The Brainbox Tutorials has especially made the chapter wise PYQs series of all important subjects. Get access to ICSE Class 10 Physics Chapter Force PYQs + Solution here. All the questions which came in Board exams from the chapter Force from the year 2014 to 2024 are provided below. Chapter wise PYQs will be very beneficial for the students preparing for the upcoming board exams.

## ICSE Class 10 Physics Chapter Force PYQs + Solution

## 2024

**Q1.** A door lock is opened by turning the lever (handle) of length 0.2 m. If the moment of

force produced is 1Nm, then the minimum force required is:

(a) 5N

(b) 10N

(c) 20N

(d) 0.2N

**Ans:** (a) 5N

**Q2.** Fill in the blanks:**(a)** When a stone tied to a string is rotated in a horizontal plane, the tension in the

string provides ____________ force necessary for circular motion.

**Ans:** centripetal

**Q3.** A non uniform beam of weight 120N pivoted at one end is shown in the diagram below.

Calculate the value of F to keep the beam in equilibrium.

**Ans:** The moment of force due to the weight of the beam is

120 N × 20 cm,

and the moment due to force F is

-F×80 cm

Since the beam is in equilibrium, the sum of the moments must be zero. Therefore, the equation is

-F×80 + 120×20 = 0

⇒ -F×80 + 2400 = 0

⇒ 2400 = 80 F

⇒ F = 30 N.

**Q4. (a)** Define Centre of Gravity.**(b)** A hollow ice cream cone has height 6 cm.

(i) Where is the position of its centre of gravity from the broad base?

(ii) Will its position change when it is filled completely with honey? Write Yes or No.

**Ans: (a)** The centre of gravity of a body is the point at which the entire weight of the body acts, irrespective of the position and orientation of the body.

**(b) (i)** The centre of gravity from the broad base is h/3 = 6/3 = 2 cm

(ii) Yes

## 2023

**Q 1.** Clockwise moment produced by a force about a fulcrum is considered to be:

a. Positive

b. Negative

c. Zero

d. None of these

**Ans:** b. Negative

**Q 2 (a)** what is the position of centre of gravity of a triangular lamina?**(b)** When this triangular lamina is suspended freely from any one vertex, what is the moment of force produced by its own weight in its rest position?

**Ans: **(a) The centre of gravity for a triangular lamina is at the point of intersection of the medians.

(b) When the triangular lamina is suspended freely from any one vertex, it will rotate until its center of gravity is directly below the point of suspension at the point where its median intersect and the triangle is in rotational equilibrium. Hence, we get zero moment of force.

**Q 3.** The diagram shows wheel O pivoted at point A. Three equal forces F1, F2 and F3 act at point B on the wheel.

**(a)** Which force will produce maximum moment about A?**(b)** Give a reason for your answer in (a).

**Ans: (a)** F1

**(b)** The moment produced by a force is given by the magnitude of the force multiplied by the perpendicular distance from the point of application of the force to the pivot point.

In this case, the perpendicular distance from the line of action of F2 and F3 is less than the diameter.

Therefore, the moment produced by force F1 will be greater than the moments produced by forces F2 and F3, since F1 has the largest perpendicular distance from the pivot point A.

**Q 4.** A metre scale of weight 50 gf can be balanced at 40 cm mark without any weight suspended on it.**(a)** If this ruler is cut at its centre then state which part [0 to 50 cm or 50 to 100 cm] of the ruler will weigh more than 25 gf?

**(b)** What minimum weight placed on this metre ruler can balance this ruler when it is pivoted at its centre?

**Ans: (a)** Let weight from ( 0 to 40 )cm be W1 and (40 to 100)cm be W2.

Hence W1 + W2 = 50

W1 = 50 – W2

By principle of moments,

Anti clockwise moment = clockwise moment

W1 x (40 – 0) = W2 x (100 – 40)

(50 – W2) x 40 = W2 x 60

2000 – 40W2 = 60W2

100W2 = 2000

W2 = 20 gf

So, W1 = 50 – 20 = 30 gf

Hence the part (0 – 50 cm) of the ruler will weigh more than 25 gf.

**(b)** Let the minimum weight needed be m.

In equilibrium,

Anti clockwise moment = clockwise moment

50 x (50 – 40) = m x (100-50)

500 = 50 m

m = 10 gf

## 2021 Semester 1

**Q 1.** The relation between CGS and S.I. unit of moment of force is: [1]

(a) 1 Nm= 10^{5} dyne cm

(b) 1 Nm = 10^{5} dyne

(c) 1 Nm= 10^{7} dyne cm

(d) 1 dyne cm = 10^{7} N m

**Ans:** (c) 1 Nm= 10^{7} dyne cm

**Q 2.** A stone tied at the end of a string is whirled by hand in a horizontal circle with uniform speed.**(i)** Name the force required for this circular motion: [1]

(a) Centrifugal force.

(b) Centripetal force.

(c) Force of gravity.

(d) Frictional force.

**(ii)** What is the direction of the above-mentioned force? [1]

(a) Towards the centre of the circular path.

(b) Away from the centre of the circular path.

(c) Normal to the radius at a point where the body is present on the circular path.

(d) Direction of this force keeps on changing alternately towards and away from the centre.

**Ans: (i)** (b) Centripetal force.

**(ii)** (a) Towards the centre of the circular path.

**Q 3.** The diagram below shows the balanced position of a metre scale. [1]

Which one of the following diagrams shows the correct position of the scale when it is supported at the centre?

**Ans:** option a.

## 2020

**Q 1. (i) **With reference to the direction of action, how does a centripetal force differ from a centrifugal force during uniform circular motion?**(ii)** Is centrifugal force the force of reaction of centripetal force?**(iii)** Compare the magnitudes of centripetal and centrifugal force.

**Ans: (i)** Centripetal force always acts towards the centre of circle along it’s radius. It is this force which compels the body to move along the circular path. A centrifugal force appears to act away (outward) from the centre of circular path. Thus, the two forces act in opposite directions.

**(ii)** No, the centrifugal force is not the force of reaction of centripetal force.

**(iii)** Centripetal force and centrifugal force have equal magnitudes.

**Q 2. (i) **define moment of force**(ii)** Write the relationship between SI and CGS unit of moment of force

**Ans: (i)** 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.

**(ii)** The S.I. and C.G.S. unit of moment of force are related as follows —

1 N m = 10^{5} dyne x 10^{2} cm = 10^{7} dyne cm

## 2019

**Q 1. **A uniform meter scale is in equilibrium as shown in the diagram.

**(i)** Calculate the weight of the metre scale**(ii)** Which of the following option is correct to keep the ruler in equilibrium when 40 gf weight is shifted to 0 cm mark ?

F is shifted towards O cm

OR

F is shifted towards 100 cm

**Ans: (i)** Let W gf be the weight of metre rule which acts downwards from 50 cm mark.

In equilibrium position, taking moments about fulcrum.

Anticlockwise moment = 40 x (30 – 5) = 1000 gf cm

Clockwise moment = W (50 – 30) = 20 W gf cm

∴ Using principle of moments:

20 W = 1000

⇒ W =1000/20

= 50 gf

∴ Weight of the metre scale = 50 gf

(ii) In order to keep the ruler in equilibrium, F is shifted towards 0 cm mark.

## 2018

Q 1. A half metre rod is pivoted at the centre with two weights of 20 g f and 12 g f suspended at a particular distance of 6 cm and 10 cm from the pivot respectively as shown below.

**(i)** Which of the 2 forces acting on the rigid rod causes clockwise moment?**(ii)** Is the rod in equilibrium?**(iii)** The direction of 20 g f forces reversed. What is the magnitude of the resultant moment of force on the rod?

**Ans: (i)** The force 12 gf rotates the rod clockwise.

(ii) Clockwise moment = 20 gf x 6 cm = 120 gf cm

Anticlockwise moment = 12 gf x 10 cm = 120 gf cm

Since, the clockwise moment is equal to the anticlockwise moment, hence the rod is in equilibrium.

(iii) On reversing the direction of force, 20 gf will also produce clockwise moment, total clockwise moment will then be

(20 gf x 6 cm) + (12 gf x 10 cm) = 240 gf cm

## 2017

**Q 1. (a)** A brass ball is hanging from a stiff cotton thread. Draw a neat labelled diagram showing the forces acting on the brass ball and the cotton thread.**(b)** The distance between 2 bodies is doubled. How is the magnitude of gravitational force between them affected?

**Ans: (a)** In the figure below, the force on the ball is the weight W acting vertically downwards and the force on the thread is the tension T acting upwards.

**(b) **The gravitational force becomes one-fourth since it is inversely proportional to the square of the distance of separation (i.e., F ∝ 1/d^{2} where F is the force and d is the distance between centers)

**Q 2.** A uniform half meter rule balances horizontally on a knife edge at 29 cm mark when a weight of 20 gf is suspended from one end.**(i)** Draw a diagram of the arrangement.**(ii)** What is the weight of the half metre rule?

**Ans: (i)** Diagram of the arrangement is shown below:

(ii) If W gf is the weight of half meter rule, by the principle of moments

W x (29 – 25) = 20 x (50 – 29)

⇒ W x 4 = 20 x 21

⇒ W = 105 gf

**Q 3.** How does uniform circular motion differ from uniform linear motion?

**Ans:** The difference between uniform circular motion and uniform linear motion is as follows:

Uniform circular motion | Uniform linear motion |

In a uniform circular motion the velocity is variable (although the speed is uniform), so it is an accelerated motion. | In uniform linear motion the speed and velocity are constant and acceleration is zero i.e., the uniform linear motion is an unaccelerated motion. |

## 2016

**Q 1.** A stone of mass m is rotated on a circular path with uniform speed by tying a strong string with the help of your hand. Answer the following questions:**(i)** Is the stone moving with uniform or variable speed?**(ii)** Is the stone moving with uniform acceleration? In which direction does the acceleration act?**(iii)** What kind of force act on the hand and state its direction?

**Ans: (i)** The stone is moving with uniform speed.

**(ii)** Yes, the stone is moving with a uniform acceleration. The acceleration acts towards the centre.

**(iii)** Centrifugal force acts on the hand. It is directed radially outwards away from the centre.

## 2015

**Q 1.** When a body is placed on a table top, it exerts a force equal to its weight downwards on the table top but does not move or fall.

(i)** **Name the force exerted by the table top.

(ii)** **What is the direction of the force? [2]

**Ans: (i)** Normal reaction force

**(ii)** It acts normal to the surface of the table top, i.e. opposite to the direction of the force due to the weight of the body.

**Q 2.(i)** On what factor does the position of the centre of gravity of a body depend ?**(ii)** What is the S.I. unit of the moment of force? [2]

**Ans: (i)** The position of the centre of gravity of a body depends on its shape, i.e. on the distribution of mass in it.

**(ii)** S.I. unit of the moment of force is newton x meter(N m).

**Q 3.** Name the factors affecting the turning effect of a body. [2]

**Ans:** The factors affecting the turning effect of a body are:

**(i)** The magnitude of force applied.

**(ii)** The distance of line of action of the force from the axis of rotation.

**Q 4. (i)** Define equilibrium.**(ii)** In a beam balance when the beam is balanced in a horizontal position, it is in ……….. equilibrium. [2]

**Ans: (i)** When a number of forces acting on a body produce no change in its state of rest or state of motion, the body is said to be in equilibrium.

**(ii)** static

**Q 5.** Explain the motion of a planet around the sun in a circular path. [2]

**Ans:** A planet moves around the Sun in a circular path for which the gravitational force of attraction on the planet by the Sun provides the necessary centripetal force. This centripetal force is always directed towards the centre of the Sun at each point of its path which is responsible for circular motion of the planet.

## 2014

**Q 1.(a)** A force is applied on (i) a non-rigid body and (ii) a rigid body. How does the effect of the force differ in the above two cases? [2]**(b)** A metallic ball is hanging by a string from a fixed support. Draw a neat labelled diagram showing the forces acting on the bail and the string. [2]**(c) **What is the weight of a body placed at the centre of the earth ?**(d)** Is it possible to have an accelerated motion with a constant speed? Explain. [2]

**Ans: (a)** A force when applied on a rigid object does not change the inter-spacing between its constituent particles and therefore it does not change the dimensions of the object, but causes motion in it. On the other hand, a force when applied on a non-rigid object, change the inter-spacing between its constituent particles and therefore cause a change in its dimensions.

**(b)**

**(c)** The weight of a body placed at the centre of the earth is zero.

**(d)** Yes. The velocity of particle in circular motion is variable or the circular motion is accelerated even though the speed of particle is uniform.

**Q 2.** Two forces each of 5N act vertically upwards and downwards respectively on the two ends of a uniform metre rule which is placed at its mid-point as shown in the diagram. Determine the magnitude of the resultant moment of these forces about the midpoint.

Ans: Given F1 = F2 = 5N, d1 = d2 = 0.5 m

Torque 1 = F1 x d1

= 5N x 0.5m = 2.5 N m (anti clockwise)

Torque 2 = F2 x d2

= 5N x 0.5m = 2.5 N m (anti clockwise)

The resultant moment of force = 2.5 + 2.5 = 5 N m (anti clockwise)

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