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Chapter 1 : Force – Answers to Exercise 1(A) (Q1 – Q25) from Selina for Class 10 Physics ICSE


Chapter 1 : Force
Answers to Exercise 1(A) (Q1 – Q25) from Selina for Class 10 Physics ICSE

Question 1 :
State the condition when on applying a force, the body has:
(a) the translational motion,
(b) the rotational motion.

Answer :
(a) The condition when on applying a force, a body has translational motion is that –
(i) The body must be free to move.
(ii) All the particles of the body must share the same displacement in the same direction in the same time interval.

(b) The condition when on applying a force, a body has rotational motion is that –
(i) The body must be pivoted or fixed at a point.
(ii) All the particles of the body must be doing circular motion about a line passing through that fixed point.

Question 2 :
The moment of a force about a given axis depends:
(a) Only on the magnitude of force
(b) Only on the perpendicular distance of force from the axis
(c) Neither on the force nor on the perpendicular distance of force from the axis
(d) Both on the force and its perpendicular distance from the axis.

Answer :
As we know, that the
Moment of force = Force x Perpendicular distance
\therefore The moment of a force about a given axis depends on both :
(i) on the force and
(ii) its perpendicular distance from the axis.
Hence, Correct Option (d)

Question 3 :
The moment of a force of 10 N about a fixed point O is 5 N-m . Calculate the distance of the point O from the line of action of the force.

Answer :
Given :
Moment of force about O = 5 N-m
Force = 10 N
Let r be the perpendicular distance of line of action of the force from point O

\because Moment of force about O = force x perpendicular distance of force from point O
\Rightarrow Moment of force about O = F x r
Substituting the given values,
\Rightarrow 5 N-m = 10 x r
\Rightarrow R = 5/10 m
\Rightarrow R = 0.5 m

So, the distance of the point O from the line of action of the force is 0.5 m

Question 4 :
Define moment of force and state its S.I. unit.

Answer :
Definition: The moment of force is equal to the product of the magnitude of the force and the perpendicular distance of the line of action of force from the axis of rotation.
S.I. unit of moment of force is newton-metre (N-m).

Question 5 :
A body is acted upon by two unequal forces in opposite directions, but not in same line. The effect is that
(a) The body will have only the rotational motion
(b) The body will have only the translational motion
(c) The body will have neither the rotational motion nor the translational motion
(d) The body will have rotational as well as translational motion.

Answer :
Correct Option (d) : The body will have rotational as well as translational motion.

[Answer Logic (Not to be written in exam) –
Whenever a body is acted upon by two unequal forces in opposite directions, but not in the same line, then the body will have both rotational as well as translational motion.]

Top view : To show the effect of two forces acting along two different lines of actions

Question 6 :
A nut is opened by a wrench of length 10 cm. If the least force required is 5.0 N, find the moment of force needed to turn the nut.

Answer :
Given:
Length of wrench, r=10 cm = 0.1 m
Force required to open the nut, F= 5 N
\therefore Moment of force needed to turn the nut = F x r = 5 x 0.1= 0.5 N-m

Question 7 :
State whether the moment of force is a scalar or vector quantity?

Answer :
Moment of a force is a vector quantity.

Question 8 :
A wheel of diameter 2 m is shown with axle at O. A force F = 2 N is applied at B in the direction shown in figure. Calculate the moment of force about (i) the centre O, and (ii) the point A.

Answer :

Given :
Force applied, F= 2 N
Diameter of the wheel, d = 2 m
Perpendicular distance between B and O, r = 1 m
\therefore Radius of the wheel , R = d/2 = 2/2 = 1 m
\Rightarrow OB = r1 = 1 m and AB = r2 = 2 m
(i) The moment of force about point O = F x r1
= 2 x 1= 2 N-m (clockwise)
(ii) The moment of force about point A = F x r2
= 2 x 2= 4 N-m (clockwise)

Question 9 :
State two factors affecting the turning effect of a force.

Answer :
The turning effect of a force is also known as the moment of a force.
About a point or an axis, it depends on the following two factors:
(a) The magnitude of the force applied and,
(b) The distance of line of action of the force from the axis of rotation.

Question 10 :
The diagram shows two forces F1= 5 N and F= 3 N acting at points A and B of a rod pivoted at a point O, such that OA = 2 m and OB = 4 m
Calculate:
(i) the moment of force F1 about O.
(ii) the moment of force Fabout O.
(iii) total moment of the two forces about O.

Answer :

Given :
AO = 2 m and OB = 4 m
(i) Moment of force F1( = 5 N) at A about the point O
= F1 x OA
= 5 x 2 = 10 N-m (anti-clockwise)
(ii) Moment of force F2 ( = 3 N) at B about the point O
= F2 x OB
= 3 x 4 = 12 N-m (clockwise)
(iii) \because Clockwise moment of F2 is more that the anti-clockwise moment of F1
\therefore The total moment of forces about the point O =
= 12 – 10 = 2 N-m (clockwise)

Question 11 :
When does a body rotate? State one way to change the direction of rotation of a body. Give a suitable example to explain your answer.

Answer :
When a body is pivoted at a point, the force applied on the body at a point other than the point of pivot rotates the body about an axis passing through the pivoted point.
The only condition on the force acting is that the line of action of the force should not pass through the pivot.

The direction of rotation can be changed by changing the point of application of force.

For example, The given figure shows the anticlockwise and clockwise moments produced in a disc pivoted at its centre by changing the point of application of force F from A to B.

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Published on September 28th, 2019 | by Abhishek Mandal


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