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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.
| 151. |
A block of mass mis kept in an elevation which starts1 moving downward with an acceleration aas shown in figure. The block is observed by two observers A and B for a time interval `t_(0)` The observer B finds that the work done by pseudo-force on the block is :A. zeroB. `-ma^(2) t_(0)`C. `+ ma^(2) t_(0)`D. `- mg a t_(0)` |
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Answer» Correct Answer - A |
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| 152. |
A block of mass mis kept in an elevation which starts1 moving downward with an acceleration aas shown in figure. The block is observed by two observers A and B for a time interval `t_(0)` According to observer B, the net work done on the block is:A. `- (1)/(2) ma^(2) t_(0)^(2)`B. `(1)/(2) ma^(2) t_(0)^(2)`C. `(1)/(2) mg a t_(0)^(2)`D. `- (1)/(2) mga t_(0)^(2)` |
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Answer» Correct Answer - B |
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| 153. |
If KE of a body increases by `300%`, by what `%` will the linear momentum of the body increase?A. 0.2B. 0.5C. 1D. 2 |
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Answer» (c ) Final kinetic energy `(K_f) = "initial kinetic energy" (K_i) + 300 %` of` K_i` ` = K_i + ((300)/(100))K_i = 4 K_i` Clearly, `K_f| K_i = 4` As `P prop sqrtK, (P_f)/(P_i) = sqrt((K_f)/(K_i)) = sqrt4 = 2 or P_f = 2p_i or p_f - p_i = 2p_i - p_i = p_i` Thus % age increases in momentum `= (P-f - p_i)/(p_i)xx 100 = 100%` |
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| 154. |
A block of mass M is hanging over a smooth and light pulley through a light string. The other end of the strinig is pulled by a constant force F. The kinetic energy of the block increases by 20J n 1s.A. the tension in the string is MgB. The tension is the string is FC. The work done by the tension on the block is 20 J in the above 1s.D. the work done by the force of gravity is -20J in the above 1s. |
| Answer» Correct Answer - B | |
| 155. |
A player kicks a ball of mass 250g at the centre of a field. The ball haves his foot with a speed of 10m//s Find the work done by the player on the ball. |
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Answer» The ball, which is initially at rest, gains kinetic energy due to work done on it by the player. Thus, the work done by the player on the ball, W = kinetic energy `(E_k)` of the ball as it leaves as it leaves his foot, i.e., `W= E_k l= (1)/(2)mupsilon^2` Here, `m = 250g =0.25kg, upsilon = 10//s` Thus, `W =(1)/(2) (0.25kg)(10m//s)^2 = 12.5J` |
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| 156. |
A spring block system is placed on a rough horizontal floor. The block is pulled towards right to give spring an eleongation less than `(2 mu mg)/(K)` but more than `(mu mg)/(K)` and released. Which of the following Jaws/principles of physics can be applied on the spring block system ?A. Conservation of mechanical energyB. Conservation of momentumC. Work energy principleD. None |
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Answer» Correct Answer - C |
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| 157. |
A block of mass m is pulled by a constant powert `P` placed on a rough horizontal plane. The friction coefficient the block and surface is `mu`. The maximum velocity of the block is.A. `(P)/(mg)`B. `(muP)/(mg)`C. `(P)/(mu mg)`D. `(P)/(mu^(2) mg)` |
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Answer» Correct Answer - C |
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| 158. |
An electric heater is rated 1500W. How much energy does it use in 10 hours ? |
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Answer» Correct Answer - 15 kWh (Unit) Energy consumed by an electric heater can be obtained with the help of the expression, P = W/T Where, Power rating of the heating, `P= 1500 W = 1.5 W` Time for which the heater has operated, `T = 10 h` Work done = Energy consumed by the heater Therefore, energy consumed = Power `xx` Time `= 1.5 xx 10 = 15 kWh` Hence, the energy consumed by the heater in 10 h is 15 kWh. |
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| 159. |
What is the gravitational potential energy of a body of mass m at a height h ? |
| Answer» Answer: - mgh. | |
| 160. |
Is gravitational potential energy of an object path dependent ? |
| Answer» No as it depends only on the differnece in the vertical heights of the intial and the final positions of the object. | |
| 161. |
What is the work done by the force of gravity on a satellite moving round the Earth ? Justify your answer. |
| Answer» When a satellite moves around the earth in a circular path, its displacement in any short interval of time is along the tangent to the around the circular path of satellite. The force of gravity acting on the satellite is along the radius of the earth at that point. Since a tangent is always at right angles to the radius, therefore, the motion of a satellite and force of gravity are at right angles (`90^(@)` angle) to each other. Now, work done `W = F cos 90^(@) xx s`. Since `cos 90^(@) = 0`, therefore, work done `W = F xx 0 xx s = 0`. Thus, the work done by the force of gravity on a satellite moving the earth is zero. | |
| 162. |
Can any object have mechaincal energy even if it momentum is zero ? Exaplain. |
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Answer» Yes. We know that mechaincal energy = potential energy + kinetic energy Momentum of a body `= m upsilon` When momentum is zero, `upsilon = 0 (as m != 0)` and as such KE. =` (1)/(2) m upsilon^2 = 0` Thus, mechanical energy = potentail energy. |
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| 163. |
A block of mass 30 kg is pulled up by a rope as shown in Fig. with a constant speed by applying a force of 200 N parallel to the slope. A and B are initial and the final positions of the block. Calculate: (a) the work done by the force in moving the block A to B, (b) the potential energy gained by the block, (c) account for the difference in work done the force and the increase in potential energy of the block. |
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Answer» Correct Answer - A::B::C::D Here, m = 30kg, F = 200N, s =AB =3m, h =1.5m (a) Work done by the force in moving the block form A to B, i.e., W =F s =(200N)(3m) =60J (b) Potential energy gained by the block `E_p =mgh = (30kg)(10m//s^2)(1.5m) =450J` (c ) The difference (600J - 450 J =150J) is used in doing work against friction between the block and the slope. This energy is converted into heat. |
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| 164. |
How do you differentiate between energy and power ? |
| Answer» The energy possessed by an object is a measure of the total amount of work it can perform without any reference to the time in which this work has been performed. Power, on the other hand, is the rate at which work is done and not the total amount of work done. | |
| 165. |
Can a body have energy without having momentum ? |
| Answer» Yes. A body at rest has no momentum, i.e., p = m upsilon =0.` But a body at rest can have potential energy (due to position or shape). | |
| 166. |
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not ? Justify your answer. |
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Answer» Work is done whenever the given two conditions are satisfied `rarr` A force acts on the body `rarr` There is a displacement of the body by the application of force in or opposite to the direction of force When a person holds a bundle of hay over his head, then there is no displacement in the bundle of hay. Althrough, force of gravity is acting on the bundle, the person is not applying any force on it. in the absence of force, work done by the person on the bundle is zero. |
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| 167. |
What does the gravitational potential energy imply ? |
| Answer» it implies that the body has the potential or capacity of gaining kinetic energy when relased from some point under gravity. | |
| 168. |
When is work done by a force negative ? |
| Answer» Work done by a force is negative is force and displacement are in opposite direction. | |
| 169. |
The potential energy at a point, relative to the reference point is always defined as the negative of work done by the force as the object moves from the reference point to the point considered. The value of potential energy at the reference point itself can be set equal to zero because we are always concerned only with differences of potential energy between two points and the associated change of kinetic energy. A particles A is fixed at origin of a fixed coordinate system. A particle B which is free to move experiences an force `vec(F) = (-(2a)/(r^(3)) + (beta)/(r^(2))) hat(r)` due to particle A where `hat(r)` is the position vector of particle B relative to A. It is given that the force is conservative in nature and _potential energy at infinity is zero. If B has to be removed from the influence of A, energy has to be supplied for such a process. The ionization energy E 0 is work that has to be done by an external agent to move the particle from a distance `r_(0)` to infinity slowly. Here `r_(0)` is the equilibrium position of the particle What is potential energy function of particle as function of r.A. `(alpha)/(r^(2)) - (beta)/(r)`B. `- (alpha)/(r^(2)) + (beta)/(r)`C. `- (alpha)/(r^(2)) - (beta)/(r)`D. `(alpha)/(r^(2)) + (beta)/(r)` |
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Answer» Correct Answer - B |
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| 170. |
In the Q. No. 26, if `M = 2 m` and friction exists between the circular track and the horizontal surface then, which of the following lot best represents the variation of frictional force versus the angle `theta`:A. B. C. D. |
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Answer» Correct Answer - B |
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| 171. |
Two students who weigh the same start at the same ground floor location at the same time to go to the same classrooms on the third floor by different routes. If they arrive at different times, which student will have expended more power ? Explain. |
| Answer» They are doing the same amount of work (same mass, same height). So the one that arrives first have expended more power due to shorter time interval. | |
| 172. |
A bullet of mass 20 g moving with a velocity of 500 m/s, strikes a free and goes out from the other side with a velocity of `400 m//s.` Calculate the work done by the bullet in joule in passing through the tree. |
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Answer» (a) Kinetic energy. It can be utilized for generatin hydroelectricity. (b) As `K_i = (1)/(2) (10xx10^(-3)) (500)^2 J = 1250 J` and `K_f= (1)/(2) (10 xx10Z^(-3)) (400)^2 J = 800 J,` work done by the bullet, `W = K_i - K_j = 450J` |
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| 173. |
A car weighing 1000 kg and travelling at `30 m//s` stops at a distance of 50 m decelerating uniformly. What is the force exerted on it by the breaks ? What is the work done by the brakes? |
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Answer» In order to calculate the force, we have to find out the acceleration (or rather retardation) first. Now, Initial speed, u = 30 m/s Final speed, v = 0 (The car stops) Acceleration, a = ? (To be calculated) And, Distance, s = 50 m Now, we know that : `v^(2) = u^(2) + 2as` So, `(0)^(2) = (30)^(2) + 2 xx a xx 50` 100 a = - 900 `a = -(900)/(100)` Thus, Acceleration, `a = - 9 m//s^(2)` The force exerted by the brakes can now be calculated by using the formula: `F = m xx a` Here, Mass, `m = 1000 kg " "` (Given) And, Acceleration, `a = -9 m//s^(2)" "` (Calculated above) So, Force, `F = 1000 xx (-9)` F = - 9000 N Thus, the force exerted by the brakes on the car is of 9000 newtons. The negative sign shows that it is a retarding force. The work done by the brakes can be calculated by using the relation : `W = F xx s` Here, Force, F = 9000 N Distance, s = 50 m So, Work done, `W = 9000 xx 50 J` = 450000 J `= 4.5 xx 10^(5) J` Thus, the work done by the brakes is `4.5 xx 10^(5)` joules. |
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| 174. |
(a) A student throws a ball vertically upward so tha it just reaches the height of window on th second floor of a dormitory. At the same time the ball is thrown upward, a student at the windoe drops a ball. Are the mechanical energies of the ball the same at half the height of the window ? Exaplain (b) What does work- energy theorem state ? |
| Answer» (a) Yes. When the thrown-up ball is at maximum height, its velocity is zero so it has same energy as the dropped ball. Since the total energy of each ball is conserved, both the balls will have same mechanical energy at half the height of the window. As a matter of fact, both the balls will have the same kinetic and potential energy at half height position. (b) Work done = change in KE (or PE) | |
| 175. |
The string of a simple pendulum can with stand a maximum tension equal to 4 times the weight of bob suspended to it. The string is made horizontal and bob is released from rest then:A. String will break somewhere during the motion and will then follow straight line pathB. String will break somewhere during the motion and then follow parabolic pathC. It will complete the vertical circleD. Motion will be oscillatory and string will not break |
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Answer» Correct Answer - D |
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| 176. |
A pendulum of mass 1 kg and length = 1m is released from rest at angle = 60º. The power delivered by all the forces acting on the bob at angle = 30º will be: (g = 10 m/s2)A. `1.34 omega`B. `13.4 omega`C. `0.67 omega`D. `5 omega` |
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Answer» Correct Answer - D |
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| 177. |
A simple pendulum consisting of a mass M attached to a string of length L is released from rest at an angle `alpha`. A pin is located at a distance l below the pivot point. When the pendulum swings down, the string hits the pin as shown in figure. The maximum angle `theta` which the string makes with the vertical after hitting the pin is A. `cos^(-1) ((L cos alpha + l)/(L + l))`B. `cos^(-1) ((L cos alpha + l)/(L - l))`C. `cos^(-1) ((L cos alpha - 1)/(L - l))`D. `cos^(-1) ((L cos alpha - l)/(L + l))` |
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Answer» Correct Answer - C |
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| 178. |
A block of mass m released from rest from point O as shown below. The velocity of the block at the lowest points are v O , v E, v F respectively. Assume coefficient of kinetic friction between surface and the block is same in all cases. Then, (a) A. `v_(D) gt v_(E) gt v_(F)`B. `v_(f) gt v_(E) gt v_(D)`C. `v_(D) = v_(E) = v_(F)`D. `v_(D) = v_(E) v_(F) = 0` |
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Answer» Correct Answer - C |
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| 179. |
A block of mass 100 g moved with,a speed of 5 m/s at the highest point in a closed circular tube of radius 10 ,cm kept in a vertical plane. The cross-section of the tube is such that the block just fits in it., The block makes several oscillations inside the tube and finally steps at the lowest point. The work done by tIe tube on the block during the process is:A. 1.45 JB. `- 1.45 J`C. 0.2 JD. zero |
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Answer» Correct Answer - B |
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| 180. |
When is gravitational potential energy of a body positive ? |
| Answer» Positive gravitational potential energy implies that the body will perform work while returing from its present position (level) to zero level. | |
| 181. |
Match the following : `{:(,"Column-1",," Column-2"),("(A)","Electrostatic potential energy",,"(P) Positive"),("(B)","Gravitational potential energy",,"(Q) Negative"),("(C)","Elastic potential energy",,"(R) Zero"),("(D)","Magnetic potential energy",,"(S) Not defined"):}` |
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Answer» Correct Answer - A - P,Q, R; B - Q; C- P,R; D - P, Q, R |
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| 182. |
(a) Why can we not associate a potential energy with the frictional force as we did with the gravitational force ? (b) A uniform chain of mass m and length I is lying on a table with `(1//4)` of its length hanging freely from the edge. Find the amount of work required to be done in dragging the chain on the table completely. |
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Answer» (a) We cannot recover the energy lost due to frictional effects. (b) Mass of `(1//4)` length of the chain =` m//4` the weith `(mgh//4)` of this part of the chain acts an its CG which is at a distance `(i//8)` from the edga. Work done in dragging the chain on the table completely = force`xx` distacne = weight of the `(I//4)` length of the chain `xx` distance through its CG is to be moved = `(mg//4) (I/8) = mgl//32`. |
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| 183. |
A certain household has consumed 250 units of energy during a month. How much energy is this in joules ? |
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Answer» Energy consumed = `250 units = 250 kWh = 250 xx 1000 Wh = 250 xx (1000 W) (3600 s)` `=250 xx 1000 xx 3600 J (1 W = 1 j//s)` ` = 900 xx 10^6 = 9 xx10^8 J` |
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| 184. |
What happens to their gravitational potential energy when firefighters slide fire pole ? |
| Answer» Most of it changes to thermal energy due to friction of the bodies of the firefighters with the pole. | |
| 185. |
An electric bulb of 60 W is used for 6 h per day. Calculate the units of energy consumed in one day by the bulb. |
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Answer» Correct Answer - C Energy, `W = Pxxt = 60Wxx6h = 360Wh = 0.36kWh = 0.36 units` |
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| 186. |
A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy ? |
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Answer» When a freely falling object having kinetic energy hits the ground and eventually stops, then : (i) Some of the kinetic energy is converted into sound energy (due to which a sound is produced when the falling object hits the ground). (ii) Some of the kinetic energy is changed into heat energy (due to which the falling object and the ground where it falls become slightly warm). (iii) Some of the kinetic energy is transformed into potential energy of configuration of the object and the ground (because the object and the ground get deformed a little bit at the point of their collison). |
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| 187. |
What happens to the kinetic energy of the molecules of a substance when it is cooled ? |
| Answer» When a substance is cooled, the velocity of its molecules decreases. As a result of this, the kinetic energy of the molecule also decreases. | |
| 188. |
The Earth moving round the sun in a circular orbit is acted upon by a force and hence work must be done on the Earth by the force. Do you agree with this statement ? |
| Answer» The statement is wrong as the centripetal force acting on the Earth is at right angles to the direction of its displacement i.e., centripetal force is along the radius of the circular orbit and the displacement is along its tangent. (Radius and tangent are perpendicular to each other). | |
| 189. |
In each of the following, a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is gegative, positive or zero. |
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Answer» (i) in (a) the direction of force, F and displacment are perpendicular to each other. There is no displacement in the direction of force and as such work done is zero. (ii) (b) the direction of force F and displacement are in the some direction and as such work done by the force is positive. (iii) (c ), the force, F acts in a direction opposite to the displacement and as such work done is negative. |
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| 190. |
In each of the following a force F is acting on an object of mass m. The direction of displacement is form west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero. |
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Answer» (a) In the first case [Figure (i)],the direction of force F and the direction of displacement are at right angles to each other. So, the work done in the first case is zero. (b) In the second case [Figure (ii)], the displacement is in the direction of force F, so the work done is positive. (c) In the third case [Figure (iii)], the force F acts in a direction opposite to the direction of displacement, so the work done is negative. |
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| 191. |
A spring is compressed by a toy cart of mass 150g. On releasing the cart, it moves with a speed of `0.2 m//s`. Calculate the elastic potential energy of the spring. |
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Answer» Correct Answer - A::C Elastic potentail energy of the spring = kinetic energy gained by the toy cart, i.e., `E_p = E_k = (1)/(2)mv^(2) = (1)/(2)(0.150).(0.2m//s)^2 = 3xx10^(-3)J` |
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| 192. |
A body is moved along the sides of a triangle ABC once in anticlockwise direction by applying different forces along different sides as represented in the figure . The total displacement of the body and work done by the forces are `"______"` and "`_______"` respectively. A. 23 m , 26 JB. 15 m , 21 JC. 25 m , 32 JD. 0 m , 258 J |
| Answer» Correct Answer - D | |
| 193. |
A box is pushed through 4.0 m across floor offering 100 N resistance.How much work is done by the resisting force? |
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Answer» Correct Answer - D `F=100N,S=4m, theta=0^0` `W=vecF.vecS=100xx4=400J` |
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| 194. |
Rohit pushed a toy truck and it moved a distance of 10 m . How much force did Rohit apply through the push , if the work done by him is 1000 J ? |
| Answer» Correct Answer - 100 N | |
| 195. |
The KE acquired by a mass m in travelling a certain distance s, starting from rest, under the action of a constant force is directly proportional to :A. mB. `sqrtm`C. `(1)/(sqrtm)`D. none of these |
| Answer» (d) For a constant force (F) and a fixed distance (s), kinetic energy acquired = work done = F s, i.e., kinetic energy is independent of m. | |
| 196. |
A truck moving at `15m//s` has KE of `4.2xx10^5` J (a) what is the mass of the truck ? (b) By what multiplicative factor does the kinetic energy of the truck increases if its speed is doubled? |
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Answer» (a) `K = (1)/(2) m upsilon^2, m = (2K)/(upsilon^2) = (24.2xx10^5J)/((15m^2) )= 3733 kg` (b) Kinetic energy depends on the speed squared, and hence doubling the speed increases the kinetic energy by a factor of four. |
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| 197. |
A boy of mass 55kg rus up a filight of 40 stairs, each measuring 0.15m Calculate the work done by the boy. |
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Answer» Correct Answer - C `W = Fs =(550 N) (6m) = 3300 J (F = 55 kg wt = 550 N, s = 40 xx 0.15m = 6m)` |
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| 198. |
When do we say that work is done ? |
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Answer» Work is done whenever the given conditions are satisfied `rarr` A force acts on the body `rarr` There is a displacement of the body caused by the applied force along the direction of the applied force. |
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| 199. |
What is the other name of power ? |
| Answer» Answer:- Activity. | |
| 200. |
Write an expression for the work done when a force is acting on an object in the direction of its displacement. |
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Answer» When a force F displaces a body through a distance S in the direction of the applied force, then the work done W on the body is given by the expression : Work done = Force `xx` Displacement `W= F xx s`. |
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