

InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
651. |
The direction of friction and the direction of motion are always A) same B) opposite C) A or B D) we cannot say |
Answer» Correct option is B) opposite |
|
652. |
The forces acting on a wooden block are shown in the figure given alongside. Find the resultant of the forces and the direction of the resultant force. |
Answer» Total forces acting downwards = 2 N + 22 N = 24 N Total forces acting upwards = 8 N + 14 N = 22 N Resultant force = Difference between the magnitude of forces = 24 N – 22 N = 2 N . The wooden block will move in the downward direction with the resultant force of 2 N acting on it. |
|
653. |
Friction is offered by 1. solids 2. liquids 3. gases 4. all of these |
Answer» 4. all of these |
|
654. |
Fill in the blanks. 1. Rolling friction is.......... than sliding friction. 2. Smoother the surface, ........... is the frictional force offered by it. 3. Friction depends upon the ............ of an object. 4. Friction is a type of ............5. Bicycles use ............... to reduce friction between the pedals and the crank. |
Answer» Fill in the blanks. 1. Rolling friction is less than sliding friction. 2. Smoother the surface, lesser is the frictional force offered by it. 3. Friction depends upon the weight of an object. 4. Friction is a type of contact force. 5. Bicycles use ball bearings to reduce friction between the pedals and the crank |
|
655. |
Fill in the blanks.1. More the roughness of a surface, greater is the ..................2. Rolling friction offers ............. resistance than sliding friction. 3. Weight remaining the same, friction is independent of ........... of surface in contact. 4. The force of friction acting between two bodies in contact while at rest is known as ......... friction. 5. Sliding friction and rolling friction are types of ......... friction. |
Answer» 1. More the roughness of a surface, greater is the friction. 2. Rolling friction offers less resistance than sliding friction. 3. Weight remaining the same, friction is independent of area of surface in contact. 4. The force of friction acting between two bodies in contact while at rest is known as static friction. 5. Sliding friction and rolling friction are types of kinetic friction. |
|
656. |
Friction always 1. supports motion 2. starts motion 3. encourages motion 4. opposes motion |
Answer» 4. opposes motion |
|
657. |
How can sliding friction be changed into rolling friction easily ? |
Answer» Sliding friction can be changed into rolling friction easily by providing wheels or ball bearings to the objects to reduce friction and facilitate easy movement. |
|
658. |
When you push something heavy, the resistance you feel decreases as soon as the object starts moving. Explain the reason. |
Answer» It is because when the object is at rest, static friction occurs. The static friction is maximum when the applied force is just short of the force required to start motion between the two bodies in contact. Once the object starts moving, friction (resistance) we feel decreases. |
|
659. |
Where do we apply a force while walking? |
Answer» While walking we apply force on the ground. |
|
660. |
A ball of dough is rolled into a flat chapatti. Name the force exerted to change the shape of the dough. |
Answer» Muscular force. |
|
661. |
Define rolling friction and give an example where rolling. friction comes into play. |
Answer» When we roll an object on a surface, the friction produced between the object and the surface is called rolling friction. Example : when we pull our large suitcases on wheels, rolling friction comes into play. The wheels reduce friction and cut down a lot of effort required to move the suitcase (Rolling friction is less than sliding friction). |
|
662. |
Force of friction comes into play when a/an 1. object attempts motion 2. moving object tries to stop 3. object is at rest 4. all of the these |
Answer» 4. all of the these |
|
663. |
A chapati maker is a machine which converts balls of dough into chapati’s. What effect of force comes into play in this process? |
Answer» The force changes the shape of the dough. | |
664. |
State the formula for pressure, Hence, determine the unit of pressure. |
Answer» Pressure = \(\cfrac{force}{area\,on\,which\,the\,force\,is\,applied}\) ∴ The unit of pressure = \(\cfrac{the\,unit\,of\,force}{the\,unit\,of\,area}\) The SI unit of force is the newton (N) and that of area is m . Therefore, the SI unit of pressure is N/m2 . It is called the pascal (Pa). 1 Pa = 1 N/m2 . [Note: The unit pascal is named in honour of Blaise Pascal (1623-62), French mathematician, physicist and philosopher.] |
|
665. |
What is a non contact force? Give one example. |
Answer» A force that acts between two objects even if the two objects are not in contact, is called a non contact force. Example: The earth revolves around the Sun. |
|
666. |
What is a contact force? Give one example. |
Answer» A force that acts through a direct contact of two objects or via one more object, is called a contact force. Example: If a ball at rest on the ground is kicked, it starts moving. contact force is a type of force which requires contact to occur example pushing a car up or kicking a ball across the room etc. |
|
667. |
Why do porters place a round piece of cloth on their heads? |
Answer» Porter place a round piece of cloth on their heads to increase the area of contact of the load with the head. So the pressure on their head is reduced and they can carry heavy load easily |
|
668. |
How much hydro-electric power is generated in India? |
Answer» At present, 23% of total electricity is generated in India using hydro-energy. |
|
669. |
A machine raises a load of 750N through a height of 16m in 5 seconds. Calculate the power at which the machine works. |
Answer» Power = (Force x Distance)/Time = (750 x 16)/5watt = 2400watt. = 2.4kw. |
|
670. |
Can two agents, doing the same work, have different power? |
Answer» Yes, since power = work/time the two agents can have different powers if they take different amounts of time to do the same work. |
|
671. |
State the amount of work done by an object when it moves in a circular path for one complete rotation. Give a reason to justify your answer. |
Answer» Amount of work done is equal to zero. Work is said to be done only when there is displacement produced. In case of a body moving in a circular path, the body comes to its original place, therefore, there is no displacement and hence work is zero. |
|
672. |
When does a force do work? What is the work done by the moon when it revolves around the earth? |
Answer» Work is said to be done if a body undergoes displacement due to applied force. Work done = Force × Displacement Work done is zero by the moon, as there is no displacement since it is moving in a circular path. |
|
673. |
What is Gravitational force? Give gravitational units of force. |
Answer» Earth exerts a force of attraction on each object. This force is due to gravity and its magnitude is directly proportional to the mass of the object. This force is known as gravitational force and its units are gravitational units. In MKS system the gravitational unit of force is kilogramme force i.e., kgf. In CGS system it is gramme force, gf. |
|
674. |
A bullet shot with high velocity pierces through a window pane, whereas, the pane is shattered by a pebble thrown at it. Why? |
Answer» A bullet moving with high velocity has a large momentum. When it strikes the window pane, the velocity decreases and momentum reduces to a very small value. This sudden change in momentum exerts a large impulsive force enough to pierce through the window pane, making a hole. On the other hand, the velocity of a pebble thrown at it is not very high, its momentum is less. So. change in momentum (or force) is less which only make cracks in the window pane. |
|
675. |
A force is applied on a body of mass 20kg moving with a velocity of 40ms-1 . The body attains a velocity of 50ms-1 in 2 seconds. Calculate the work done by the body. |
Answer» Work done, W = 1/2m(v22 - v12) = 1/20x 20(502 - 402) = 9000J |
|
676. |
What is the work done by the gravitational force on the moon that revolves around the earth? |
Answer» The moon is revolving around the earth. In this process, the work done by the gravitational force is zero because the displacement of the moon is at right angle to the gravitational force. |
|
677. |
A bullet fired against a window glass pane makes a hole in it without any cracks while a stone striking the same glass pane smashes it; explain with reason. |
Answer» When the bullet strikes the glass, the part of the glass coming in contact with the bullet, immediately shares the large velocity of the bullet and makes a hole in it while the rest of the glass pane, due to inertia of rest, remains at rest and so is not smashed. But when a stone strikes the same glass pan, the surrounding part of the glass gets time to share the comparatively low velocity of stone and so it is smashed. |
|
678. |
A coolie, with a load of 30kg on his head, walks on the platform. If he walks a distance of 60m, what is the work done by him? |
Answer» The work done by the coolie is zero because there is no displacement of’ the load in the direction of force, which he is applying vertically upwards. |
|
679. |
A coolie carrying a load on his head and moving on a frictionless horizontal platform does no work. Explain the reason why. |
Answer» We know w = Fdcosθ. Since force is normal to displacement. so θ = 90° Hence work done W = Fdcosθ 90 = 0. |
|
680. |
Observe the figure given and name the force needed for the motion :Fruit falling from tree. |
Answer» Gravitational force. |
|
681. |
Observe the figure given and name the force needed for the motion :A ramp. |
Answer» Gravitational force. |
|
682. |
Observe the figure given and name the force needed for the motion :A man carrying heavy sac. |
Answer» Gravitational force. |
|
683. |
(i) Where is the centre of gravity of a uniform ring situated? (ii) ‘The position of the centre of gravity of a body remains unchanged even when the body is deformed.’ State whether the statement is true or false. |
Answer» (i) Centre of gravity of a uniform ring is at its centre. (ii) False. |
|
684. |
(i) Where is the centre of gravity of a uniform ring situated. (ii) ‘The position of centre of gravity of a body remains unchanged even when the body is deformed. ’State whether the statement is true or false. |
Answer» (i) At the centre of ring, (where there is no mass) (ii) False |
|
685. |
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» (a) False (b) True |
|
686. |
Observe the figure given and name the force needed for the motion :A pin holder. |
Answer» Magnetic force. |
|
687. |
Define the term ‘centre of gravity of a body’. |
Answer» Centre of gravity is the point about which the algebraic sum of moments of weights of particles constituting the body is zero and the entire weight of the body is considered to act at this point. |
|
688. |
On whatfactor does the position of the centre of gravity of a body depend? |
Answer» The whole weight of body should act at that point. |
|
689. |
On what factors does the effect of force depend? |
Answer» The effect of force depends on two things 1. The amount of force applied the more the force applied the greater the effect will be. 2. The area on which the force is applied. The smaller the area, the force is applied on the greater the effect will be. |
|
690. |
State one factor on which the magnitude of a non – contact force depends. How does it depend on the factor stated by you? |
Answer» The magnitude of a non-contact force depends on distance of separation of two bodies. Magnitude of force decreases as the distance increases. |
|
691. |
Mention the factors that the effect of a force depend. |
Answer» 1. Magnitude of the force 2. The area over which it acts. |
|
692. |
State whether the following statements are true or false.(i) ‘The position of centre of gravity of a body remains unchanged even when the body is deformed’(ii) ‘Centre of gravity of a freely suspended body always lies vertically below the point of suspension’ |
Answer» (i) False. The position of centre of gravity of a body of given mass depends on its shape i.e., on the distribution of mass in it. (ii) True. |
|
693. |
Can the centre of gravity be situated outside the material of the body? Give an example. |
Answer» Yes, the centre of gravity can be situated outside the material of the body. For example, centre of gravity of ring. |
|
694. |
On what factor does the position of centre of gravity of a body depend? Explain your answer with an example. |
Answer» The position of centre of gravity of a body of given mass depends on its shape i.e., on the distribution of mass in it. For example: the centre of gravity of a uniform wire is at its mid-point. But if this wire is bent into the form of a circle, its centre of gravity will then be at the centre of circle. |
|
695. |
What is the position of centre of gravity of a:(a) rectangular lamina(b) cylinder? |
Answer» The position of centre of gravity of a (a) rectangular lamina is at the point of intersection of its diagonals. (b) cylinder is at the mid point on the axis of cylinder. |
|
696. |
Friction 1. causes wear and tear 2. produces heat 3. stops a moving body 4. has all the above disadvantages |
Answer» 4. has all the above disadvantages |
|
697. |
The position (x) of a particle of mass 1 kg moving along X-axis at time t is given by `(x=1/2 t^(2))` metre. Find the work done by force acting on it in time interval from `t=0` to `t=3 s`.A. 4.2 JB. 5.4 JC. 4.6 JD. 4.5 J |
Answer» Correct Answer - D `x=1/2 t^(2)` `implies v=(dx)/(dt)=1/2 (2t)=t` `:.` At `t=0, v_(i)=0 implies` At `t=3 s, v_(f)=3 ms^(-1)` According to work -energy theorem, `W=Delta K=K_(f)-K_(i)` `=1/2 mv_(f)^(2)-1/2 mv_(i)^(2)=1/2xx1xx3^(2)=4.5 J` |
|
698. |
A body of mass 6 kg is acted on a by a force so that its velocity changes from `3 ms^(-1)` to `5ms^(-1)`, then change in momentum isA. 48 NsB. 24 NsC. 30 NsD. 12 Ns |
Answer» Correct Answer - D Given, mass `=6` kg velocity `v=v_(2)-v_(1)=5-3=2` `:.` Momentum `p=mv=6xx2=12 N-s` |
|
699. |
Force acting on a particale is `(2hat(i)+3hat(j))N`. Work done by this force is zero, when the particle is moved on the line `3y+kx=5`. Here value of k is `(`Work done `W=vec(F).vec(d))`A. 2B. 4C. 6D. 8 |
Answer» Correct Answer - A Given, `F=2hat(i)+3hat(j)` displacement `ds=dx hat(i)+dyhat(j)+dz hat(k)` work done, `w=int F.ds =int (2 dx+3dy)` Also `3y+kx=5 implies (3dy)/(dx)+k=0` `implies 3dy=-kdx implies W=int (2dx-kdx)=0` `implies 2x=kx implies k=2` |
|
700. |
Find the torque of a force `F=(hat(i)+2 hat(j)-3hat(k))N` about a point O. The position vector of point of application of force about O is `r=(2hat(i)+3hat(j)-hat(k))m`.A. `tau=(-7 hat(i)+5hat(j)+hat(k))Nm`B. `tau=(7hat(i)-5hat(j)+hat(k)) Nm`C. `tau=(-7hat(i)+8hat(j)+hat(k))Nm`D. `tau=(-5hat(i)+7hat(j)+hat(k))Nm` |
Answer» Correct Answer - A Torque, `tau =r xx F=|(hat(i),hat(j),hat(k)),(2,3,-1),(1,2,-3)|` `tau=hat(i)(-9+2)+hat(j)(-1+6)+hat(k)(4-3)` or `tau=(-7hat(i)+5hat(j)+hat(k))Nm` |
|