Presence of beak
   • Forelimbs are modified into wings.
   • Warm blooded
   • Respiration by lungs with air sacs.
   • Oviparous
2. Flight adaptations:
   • Forelimbs are modified into wings
   • Long bones are hollow with air cavities (Pneumatic)
   • Air sacs connected to lungs supplement respiration
   • Constant body temperature

Plus One Animal Kingdom NCERT Questions and Answers

Question 1.
What is the difference between direct and indirect development?
Answer:
1. Direct Development:
In direct development the young animal resembles an adult. There is no intermediate stage.

2. Indirect Development:
In indirect development there is intermediate stage, like larval stage. For example, frog before being developed into adult passthrough a tadpole stage. This is the case of indirect development.

Question 2.
What are the peculiar features that you find in parasitic platyhelminthes?
Answer:
In parasitic platyhelminthes hooks and suckers are present. Suckers help the parasite, in sucking the blood from the host.

Question 3.
What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
Answer:
Arthropods are the first phylum to have well developed systems to carry out different activities. There is distinct system for respiration, locomotion and reproduction. Their survival capacity is great because of elaborate system. This has helped them survive in diverse conditions. They can live in water, on land and in air.

This can be one of the reasons why arthropods are the largest group among the animal kingdom. Another reason is their early development compared to animals of higher phylum.

Question 4.
Water vascular system is the characteristic of which group of the following:
(a) Porifera
(b) Ctenophora
(c) Echinodermata
(d) Chordata
Answer:
(c) Echinodermata

Question 5.
“All vertebrates are chordates but all chordates are not vertebrates”. Justify the statement.
Answer:
All chordates have notochord present in some stage of life.
The difference between vertebrates and nonvertebrates is as follows:
In vertebrates the notochord is present in the embryonic stage. This is replaced by a vertebral column during the adult stages.

Question 6.
How important is the presence of air bladder in Pisces?
Answer:
Presence of air-bladder in Pisces helps in buoyancy. This means that members of pisces don’t have to keep on swimming to remain floating.

Question 7.
What are the modifications that are observed in birds that help them fly?
Answer:
Following modification in birds help them fly:

1. Pneumatic or hollow bones make for a light weight skeleton.
2. Fore limbs are modified into wings to assist in flight.
3. Excertion of urine and faeces is through single opening facilitating weight reduction.
4. Aerodynamic body helps in flying.

Question 8.
Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
Answer:
Usually number of eggs produced by oviparous mothers is greater than number of young ones produced by viviparous mothers. The main reason for this is the need of resources required for development of the embryo.
In oviparous the major part of development of the embryo takes place outside the uterus. This makes lesser burden on the mother.

On the other hand in viviparous animals the development takes place inside the uterus so lesser number of young ones can be successfully incubated. Moreover, once eggs are outside they are at risk of getting eaten by some predator because of their immobility, so need of more eggs is there to ensure continuity of progeny.

Plus One Animal Kingdom Multiple Choice Questions and Answers

Question 1.
Calcareous skeleton is found in
(a) echinoderms
(b) some sponges
(c) mollusca
(d) all the above
Answer:
(d) all the above

Question 2.
Which cannot be the character of cnidaria
(a) musculoepithelial cells
(b) gastrovascular cavity
(c) nerve cells and process
(d) organ grade organization
Answer:
(d) organ grade organization

Question 3.
A non-matching set in the following is
(a) sepia – cuttle fish
(b) octopus – devilfish
(c) limulus – king crab
(d) ancylostoma – pinworm
Answer:
(d) ancylostoma – pinworm

Question 4.
A character common to Echinoderms and chordates
(a) marine
(b) benthonic
(c) deuterostome
(d) none of the above
Answer:
(c) deuterostome

Question 5.
Largest animal in the world that feed on smallest plankton is the
(a) dolphin
(b) killer whale
(c) blue whale
(d) sea cow
Answer:
(c) blue whale

Question 6.
Which is common to amphibian, reptelia and fishes
(a) nucleated RBC
(b) dermal scales
(c) poikelothermic condition
(d) both a and d
Answer:
(d) both a and d

Question 7.
Which of the following has pseudocoelomate tube within a tube body plan
(a) hydra
(b) planaria
(c) ascaris
(d) pheretima
Answer:
(c) ascaris

Question 8.
Ink gland associated with alimentary canal is found in
(a) sepia
(b) earthworm
(c) starfish
(d) cockroach
Answer:
(a) sepia

Question 9.
Which is common to all tetrapods
(a) epidermal scales
(b) red coloured blood
(c) 12 pairs of cranial nerve
(d) ureotelism
Answer:
(b) red coloured blood

Question 10.
Which one of the following is not a characteristic of phylum annelida?
(a) closed circulatory system
(b) segmentation
(c) pseudocoelom
(d) ventral nerve cord
Answer:
(c) pseudocoelom

Question 11.
Respiratory pigment of mollusc is
(a) haemocyanin
(b) haemoglobin
(c) haemoerythrin
(d) both a and b
Answer:
(a) haemocyanin

Question 12.
Select the character that can be attributed to chondrithytes
(a) persistened notochord
(b) placoid scales
(c) poikelothermic body
(d) all the above
Answer:
(d) all the above

Question 13.
The central cavity of sponge is called
(a) spongocoel
(b) coelocentron
(c) canal system
(d) spongilla
Answer:
(c) canal system

Question 14.
Flame cell are excretory organ of
(a) hydra
(b) cockroach
(c) planaria
(d) frog
Answer:
(c) planaria

Question 15.
Pneumatic skeleton is a feature of
(a) amphibians
(b) reptiles
(c) fishes
(d) birds
Answer:
(d) birds

Question 16.
The number of gills present in osteichthyes is
(a) 2 pairs
(b) 6-15 pairs
(c) 5 pairs
(d) 4 pairs
Answer:
(d) 4 pairs