# Force Class-10 Goyal Brothers ICSE Physics Solutions Ch-1

**Force ICSE Class-10 Goyal Brothers** Physics Solutions Chapter-1 **TURNING FORCES AND UNIFORM CIRCULAR MOTION**. We Provide Step by Step Answer of Exercise, MCQs, Numericals Practice Problem Questions of Exercise-1 **Force TURNING FORCES AND UNIFORM CIRCULAR MOTION ICSE Class-10**** **. Visit official Website **CISCE ** for detail information about ICSE Board Class-10 Physics.

## Force Class-10 Goyal Brothers ICSE Physics Solutions Ch-1

**-: Select Exercise :-**

**Exe-1 Force Class-10 Goyal Brothers ICSE Physics Solutions Ch-1**

Page-9

**Question 1.**

**(a) Define the following :**

- Rigid body
- point of action of force
- line of action of force
- principle of transmissibility of force.

**(b)** Name four bodies which can be called rigid bodies for practical purposes.

**Answer 1:**

**(a)**

**(i) Rigid Body :** “A body which does not get deformed under the action of a force or a number of forces is called rigid body,

**(ii) Point Of Action Of Force :** “The point on the rigid body where a force acts is called the point of action of force.”

**(iii) Line Of Action Of Force :** “An imaginary line passing through the point of action of force and drawn in the same direction in which the force acts is called line of action of force. LM is the line of action of force.

**(iv) Principle Of Transmissibility Of Force :** “It states that the point of action of force can be transmitted anywhere along the line of action of the force without causing any material difference in the movement of the body.”

**(b)** Wood, iron rod, stone and glass are rigid bodies.

**Question 2. Force Class-10 Goyal Brothers**

Under what condition (s) a body describes a motion of :

(a) translation

(b) rotation.

**Answer 2:**

**Conditions for**

**(a) Translational motion :** Motion of rigid body in straight line and in the direction of applied force.

**(b) Rotational motion :** Rigid body should rotate about axis (or pivot).

**Question 3.**

**(a)** What do you understand by the term moment of force

**(b)** State two factors which determine the moment of force.

**Answer 3:**

**(a) Moment of force :** “The turning effect of force acting on a body about an axis is called the moment of force.”

**(b) Factors affecting the moment of force :**

- The magnitude of the force applied
- The distance of line of action of the force from the axis of rotation.

**Question 4.**

State one way of

**(a)** reducing moment of force

**(b)** Increasing moment of force for a given force acting on a body capable of turning around a fixed point.

**Answer 4:**

**(a)** Moment of force can be reduced by reducing the distance between force and the turning point.

**(b)** Moment of force can be increased by increasing the distance between force and the turning point.

**Question 5.**

State the law of moments.

**Answer 5:**

**Law of moments :** In equilibrium sum of anticlockwise moments is equal to sum of clock-wise moments.

Sum of anticlock wise moments = sum of clockwise moments.

**Question 6.**

**(a)** What do you understand by the terms (i) positive moments (ii) negative moments

**(b)** State the absolute units of moment of force in (i) CGS system (ii) S.I. system.

**Answer 6:**

**(a)**

**Positive moments :**If the force produces anticlock wise motion in the rigid body about turning point, it is positive moment.

**Negative moment :**If the force produces clock wise motion in the body about turning point, it is negative moment.

**(b) Absolute units of moment of force in**

- CGS system is dyne-cm.
- S.I system is newton-metre (N-m)

**Question 7.**

(a) What do you understand by the following terms ? (i) couple, (ii) arm of couple, (iii) moment of couple

(b) State the units of the moment of couple in (i) CGS system, (ii) S.I. system.

(c) Give four examples of couple in everyday life.

(d) State the mathematical expression for the moment of a couple.

**Answer 7:**

**(a)**

**Couple :**“Two forces acting on a rigid body are equal in magnitude opposite parallel and not along the same straight line constitute a couple.”**Arm of the couple :**“Perpendicular distance between two forces is called arm of the couple.” i.e. AB is arm of couple.**Moment of the couple :**“Is the product of any one of the force and the arm of couple. i.e. F × AB is moment of couple.

**(b)**

- In CGS system unit of moment of couple-dyne-cm
- In S.I. system unit of moment of couple-N-m.

**(c) Examples of couple in everyday life :
**

- Steering wheel of a car.
- Turning of a screw driver.
- Opening and closing of water tap.
- Winding the watch.

**(d) Mathematical expression :** Moment of couple = force × Arm of couple.

**Question 8.**

**Explain the following :**

**(a)** Jack screw is provided with a long arm.

**(b)** It is easier to open a door by holding it from its edge.

**(c)** A small boy can balance a stout man on a see-saw.

**(d)** The handles of a hand flour grinder (chakki) is provided near its edge.

**(e)** It is easier to turn a steering wheel of larger diameter than a steering wheel of smaller diameter.

**(f)** A wrench or a spanner has a long handle.

**Answer 8:**

**(a)** So that there is maximum distance between force and turning point and minimum force is needed.

**(b)** Holding a door from edge increases the distance between force and turning point (hinge) and hence minimum force is needed.

**(c)** By making the smaller boy sit at longer distance from axis of rotation. Hence turning effect of force increases.

**(d)** To increase the distance between force and turning point and hence to apply minimum force.

**(e)** In this way we increase the ⊥ distance between force and point of rotation and minimum force is needed to rotate the steering.

**(f)** Moment of force increases by increasing the length of handle and minimum force is required.

**Question 9. Force Class-10 Goyal Brothers**

The diagram along side shows a heavy roller, with its axle at O which is to be pulled on to the pavement XY by applying the minimum possible force. Draw the diagram and mark on it the point and direction in which the force should be applied?

**Answer 9:**

Force should be applied at A near the rim to increase the moment of force by increasing perpendicular distance OA (between force and point of rotation).

**Question 10.**

**(a)** What do you understand by the term equilibrium of a body?

**(b)** State the condition when the body is in (i) dynamic equilibrium (ii) static equilibrium support your answer with one example each.

**(c)** A body is acted upon by a number of forces acting in different directions. State two conditions for a body to be inequilibrium.

**Answer 10:**

**(a) Equilibrium :** “When two or more forces act on a rigid body such that the state of body (rest or of uniform motion) does not change, the body is said to be in equilibrium”.

**(b) Condition when the body is in**

**(i) Dynamic equilibrium** …. the body should remain in its state of uniform motion when it is under the action of forces.

**Example :** A train running with a constant speed is in dynamic equilibrium, because the force generated by its engine is equal to the force of friction due to rails and force of friction due to air.

**(ii) Static equilibrium :** The body should remain in state of rest when it is under the action of forces.

**Example :**

When a body is at rest on the ground the various forces are :

W = R and F = F

**(c)** When a body is acted upon by a number of forces the body is to be in equilibrium.

**The two conditions are :**

**(i)** The resultant of all the translational forces should be zero.

**(ii**) All the clockwise moments should be equal to all the anticlock wise moments.

**Question 11.**

(a) What do you understand by the term centre of gravity?

(b) State the position of CG in case of the following regular bodies.

- A triangular lamina
- a rectangular lamina
- a circular lamina
- a cylinder
- a sphere
- a square lamina.

**Answer 11:**

**(a)** Centre of gravity : “Is a point with in a body where the whole weight of body is supposed to act.”

**(b)** Position of C.G.

- A traingular lamina is the point of intersection of medians.

- A rectangular lamina: is the point of intersection of its diagonals

- A circular lamina is its geometric centre O

- A cylinder is a point at the centre of its axis.

- A sphere is its geometric centre.

- A square is the point of intersection of its diagonals.

**Question 12. Force Class-10 Goyal Brothers**

**(a)** Is it possible to have a body whose centre of gravity is outside the body ? If so explain ?

**(b)** How will you determine the centre of gravity of an irregular piece of a cardboard ?

**Answer 12:**

**(a)** Yes, the body with L shape or a Boomrang has its centre of gravity out side the body.

Medians AB, CD and EF meet at point O out side the body is C.G.

**(b)** To find C.G of irregular body with the help of plumb line. Make holes at A, B, C, D, E sharp corners and suspand the cardboad at each hole in turn with the help of thread of plumb line draw the line below the thread.

The centre of gravity is point where maximum number of lines meet.

**Question 13.**

A flat triangular cardboard equilateral in shape is suspended by passing a common pin through a narrow hole at its one corner. Draw a diagram to show its position in the state of rest. In the diagram mark the position of suspension by the letter A and centre of mass (cenre of gravity) by the letter B.

**Answer 13:**

Point of suspension A and CG (B) is shown. B is the point where medians of ∆AX, DZ and CY intersect.

**Question 14. Force Class-10 Goyal Brothers**

A stone of mass ‘m’ is rotated in circular path with uniform speed by tying a strong string with the help of your hand. Answer the following questions.

**(a)** Is the stone moving with a uniform or variable speed ?

**(b)** Is the stone moving with a uniform acceleration ? What is the effect of acceleration ? In which direction does the acceleration act ?

**(c)** What kind of force acts on the stone and state its direction ?

**(d)** What kind offorce acts on the hand and state its direction?

**Answer 14:**

**(a)** Uniform speed.

**(b)** Yes, the stone is moving with uniform acceleration.

Effect of accelerated motion is it makes the body move in circular path. Acceleration acts at right angle to the radius of circle.

**(c)** Centripetal force acts on stone. Its direction is towards the centre of circle.

**(d)** Centrifugal force acts on hand. Its direction is opposite to the direction of centripetal force i.e. away from centre of circle along the radius.

**Question 15.**

**State whether the following statements are true or false :**

**(a)** on deformation of a body,Jhe position of its centre of gravity does not change.

**(b)** The centre of gravity of freely suspended body is always vertically below the point of suspension.

**Answer 15:**

**(a)** False

**(b)** True

**Question 16.**

**Define or explain**

**(a)** circular motion

**(b)** centripetal force

**(c)** centrifugal force

**Answer 16:**

**(a) Circular Motion :** “Motion of a body around a circular path with uniform speed, but variable velocity, such that it is acted upon by a uniform acceleration is called circular motion.”

**(b) Centripetal Force :** “A force which is directed towards the centre of a circular path and always acts at right angles to the direction of motion, along the circular path is called centripetal force.”

**(c) Centrifugal Force :** “The force acting on a body away from centre of its circular path is called Centrifugal Force.”

**Question 17. Force Class-10 Goyal Brothers**

Give an example of a body moving with a uniform speed, but has an accelerated motion.

**Answer 17:**

Motion of moon around the earth is example of moon (with uniform speed and accelerated motion). Gravitational force of the earth provides necessary centripetal force.

**Question 18.**

Compare uniform circular motion and uniform linear motion.

**Answer 18:**

**Uniform Circular motion :**

- Speed of body is uniform.
- It is accelerated motion.
- Velocity of body is variable.
- External force is needed.

**Uniform Linear motion :**

- Speed is uniform.
- Acceleration is zero.
- Velocity is constant.
- No external force is needed.

**Question 19.**

Explain the motion of moon around the earth.

**Answer 19:**

The moon moves around the earth with uniform speed in a circular path. The centripetal force for its motion is provided by earth.

**Question 20.**

With reference to the magnitude of force and its direction, how does centripetal force differ from centrifugal force.

**Answer 20:**

Both centripetal force and centrifugal force have same magnitude but they are opposite in direction.

** ****Multiple Choice Questions**

**Force Class-10 Goyal Brothers ICSE Physics Solutions Ch-1 **

Page-11

** Tick ( ✓ ) the most appropriate option**

**1. The point of action of force on a rigid body is :**

(a) Fixed point on rigid body

(b) Fixed point but can be transferred any where along the line of action of force.

(c) Fixed point but can be transferred anywhere along the direction of force.

(d) Fixed point, but can be transferred anywhere opposite to the direction of force.

**Answer 1:**

(b) Fixed point but can be transferred any where along the line of action of force.

**2. The turning effect produced in a rigid body around a fixed point by the application of force is called;**

(a) turning force

(b) movement of force

(c) moment of couple

(d) none of these

**Answer 2:**

(b) movement of force

**3. The unit of moment of force in SI system is :**

(a) Nm

(b) dynecm

(c) dynem

(d) Ncm

**Answer 3:**

(a) Nm

**4. The moment of couple is mathematically the :**

(a) product of one force and the perpendicular distance between two forces

(b) product of both forces and the perpendicular distance between them

(c) product of one force and tKe perpendicular distance between the point of application of force and turning point.

(d) None of the above.

**Answer 4:**

(a) product of one force and the perpendicular distance between two forces

**5. The condition for equilibrium is:**

(a) the resultant of all the forces acting on the body be zero only.

(b) the resultant of moments of all the forces acting on the body about the turning point should be zero.

(c) both (a) and (b)

(d) none of the above

**Answer 5:**

(b) the resultant of moments of all the forces acting on the body about the turning point should be zero.

**6. A body is acted upon by two unequal and opposite forces along different lines of action of force. The body will have**

(a) only rotatory motion

(b) only translatory motion

(c) both (a) and (b)

(d) neither (a) nor (b)

**Answer 6:**

(a) only rotatory motion

**7. A force F acts on a rigid body capable of turning around a fixed point. The moment of force depends upon**

(a) magnitude of force F

(b) magnitude of perpendicular distance between the point of action of force and the turning point

(c) both (a) and (b)

(d) none of these

**Answer 7:**

(c) both (a) and (b)

**8. The centre of gravity of a cricket ball is at:**

(a) its geometric centre

(b) at its bottom touching the ground

(c) its top most point

(d) at any point on its surface

**Answer 8:**

(a) its geometric centre

**9. A body is describing a uniform cicular motion. Which of the following quantities is/are constant**

(a) speed

(b) acceleration

(c) velocity

(d) both (a) and (b)

**Answer 9:**

(a) speed

**10. In a uniform circular motion :**

(a) speed of body continuously changes because the direction of motion changes

(b) velocity of body continuously changes because the direction of motion changes

(c) the motion of body is accelerated

(d) both (b) and (c)

**Answer 10:**

(d) both (b) and (c)

**Numerical Problems on Moment of Force**

**Force Class-10 Goyal Brothers ICSE Physics Solutions Ch-1**

**(Page-11,12,13,14,15)**

**Practice Problems 1 **

**Question 1.**

A force of 50 dynes acts on a rigid body, such that the perpendicular distance between the fulcrum and the point of application of force is 75 cm. Calculate the moment of force.

**Answer 1:**

Force F = 50 dynes

⊥ distance = 75 cm

∴ Moment of force = F × ⊥ distance

= 50 × 75 = 3750 dynecm

**Question 2.**

The perpendicular distance between the point of application of force and the turning point is 1.75 m, when a force of 80 N acts on a rigid body. Calculate the moment of force.

**Answer 2:**

⊥ distance = 1.75 m

Force F = 80 N

∴ Moment of force = F × ⊥ distance

= 80 × 1.75 = 140.00

= 140 Nm

**Practice Problems 2**

**Question 1.**

A force of 50 N produces a moment of force of 10 Nm in a rigid body. Calculate the perpendicular distance between the point of application of force and the turning point is 45 cm.

**Answer 1:**

F = 50 N

Moment of force =10 Nm

⊥ distance = ?

Moment of force = F × ⊥ dist.

10 = 50 × ⊥ distance.

∴ ⊥ distance 10/50 = 0.2 m

**Question 2. Force Class-10 Goyal Brothers**

Calculate the force which will produce a moment of force of 1575 dyne cm, when the perpendicular distance between point of application of force and turning point is 45 cm.

**Answer 2:
**

**Practice Problems 3**

**Question 1.**

A couple of 15 N force acts on a rigid body, such that the arm of couple is 85 cm. Calculate the moment of couple in S.I. system.

**Answer 1:
**

**Question 2.**

Calculate the length of the arm of couple, if a force of 13 N produces a moment of couple of 14.3 Nm.

**Answer:
**

**Question 3. Force Class-10 Goyal Brothers**

Two forces each of magnitude 2 N act vertically upward and downward respectively on two ends of a uniform rod of length l m, freely pivoted at its centre. Determine the resultant moment of forces about the mid-point of the rod.

**Answer 3:**

As the two equal forces 2 N are acting at the ends of pivoted rod AB. These constitute a couple in anti-clockwise direction.

**Practice Problems 4**

**Question 1.**

The diagram along side shows a force F = 5 N acting at point A produces a moment of force of 6 Nm about point O. What is the diameter of the wheel

**Answer 1:
**

**Question 2. **

The diagram alongside shows a force F acting at point A, such that it produces a moment of force of 20 Nm in clockwise direction. Calculate the magnitude of force F

**Answer 2:
**

**Practice Problems 5**

**Question 1.**

Study the diagram alongside and calculate the moment of couple.

**Answer 1:
**

**Question 2. Force Class-10 Goyal Brothers**

Two forces F_{1} and F_{2} are applied on a circular body such that moment of couple is 9 Nm in CWD. Calculate the radius of circular body.

**Answer 2:
**

**Question 3. Force Class-10 Goyal Brothers**

Two forces F_{1} = F_{2} are applied on a wheel of 1.5 m radius, such that moment of couple is 30 Nm. Calculate the magnitude of each of the force.

**Answer 3:**

Anticlockwise direction

F_{1} = F_{2} = F in opp. direction and are parallel

Hence constitute a couple

moment of couple = F × diameter AB

30 = F

30 = F × (2 × 1.5)

F= 30/3 = IO N

**Practice Problems 6**

**Question 1.**

A uniform metre scale is balanced at 60 cm mark, when weights of 5 gf and 40 gf are suspended at 10 cm mark and 80 cm mark respectively. Calculate the weight of the metre scale.

**Answer 1:
**

Let w be the mass of metre scale acting a mid point 50 cm

Clock wise moment = 40 × (80 – 60)

= 40 × 20 = 800 g f cm …….(i)

Anticlockwise moments

= 5 × (60 – 10) + w × (60 – 50)

(250 + 10 w) gf cm ……..(ii)

Anticlockwise moment = clock wise moment

250+ 10 w = 800

10 w= 800 -250 = 550

w = 550/10 = 55 gf

**Question 2.**

A uniform metre scale is balanced at 20 cm mark, when a weight of 100 gf is suspended from one end.Where must the weight be suspended ? Calculate the weight of the metre scale.

**Answer 2:
**

Let w be the weight of metre scale.As the scale is balanced at 20 cm and 100 gf is suspended on one end (0 mark). The weight of longer arm i.e. BC is balanced by 100 gf

∴ CW moment = A.C.W moment

W × 30 = 20 × 100

W = 2000/30 = 66.66 gf

wt. of 100 gf should be suspended at zero mark.

**Question 3. Force Class-10 Goyal Brothers**

A uniform metre scale balances horizontally on a knife edge placed at 55 cm mark, when a mass of 25 g is supported from one end. Draw the diagram of the arrangement. Calculate mass of the scale.

**Answer 3:
**

As the metre scale is balanced at 55 cm mark i.e. large arm is balanced by arm BC and 25 g is suspended at one end i.e. at 100 cm mark

∴ Anticlockwise moment = clockwise moment

W × (55 – 50) = 25 × (100 – 55)

W = 25 × 45 / 5 = 225g

**Practice Problems 7**

**Question 1.**

A uniform metre scale of weight 50 gf is balanced at the 40 cm mark, when a weight of 100 gf is suspended at the 5 cm mark. Where must a weight of 80 gf be suspended to balance the metre scale ?

**Answer 1:
**

Let a wt. of 80 gf be placed at a distance x from 40 cm mark.

Clock wise moments = Anticlock wise moment

(80 × x) + 50 × (50 – 40) = 100 × (40 – 5)

80 x +500 = 3500

80 x = 3500 – 500 = 3000

x = 3000/80 = 37.5 cm

∴ A is at 40 + 37.5 = 77.5 cm mark

**Question 2.**

A see-saw 8 m long is balanced in the middle. Two children of mass 30 kgf and 40 kgf are sitting on the same side of the fulcrum at a distance of 1.5 m and 3.5 from the fulcrum respectively. Where must a lady weighing 60 kgf sit from the fulcrum, so as to balance the see-saw ?

**Answer 2:
**

**Question 3. Force Class-10 Goyal Brothers**

A uniform wooden beam AB, 80 cm long and weighing 250 gf, is supported on a wedge as shown in the figure. Calculate the greatest weight which can be placed on end A without causing the beam to tilt.

**Answer 3:
**

**Question 4.**

Figure shows a uniform metre rule weighing 100 gf, pivoted at its centre ‘O’. Two weights of 150 gf and 250 gf hang from the metre rule as shown. Calculate :

**(a)** Total C.W. moment about ‘O’.

**(b)** TotalA.C.W. moment about ‘O’.

**(c)** Differentiate of C.W. andA.C.W. moments.

**(d)** The distance from ‘O’ where a 100 gf weight should be suspended to balance the metre scale.

**Answer 4:
**

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