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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.
| 1. |
A spring which is kept comprassed by tying its ends together is allowed to be dissolved in an acid. What happens to the potnetial energy of the spring ? |
| Answer» The potential energy of the spring gets converted into heat energy (kinetic energy of acid molecules). Due to this heat, the temperature of the acid rises. | |
| 2. |
An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy ? If the object is allowed to fall, find its kinetic energy when it is half - way down. |
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Answer» Here, Mass, m = 40 kg Acceleration due to gravity, `g = 10 m//s^(2)` And Height, h = 5 m So, Potential energy `m xx g xx h` (at 5 m height) `= 40 xx 10 xx 5 J` = 2000 J `" "` (or 2000 joules) Initial, the object is at a height above the ground will be half of 5 m (and has a potential energy of 2000 J). When this object is allowed to fall and it is half way down, its height above the ground will be half of 5 m which is `(5)/(2) = 2.5 m`. Now, the potential energy of this object of 40 kg when it is at a height of 2.5 m will be : Potential energy `= m xx g xx h` (at 2.5 m height) `= 40 xx 10 xx 2.5 J` = 1000 J `" "` (or 1000 joules) Now, according to the law of conservation of energy : Total potential energy = Potential energy at half way down + Kinetic energy at half way down or 2000 = 1000 + Kinetic energy at half way down So, Kinetic energy at = 2000 - 1000 half way down = 1000 J `" "` (or 1000 joules) |
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| 3. |
A body dropped from a height H reaches the ground with a speed of 1.2 `sqrt(gH)`. Calculate the work done by air friction. |
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Answer» The forces acting on te body are the force of gravit and the air friction. By work energy theorem, the total work done on the body is `W=1/2m(1.2sqrt(gH))^2-0=0.72mgH` The work done by the force of gravity is mgh. Hence, the work done by the air friction is `0.72mgH-mgH=00.28mgH. |
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| 4. |
Two idential objects made of iron and wood, are allowed to fall from the same height on a heap of sand. It is found that the iron object penetrates more in the wooden object. Which of the two objects has more potential energy ? |
| Answer» Since the iron object penderates more in the sand (then the wooden object), it does more work. As potential energy of an object has more potential energy then the wooden object. | |
| 5. |
When an arrow is shot from a bow, it has kinetic energy. From where does it get the kinetic energy ? |
| Answer» A bow with its string stretched, possesses potential energy on account of a change in its shape. To shoot an arrow, the string is relased. The potential energy of the bow in converted into kinetic energy of the arrow. | |
| 6. |
A bullet of mass 5 g travels with a speed of 500m/s. If it penetrates a fixed target which offers a constant resistive force of 1000N to the motion of the bullet, find (a) the initial kinetic energy of the bullet (b) the distance through which the bullet has penetrated. |
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Answer» Correct Answer - A::B Here, `m = 5g =0.005kg, upsilon =500m//s, F =1000N` (a) `E_k = (1)/(2)mupsilon^2 = (1)/(2) = (1)/(2)(0.005kg)(500m//s)^2 =625J` (b) As `W= E =F_s,s =(E_k)/(F) = (625J)/(1000N) = 0.625m` |
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| 7. |
(a) Which of the following does the most work : a force of 3N acting through a distance of 3 m or a force of 4N acting through a distnace of 2m? (b) How much work is done by the centripetal force? |
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Answer» (a) Force of 3N acting through 3m as 3 N xx3m = 9J and 4N xx2m = 8J. (b) Zero as the cantripetal force is always perpendicular to displacement. |
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| 8. |
Certain force acting on a 20 kg mass changes its velocity from `5m//s to 2m//s`. calculate the work done by the force. |
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Answer» The work done by the force will be equal to the change in kinetic energy when the velocity changes from `5 m s^(-1)` to `2 m s^(-1)`. (i) In the first case : Mass, m = 20 kg And Velocity, `v = 5 m s^(-1)` `K.E. = (1)/(2) mv^(2)` `= (1)/(2) xx 20 xx (5)^(2)` `= (1)/(2) xx 20 xx 25` `= 250 J" "...(1)` (ii) In the second case : Mass, m = 20 kg And Velocity, `v = 2 m s^(-1)` So, `K.E. = (1)/(2) mv^(2)` `= (1)/(2) xx 20 xx (2)^(2)` `= (1)/(2) xx 20 xx 4` `= 40 J" "...(2)` Work done by force = Change in kinetic energy `= 250 - 40 J` = 210 J |
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| 9. |
A force of 5 N is acting on an object. The object is displaced through 2m in the direction of the force, if the forec acts all through the displacement, find the work done by the force. |
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Answer» Correct Answer - A `W = Fs = (5N)xx(2m) = 10J` |
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| 10. |
A bullet of mass 20g is found to pass two points 30m apart in 4 s? Assuming the speed to be constant, find its kinetic energy? |
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Answer» Correct Answer - B Uniform velocity of the bullet, `upsilon = ("distance convered")/("time taken") = (30m)/(4s) =.5m//s` `"mass of bullet"`, `m =20g =0.02kg` `KE "of the bullet" =(1)/(2)mupsilon^2 =(1)/(2)(0.02kg)(7.5m//s)^2 =0.5625J` |
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| 11. |
A car of mass 2000 kg is lifted up a distance of 30m by a crane in 1 min. A second crane does the same job is 2 min. What is the power applied by each crane? Do the crnae? Do the caranes consume the same or different amount of fule? Neglect power dissipation against friction. |
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Answer» Here, mass of the car to be lifted, `m = 2000kg` height through which the car is to be lifted, `h = 30m` time taken by second crane, `t_1 = 1 min =60s` time taken by second crane, `t_2 =2min =120s` Amount of work done by each crane, i.e.,` W = mgh = (2000kg)(10m//s^2)(30m) = 6xx10^5J` Power of the first crane, `P_1 = ("work done"(W))/("time taken"(t_1)) = (6xx10^5J)/(60S) = 10000 W =10kW` Power of the second crane, `P_2 = ("work done("W)/("time taken"(t_2)) = (6xx10^5J)/(120s) = 5000 W =5kW` The two cranes would consume the same amount of fuel as these have to perform the same amount of work `(i.e., 6xx10^5J)`. |
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| 12. |
Calculate the increase in potential energy as a block of 2kg is lifted through 2m. |
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Answer» Correct Answer - D `E_p =mgh =(2kg)(10m//s^2)(2m) = 40J` |
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| 13. |
How is the power related to the speed at which a body can be lifted ? How many kilograms will a man working at the power of 100 W, be able to lift at constant speed of `1 m//s vertically ? (g= 10 m//s^2)` |
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Answer» Let a body of mass m be lifted to a height h is time t. Work done in time t, i.e., `W = mgh` Thus, power = `(W)/(t) = (mgh)/(t) = mg (h//t) = mg xx` speed where `h//t` is the speed at which the body is being lifted. Aliter : `P = F upsilon = mg xx` speed. Since power = `mg xx` speed, `m = (power)/(g xx speed) = (100W)/((10 m//s^2) (1m//s)) = 10 kg (as power = 100 W and speed = 1m//s)` |
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| 14. |
A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km, he forgot the correct path at a round about, of radius 100m. However, he moves on the circular path for one and half cycle and then he moves forward upto 2.0 km. Calculate the work done by him. |
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Answer» Frictional force, F = 5 N Total distance covered by the boy, `s = 1.5 km + 1.5 (2 pi xx 100m) + 2.0 km` = 1500 m + 942m + 2000m = 4442 m work done, W= Fs = (5N) (4442 m) = 22210J |
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| 15. |
A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. If the value of g be `10 m//s^(2)`, the work done by the girl against the gravitational force will be :A. `6 xx10^3 J`B. 6 JC. 0.6 JD. zero |
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Answer» Correct Answer - D The gravitational force (F) acting on the bag, i.e., weight of the bag (3 kg) and displacement (s = 200 m) are parependicular to each other. We know no work is done when F and s are perpendicular to each other. |
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| 16. |
An automobile engine propels a 1000 kg car (A) along a levelled road at a spedd of 36 km//h. Find the power if the opposing frictional force is 100 Now, suppose after travelling a distance of 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stop at the same time. Now car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just after the collision. |
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Answer» Mass of the car `A, m_A = 1000kg` Mass of the car `B, m_B = 1000kg` initial velocity of car `A, u_A= 36 km//h = 36 ((5)/(18))m//s = 10m//s (as 1km//h = (5)/(18)m//s)` Opposing force of friction, `F = 100 N` Power of the engine of car `A = F u_A = (100N) (10m//s) = 1000 W` `("as power" ="force" x "velocity"` and `N m//s = J//s =W)` Initial velocity of car `B, u_B = 0` Final velocity of car `A, upsilon_A = 0 ("as it comes to rest after colliding with car B")` If `upsilon_B` is final velocity of car B when car A collides with it applying the law conservation of momentum, initial momentum of cars A and B = final momentum of cars A and B `i.e., m_A u_A + u_B u_B = m_A upsilon__B` or `1000 xx 10 + 1000 xx 0 = 1000 xx 0 + 1000 xx upsilon_B` or `upsilon_B = 10m//s` |
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| 17. |
A tank of size 5 m`xx` 5m `xx` 5m is full of water and built on ground. Find the potential energy of the water in the tank. |
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Answer» PE = mgh= `(V rho)gh` (as mass = volume xx density) `=(125 xx 1000) (10) (5//2) J= 3125 xx 10^3 J` [Volume of water `=5 m xx 5m xx5m = 125 m^3,` distance of CG of water in the tank from the ground = (5//2)m] |
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| 18. |
A self-propelled vehicle of mass m, whose engine delivers a constant power P, has an acceleration `a = (P//mv)`. (Assume that there is no friction). In order to increase its velocity from `v_(1)` to `v_(2)`, the distan~e it has to travel will be:A. `(3P)/(m) (v_(2)^(2) - v_(1)^(2))`B. `(m)/(3P) (v_(2)^(3) - v_(1)^(3))`C. `(m)/(3P) (v_(2)^(2) - v_(1)^(2))`D. `(m)/(3P) (V_(2) - v_(1))^(3)` |
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Answer» Correct Answer - B |
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| 19. |
A scooter company gives the following specifications about its product. Weight of the scooter -95 kg Maximum speed -60 km/h Maximum engine power -3.5 hp Pick up time to get the maximum speed -5s Check the validity of these specifications. |
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Answer» Correct Answer - A::B::C::D The specificatiion givenby the compane are, u=0, m=95m, P_m=3.5hp` V_m=60km/h` `=50/3 m/sec` `t_m=5sec` So, the maximum acceleration that can be producd is given by, `=50/(3xx5(=10/3m/sec^2` So the driving force is given by `F=ma=95xx(10/3)` `=950/3N will be` `v=p/F` `rarr v=3.5xx746xx3/950` `=8.2m/sec` Because teh scooter can reach a maximum of 8.2 m/sec while producing a force of 950/3N. The specification given are some what over claimed. |
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| 20. |
An automobile travelling with a speed `60 km//h` , can brake to stop within a distance of `20 m` . If the car is going twice as fast i. e. , `120 km//h`, the stopping distance will be |
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Answer» Let us suppose that when the brakes are applied, the automobile does not skid. To get the maximum stopping distacne (s), we take the ffirctional force (f) between the tyres and the road to a maximum. The work done by the frictiional force must be equal to the change in the kinetic energy of the automobile. if `upsilon` is the initial speed of the automobile, `fs = (1)/(2) mupsilon^2 or s = (m upsilon^2)/(2f)` If we assume f to be the same for two initial speeds `(upsilon_1 and upsilon_2)`, `s prop upsilon^2 or (s^2)/(s_1) =((upsilon_2)/(upsilon_1))^2 or s_2 = ((upsilon_2)/(upsilon_1))^2 s_1` As `upsilon_2 = 96km//h, upsilon_1 = 48 km//h and s_1 = 40 m, s_2 = ((96km)/(48))^2 xx40` =160m |
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| 21. |
Find the amount of work donw by a labourer who carrier n bricks of m kilogram each to the roof of a house h metre high by climbing a ladder. |
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Answer» Here, force exerted by the labourer in carrying n briks `("each of mass m kg")`, `F = (nm) g =`" newton"` displacement of the bricks, s =h metre work done by the labourer, `W =Fxxs = (mng newton)xx(h metre)` = mngh joule |
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| 22. |
A person is painting his house walls. He stands on a ladder with a bucket contaiing paint in one hand and a brush inother.Suddenly the bucket slips from his hand and falls down on the floor. If the bucket with the paint had a mass of 6.0 kg andwas ast a height of 2.0 m at the time it slipped, how much gravirtational potential energy is lost together with the paint? |
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Answer» Correct Answer - A `m=6, h=2m` `P.E. at a height, h=mgh` `=6xx(9.8)xx2` `=117.6J` P.E. at floor =0 Loss in P.E. `=117.6-0` `=117.6J~~118J` |
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| 23. |
This diagram depicts a block sliding along a, frictionless ramp in vertical plane. The eightl numbered arrows in the diagram represent directions, to be referred to when answering the questions. The direction of the acceleration of the block, when in position I, is best represented by which of the arrows in the diagram ?A. 2B. 4C. 5D. None of the arrows, the acceleration is zero |
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Answer» Correct Answer - B |
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| 24. |
A projectivle is fired form the top of a 40 m hnigh cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground. |
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Answer» `h=40m, u=50m/sec` let the speed be v when it strikes the ground. Applying law of conservation of energy, `mgh+1/2mu^2=1/2mv^2` `rarr 10xx40+(1/2)xx2500=1/2v^2` `rarr v^2=3300` `rarr v=57.4m/sec=58m/sec` |
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| 25. |
figure depicts a block sliding along a frictionless ramp in vertical plane Eight numbered arrows in the diagram repreesent direction to be referred to when answering the questions Position - i : Starts sliding on curve path Position - ii : Lowest position of the curve path Position - iii : Just outside of the curve path The direction of the acceleration of the blocks when in position ii is best represented by which of the arrows in the diagram?A. 1B. 3C. 5D. 8 |
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Answer» Correct Answer - A |
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| 26. |
This diagram depicts a block sliding along a, frictionless ramp in vertical plane. The eightl numbered arrows in the diagram represent directions, to be referred to when answering the questions. The direction of the acceleration of the block (after leaving the ramp) at position III is best represented by which of the arrows in the diagram ?A. 2B. 5C. 6D. None of the arrows, the acceleration is zero |
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Answer» Correct Answer - B |
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| 27. |
Water is pumped out of a well 10m deep by means of a pump rated at 10 kW. Find the efficiency of the motor if 4200 kg of water is pumped out every minute. Take `g = 10m//s^2.` |
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Answer» Work done by the pump per minute, `W =mgh = (4200) (10 (10)) = 4.2 xx 10^5 J` Output power of the pump `= (W)/(t) = (4.2xx 10^5J)/(60s) = 7 xx 10^3 W = 7kW` Efficiency of the pump `= ("output power")/("input power" (i.e. "rated power")) = (7kW)/(10kW) = 0.7 = 70%` |
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| 28. |
Find the power of an engine which lifts 90 metric tonnes of coal per hour from a depth of 200m. |
| Answer» `P = (mgh)/(t) = ((90xx1000) (10) (200)J)/((60 xx 60)s) = 5xx10^4 W = 50 kW` | |
| 29. |
Calculate the power of an engine required to lift `10^5`kg of coal per hour from a mine 360 m deep. |
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Answer» Correct Answer - A Power of the engine required, `P = (W)/(t) = (mgh)/(t) = ((10^5kg)(10m//s^2)(360m))/(60xx60s) = 10^5W =100kW` |
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| 30. |
The potential energy `U` in joule of a particle of mass `1 kg` moving in `x-y` plane obeys the law`U = 3x + 4y`, where `(x,y)` are the co-ordinates of the particle in metre. If the particle is at rest at `(6,4)` at time `t = 0` then :A. the particle has constant accelerationB. the particle has zero accelerationC. the speed of the particle when it crosses y-axis is `10 m//s`D. co-ordinate of particle at `t = 1` sec is (4.5, 2) |
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Answer» Correct Answer - A::C::D |
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| 31. |
A heavy particle is suspended by a stirng of length l. The particle is given a horizontal velocity `v_0`. The stirng becomes slack at some angle and the particle proceeds on a parabola.find the value of `v_0` if the prticle psses through the ponit of suspension |
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Answer» Suppose the string becomes slack when the particle reaches the point P ure. Suppose the string OP makes an angle `theta` with the upward vertical. The only force actig on the particle at P is its weight mg.The radia component of the force is `mg costheta`. As the particle moves on the circle upto P, `mg costheta=m(v^2/l)` or, `v^2=glcostheta` ...........i where v is its speed at P. Using conservation of energy, `1/2 mv_0^2=1/2mv^2+mgl(1+costheta)` or `v^2=v^2_0-2gl(1+costheta)`......ii From i and ii `v_0^2-2gl(1+costheta)=glcostheta` or, `v_0^2=gl(2+3costheta)` .........iii Now onwards the particle goes in a parabola under the action of gravity. As it passes through the point of suspension O, the equations for horizontal and verticla motions give, `lsintheta=(v costheta)t` and `-lcostheta=(vcostheta)t-1/2gt^2` or, `-lcostheta=(vsintheta((lsintheta)/(vcostheta))-1/2g((lsintheta)/(vcostheta))^2` or `-cos^2theta=sin^2theta-1/2g(lsin^2theta)/(v^2costheta)` or `-cos^2thet=1-cos^2theata-1/2 (glsin^2theta)/(2glcos^2theta) [from i]` or, `=1/2tan^2theta` `or tantheta=sqrt2` From iii `v_0=[gl(2+sqrt3)]^(1/2)` |
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| 32. |
Consider two observers moving with respect to each other at a speed v along·a straight line. They observe a block of mass m moving a distance 1 on a rough surface. The following quantities will be same as observed by the observersA. Kinetic energy of the block at timeB. Work done by frictionC. Total work done on the blockD. Acceleration of the block |
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Answer» Correct Answer - D |
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| 33. |
In the shown diagram mass of A is m and that of B is 2 m. All the surfaces are smooth. System is released from rest with spring unstretched. Then, the maximum extension `(x_(m))` in spring will be: A. `(mg)/(k)`B. `(2mg)/(k)`C. `(3 mg)/(k)`D. `(4 mg)/(k)` |
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Answer» Correct Answer - D |
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| 34. |
An elastic spring of unstretched length `L` and force constant `K` is stretched by amoun t `x` .It is further stretched by another length `y` The work done in the second streaching isA. `((1)/(2)) ky^(2)`B. `((1)/(2)) k (x^(2) + y^(2))`C. `((1)/(2)) k (x + y)^(2)`D. `((1)/(2)) ky (2 x + y)` |
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Answer» Correct Answer - D |
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| 35. |
Consider two observers moving with respect to each other at a speed v along a straighline. They observe a block of mass m movig a distasnce l on a rough surface. The following quantities will be same as observed by the two observersA. kinetic energy of the block t tiem tB. work doen by frictionC. tota work done the blockD. accelertion of the block |
| Answer» Correct Answer - D | |
| 36. |
A particle is acted upon by a force of constant magnitude which is always perpendiculr to the velocity of the particle. The motion of the particle takes place in a plane. It follows thatA. its velocity is constantB. its acceleration is constantC. its kinetic energy is constantD. it moves in a circular path |
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Answer» Correct Answer - C::D |
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| 37. |
A block of mass 2.00 kg moving at a speed of 10.0 m/s accelerates at 3.0 `m/s^2` for 5.00 s. Compute its final kinetic enegy. |
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Answer» Correct Answer - B `M_b=2kg,u=10m/sec` `a=m/sec^2,t=5sec` `v=u+at` `=10+3xx5=25 m/sec` `:. FiN/Al K.E. =1/2 mv^2` `=1/2xx2x625=625J` |
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| 38. |
A particle is acted upon by a force of constant magnitude which is always perpendiculr to the velocity of the particle. The motion of the particle takes place in a plane. It follows thatA. it velocity is constantB. its acceleration is constantC. its kinetic energy is constantD. it moves in a circular path |
| Answer» Correct Answer - C::D | |
| 39. |
A 250 g block slides on aeroug horizontal table. Find the work donen by the frictinal force in bringing the block to rest if it is initially moving at a speed of 40 cm/s. If the friction coefficient between te table and the block is 0.1 how far does the block move before coming to rest? |
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Answer» Correct Answer - B::C Given `m=250gm=0.250kg `u=40 cm/sec` =0.4m/sec, mu=0, v=0` here, `muR=ma` (where a is deceleration)` `a=((muR)/m)=((mumg)/m)=mug` `=0.1xx9.8=098m/s^2` `s=(v^2-u^2)/(2a)=0.082m` Again work done against friction is given by `-W=muRscostheta` `=1xx2.5xx0.082xx1` `rarr+0.02J` `rarrW=-0.02J` |
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| 40. |
The total work done on a particle is equal to the change in its kinetic energyA. alwaysB. only if the forces acting on ilt are conservativeC. only if gravitational force alone acts on itD. only if elastic force alone acts on it |
| Answer» Correct Answer - A | |
| 41. |
A particle of maas m is attched to a light string of length l, the other end of which is fixed. Initially the string is kept horizontal and the particle is given an upwrd velocity v. The particle is just able to complete a circleA. The string becomes slack when the partice reaches its highest pointB. The velocity of the particle becomes zero at the highest point.C. The kinetic energy of the ball in initial position was `1/2mv^2=mg/`D. The particle again passes through the initial position. |
| Answer» Correct Answer - A::D | |
| 42. |
The kinetic energy of a particle continuously icreses with timeA. the resultant force on the particle must be parallel to the velocity at all instants.B. the resultant force on the particle must be at an angle less than `90^0` all the timeC. Its height above the ground level must continuously decreaseD. the magnitude of its linear momentum is increasing continuously. |
| Answer» Correct Answer - B::D | |
| 43. |
A particle is rotated in a vertical circle by connecting it to a string of length `l` and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle isA. `sqrt(gl)`B. `sqrt(2g l)`C. `sqrt(3 gl)`D. `sqrt(5g l)` |
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Answer» Correct Answer - C |
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| 44. |
A particle is rotated in verticla circle by connecting it to string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle isA. `sqrt(gl)`B. `sqrt(2ghl)`C. `sqrt(3gl)`D. `sqrt(5gl)` |
| Answer» Correct Answer - C | |
| 45. |
Two cylindrical vessels of equal cross-sectional area. A contain water upto height `h_(1)` and `h_(2)`. The vessels are interconnected so that the levels in them becomes equal. The work done by the force of gravity during the process is:A. zeroB. `rho A ((h_(1) + h_(2))/(2))^(2) g`C. `rho A ((h_(1) - h_(2))/(2))^(2) g`D. `(rho A h_(1) h_(2))/(2) g` |
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Answer» Correct Answer - C |
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| 46. |
A person pushes a 72 kg patient on a 15 kg trolley, producing an acceleration of `0.60m//s^2` (a) How much work does the person do by pushing the patient and the trolley through a distance of 2.5m ? Assume the trolley moves without friction. (b) How far must the person push the trolley to do 140J of work. (c) Does the work done by the person depends on the speed of the trolley? |
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Answer» (a) `F = ma = (72 kg + 15kg)(0.60 m//s^2) = 52N` W = F s =(52N)(2.5m) =130J (b) As W = F s, `s = (W)/(F) = (140J)/(52N) = 2.7m ` (c ) No, the work done on an object W = F s, does not depends on whether the object moves through the distance s quickly or slowly. |
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| 47. |
Figure shows a smooth curved track terminating in a smooth horizontal part. A spring of sprng constant 400 N/m is asttached at one end ot a wedge fixed rigidly with the horizontal part. A 40 g mas is released from rest at a height of 4.9 m n the curved track. Find the maximumcompression of the spring. |
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Answer» At the instant of maximum compressin the speed of the 40 g mass reduces to zero. Taking the gravitatioN/Al potential energy to be zero at the horizontal part the conservation of energy shows, `mgh=1/2kx^2` where m=0.04 k, h=4.9 m, k=400 N/m and x is the maximum compression. Thus, `x=(sqrt(2mgh))/k` `=(sqrt(2xx(0.04kg)xx(9.8m/s^2)xx(4.9m)/((400N/m)))` `9.8cm` |
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| 48. |
A block hangs freely from the end of a spring. A boy then slowly pushes the block upwards so that the spring becomes strain free. The gain in gravitational potential energy of the block during this process is not equal toA. The work done by · the boy , against the gravitational force acting on the block.B. The loss of energy stored in the spring minus the work done by the tension in the springC. The work done on tlie block by the boy plus the loss of energy stored in the spring.D. The work done on the block by the boy minus the work done by the tension in the spring plus the loss of energy stored in the spring. |
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Answer» Correct Answer - C |
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| 49. |
The mass m slides down the track and completes the vertical circle on the smooth curved surface. The minimum value of h will be: A. RB. 2 RC. 2.5 RD. 3 R |
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Answer» Correct Answer - C |
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| 50. |
A particle of mass m is kept at the top of a smooth fixed sphere. It is given a horizontal velocity v then:A. it will start moving along a circular path if `v lt sqrt(gR)`B. it will start moving along a circular path if `v gt sqrt(gR)`C. it will start moving along a parabolic path if `v lt sqrt(gR)`D. it will start moving along a parabolic path if `v gt sqrt(gR)` |
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Answer» Correct Answer - A::D |
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