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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.
| 1351. |
Let the mass of the flower pot 15kg and the height of the sunshade 4m. (a) When the flower pot is on the sunshade, what is its potential energy? (g = 10m/s2). (b) When it is on the sunshade, what is its kinetic energy? (c) If so, what is its total energy? |
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Answer» (a) U = mgh = 15 × 10 × 4 = 600J (b) Kinetic energy will be zero (c) Total energy = PE + KE = 600J + 0 = 600J |
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| 1352. |
Identify more situation in which potential energy is acquired by virtue of position |
Answer»
Inference: Height increases, potential energy increases |
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| 1353. |
Given below is the information regarding the working of pumps in three neighboring houses. Complete the table (g = 10 m/s2). |
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Answer» (b) 150000J (c) 150000J |
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| 1354. |
If a force of 50N is applied on a body and it under goes a displacement of 2m in the direction of the force, calculate the amount of work done. |
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Answer» F = 50N s = 2m W = Fs = 50 x 2 = 100J |
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| 1355. |
The potential energy of a system increased if work is doneA. by the system against a conservative forceB. by the system against a nonconservative forceC. upon the system by a conservative forceD. upon the system by a nonconservative force |
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Answer» The potential energy of a system increase if work is done by the system against a conservative force. `- Delta U = W_("conservative force")` |
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| 1356. |
A particle of mass `0.2 kg` is moving in one dimension under a force that delivers constant power `0.5 W` to the particle.If the initial speed =0 then the final speed (in `ms^(-1)`) after `5s` is. |
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Answer» Correct Answer - `(5)` `Power=(dW)/(dt)impliesW=0.5xx5=2.5=KE_f-KE_i` `2.5=M/2(v_f^2-v_i^2)rarrv_r=5` |
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| 1357. |
A particle A of mass `10//7kg` is moving in the positive direction of `x-axis`. At initial position `x=0`, its velocity is `1ms^-1`, then its velocity at `x=10m` is (use the graph given) A. (a) `4ms^-1`B. (b) `2ms^-1`C. (c) `3sqrt2ms^-1`D. (d) `100/3ms^-1` |
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Answer» Correct Answer - A Area under `P-x` graph `=int pdx=(m(dv)/(dt))vdx=underset1oversetvintmv^2dV` `=[(mv^3)/(3)]_1^v=(10)/(7xx3)(v^3-1)` From the graph, `area=1/2(2+4)xx10=30` or `(10)/(7xx3)(v^3-1)=30` or `v=4ms^-1` Alternative method: From graph `P=0.2x+2` `mv(dv)/(dx)v=0.2x+2` `mv^2dv=(0.2x+2)dx` Now integrating both sides, we get `underset1oversetvintmv^2dv=underset0overset10int(0.2x+2)dx` `v=4ms^-1` |
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| 1358. |
The teeter toy consists of two identical weights hanging from a peg on dropping arms as shown. The arrangement is surprisingly stable. Let us consider only oscillatory motion in the vertical plane. Consider the peg and rods (connecting the weights to the peg) to be very light. The length of each rod is l and length of the peg is L. In the position shown the peg is vertical and the two weights are in a position lower than the support point of the peg. Angle `alpha` that the rods make with the peg remains fixed. (a) Assuming the zero of gravitational potential energy at the support point of the peg evaluate the potential energy (U) when the peg is tilted to an angle `theta` to the vertical. The tip of the peg does not move. (b) Knowing that U shall be minimum in stable equilibrium position prove that `theta = 0` is the stable equilibrium position for the toy if the two weights are in a position lower than the support point of the peg |
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Answer» Correct Answer - (a) `2mg cos theta[L-l cos alpha]` |
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| 1359. |
A physics student writes the elastic potential energy stored in a spring as `U=1/2KL^(2)+1/2Kx^(2),` where L is the natural length of the spring, x is extension or compression in it and K is its force constant. A block of mass M travelling with speed V hits the spring and compresses it. Find the maximum compression caused. |
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Answer» Correct Answer - `sqrt(M/K) .V` |
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| 1360. |
A heavy particle is attached to one end of a light string of length l whose other end is fixed at O. The oarticle is projectyed horizontally with a velocity `v_0` from its lowest position A. When the angular displacement of the string is more than`90^(@)`, the particle leaves the circular path at B. The string again becomes taut at C such that B,O,C are collinear. Find `v_0` in terms of l and g |
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Answer» Correct Answer - `((4+3sqrt(2))/2 gl` |
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| 1361. |
A small ball is attached to an end of a light string of length R. It is suspended in vertical plane supported at point A. B and C are two nails (of negligible thickness) at a horizontal distance 0.3 R from A and a vertical distance 0.4 R above A respectively. The ball is given a horizontal velocity `u = sqrt5gR)` at its lowest point. Subsequently, after the string hitting the nails, the nails become the centre of rotation. Assume no loss in kinetic energy when the string hits the nails. It is known that the string will break if tension in it is suddenly increased by 200% or more. Will the string break during the motion? If yes, where? What is tension in the string at the instant the string breaks? |
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Answer» Correct Answer - The string will break on hitting the second nail at C.T = 8.6 mg |
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| 1362. |
A mass m = 0.1 kg is attached to the end B of an elastic string AB with stiffness k = 16 N/m and natural length `l_0 = 0.25 m`. The end A of the string is fixed. The mass is pulled down so that AB is `2l_0` = 0.5 m and then released. (a) Find the velocity of the mass when the string gets slack for the first time. (b) At what distance from A the mass will come to rest for the first time after being released. |
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Answer» Correct Answer - (a) `sqrt(5) m//s` (b) zero |
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| 1363. |
A person weighing 70kg runs up a flight of 30steps in 35seconds. What is the power of the person if each step is 20cm high? |
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Answer» Here, `M=70kg, t=35s` `h=30xx20cm=6m. P=? P=(W)/(t)=(Mgh)/(t)=(70xx9*8xx6)/(35)=117*6watt` |
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| 1364. |
According to work energy principle, work done by …………………..in displacing body …………………to…………………… in………………. . |
| Answer» net force `,` is equal `,` change `,` kinetic energy of the body. | |
| 1365. |
A spring is compressed by tieing its ends together tightly. It is then placed in acid and dissolves. What happens to its stored potential energy. |
| Answer» When the spring dissolves in acid, the elastic potential energy stored in the spring passes to the acid. The internal energy of the acid increases and hence its temp. rises. | |
| 1366. |
The linear momentum of a body is increased by `10%`. What is the percentage change in its KE? |
| Answer» Correct Answer - `21%` | |
| 1367. |
Two springs have spring constants `k_(1)and k_(2) (k_(1)nek_(2)).` Both are extended by same force. If their elastic potential energical are `U_(1)and U_(2),` then `U_(2)` isA. `(k_(2))/(k_(1))U_(1)`B. `(k_(1))/(k_(2))U_(1)`C. `sqrt((k_(1))/(k_(2)))U_(1)`D. `sqrt((k_(2))/(k_(1)))U_(1)` |
| Answer» Correct Answer - B | |
| 1368. |
A spring with spring constaant k when compressed by 1 cm the PE stored is U. If it is further compressed by 3 cm, then change in its PE isA. 3 UB. 9 UC. 8 UD. 15 U |
| Answer» Correct Answer - D | |
| 1369. |
KE of a body is increased by `44%.` What is the percent increse in the momentum ?A. `10%`B. `20%`C. `30%`D. `44%` |
| Answer» Correct Answer - B | |
| 1370. |
Momentum of a body increae by `0.02%` percent increase in KE isA. `0.02%`B. `0.04%`C. `0.01%`D. `0.08%` |
| Answer» Correct Answer - B | |
| 1371. |
The PE of a 2 kg particle, free to move along x-axis is given by `V(x)=((x^(3))/(3)-(x^(2))/(2))J.` The total mechanical energy of the particle is 4 J. Maximum speed (in `ms^(-1)`) isA. `(1)/(sqrt2)`B. `sqrt2`C. `(3)/(sqrt2)`D. `(5)/(sqrt6)` |
| Answer» Correct Answer - D | |
| 1372. |
A body loses half of its velocity on penetrating 6 cm in wooden block. How much will it penetrate more before coming to rest ?A. `1 cm`B. `2 cm`C. `3 cm`D. `4 cm` |
| Answer» Correct Answer - B | |
| 1373. |
A block of mass `m` is lying at rest at point P of a wedge having a smooth semi-circular track of radius R. What should be the minimum value of `a_0` so that the mass can just reach point Q A. (a) `g/2`B. (b) `sqrtg`C. (c) `g`D. (d) Not possible |
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Answer» Correct Answer - C Work done by pseudo-force=change in PE `ma_0R=mgR` `a_0=g` |
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| 1374. |
Two spring P and Q having stiffness constants `k_1` and `k_2(ltk_1)`, respectively are stretched equally. ThenA. (a) More work is done on QB. (b) More work is done on PC. (c) Their force constants will become equalD. (d) Equal work is done on both the springs |
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Answer» Correct Answer - B Since `k_1gtk_2`. Therefore, P will develop greater restoring force than Q. So the applied force on P will be more than the applied force on Q. Extension is same in both the cases. So, more work will be done on P. |
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| 1375. |
A car of mass `500kg` moving with a speed `36kmh` in a straight road unidirectionally doubles its speed in `1min`. Find the power delivered by the engine. |
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Answer» Its initial speed `v_1=(36000)/(3600)=10ms^-1` If the car doublese its speed, finally its speed becomes `v_2=20ms^-1`. Change in KE of the car `=DeltaKE=(1//2)mv_2^2-(1//2)mv_1^2` During `Deltat=1min=60s` the power delivered by the engine, P=Work done/Time Taken=`(|DeltaKE|)/(Deltat)` `=(1/2xxm[v_2^2-v_1^2])/(Deltat)` `=(1//2xx500[20^2-10^2])/(60)=1250W` |
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| 1376. |
A 1 kg block situated on a rough incline is connected to a spring constant 100Nm-1.The block is released from rest with the spring in the unstretched position.The block moves 10cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. |
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Answer» The normal reaction force of block on incline R = mg cosθ Let the coefficient of friction be M ∴ F = MR = μ mg cosθ ∴ Net force on block = mg sinθ – F = mg (sinθ – μ cosθ) we know that the block moves by distance x = 10cm = 0.1 m. The work done by net force in moving block 0.1m is = Energy stored in the spring. ∴ mg (sin θ – μ cos θ) x = \(\frac{1}{2}\) kx2 ∴ 2 mg (sin 37° – μ cos 37°) = kx ⇒ 2 × 1 × 9.8 ms-2 (sin37°- μ cos37°) = 100 × 0.1m ⇒ 19.6 (0.601 – μ × 0.798) = 10 0.601 – μ × 0.798 = 0.5102 ⇒ μ × 0.798 = 0.09079 ∴ μ = \(\frac{0.09079}{0.798}\)= 0.1137 |
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| 1377. |
A `500-kg` car, moving with a velocity of `36kmh^-1` on a straight road unidirectionally, doubles its velocity in `1min`. The average power delivered by the engine for doubling the velocity isA. (a) `750W`B. (b) `1050W`C. (c) `1150W`D. (d) `1250W` |
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Answer» Correct Answer - D `u=10ms^-1`, `v=20ms^-1` Work done=Increase in kinetic energy `=1/2xx500[20^2-10^2]=(500xx30xx10)/(2)` `Power=(500xx30xx10)/(2xx60)=1250W` |
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| 1378. |
Two inclined frictionless tracks, one gradual and the other steep meet at A from where to stones are allowed to slide down from rest, one on each track (fig.) Will hte stones reach the bottom at the same time? Will they reach there with the same speed? Explain, given `theta_(1)=30^(@)`, `theta_92)=60^(@)` and h=10m. What are the speeds and time taken by the two stones? |
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Answer» `1/2 mv^(2) = mgh, v=sqrt(2gh)` `=sqrt(2 xx 10 xx 10) ms^(-1) = 14.4 ms^(-1)` `v_(B) = v_(C) = 14.14ms^(-1), l=1/2(gsintheta)t^(2)` `sintheta = h/l, l=h/sintheta)` `h/sintheta = 1/2gsinthetat^(2)` or `t=sqrt((2h)/g.1/sintheta` `t_(B) = sqrt((2 xx 10)/10 1/(sin30^(@)) = 2sqrt(2)s` `t_(C) = sqrt((2 xx 10)/(10) . 1/(sin60^(@)) = (2sqrt(2))/(sqrt(3))s` |
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| 1379. |
A large family uses `8KW` of power. Direct solar energy is incident on the horizontal surface at an average rate of `200Wm^-2`. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply `8kW`? |
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Answer» If `Am^2` is the area, then power `=200xxAW` Useful electrical energy produced per second is `200/100(200A)=40xxAW` But `40A=8000` or `A=200m^2` |
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| 1380. |
Two inclined frictionless tracks, one gradual and the other steep meet at `a` from where two stones are allowed to slide down from rest, one on each track as shown in Figure. Which of the following statement is correct ? .A. Both the stones the bottom at the same time but not with the same speedB. Both the stones reach the bottom with the same speed and stone i reaches the botom earlier than stone IIC. Both the stones reach the bottom with the same speed and stone II reaches the botom earlier than stone ID. Both the stones reach the bottom different times and with different speeds |
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Answer» (c ) As the given tracks are frictionless hence, mechaical energy will be conserved. As both the tacks having common height, h From conservation of mechanical energy `(1)/(2)mv^(2)=mgh` (for both tracks I and II) Hence, speed is same for both stone. For stone I, `a_(1)`=acceleration along inclined planed `=g sin theta_(1)` Similarly, for stone II `a_(2)=g sin theta_(2) as theta_(2) gt theta_(1) hence, a_(2) gt a_(1)`, And both length for track II is also less hence, stone II reaches earlier than stone I. |
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| 1381. |
A family uses 8kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square metre. If `20%` of this energy can be converted to useful electrical energy, how large an area is needed to supply 8kW? (a) Compare this area to that of the roof of a typical house. |
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Answer» Let the area be A square metre. `:.` Total power `=200A`. Useful electrical energy produced `//` sec `=(20)/(100)(200A)=40A=8000(2wat t)` Therefore, `A=(8000)/(40)=200sq. m` This area is comparable to the roof of a large house of 250sq. metre. |
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| 1382. |
A bullet of mass 0.012 kg and horizontal speed `70ms^(-1)` strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by thin wire. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block. |
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Answer» Here, `m_(1)=0.012kg, u_(1)=70m//s,m_(2)=0.4kg, u_(2)=0` As the bullet comes to rest with respect to the block, the two behave as one body. Let `upsilon` be the velociy acquired by the combination. Applying principle of conservation of linear momentum, `(m_(1)+m_(2))upsilon=m_(1)u_(1)+m_(2)u_(2)=m_(1)u_(1)` `upsilon=(m_(1)u_(1))/(m_(1)+m_(2))=(0.012xx70)/(0.012+0.4)=(0.84)/(0.412)=2.04ms^(-1)` Let the block rise to a height h. P.E. of the combination `=` K.E. of the combination `(m_(1)+m_(2))gh=(1)/(2)(m_(1)+m_(2))upsilon^(2):. h =(upsilon^(2))/(2g)=(2.04xx2.04)/(2xx9.8)=0.212m` For calculating heat produced, we calculate energy energy lost (W) , where `W=` initial of bullet `-` final K.E. of combination `=(1)/(2)m_(1)u_(1)^(2)-(1)/(2)(m_(1)+m_(2))=(1)/(2)xx0.012(70)^(2)-(1)/(2)(0.412)(2.04)^(2)` `W=29.4-0.86=28.54jou l e:. ` Heat produced, `H=(W)/(J)=(28.54)/(4.2)=6.8cal` |
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| 1383. |
In case of a moving body as force of friction is μmgx as potential energy similar to mgh? |
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Answer» No, potential energy is defined only for conservation forces. |
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| 1384. |
A body is being raised to a height h from the surface of earth. What is the sign of work done by (a) applied force (b) gravitational force ? |
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Answer» (a) Force is applied on the body to lift it in upward direction and displacement of the body is also in upward direction, therefore angle between the applied force and displacement is `theta==0^(@)` Work done by the applied force `" " "W=Fs cos theta=Fs cos 0^(@)=Fs (therefore cos 0^(@)=1)` `i.e. " " "W=Positive"` (b) The gravitational force acts inm download direction and displacement in upward direction therefore angle between them is `theta=180^(@)` Work done by the gravitional force W=Fs cos `180^(@)=-Fs` |
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| 1385. |
A bob of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of `7ms^(-1)`. If hits the floor of the elevator (length of the elevator `=` 3m) and does not rebound. What is the heat produced by the impact ? Would your answer be different if the elevator were stationary ? |
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Answer» P.E of bolt= mgh = `0.3 xx 9.8 xx 3 xx 8.82 J` The bolt does not rebound. So the whole of the energy is converted into heat. Since the value of the acceleration due to gravity is the same in all inertial system, therefore the answer will not change even if the elevator is stationary. |
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| 1386. |
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s–1 . It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary? |
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Answer» Mass of the bolt, m = 0.3 kg Speed of the elevator = 7 m/s Height, h = 3 m Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy. Heat produced = Loss of potential energy = mgh = 0.3 × 9.8 × 3 = 8.82 J The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero. |
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| 1387. |
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty? |
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Answer» The sand bag is placed on a trolley that is moving with a uniform speed of 27 km/h. The external forces acting on the system of the sandbag and the trolley is zero. When the sand starts leaking from the bag, there will be no change in the velocity of the trolley. This is because the leaking action does not produce any external force on the system. This is in accordance with Newton’s first law of motion. Hence, the speed of the trolley will remain 27 km/h. |
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| 1388. |
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s-1. What is the speed of the trolley after the entire sandbag is empty? |
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Answer» The system of trolley and sandbag is moving with a uniform speed. Clearly, the system is not being acted upon by the external force. If the sand leaks out, even then no external force acts. So there shall no change in the speed of the trolley. |
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| 1389. |
A body of mass ‘M’ collides against a wall with a velocity v and retraces its path with the same speed. The change in momentum is (take initial direction of velocity as positive)A. zeroB. 2MvC. MvD. `-2Mv` |
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Answer» Correct Answer - D |
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| 1390. |
Momentum of a particle is increased by `50%`. By how much percentage kinetic energy of particle will increase?A. `25%`B. `50%`C. `100%`D. `125%` |
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Answer» Correct Answer - B We know that `K prop P^(2)` or `(K_(2))/(K_(1)) = (P_(2)^(2))/(P_(1)^(2))` `(K_(2) - K_(1))/(K_(1)) xx 100 = ((P_(2)^(2) - P_(1)^(2))/(P_(1)^(2))) xx 100` `= [((150)^(2) - (100)^(2))/((100)^(2))] xx 100 = 125%` |
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| 1391. |
Assertion: A light body and a heavy body have same momentum. Then they also have same kinetic energy. Reason: Kinetic energy does not depand on mass of the body.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason are false. |
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Answer» Kinetic energy of body of mass `m_(1)` `K_(1) = (1)/(2) m_(1)v_(1)^(2) = (P_(1)^(2))/(2 m_(1))` Again, kinetic energy of a body of mass `m_(2)`. `K_(2) = (1)/(2) m_(2)v_(2)^(2) = (P_(2)^(2))/(2 m_(2))` If `P_(1) = P_(2), (K_(1))/(K_(2)) = (m_(2))/(m_(1))` If `m_(2) gt m_(1)` Then , `K_(1) gt K_(2)` i.e., the kinetic energy of light body will be more than the kinetic energy of heavy body when both have same momentum. |
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| 1392. |
Mark out the correct statement(s).A. Total work done by internal forces on a system is always zeroB. Total work done by internal forces on a system may sometimes be zeroC. Total work done by friction can never be zeroD. Total work done by friction is always zero |
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Answer» Correct Answer - B (b) Work done by friction mayt be positive, negative and zero. Work done by conservative, internal forces may be zero in a round trip but it is non-zero for non-conservative internal forces. |
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| 1393. |
If the speed of a vehicle is increased by `1 ms^(-1)`, its kinetic energy is doubled , then original speed of the vehicle isA. `(sqrt2+1)ms^(-1)`B. `2(sqrt2-1)ms^(-1)`C. `2(sqrt2+1)ms^(-1)`D. `sqrt2(sqrt2+1)ms^(-1)` |
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Answer» Correct Answer - A (a) It is given that, `K_(f)=2K_(i)` or `" "(1)/(2)m(v+1)^(2)=((1)/(2)mv^(2))(2)" or " v+1=sqrt2v` `v=(1)/(sqrt2-1)=sqrt2+1ms^(-1)` |
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| 1394. |
A gas filled in a cylinder fitted with movable piston is allowed to expand. What is the nature of the work done by the gas? |
| Answer» Positive as force due to gaseous pressure and displacment of piston are in the same direction. | |
| 1395. |
A vehicle needs an engine of 7500 W to keep it moving with a constant velocity of 20 `ms^(-1)` on a horizontal surface. The force resisting the motion isA. 375 dyneB. 375 NC. 150000 dyneD. 150000 N |
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Answer» Correct Answer - B (b) Power required to move the vehicle `P=F.v rArr F=(P)/(v)=(7500)/(20)=375 N` |
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| 1396. |
A body is pushed through 5 m across a sirface affering 75 N resistance How much work is done by the (1) applied force (ii) resisting force ? |
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Answer» (i) `F=75N, d=5 m` Work done by applied force `=W=Fdcos0^(@)=75xx5xx1=375J` (ii) Work done by resisting force `=Fdcos180^(@)` `=75xx5xx(-1)=-375J` |
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| 1397. |
A box is pushed through 4-0m across a floor offereing 100N resistance. How much work is done by the (i) applied force (ii) resisting force? |
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Answer» Here, `s=4.0m, F=100N,W=?` Work done by applied force `W=F s cos0^(@)=Fscos0^(@)=100xx4xx1=400J` Work down by resistinig force `W=Fscos180^(@)=100xx4(-1)=-400J` |
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| 1398. |
In a shotput event an athlete throws the shotput of mass `10 kg`with an initial speed of `1 ms^(-1)` at `45^@` from a height `1.5 m` above ground. Assuming air resistance to be negligible and acceleration due to gravity to be `10 ms^(-2)`, the kinetic energy of the shotput when it just reaches the ground will beA. 2.5 JB. 5 JC. 52.5 JD. 155 J |
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Answer» Correct Answer - D (d) Given, h=15 m, `v=1ms^(-1), m=10 ms^(-2)` From conservation of mechanical energy, `(PE)_(i)+(KE)_(i)=(PE)_(f)+(KE)_(f)` `rArr " " mgh+(1)/(2)mv^(2)=0+(KE)_(f)` `rArr" "(KE)_(f)=mgh+(1)/(2)mv^(2)` `rArr" "(KE)_(f)=10xx10xx15+(1)/(2)xx10(1)^(2)` =150+5=155J |
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| 1399. |
In a shotput event an athlete throws the shotput of mass 20 kg with an initial speed of `2 m s^(-1)` at `45^@` from height 3 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be `10 m s^(-2)`, the kinetic energy of the shotput when it just reaches the ground will beA. 2.5JB. 5JC. 525JD. 640 J |
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Answer» Correct Answer - D Using law of energy conservation, `(TME)_(i) = (TME)_(f)implies U_i+ K_i = U_f + K_f` `implies 20 xx 10 xx 3 + 1/2 xx 20 xx 2^(2)= 0+ K_fimplies K_f= 640 J` |
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| 1400. |
The decrease in the potential energy of a ball of mass 20 kg which falls from a height of 50 cm isA) 968 JB) 98 JC) 1980 JD) None of these |
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Answer» B) 98 J Explanations: \(\Delta U = mgh\) \(20 \times 9.8 \times0.5 = 98 J\) |
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