174 + Interview Questions in WORK-POWER-AND-ENERGY IN GENERAL AWARENESS Page 4 InterviewSolution

151.	A tennis ball dropped on a barizoontal smooth surface , it because back to its original postion after hiting the surface the force on the bell during the collision is propertional to the length of compression of the bell . Which one of the following skethes desches discribe the variation of its kinetic energy `K` with time `1` mass apporiandly ? The figure as only illistrative and not to the scale .A. B. C. D.
Answer» Correct Answer - B `v prop t , v^(2) prop t^(2) , K.E prop t^(2)`.

Discussion

152.	An engine is hauling a train of mass `m` on a level track at a constant speed `v`. The resistance due to friction is `f`. What power is the engine producing? What extra power must the engine develop to maintain the speed up a gradient 1 in l. What is the new total power developed by the engine develop to maintain the speed up a gradient 1 in l. What is the new total power developed by the engine ?A. Power expended by the engine is "mfu".B. The extra power developed by the engine to maintain a speed `u` up a gradient on of `h` in `s` is `(mghu)/(s)`.C. The frictional force exerting on the train is `mf` on the level trackD. None of above is correct
Answer» Correct Answer - B

Discussion

153.	A car of mass `m` is driven with acceleration `a` along a straight level road against a constant external resistive force R. When the velocity of the car V, the rate at which the engine of the car is doing work will beA. RVB. maVC. (R+ma)VD. (ma-R)V
Answer» Correct Answer - C F-R=ma F=ma+R P=FxV P=(ma+R)V

Discussion

154.	A particle is taken from point A to point B under the influence of a force field. Now it is taken back from B to A and it is observed that the work done in taking the particle from A to B is not equal to the work done in taking it from B to A. If `W_(nc)` and `W_c` are the work done by non-conservative and conservative forces present in the system, respectively, `DeltaU` is the change in potential energy and `Deltak` is the change in kinetic energy, thenA. `W_(nc) - Delta U = Delta D`B. `W_(c) = - Delta U`C. `W_(nc) + W_(c) = Delta k`D. `W_(nc) - Delta U = - Delta k`
Answer» Correct Answer - A::B::C

Discussion

155.	The alternative that gives the conservative force of the following is.A. `vec F_(1)=2xy hat i + x^(2) hatj`B. `vecF_(2) =y^(3) hati + xy^(2) hat j`C. `vec(F_(3))= y hat i + x hat j`D. `vec(F_(4)) = xy^(2) hat i + x^(2) hatj`
Answer» Correct Answer - A::C

Discussion

156.	An engine is pulling a train of mass `m` on a level track at a uniform speed `u`. The resistive froce offered per unit mass is `f`.A. Power produced by the engine is `mfu`B. The extra power developed by the engine to maintain a speed `u` up a gradient on of `h` in `s` is `(mghu)/(s)`.C. The frictional force exerting on the train is `mf` on the level trackD. None of above is correct
Answer» Correct Answer - A::B::C

Discussion

157.	At high altitude , a body explodes at rest into two equal fragments with one fragment receiving horizontal velocity of `10 m//s`. Time taken by the two radius vectors connecting of explosion to fragments to make `90^(@)` isA. 10 sB. 4 sC. 2 sD. 1 s
Answer» Correct Answer - D `t=(sqrt(u_(1)u_(2)))/(g)`.

Discussion

158.	Figure shows a light, inextensible string attached to a cart that can slide along a frictionless horizontal rail aligned along an `x` axis. The left end of the string is pulled over a pulley, of negligible mass and friction and fixed at height `h = 3m` from the ground level. The cart slides from`x_(1) = 3 sqrt(3) m` to `x _(2) = 4 m` and during to move, tension in the string is kept constant `50 N`. Find change in kinetic energy of the cart in joules. `(Use sqrt(3) = 1.7)` in form of `10 x n`, where `n=` .
Answer» Correct Answer - `5` Change in kinetic energy = Work done by the force , so `W = 50 xx 1` (along the string) , so `W = 50 "Joule"`.

Discussion

159.	A tank on the roof of a `20 m` high building can hold `10 m^(3)` of water. The tank is to be filled form a pond on the ground in `20` minutes. If the pump has an efficiency of `60 %`, then the input power in `kW` isA. `1.1`B. `2 .74`C. `5.48`D. `7.0`
Answer» Correct Answer - B `eta = (P_(out))/(P_(i n)) = ((mgh)//t)/(P) , (x)/(100) p = (mgh)/(t)`.

Discussion

160.	A tank of size `10 m xx 10 m xx 10 m` is full of water and built on the ground. If `g = 10 ms^(-2)`, the potential energy of the water in the tank isA. `5 xx 10^(7) J`B. `1 xx 10^(8) J`C. `5 xx 10^(4) J`D. `5 xx 10^(5) J`
Answer» Correct Answer - A `P.E = mgh_(1)`, here `h_(1) = (h)/(2)` and `m = rho xx V`.

Discussion

161.	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 ?A. 8.82 JB. 7.72 JC. 6.62 JD. 5.52 J
Answer» Correct Answer - A Heat produced = loss of potential energy = mgh.

Discussion

162.	A spring of force constant `800N//m` has an extension of 5cm. The work done in extending it from 5cm to 15cm isA. 16JB. 8JC. 32JD. 24 J
Answer» Correct Answer - B `W_(C)=-DeltaU` `=(U_("final")-U_("initial"))` `=-(1/2xxkxx15^(2)-1/2xxkxx5^(2))` ltBrgt `W_(c)=8` joule

Discussion

163.	A motor boat is going in a river with a velocity `vec(V) = (4 hat i-2 hat j + hat k) ms^(-1)`. If the resisting force due to stream is `vec (F)=(5 hat i-10 hat j+6 hat k)N`, then the power of the motor boat isA. 100 WB. 50 WC. 46 WD. 23 W
Answer» Correct Answer - C `P vec F. vec V`.

Discussion

164.	A block of mass `m = 25 kg` on a smooth horizontal surface with a velocity `vec v = 3 ms^(-1)` meets the spring of spring constant `k = 100 N//m` fixed at one end as shown in figure. The maximum compression of the spring and velocity of block as it returns to the original position respectively are. .A. `1.5m, -3 ms^(-1)`B. `1.5m, 0 ms^(-1)`C. `1.0m, 3 ms^(-1)`D. `0.5 m, 2 ms^(-1)`
Answer» Correct Answer - A `K.E` of block is converted into elastic `P.E` in spring i.e. `(1)/(2) mv^(2) = (1)/(2) kx^(2)`.

Discussion

165.	The blocks A and B shown in figure have masses `M_A=5kg` and `M_B=4kg`. The system is released from rest. The speed of B after A has travelled a distance `1m` along the incline is A. `(sqrt(3))/(2) sqrt(g)`B. `(sqrt(3))/(4)sqrt(g)`C. `(sqrt(g))/(2 sqrt(3))`D. `(sqrt(g))/(2)`
Answer» Correct Answer - C If `A` moves down the incline by `1` metre, `B`shall move up by `(1)/(2)` metre. If the speed of `B` is `v` then the speed of `A` be `2v`. From conservation of energy : Grain in `K.E` = loss in `P.E`. `(1)/(2) m_(A)(2v)^(2)+(1)/(2) m_(B)v^(2) = m_(A)g xx (3)/(5) -m_(B) g xx (1)/(2)` Solving we get `v = (1)/(2) sqrt((g)/(3))`.

Discussion

166.	Two blocks A and B, each of mass m, are connected by a masslesss spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in fig. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides elastically with A. Then A. `v sqrt((m)/(2k))`B. `m sqrt((v)/(2k))`C. `sqrt((mv)/(2k))`D. `(mv)/(2k)`
Answer» Correct Answer - A `mv = (m + m) v^(1)` and `(1)/(2) (m+m) (v^(1))^(2) = (1)/(2) kx^(2)`.

Discussion

167.	Two blocks, of masses `M` and `2 M`, are connected to a light spring of spring constant `K` that has one end fixed, as shown in figure. The horizontal surface and the pulley are frictionless. The blocks are released from when the spring is non deformed. The string is light. .A. Maximum extension in the spring is `(4Mg)/(K)`.B. Maximum kinetic energy of the system is `(2M^(2)g^(2))/(K)`.C. Maximum energy stored in the spring is four times that of maximum kinetic energy of the system.D. When kinetic energy of the system is maximum energy stored in the spring is `(4M^(2)g^(2))/(K)`.
Answer» Correct Answer - A::B::C `(4 Mg)/(K)` System will have maximum `KE` when net force on the system becomes zero. Therefore `2 Mg = T` and `T = kx rArr x = (2Mg)/(K)` Hence `KE` will be maximum when `2M` mass has gone down by `(2Mg)/(K)`. Applying `W//E` theorem `k_(f)-0 =2Mg.(2Mg)/(K)-(1)/(2)K.(4M^(2)g^(2))/(K^(2)) , k_(f) = (2M^(2)g^(2))/(K^(2))` Maximum energy of spring `(1)/(2) K.((4Mg)/(K))^(2) = (8 M^(2) g^(2))/(K)` Therefore Maximum spring energy `= 4 xx "maximum" K.E`. When `K.E` is maximum `x = (2Mg)/(K)`. Spring enery ` = (1)/(2).K.(4M^(2)g^(2))/(K) = (2M^(2)g^(2))/(K)`. i.e. `(D)` is wrong.

Discussion

168.	Displacement time graph of a particle moving in a straight line is as shown in figure. Select the correct alternative (s) : .A. Work done by all the forces in region `OA` and `BC` is positiveB. Work done by all the forces in region `AB` is zeroC. Work done by the forces in region `BC` is negativeD. Work done by all the forces in region `OA` is negative.
Answer» Correct Answer - B In region `OA` particle is acceleratedin region `AB` particle has uniform velocity while in region `BD` particle is deceleration, Therefore, work done is positive in region `OA`, zero in region `AB` and negative in region `BC`.

Discussion

169.	The kinetic energy of a particle moving along a circle of radius `R` depends on the distance covered `s` as `K=lambdas^(2)`, where `lambda` is a constant. Find the force acting on the particle as a function of `s`.
Answer» `KE = (1)/(2) mv^(2) = cs^(2) rArr v = (sqrt((2c)/(m))) s` `a_(t) = (dv)/(dt) = sqrt((2c)/(m)) xx (ds)/(dt) = v sqrt((2c)/(m))` `F_(t) =ma_(t) = mv sqrt((2c)/(m)) = [m sqrt((2c)/(m))s] sqrt((2c)/(m)) = 2cs` Total force `F = sqrt(F_(t)^(2)+ F_(c)^(2))= sqrt((2cs)^(2) + ((mv^(2))/(r))^(2))` `F = 2cs sqrt(1+ (s^(2))/(r^(2)))`.

Discussion

170.	Two equal sphere `A and b` lie on a smooth horizontal circle groove at opposite ends of a diameter. At time `t= 0,A` is projected along the groove and tis first implings on`B` at time `t = T_(1) and again at time `t = T_(2)`. If `e` is the coefficient of restitution, the ratio `T_(2)//T_(1)` is
Answer» `T_(1) = (pi R)/(u_(1)` ….(1) `(v_(2)-v_(1))/(u_(1)) =e rArr v_(2) -v_(1) = eu_(1)` Time taken for `A` to collide with `B` again us `T_(2)-T_(1)=(2pi R)/(v_(2)-v_(1)) rArr T_(2)-T_(1) =(2 piR)/(eu_(1))` ….(2) from (1) and (2), `(T_(2))/(T_(1)) = (2+e)/( e)`.

Discussion

171.	Ball 1 collides directly with another identical ball 2 at rest. Velocity of second ball becomes two times that of 1 after collison. Find the coefficient of restitution between the two balls?
Answer» Here `m_(1) = m_(2)` and `u_(2) = 0` After collision, `v_(2) = ((1+e)/(2))u` & `v_(1) = ((1-e)/(2))u` Given `v_(2) = 2v_(1) , ((1+e)/(2)) u=2((1-e)/(2))u` `1 +e =2 - 2e , 3e = 1, e = (1)/(3)`.

Discussion

172.	Two identical bodies moving in opposite direction with same speed, collide with each other. If the collision is prefectly elastic thenA. after the collision both comes to restB. after the collision first comes to rest and second moves in the opposite direction with same speed.C. after collision they recoil with same speedD. both and `1` and `2`
Answer» Correct Answer - C

Discussion

173.	A body `X` with a momentum `p` with another identical stationary body `Y` one dimensionally. During the collision `Y` gives an impulse `J` to body `X`. Then coefficient of restitution is:
Answer» From the law of conservation of linear momentum, `m_(1)u_(1) +m_(2)u_(2) = m_(1)v_(1)+m_(2)v_(2)` `m u +m(0) = mv_(1) + mv_(2)` `rArr P -P_(1) = P_(2)` where `P_(2) = J`, (given) `:. e = (v_(2) -v_(1))/(u_(1) -u_(2)) = (mv_(2) - mv_(1))/("mu" -0) = (P_(2)-P_(1))/(P)` =`(P_(2)-(P-P_(2)))/(P) = (2P_(2)-P)/(P) =(2J -P)/(P) =(2J)/(P) -1`.

Discussion

174.	In the figure (a) and (b) `AC` and `GF` are fixed inclined planes `BC = EF = x` and `AB = DE = y` A small block of mass `M` is rdeased from the point `A` it sides down `AC` and maches `C` with a speed `V_(C) ` The same block is relessed from rest from the point `D` it sides down `DGF` and reached the point the `F` with `V_(p) THe coefficients of kiletic friction between the block and the sarface `AC` and `DGF` are `mu` colculate `V_(C) ` and V_`(p)`A. `1.7 m//s`B. `2.7 m//s`C. `3.7 m//s`D. `0.7 m//s`
Answer» Correct Answer - A In both the cases work done by friction will be `mu Mgx` `:. (1)/(2)MV_(c)^(2) =(1)/(2)MV_(F)^(2) =Mgy -mu gx` `:. V_(c) =V_(F)= sqrt(2gy-2 mu gx)`.

Discussion

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A car of mass `m` is driven with acceleration `a` along a straight level road against a constant external resistive force R. When the velocity of the car V, the rate at which the engine of the car is doing work will beA. RVB. maVC. (R+ma)VD. (ma-R)V

The alternative that gives the conservative force of the following is.A. `vec F_(1)=2xy hat i + x^(2) hatj`B. `vecF_(2) =y^(3) hati + xy^(2) hat j`C. `vec(F_(3))= y hat i + x hat j`D. `vec(F_(4)) = xy^(2) hat i + x^(2) hatj`

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A tank on the roof of a `20 m` high building can hold `10 m^(3)` of water. The tank is to be filled form a pond on the ground in `20` minutes. If the pump has an efficiency of `60 %`, then the input power in `kW` isA. `1.1`B. `2 .74`C. `5.48`D. `7.0`

A tank of size `10 m xx 10 m xx 10 m` is full of water and built on the ground. If `g = 10 ms^(-2)`, the potential energy of the water in the tank isA. `5 xx 10^(7) J`B. `1 xx 10^(8) J`C. `5 xx 10^(4) J`D. `5 xx 10^(5) J`

A spring of force constant `800N//m` has an extension of 5cm. The work done in extending it from 5cm to 15cm isA. 16JB. 8JC. 32JD. 24 J

A motor boat is going in a river with a velocity `vec(V) = (4 hat i-2 hat j + hat k) ms^(-1)`. If the resisting force due to stream is `vec (F)=(5 hat i-10 hat j+6 hat k)N`, then the power of the motor boat isA. 100 WB. 50 WC. 46 WD. 23 W

The blocks A and B shown in figure have masses `M_A=5kg` and `M_B=4kg`. The system is released from rest. The speed of B after A has travelled a distance `1m` along the incline is A. `(sqrt(3))/(2) sqrt(g)`B. `(sqrt(3))/(4)sqrt(g)`C. `(sqrt(g))/(2 sqrt(3))`D. `(sqrt(g))/(2)`

The kinetic energy of a particle moving along a circle of radius `R` depends on the distance covered `s` as `K=lambdas^(2)`, where `lambda` is a constant. Find the force acting on the particle as a function of `s`.

Two equal sphere `A and b` lie on a smooth horizontal circle groove at opposite ends of a diameter. At time `t= 0,A` is projected along the groove and tis first implings on`B` at time `t = T_(1) and again at time `t = T_(2)`. If `e` is the coefficient of restitution, the ratio `T_(2)//T_(1)` is

Ball 1 collides directly with another identical ball 2 at rest. Velocity of second ball becomes two times that of 1 after collison. Find the coefficient of restitution between the two balls?

A body `X` with a momentum `p` with another identical stationary body `Y` one dimensionally. During the collision `Y` gives an impulse `J` to body `X`. Then coefficient of restitution is: