Selina Concise Class 10 Physics Solution Chapter No. 1- ‘Force’ For ICSE Board Students.
Selina Concise Class 10 Physics Chapter 1 Force Exercise All Questions and Answers by Physics Teacher here in this post.
Exercise 1 (A)
Question: 1
State the condition when on applying a force, the body has:
(a) The translational motion
(b) The rotational motion.
Solution: (a) If the rigid body is free to move then that body can undergo translational motion.
(b) if the body is rigid and pivoted at certain location and force is applied at certain point.
Question: 2
Define moment of force and state its S.I unit.
Solution: Turning effect caused by force is defined asMoment of force . It is the product of force applied and shorted perpendicular distance of the line of action of force from axis of rotation.
Question: 3
State whether the moment of force is a scalar or a vector quantity.
Solution: Moment of force is a vector quantity.
Question: 4
State two factors affecting the turning effect of a force.
Solution: When you are applying a force to turn an object it’s turning effect its depends upon
(a) Magnitude of the force applied
(b) Distance between pivoted point and point of force applied.
Both the quantities are directly proportional to turning effect of force.
Question: 5
When does a body rotate? State one way to change the direction of rotation of a body given a suitable example to explain your answer
Solution: when force is applied from certain direction on pivoted body causing it to rotate.
The direction of rotation of the body depends upon two things
(a) Point of application of force
(b) Direction of application of force
Consider a rotating disc having a line passing along with it’s diameter PQ. If we apply force in clockwise direction on point P and disc is rotating in clockwise direction then applying force in same direction on point Q will cause change in the direction of rotation of the disc.
Question: 6
Write the expression for the moment of force about a given axis.
Solutions: The moment of force about given axis= force * perpendicular distance from the axis of rotation.
Question: 7
What do you understand by the clockwise and anticlockwise moment of force? When is it taken positive?
Solutions: The moment of force is a vector quantity. When the effect of force on the body is to turn clockwise it’s called clockwise moment. Clockwise moment’s direction inwards along the rotation of axis, on the other hand when the effect of force on the body is to turn anticlockwise I’s called anticlockwise moment, anticlockwise moment’s direction is outward along the rotation of axis. Anticlockwise moment is taken positive.
Question: 8
State one way to reduce the moment of a given force about a given axis of rotation.
Solutions: The moment of force depends upon force applied as well as the perpendicular distance between axis of rotation, hence to reduce moment of the given force we can reduce the distance the rotation of axis as they are directly proportional decrease the distance will result in the reduced moment of a given force.
Question: 9
State one way to obtain a greater moment of a force about a given axis of rotation.
Solutions: The moment of force depends upon force applied as well as the perpendicular distance between axis of rotation, hence to obtain greater moment of the given force we can increase the distance the rotation of axis as they are directly proportional increase the distance will result in the greater moment of a given force.
Question: 10
Why is it easier to open a door by applying the force at the free end of it?
Solutions: To open the door the handle is normally provided on the free end of the door which is the farthest distance from the hinge i.e. axis of rotation. The moment of force is product of force applied and distance between rotating axis hence the door opens easier with less amount of force from free end of the door
Question: 11
The stone of hand flour grinder is provided with a handle near its rim. Give a reason.
Solutions: The stone of hand floor grinder is provided with handle near the rim because tit’s located at the farthest distance from the pivoted point. Hence it can be easily rotated about the iron pivot at its centre by applying small amount of force.
Question: 12
It is easier to turn the steering wheel of a large diameter than that of a small diameter. Give reason.
Solutions: While using the steering wheel force is applied tangentially along with perimeter of the wheel. Steering wheel is pivoted at the centre so bigger the wheel bigger the distance from the centre and smaller the force applied to rotate it. Hence It is easier to turn the steering wheel of a large diameter than that of a small diameter.
Question: 13
A spanner (or wrench) has a long handle. Why?
Solutions: The moment of force is the product of force applied and distance between axis of rotation. Hence more the distance between pivot point less force is required for the rotation this is the reason behind long handles of spanner.
Question: 14
A jackscrew is provided with a long arm. Explain why?
Solutions: Jackscrew is used to lift heavy load. It is provided with long arm to ease the effort to raise of lower the heavy objects as long handle requires less force to be applied for the actions.
Question: 15
A, B and C are the three forces each of magnitude 4N acting in the plane of the paper as shown in Figure. The point O lies in the same plane.
(i) Which force has the least moment about O? Give a reason.
(ii) Which force has the greatest moment about O? Give a reason.
(iii) Name the forces producing
(a) clockwise and
(b) anticlockwise moments.
(iv) What is the resultant torque about the point O?
Solutions:
(i) Force C has the least moment about O as the distance between force C and point O is smallest as compared to A and B.
(ii) Force A has the greatest moment on O as the distance between force A and point O is highest.
(iii) Force A and B produce clockwise moment on the other hand force C produces anticlockwise moment
(iv) The torque due to force A= 4N* 0.9m= 3.6 clockwise
The torque due to force B= 4N* 0.8m= 3.2 clockwise
The torque due to force C= 4N* 0.6m= 2.4 anticlockwise
Total torque= 3.6 clockwise + 3.2 clockwise -2.4 anticlockwise= 4.4 clockwise
Total torque= 4.4 clockwise
Question: 17
A body is acted upon by two forces each of magnitude F, but in opposite directions. State the effect of the forces if
(a)both forces act at the same point of the body.
(b)the two forces act at two different points of the body at a separation r.
Solutions:
- If both the forces act at the same point of the body as the forces are opposite to each other and as of the same magnitude causing cancelling out each other
- If the forces act at two different point as the forces are opposite and as of same magnitude it will cause adductive effective on the moment of forces causing increased total torque. The forces will cause the rotation of the body about the midpoint between two forces.
Question: 18
Draw a neat labelled diagram to show the direction of two forces acting on a body to produce rotation in it. Also mark the point O about which the rotation takes place.
Solution:
Question: 19
What do you understand by the term couple? State its effect. Give two examples in our daily life where a couple is applied to turn a body.
Solutions: Couple termed for the pair of forces applied on the body.
Daily example of the couple applied turning of water tap, opening of bottle cap, turning steering wheel etc.
Question: 20
Define the moment of a couple. Write its S.I unit.
Solutions: moment of couple id defined as product of one force and the perpendicular distance between the forces. S.I. unit of couple is newton meter (Nm)
Question: 21
Prove that
Moment of couple = Force x Couple arm.
Solutions: Consider two equal and opposite forces acting on the body of magnitude F on a pivoted point O, which rotate the bar in anticlockwise direction. The perpendicular distance between two forces is PQ called couple arm
Moment of force F at the end P= F * OP
Moment of force F at the end Q=F *OQ
Therefore,
total moment of force
= F * OP +F * OQ
= F (OP+OQ)
=F (PQ) as (OP+OQ)= PQ
Here PQ is the couple arm
Hence it is proved that moment of couple = Force * Couple arm
Question: 22
What do you mean by equilibrium of a body?
Solutions: When the number of forces acting on the body produce no change in its state of the rest or of linear or rotational motion, the body is said to be in a state of equilibrium.
Question: 23
State the condition when a body is in (i) static, (ii) dynamic equilibrium. Give one example each of static and dynamic equilibrium.
Solutions: (i) When the body remains in state of rest under the influence of several forces then the body is said to be in static equilibrium. For example, when a body lying on the table weight of the body exerted on the table vertically which is balanced by equal and opposite forces of reaction applied by table on the body
(ii) when a body remains in the same state of motion under influence of several forces then the body is said to be in dynamic equilibrium.
Question: 24
State two conditions for a body, acted upon by several forces to be in equilibrium.
Solutions: Two condition for a body acted upon by several forces to be in equilibrium are
(1) Net external force on the body must be zero
(2) Net external torque on the body must be zero
Question: 25
State the principle of moments. Name one device based on it.
Solutions: the principle of moment states that when the total clockwise moment of body is identical to the total anticlockwise moment of the body then that body is assumed to be balanced.
Most common example of the device based on the principle of moment is seesaw.
Question: 26
Describe a simple experiment to verify the principle of moments, if you are supplied with a metre rule, a fulcrum and two springs with slotted weights.
Solutions:
A metre rule is suspended horizontally from a fixed support by means of a strong thread at O. Now two spring balances A and B are suspended on the metre rule on either side of the thread. Weights W1 and W2 are the suspended on the spring balance A and B respectively. The metre rule may tilt to one side. The slotted weights are adjusted on the spring balances or the position of the spring balances on either side of the thread from O in such a way that the metre rule becomes horizontal again.
Let the weight suspended from the spring balance A on the right side of the thread be W at a distance OA= l1, while the weight suspended from the spring balance B on the left side of the thread be W, at a distance OB = l2
The weight W, tends to turn the metre rule clockwise, while the weight W, tends to turn the metre rule anticlockwise.
Clockwise moment of weight W1 about the point O = W1 * l1
Anticlockwise moment if weight W2 about point O = W2 * l2
In equilibrium when the metre rule is horizontal it is found that W1 *l1 = W2 *l2
i.e. clockwise moment = anticlockwise moment
Hence, the it proves the principle of moments.
Question: 27
Complete the following sentences:
(i) The S.I. unit of moment of force is _________.
(ii) In equilibrium algebraicsum of moments of all forces about the point of rotation is ______________.
(iii) In a beam balance when the beam is balanced in a horizontal position, it is in ____________equilibrium.
(iv) The moon revolving around the earth is in ____________ equilibrium.
Solutions:
(i) Nm
(ii) zero
(iii) Static equilibrium
(iv) Dynamic equilibrium
Multiple Choice Type:
1.) 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.
Solution:
(d) on both the forces and its perpendicular distance from the axis.
2.) A body is acted upon by two unequal forces in opposite directions, but not in the 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.
Solution:
(d) the body will have rotational as well as translational axis.
NUMERICALS
Question: 1
The moment of a force of 10N about fixed point O is 5Nm. Calculate the distance of the point O from the line of action of the force.
Solutions: Given
Force= 10N
Force of moment= 5N
Therefore,
Force of moment= force applied * perpendicular distance
10N= 5N * perpendicular distance
Perpendicular distance = 10N/5N
Perpendicular distance= 2m
The distance of the point O from the line of action of the force is 2m.
Question: 2
A nut is opened by a wrench of length 10cm. If the least force required is 5.0N, find the moment of force needed to turn the nut.
Solutions: Given
Perpendicular distance=10cm = 0.1m
Force applied= 5N
Therefore,
Force of moment= force applied * perpendicular distance
Force of moment= 5N * 0.1m
Force of moment= 50N cm
Force of moment required to turn the nut is 0.5Nm
Question: 3
A wheel of diameter 2m is shown with the axle at O. A force F = 2N is applied at B in the direction shown in the figure. Calculate the moment of force about
(i) the centre O, and
(ii) the point A.
Solutions: Given, Diameter of the wheel = 2m
Force = 2N
Therefore, Moment of force on point O = force applied * perpendicular distance OB
= 2N * 1m
= 2Nm (clockwise) And, Moment of force on point A = force applied * perpendicular distance AB
= 2N * 2m
= 4Nm (clockwise)
Question: 4
The diagram shows two forces F1= 5N and F2 = 3N acting at points A and B of a rod pivoted at a point O, such that OA = 2m and OB = 4m
Calculate:
(i) the moment of force F1 about O.
(ii) the moment of force F2 about O.
(iii) total moment of the two forces about O.
Solutions: Given, Force F1= 5N
Force F2 = 3N
OA = 2m, OB = 4m
Therefore, The moment of force F1 about O = force F1 * OA
= 5N * 2m
= 10Nm (anticlockwise)
The moment of force F2 about point O = force F2 * OB
= 3N * 4m
= 12Nm (clockwise)
And the direction of torque is opposite they will try to cancel each other
Total moment of the two forces about O = 12Nm (clockwise) – 10Nm (anticlockwise)
= 2Nm (clockwise)
Question: 5
Two forces each of magnitude 10N act vertically upwards and downwards respectively at the two ends A and B of a uniform rod of length 4m which is pivoted at its mid-point O as shown. Determine the magnitude of the resultant moment of forces about the pivot O.
Solutions: Given, Force applied = 10 N
Length of the rod = 4m
Total resultant moment of forces as the forces acting are in opposite directions will be
= force applied * total length of rod
= 10 N * 4 m
= 40 Nm (clockwise)
Question: 6
Figure shows two forces each of magnitude 10N acting at the points A and B at a separation of 50 cm, in opposite directions. Calculate the resultant moment of the two forces about the point (i) A, (ii) B and (iii) O, situated exactly at the middle of the two forces.
Solutions: Given, Distance between forces = 50cm = 0.5 m
Magnitude of forces = 10 N
Therefore, Moment of force at point A = force applied * distance
= 10 N * 0.5 m
= 5 Nm (clockwise)
Moment of force at point B = force applied * distance
= 10 N * 0.5 m
= 5 Nm (clockwise)
Moment of forces at point O = force applied * separation between forces
= 10 N * 0.5m
= 5 Nm (clockwise)
Question: 8
A uniform metre rule is pivoted at its mid-point. A weight of 50 gf is suspended at one end of it. Where a weight of 100g should be suspended to keep the rule horizontal?
Solutions: Given, W1 = 50 g
L1 = 0.5 m
W2 = 100 g
To keep the metre rule in horizontal position we have to keep the rule in equilibrium
According the principle of the moments
W1 * l1 = W2 * l2
50 g * 0.5 m = 100 g * l2
50g * 0.5m / 100g = l2
0.25m = l2
Hence, a weight of 100 g should be suspended at 0.25m or 25 cm from the midpoint.
Question: 10
The diagram shows a uniform bar supported at the middle point O. A weight of 40 g is placed at a distance 40cm to the left of the point O. How can you balance the bar with a weight of 80 g?
Solutions: Given, W1 = 40 g
L1 = 40cm = 0.4m
W2 = 80g
According to principle of moment
W1 * L1 = W2 * L2
40 g * 40 cm = 80g * L2
40 g * 40 cm /80g = L2
20 cm = L2
Hence, the weight can be balanced on the bar at 20 cm from right of O.
Question: 11
Figure shows a uniform metre rule placed on a fulcrum at its mid-point O and having a weight 40gf at the 10 cm mark and a weight of 20 gf at the 90 cm mark.
(i) Is the metre rule in equilibrium? If not, how will the rule turn?
(ii) How can the rule be brought in equilibrium by using an additional weight of 40gf?
Solutions:
(i) the metre rule is not in the equilibrium as it does not obey principle of moment hence the rule will turn in anticlockwise direction as W1 * L1 > W2 * L2
(ii) To bring rule in equilibrium using additional weight of 40g
W1 * L1 = (W2 * L2) + ( W3 + L3)
40g * 40cm = (20g * 40cm) * (40g + L3)
40g * 40cm – 20g * 40cm = (40g * L3)
1600 g cm – 800 g cm = 40g * L3
800 g cm /40 g = L3
20 cm = L3
Hence, to bring the rule into equilibrium weight should be put on 20 cm on right from the point O or at 70 cm on the rule.
Question: 12
When a boy weighing 20 kg sits at one end of a 4m long see-saw, it gets depressed at its end. How can it be brought to the horizontal position by a man weighing 40 kg.
Solutions: Given
W1 = 20 kg
L1 = 2 m (from the centre)
W2 = 40 kg
According to principle of moments
W1 * L1 = W2 * L2
20 kg * 2m = 40kg * L2
40 kg m / 40kg = L2
1 m = L2
Therefore, the seesaw can be brought horizontal when the 40kg man is seating at 1m from the centre.
Question: 13
A physical balance has its arms of length 60 cm and 40 cm. What weight kept on a pan of longer arm will balance an object of weight 100 gf kept on the other pan?
Solutions: Given
W1 = 100 g
L1 = 40 cm
L2 = 60 cm
According to principle of moments
W1 * L1 = W2 * L2
100 g * 40 cm = 60 cm * W2
4000 g cm / 60 cm = W2
66.67 g = W2
The balance the object of 100g we have to put weight of 66.67g in the pan with longer arm.
Question: 14
The diagram shows a uniform metre rule weighing 100 gf, pivoted at its centre O. Two weights 150gf and 250gf hang from the point A and B respectively of the metre rule such that OA = 40 cm and OB = 20 cm. Calculate: (i) the total anticlockwise moment about O, (ii) the total clockwise moment about O, (iii) the difference of anticlockwise and clockwise moment, and (iv) the distance from O where a 100gf weight should be placed to balance the metre rule.
Solutions: Given, Weight of rule 100g
W1 = 150g
W2 = 250g
W3 = 100g
OA = 40 cm
OB = 20 cm
Total anticlockwise moment about O = W1 * OA
= 150g * 40 cm
= 6000 g cm
Total clockwise moment about O = W2 * OB
= 250g * 20cm
= 5000 g cm
The difference between anticlockwise moment and clockwise moment
= 6000 g cm – 5000 g cm
= 1000 g cm
To balance the rule in horizontal position according to principle of moments
W1 * L1 = W2 *L2 + W3 * L3
6000 g cm = 5000 g cm + 100g * L3
1000g cm/ 100 g = L3
L3 = 10 cm
Therefore, weight of 100g must be kept on 10cm to the right.
Question: 15
A uniform metre rule of weight 10 gf is pivoted at its 0 mark.
(i) What moment of force depresses the rule?
(ii) How can it be made horizontal by applying a least force?
Solutions: (i) Given
Length of rule = 100cm
Weight of the rule = 10g
As the rule is pivoted at 0 mark the perpendicular distance will be 50 cm
Therefore,
Moment of force = force applied * perpendicular distance
= 10g * 50 cm
= 500 g cm
Hence, the moment of fore that depresses the rule is 500 gcm.
(ii) We have to apply minimum force hence the perpendicular distance must be maximum. So, maximum distance we have is 100 cm
The calculated moment of force is 500 g cm
Moment of force = force applied * perpendicular distance
500 g cm = Force applied * 100 cm
Force applied = 500 g cm / 100 cm
Force applied = 5 g
Hence, the minimum force to be applied on the rule to keep it horizontal is 5 g.
Question: 17
A uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 cm by suspending an unknown mass m at the mark 20 cm.
(i) Find the value of m.
(ii) To which side the rule will tilt if the mass m is moved to the mark 10 cm?
(iii) What is the resultant moment now?
(iv) How can it be balanced by another mass 50 g?
Solutions: (i) As the rule is suspended at 40 cm the centre of gravity is 10 cm away from the pivoted point
Therefore, distance of centre of gravity of rule = 10 cm
Mass suspended = m g
Distance of mass = 20cm
As the system is in equilibrium
20 cm * m g = 10 cm * 100 g
m g = 50 g
Value of m is 50 g
(ii) if the mass moved to 10 cm mark then the resulting moment of force will increase as the distance between point of force and pivoted point will increase. Hence, the rule will tilt in downward direction from the side of weight m.
Question: 18
In figure below, a uniform bar of length l m is supported at its ends and loaded by a weight W kgf at its middle. In equilibrium, find the reactions R1 and R2 at the ends.
Solutions: according to the condition of equilibrium to be for bar to stay in equilibrium
R1+R2 = W
But for rotational equilibrium R1= R2
Therefore,
R1+R1 = W
R1= W/2
R1= R2= W/2
Exercise 1(B)
Question: 1
Define the term ‘centre of gravity of a body’.
Solutions: The point about which the sum of moment of weights of all the particles containing the body is zero it is called as centre of gravity of the body.
Question: 2
Can the centre of gravity of a body be situated outside its material of the body? Give an example.
Solutions: Yes. The centre of gravity of a circular ring is situated outside of the body at the centre of the ring.
Question: 3
State factor on which the position of the centre of gravity of a body depends? Explain your answer with an example.
Solutions: The position of centre of gravity depends upon its shape (distribution of mass).
For example, the centre of the gravity of a straight uniform wire is at its middle but when that wire is bent in the form of circle the centre od gravity lies at the centre of that circle.
Question: 4
What is the position of the centre of gravity of a:
(a) rectangular lamina
(b) cylinder?
Solutions: (a) The centre of gravity of a rectangular lamina lies on the intersection of diagonals.
(b) The centre of gravity of a cylinder lies on the midpoint of axis of cylinder.
Question: 5
At which point is the centre of gravity situated in:
(a) A triangular lamina and
(b) A circular lamina?
Solutions: (a) The centre of gravity of a triangular lamina lies on the point of intersections of median.
(b) The centre of gravity of a circular lamina lies at he centre of the circular lamina.
Question: 6
Where is the centre of gravity of a uniform ring situated?
Solutions: centre of gravity of the uniform ring is situated at centre of the ring.
Question: 8
Explain how you will determine experimentally the position of the centre of gravity for a triangular lamina (or a triangular piece of cardboard).
Solutions: To determine the position of centre of gravity for a triangular lamina experimentally, make fine hole on the corners of the triangle. Now, suspend the lamina using plumb line through one of the hole. Make sure that lamina is free to oscillate about the point of suspension. When lamina comes to rest draw a straight line along the plumb line. Repeat the same procedure the remaining two holes. We will get three lines crossing each other at exact same point which is the centre of gravity of triangular lamina.
Question: 9
State whether the following statements are true or false.
(i) ‘The position of the centre of gravity of a body remains unchanged even when the body is deformed’.
(ii) ‘The centre of gravity of a freely suspended body always lies vertically below the point of suspension’.
Solutions:
(i) False
(ii) True
Multiple Choice Type:
Question: 1
The centre of gravity of a uniform ball is
(a) At its geometrical centre
(b) At its bottom
(c) At its topmost point
(d) At any point on its surface
Solutions: (a) at its geometrical centre
Question: 2
The centre of gravity of a hollow cone of height h is at distance x from its vertex where the value of x is:
(a) h/3
(b) h/4
(c) 2h/3
(d) 3h/4
Solutions: (c) 2h/3
Exercise 1 (c)
Question: 1
Explain the meaning of uniform circular motion. Why is such motion said to be accelerated?
Solutions: When a particle moves with a constant speed in circular path, its motion is said to be uniform circular motion. In this motion the speed of particle is constant as particle travels equal distance in equal interval of time. But the direction of motion changes at each of point of circular path hence the velocity of the particle changes. That is why uniform circular motion is said to be accelerated.
Question: 3
Is it possible to have an accelerated motion with a constant speed? Name such type of motion.
Solutions: Yes. The accelerated motion with constant speed is uniform circular motion.
Question: 4
Give one example of motion in which speed remains uniform, but the velocity changes.
Solutions: circular motion.
Question: 5
A uniform circular motion is an accelerated motion. Explain it. State whether the acceleration is uniform or variable? Name the force responsible to cause this acceleration. What is the direction of force at any instant? Draw diagram in support of your answer.
Solutions:
When a particle moves with a constant speed in circular path, its motion is said to be uniform circular motion. In this motion the speed of particle is constant as particle travels equal distance in equal interval of time. But the direction of motion changes at each of point of circular path hence the velocity of the particle changes. That is why uniform circular motion is said to be accelerated.
The acceleration is uniform.
The force responsible for this acceleration is centripetal force. The direction of this force at any instant is towards the centre.
Question: 6
Differentiate between a uniform linear motion and a uniform circular motion.
Solutions:
Uniform linear motion |
Uniform circular motion |
In uniform linear motion particle moves in straight line with constant velocity | In uniform circular motion particle moves in circle with changing velocity |
The motion s not accelerated | The motion is accelerated |
The direction of motion doesn’t change | Direction of motion changes constantly |
Example is train running on straight track with constant velocity | Example is motion of planets around the sun |
Question: 7
Name the force required for circular motion. State its direction.
Solutions: Force required for circular motion is centripetal force which acts towards the centre.
Question: 8
What is centripetal force?
Solutions: Centripetal force is the force acting on an object in circular motion directed towards the axis of rotation or centre of curve.
Question: 9
Explain the motion of a planet around the sun in a circular path.
Solutions: A planet moves around the sun in a circular path the gravitational force between the planets which acts as the required centripetal force. This centripetal force is always directed towards the centre of the sun at each point of its path which is responsible for the circular motion of a planet.
Question: 10
(a) How does a centripetal force differ from a centrifugal force with reference to the direction in which they act?
(b) Is centrifugal force the force of reaction of the centripetal force?
(c) Compare the magnitudes of centripetal and centrifugal force.
Solutions:
(a) The main difference between centripetal and centrifugal forces is about the direction they on which is exactly opposite. The centripetal force is the force pointing towards the centre of a circle that keeps an object moving in a circular path, while the centrifugal force is the sensation that an object feels when it moves in that circular path, with that sensation seeming to push it away from the centre of a circle.
(b) Centrifugal force is not the reaction to centripetal force.
(c) Centrifugal force is not a real force it is a result of inertia. Its tendency of a body to resist to the change of state of rest or motion. On the other hand centripetal force is a real force that counteracts the centrifugal force and keeps body from flying out of its path.
Question: 11
Is centrifugal force a real force?
Solutions: centrifugal force is a pseudo force.
Question: 12
A small pebble tied at one end of a string is placed near the periphery of a circular disc, at the centre of which the other end of the string is tied to a peg. The disc is rotating about an axis passing through its centre.
(a) What will be your observation when you are standing outside the disc? Explain.
(b)What will be your observation when you are standing at the centre of the disc. Explain.
Solutions:
(a) As the tension in the string provides the required centripetal force the pebble moves in a circular path.
(b) As the centrifugal force on the pebble balance the tension in the string the pebble is stationary just in front.
Question: 13
A piece of stone tied at the end of a thread is whirled in a horizontal circle with uniform speed with the help of hand. Answer the following questions.
a.) Is the velocity of stone uniform or variable?
b.) Is the acceleration of stone uniform or variable?
c.) What is the direction of acceleration of stone at any instant?
d.) What force does provide the centripetal force required for circular motion?
e.) Name the force and its direction which acts on the hand.
Solutions:
a.) Variable
b.) Variable
c.) Towards the centre of the circular path
d.) Tension in the string
e.) The reaction of tension away from the centre of the circular path
Question: 14
State two differences between the centripetal and centrifugal force.
Solutions:
Centripetal force |
Centrifugal force |
This is real force acting on the body which cause circular motion | This is pseudo force resulting due to inertia of the body |
This force acts towards the centre of the rotation | This force acts away from the centre of rotation |
Question: 15
State whether the following statements are true or false by writing T/F against them.
(a) The earth moves around the sun with a uniform velocity.
(b) The motion of the moon around the earth in circular path is an accelerated motion.
(c) A uniform linear motion is unaccelerated, while a uniform circular motion is an accelerated motion.
(d) In a uniform circular motion, the speed continuously changes because the direction of the motion changes.
(e) A Boy experiences a centrifugal force on his hand when he rotates a piece of stone tied at one end of a string, holding the other end in the hand.
Solutions:
(a) False
(b) True
(c) True
(d) False
(e) False
Multiple Choice Type:
Question: 1
Which of the following quantity remains constant in uniform circular motion:
(a) Velocity
(b) Speed
(c) Acceleration
(d) Both velocity and speed
Solutions: (b) speed
Question: 2
The centrifugal force is:
(a) a real force
(b) the force of reaction of centripetal force
(c) a fictitious force
(d) directed towards the centre of the circular path
Solutions: (c) a fictitious force