InterviewSolution
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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.
| 101. |
The weight of a person on a plannet A is about half that on the Earth. He can jump upto 0.4 m height on the surface of the Earth. How high can be jump on the planet A ? |
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Answer» Weight of the person on a planet `A, W_A = mg_A` Weight of the person on Earth, `W_E = mg_E ("as mass of the person is the same on the planet A and the Earth")` We are given that `W_A = (1)/(2)W_E, mg_A = (1)/(2)mg_E or g_A = (1)/(2)g_E` Let `h_A, h_E` be the heights to which the person can jump on the planet A and the Earth respectively. Gain in PE on jumping on planet `A = mg_A h_A` Gain in PE on jumping on Earth ` = mg_E h_E` Since gain in potential energy by the person is the same in both the cases as the muscular effort of the person remains unaffected by the change is planets, `mg_A h_A = mg_E h_E` or `h_A = (g_Eh_E)/(g_A) = (g_E h_E)/((1)/(2)g_E) = 2h_E` As `h_E = 0.4 m, h_A = 2 (0.4 m) = 0.8m` |
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| 102. |
How does the wind move the blades of a wind- mill ? |
| Answer» Air in motion is called wind. Thus, the wind possesses a lot of kinetic energy which enables it to perform the work required to turn the blades of a wind- mill. | |
| 103. |
Jasmine is running away from a monster with a speed of `5 m s^(-1)` . If her weight is 100N , then find out the energy spent by Jasmine to escape the monster. (Take g = `10 m s^(-2)` , energy possessed by a body in motion = 1/2 `mv^(2)` where m = mass of the body , v = velocity) |
| Answer» Correct Answer - 125 J | |
| 104. |
According to the Law of Conservation of Energy , as energy changes from one form to another form , the total energy of that systemA. Keeps increasingB. Keeps decreasingC. Increases for some time then remains constant .D. None of these |
| Answer» Correct Answer - D | |
| 105. |
A boy of mass 50kg runs up a staricase of 45 steps in 9s. If the height of each step is 15cm, find his power. Take `g = 10m//s^2` |
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Answer» Correct Answer - C `P = (W)/(t) = (mgh)/(t) = ((50kg)(10m//s^2)(6.75m))/(9s) (h = 45xx15cm = 675cm = 6.75m)` = 375W |
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| 106. |
Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her ? Why ? |
| Answer» The acceleration of an object can be zero even if several forces are acting on it provided the resultant force (F) is zero. As F = 0, ma = 0 and accordingly a = 0 `(as m != 0 )`. | |
| 107. |
`300 J`of work is done in slide a `2kg` block up an inclined plane of height `10m`. `Taking g = 10 m//s^(2), work done against friction is |
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Answer» Work done against friction = total work done - PE at a height of 10m = `(300 J) - (20 xx 10 xx10) J =100 J` |
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| 108. |
A 60 kg person climbs stairs of total height 20m in 2min. Calculate the power deliverd. |
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Answer» Correct Answer - A Power delivered, `P =(W)/(t) = ((60kg)(10m//s^2)(20m))/(2xx60s) = 100W` |
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| 109. |
Which one of the following is not the unit of energy ?A. jouleB. newton metreC. kilowattD. kilowatt hour |
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Answer» Correct Answer - C kilowatt is the unit of power. |
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| 110. |
The work done on an object does not depend upon the :A. displacementB. force appliedC. angle between force and displacementD. initial velocity of the object |
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Answer» Correct Answer - D Work done on an object by a force is independent of its initial velocity. |
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| 111. |
A family uses 250 units of electrical energy during a mont. Calculate this electrical energy in joules. |
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Answer» 250 units of electrical energy means 250 kWh of electrical energy. Now, we know that : 1 kWh of energy ` = 3.6 xx 10^(6) J` So, 250 kWh of energy `= 3.6 xx 10^(6) xx 250 J` `= 9 xx 10^(8)J` Thus, 250 units of electrical energy is equal to `9 xx 10^(8)` joules. |
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| 112. |
Is a person doing any work by holding a suitcase ? |
| Answer» No, as there is no displacement of the suticase. | |
| 113. |
what are the units of work and energy ? |
| Answer» The unit of work and energy is the same, i.e., joule (J). | |
| 114. |
A particle moves rom a point `vecr_1=(2m)veci(3m)vecj` to another point `vecr_2=(3m)veci+(2m)vecj` during which a certain force `vecF(5N)veci+(5N)vecj` acts on it. Fikd the work doen by the force on the particle during the displcement. |
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Answer» Given `vecr_1=2veci+3vecj` `vecr_2=3veci+2vecj` so, displacement vector is given by `vecr=vecr_2-vecr_1` ltbr `=(3veci+2vecj)-(2veci+vecj)` `=veci-vecj` Again `vecF=5veci+5vecj` So, work doe `=vecF.vecS=5xx1+5(-1)=0` |
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| 115. |
(a) How is the energy of an object measured? (b) What does work - energy theorem state? |
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Answer» (a) It is measured by the quantity of work that an object can do. (b) Net work done on an object by the external forces in equal to change in its kinectic energy. |
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| 116. |
When is the work done by gravity negative ? |
| Answer» when the object is thrown upward. | |
| 117. |
The kinetic energy of an object of mass` m moving with a velocity of 5 m//s is 25 J. What will be its kinetic energy when its velocity is doubled ? What will be its kinetic energy - when its velocity si increased three times ? |
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Answer» Here, kinetic energy of the object, `E_k = 25J` velocity of the object, `upsilon = 5 m//s` `As E_k= (1)/(2)m upsilon^2, m = (2E_k)/(upsilon^2) = (2xx25J)/((5//s)^2) = 2kg` Kinetic energy of the object when its velocity is doubled `(i.e., upsilon = 10 m//s)` `(1)/(2)m upsilon^2 = (1)/(2) (2kg) (10m//s)^2 = 100 J` Kinetic energy of the object when its velocity in increased three times `(i.e., upsilon =15 m//s)` `=(1)/(2)m upsilon^2 = (1)/(2) (2kg) (15m//s)^2 = 225 J` Aliter. Since `E_k prop upsilon^2` (as m is constant), (i) when `upsilon` is doubled, `E_k` becomes `(2)^2 = 4` times its previous value, i.e.,` 4xx25 J = 100 J` (ii) when `upsilon` is increased three times, `E_k` becomes `(3)^9 = 9` times it previous value, i.e., `9 xx 25 J = 225 J` |
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| 118. |
You can decrease the kinetic energy of an object as much as you want. You can do so by either reducing mass by half or reducing the speed by half. Which option would you pick, and why ? |
| Answer» Reducing the speed `(upsilon)` by half, KE will become `(1//4)th` of the its previous value and hecne will be reduced by `(3//4)`, whereas reducing the mass (m) by half will rreduce the kinetic energy only by half. | |
| 119. |
A man does 200 J of work in 10s and a boy does 100 J of work in 4s. (a) who is delivering more power? (b) Find the ratio of the power delivered by the man to that delivered by the boy. |
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Answer» Correct Answer - A::B::D (a) Power delivered by the man, `P_1 =(200J)/(10s) = 20W` Power delivered by the boy, `P_2 = (100J)/(4s) = 25W` Thus, the boy delivers more power. (b) `(P_1)/(P_2) =(20)/(25) 4//5` |
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| 120. |
A body of mass 2kg is projected vertically upwards with a speed of `3m//s`. The maximum gravitational potential energy of the body is :A. 18 JB. 4.5 JC. 9 JD. 2.25 J |
| Answer» (c ) Gravitational potential energy (maximum) = kinetic energy `=(1)/(2)m upsilon^2 = (1)/(2) (2) (3)^2 =9J` | |
| 121. |
By now much will the kinetic energy of a body increase if its speed is doubled ? |
| Answer» The kinetic energy `(E_k)` of the body will become four times as `E_k = (1)/(2) m upsilon^2 or E_k prop upsilon^2.` | |
| 122. |
A ball is projected vertically upwards. Air resistance and variation in g may be neglected. The ball rises to its maximum height H in a time T, the height being h after a time t : (1) The graph of kinetic energy `E_(k)` of the ball against height h is shown in figure 1 (2) The graph of height h against time t is shown in figure 2 (3) The graph of gravitational energy `E_(k)` of the ball against height h is shown in figure 3 (1) Two particles move on a circular path ( one just inside and the other just outside) with angular velocities ro and `5 omega` starting from the same point. Then :A. they cross each other at regular intervals of time `(2 pi)/(4 omega)` when their angular velocities are oppositely directedB. they cross each other at points on the path subtending an angle of `60^(@)` at the centre if their angular velocities are oppositely directedC. they cross at intervals of time `(pi)/(3 omega)` if their angular velocities are oppositely directedD. they cross each other .at points on the path subtending `90^(@)` at the centre if their angular velocities are in the same sense |
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Answer» Correct Answer - B::C::D |
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| 123. |
A single conservative `f_((x))` acts on a `m = 1 kg` particle moving along the x-axis. The potential energy `U_((x))` is given by : `U_((x)) = 20 + (x - 2)^(2)` where x is in metres. At `x = 5 m`, a particle has kintetic energy of 20 J The maximum value of potential energy is:A. zeroB. 20 JC. 29 JD. 49 J |
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Answer» Correct Answer - B |
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| 124. |
The US athlete Florence Griffith oyner won the 100 m spring gold medal at Seol Olympic 1988 setting a new Olympic record of 10.54s. Assume that she achieved her maximum speed in a very short time and then ran the race with that speed till she crossed the line. Take her mas ot be 50 kg. a. Calculate the kinetic energy of Griffith Joyner at her full speed. b. Assuming tht the track the wind etc. offered an average resistance of one tenth of her weight, cfalculate the work done by the resistsnce durng the run. c. What power Griffith Joyner had to exet ot maintain uniform speed? |
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Answer» Correct Answer - A::B::C::D `S=100m, t=10.54sec, m=50kg` Themotion can be assumed to be uniform because acceleration is minimum a. `Speed v=S/t=9.487 m/s` So K.E. `=1/2mv^2=2250J` `b. Weight `=mg=490J` `given R=(mg)/10=49J` So, work done against resistance `W_F=-RS=-49xx100` `rarr W_(Ff)=-4900J` c. To maintain her uniform speed she has to exert 4900J of energy to overcome friction. `P=W/t` `=4900/1054=465W` |
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| 125. |
What happens when a body performs work? |
| Answer» There is a transfer of energy. | |
| 126. |
The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body. |
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Answer» Les s be the distance covered by the body moving in the direction of constant force F. Work done by this force on the body, W = F s ........(i) If a is the acceleration produced in the body of mass m due to this constant force F and if its velocity changes from u to `upsilon` then F = ma.....(ii) and `upsilon^2 - u^2 = 2 as or s = (upsilon^2 - u^2)/(2a) ..... (iii)` From eqns. (i), (ii) and (iii), `W = (ma)((upsilon^2- u^2)/(2a)) = (1)/(2)mupsilon^2 - (1)/(2)mu^2` Since `(1)/(2)mupsilon^2 - (1)/(2)mu^2` = final kinetic energy - initial kinetic energy = increases in kinetic energy, Work done, W = increases in kinetic energy. |
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| 127. |
find the energy in kWh consumed in 10 hours by four devices of power 500 W each. |
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Answer» Here, total power of the four devices, `P = 4xx500 W = 2000 W = 2kW` time for which the devices are operated `t = 10h` Energy consumed by the four devices, i.e., `W = P xx t = (2kW) (10h) = 20 kWh = 20` units |
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| 128. |
What is power ? |
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Answer» Power is defined as the rate of doing work. `"Power" = ("Work done")/("Time taken")` or `P = (W)/(t)` where P = Power W = Work done and t = Time taken Power is also defined as the rate at which energy is consumed (or utilised). |
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| 129. |
The mass of a ball A is double the mass of another ball B. The ball a moves at half the speed of the ball. B. Calculate the ratio of the kinetic energy of A to the kinetic energy of B, |
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Answer» Correct Answer - A::B mass of ball `A = 2m`, mass of ball `B = 2 upsilon` velocity of ball `A = upsilon," velocity of ball" B =2upsilon` `("Kinetic energy of ball A")/("Kinetic energy of ball B") = ((1)/(2)(2m)upsilon^2)/((1)/(2)m(2upsilon)^2) =("mv"^(2))/(2"mu"v^2) = (1)/(2) = 1:2` |
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| 130. |
A truck weighing 5000 kg f and a cart weighing 500 kg f are moving with the same speed. Compare their kinetic energies. |
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Answer» Correct Answer - A `(KE of truck)/(KE of cart) = ((1)/(2)m_1 upsilon^2)/((1)/(2)m_2upsilon^2) = (m_1)/(m_2) = (5000kg f)/(500kg f) =10 =10:1` |
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| 131. |
What are the verious energy transformations that occur when you are riding a bicycle ? |
| Answer» The chemical energy of food is transformed in our body into muscular energy is muscles. When we pedal a bicycle,the muscular energy of legs is transformed in our body into muscular energy of muscles. When we pedal a bicycle, the muscular energy of legs is transformed into kinetic energy which rotates the pedals. The rotational kinetic energy of pedals is transferred by bicycle chain to rotational kinetic energy of bicycle wheels. Due to this the bicycle wheels move forward. When the bicycle is moving, then the bicycle as well as the person riding the bicycle, both have kinetic energy. | |
| 132. |
Express 1250 N m in terms of Kilojoules . |
| Answer» Correct Answer - 1.250 J | |
| 133. |
Find the work done if a force of 2000 N is applied on a load and the load moves through a distance of 100 cm . |
| Answer» Correct Answer - 2000 J | |
| 134. |
Can there be displacement of an object in the abence of any force acting on it ? Think. Discurss this question with your friends and teacher. |
| Answer» Yes. In the absence of any force on the object, i.e., F = 0, ma = 0 (as F = ma). Since `m!=0,` a =0. In such a case, the object is either at rest or in a state of uniform motion in a straingth line. In the latter case, there is a displacement of the object without any force acting on it. | |
| 135. |
`"_______"` sources of energy can be replenished with the passage of time . |
| Answer» Correct Answer - Renewable | |
| 136. |
How much work is required for a 74Kg sprinter to acclerate from rest to 2.2 m//s? |
| Answer» Since `u = 0. W = (1)/(2) m upsilon^2 = (1)/(2) (74 kg) (2.2m//s)^2 = 179J` | |
| 137. |
`{:("Across") , (3. "It is a non-conventional source of energy (Two words)") , (4."The SI unit of work is") , (6. "It is a form of chemical energy"), (7."Work done on the body if the displacement of the body is in the direction of force applied on it") , (8."Electricity produced from the stored water in a dam is called (Two words)"), (10 . "This converts light energy into bio-chemical energy"):}:|{:("Down") , (1. "It is a domestic source of energy (Three words)"), (2. "Energy stored in a compressed string (Two words)") , (5 . "Heat energy is measured in________"), (8."Gravitational potential energy of the body depends on mass and ______") , (9"It is a conventional source of energy"):}` |
| Answer» Correct Answer - `{:("Across") , (3."Wind Energy") , (4."Joule") , (6."Biomass"), (7."Positive") , (8."Hydro Electricity") , (10. "Plants"):}:|{:("Down"), (1."Liquefied Petroleum Gas") ,(2. "Mechnical Energy") , (5. "Calorie") , (8."Height") , (9."Coal"):}` | |
| 138. |
The ultimate source of energy isA. Bio-gasB. Solar energyC. Fossil fuelsD. All of these |
| Answer» Correct Answer - BB. Solar energy | |
| 139. |
Ramu is rolling a gas cylinder from A to C though B as shown in the adjoining figure by applying a force of 100 N . Calculate the total work done by him . A. 700 JB. 400 JC. 500 JD. 300 J |
| Answer» Correct Answer - A | |
| 140. |
A force F=a+bx acts on a particle in the x-directioin, where a and b are constants. Find the work done by this force during a displacement form `x=0 to x=d`. |
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Answer» Correct Answer - A::B::D Given that `F=a+bx` Where a and b are constants So, work done by this force during the displacement x=0 to x=d is given by `W=int_0^d Fdx ` `int_0^d(a+bx)dx` `=[ax+(bx^2)/2]_0^d` `=ad+(bd^2)/2` |
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| 141. |
A rocket of maxx `3xx10^6`kg takes off from a launching pad and acquires a verticle velcoity of 1km//s and an altitude of 25km. Calculate its (a) potential energy (b) kineitc energy. |
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Answer» Here, mass of the rocket `m =3xxx10^6kg` velocity acquired by the rocket, `upsilon =1km//s =1000m//s` height attained by the rocket, h =25km = 25000m (a) Potential energy of the rocket, `E_p = mgh =(3xx10^6kg)(10m//s^2) (25000m) = 7.5xx10^(11)J` (b) Kinetic energy of the rocket, `E_k = (1)/(2)m upsilon^2 =(1)/(2)(3xx10^6kg)(1000m//s)^2 = 1.5xx10^(12)J` |
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| 142. |
__________ of a two particle system depends only on the separation between the two particles. The most appropriate choice for the blank space n the above sentence isA. kinetic energyB. total mechanical energyC. potential energyD. total energy |
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Answer» Correct Answer - C |
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| 143. |
The work done by the external forces on a system equals the change inA. total energyB. kinetic energyC. potential energyD. none of these |
| Answer» Correct Answer - A | |
| 144. |
The work done by all the forces (external and internal) on a system equals the change inA. total energyB. kinetic energyC. potential energyD. none of these |
| Answer» Correct Answer - B | |
| 145. |
A single conservative force `F(x)` acts on a `1.0-kg` particle that moves along the x-axis. The potential energy `U(x)` is given by `U(x)=20+(x-2)^2` where x is in meters. At `x=5.0m`, the particle has a kinetic energy of `20J`. What is the mechanical energy of a system?A. zeroB. 20 JC. 29 JD. 49 J |
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Answer» Correct Answer - D |
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| 146. |
Two girls each of weigth 400N, climb up a rope through a height of 8m. We name one of the girls A and the other B. Girl A takes 20s while B takes 50s to accomplish this task. What is the power expneded by each girl. |
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Answer» (i) Power expended by girl A: Weight of the girl, `mg = 400 N` Displacement (height), `h = 8 m` Time taken, `t = 20 s` From Eq. (11.8), Power, P = Work done/time taken `= (mgh)/(t)` `= (400 N xx 8m)/(20 s)` `160 W` (ii) Power expended by girl B: Weight of the girl, `mg = 400 N` Displacement (height), `h = 8 m` Time taken, `t = 50 s` Power, `P = (mgh)/(t)` `= (400 N xx 8m)/(50 s)` `= 64 W` Power expended by girl A is `160 W`. Power expended by girl B is `64 W`. |
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| 147. |
Four men light a 250 kg box to a height of 1m and hold it without raising or lowering it. (a) How much work is done by the men in lifting the box ? (b) How much work do they do in just holding it ? (c ) why do they get tired while holding it ? `(g = 10m//s^2)` |
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Answer» (a) Mass of the box, m = 250 kg Height through which it is lifted, h = 1 m Work done in lifting the box = increases in the PE = of the box= mgh = (250) (10) (1) = 2500 N (b) When the men just hold the box, s = 0 and as such W = F s = 0 (c ) To hold the box, the men have still to apply a force equal to the weight of the box in the opposite direction To apply this force, men have to exert muscular effort which is the cause of their tiredness. |
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| 148. |
In above question, speed of block A, when the extension in spring is `(x_(m))/(2)`, is:A. `2g sqrt((m)/(k))`B. `2 g sqrt((2m)/(k))`C. `2g sqrt((2m)/(3k))`D. `g sqrt((4m)/(3k))` |
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Answer» Correct Answer - D |
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| 149. |
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 the observer A :A. the work done by gravity is zeroB. the work done by normal reaction is zeroC. the work done by pseudo-force is zeroD. all of the above |
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Answer» Correct Answer - D |
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| 150. |
A chain of length L and mass Mis arranged as shown in following four cases. The correct decreasing order of potential energy (assumed zero at horizontal surface) is: (i) A. `i gt ii gt iii gt iv gt v`B. `i= ii gt iii gt iv gt v`C. `i= ii gt iv gt iii gt v`D. `i= ii gt iv gt v gt iii` |
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Answer» Correct Answer - C |
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