1582 + Interview Questions in WORK IN GENERAL AWARENESS Page 6 InterviewSolution

251.	A lead ball strikes a wall and falls down, a tennis ball having the same mass and velocity strikes the wall and bounces back. Check the correct statementA. The momentum of the lead ball is greater than that of the tennis ballB. The lead ball suffers a greater change in momentum compared with the tennis ballC. The tennis ball suffers a greater change in momentum as compared with the lead ballD. Both suffer an equal change in momentum
Answer» Correct Answer - C

Discussion

252.	A ball of mass `m` moves with speed `v` and stricks a wall having infinite mass and it returns with same speed then the work done by the ball on the wall isA. zeroB. mv JC. m/v. JD. v/m J
Answer» Correct Answer - A

Discussion

253.	A particle of mass `0.5kg` travels in a straight line with velocity `v=ax^(3//2)` where `a=5m^(-1//2)s^-1`. What is the work done by the net force during its displacement from `x=0` to `x=2m`?
Answer» Here, `m=0.5kg, upsilon=ax^(3//2), a=5m^(-1//2)s^(-1), W=?` Initial vel. At `x=0, upsilon=axx0=0, ` Final vel. At `x=2, upsilon_(2)=a2^(3//2)=5xx2^(3//2)` Work done `=` increase in K.E. `=(1)/(2)m(upsilon_(2)^(2)-upsilon_(1)^(2))` `W=(1)/(2)xx0.5[(5xx2^(3//2))^(2)-0]=50J`

Discussion

254.	A force `F=(10+0.50x)` acts on a particle in the x direction, where F is in newton and x in meter./find the work done by this force during a displacement form x=0 tox=2.0m
Answer» Here, `F=(10+0.50x)` Small amount of word done in moving the particle through a samll distance dx is `dW=hat(F).d hatx=(10+0.5x)dx` Total work done, `W=int_(x=0)^(x=2)(10+00.5x)dx` `W=[10x +0.5(x^(2))/(2)]_(0)^(2)` `=10(2-0)+(0.5)/(2)(2^(2)-0)=20+1` `=21joule`

Discussion

255.	A particle of mass `0.5kg` travels in a straight line with a velocity `v=(5x^(5//2))m//s`. How much work is done by the net force during the displacement from `x=0` to `x=2m` ?
Answer» Here,`m=0.5kg` `v=5x^(5//2),W=?` When `x=0,v_(1)=0` When `x=2m, v_(2)=5(2)^(5//2)` Using work energy theorem, `W=` change in K.E. `=(1)/(2)mv_(2)^(2)-(1)/(2)mv_(1)^(2)` `W=(1)/(2)m(v_(2)^(2)-v_(1)^(2))=(1)/(2)xx0.5[{5(2)^(5//2)}^(2)-0]` `=(1)/(2)xx(1)/(2)xx25xx2^(5)=200J`

Discussion

256.	A force `F=(10+0.50x)` acts on a particle in the x direction, where F is in newton and x in meter./find the work done by this force during a displacement form x=0 tox=2.0mA. 31.5 JB. 63 JC. 21 JD. 42 J
Answer» Correct Answer - C (c ) Given, force, `F=10+0.5x=10+(1)/(2)x` Let during small displacement the workdone by the force is dW=Fdx So, work done during displacement from x=0 to x=2 is `W=int_(0)^(w)dW=int_(0)^(2)Fdx=int_(0)^(2)(10+(1)/(2)x)dx` `=10[x]_(0)^(2)+[(x^(2))/(4)]_(0)^(2)=21J`

Discussion

257.	A raindrop falling from a height `h` above ground, attains a near terminal velocity when it has fallen through a height `(3//4)h`. Which of the diagrams shown in figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground ?A. B. C. D.
Answer» Correct Answer - B At a height h above the ground, PE of raindrop is maximum and KE = 0. As the raindrop falls its PE goes on decreasing and KE goes on increasing up to a point `h/4`above the ground. At this stage, rain drop has acquired near terminal velocity (= constant). Therefore, at this stage. KE tends to be constant. PE becomes zero when raindrop falls to the ground. Hence, option (b) is most appropriate

Discussion

258.	A raindrop falling from a height `h` above ground, attains a near terminal velocity when it has fallen through a height `(3//4)h`. Which of the diagrams shown in figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground ?A. B. C. D.
Answer» (b) When drop falls first velocity increases, hence first KE also increase. After sometime speed (velocity) is constant this is called terminal velocity, hence, KE also become constant. PE decrease continously as the drop is falling continously. The variation in PE and KE is best represented by (b).

Discussion

259.	A `20 g` bullet pierces through a plate of mass `M_(1) = 1 kg` and then comes to rest inside a second plate of mass `M_(2) = 2.98 kg` as shown in Fig. It is found that the two plates, initially at rest, now move with equal velocities. Find the percentage loss in the initial velocity of the bullet when it is between `M_(1)` and `M_(2`. Neglect any loss of material of the plates due to the action of bullet.
Answer» Correct Answer - B

Discussion

260.	A mass of `5kg` is moving along a circular path or radius `1m`. If the mass moves with 300 revolutions per minute, its kinetic energy would beA. `250pi^2 J`B. `100 pi^2 J`C. `5pi^2 J`D. 0 J
Answer» Correct Answer - A Here, Mass, m=5 kg, Radius ,R=1 m `upsilon`=300 rpm =`300/60` rps = 5 rps The angular speed is given by `omega=2piupsilon =2pixx5=10 pi "rad" s^(-1)` The linear speed is `v=omegaR=(10pi s^(-1)) (1 m) = 10 pi m s^(-1)` Kinetic energy, `K=1/2mv^2` `K=1/2xx(5 kg) (10 pi m s^(-1))^2 =250pi^2 J`

Discussion

261.	A mass of `5kg` is moving along a circular path or radius `1m`. If the mass moves with 300 revolutions per minute, its kinetic energy would beA. `250pi^(2)`B. `100pi^(2)`C. `5pi^(2)`D. 0
Answer» (a) Given, mass=m=5kg Radius=1m=R Revolution per minute `omega=300"rev/min"` `(300xx2pi)"rad/min"` `=(300xx2xx314)"rad/60s"` `(300xx2xx3.14)/(60)"rad/s"=10pi"rad/s"` linear speed=v=`omegaR` `((300xx2pi)/(60))(1)=10pim//s` `KE=(1)/(2)xx5xx(10pi)^(2)` `100pi^(2)xx5xx(1)/(2)` `=250pi^(2)J`

Discussion

262.	AB is a quarter of smooth circular track of radius `R=6m`. A particle P of mass `0.5kg` moves along the track from A to B under the action of the following forces. a. A force `F_1` directed always towards the point B, its magnitude is constant and is equal to `20N`. b. A force `F_2` directed along the instantaneous tangent to the circular track, its magnitude is `(15-10S)N`, where S is the distance travelled in metre. c. A horizontal force of magnitude `30N`. Find the work done by forces mentioned in (a), (b) and (c)
Answer» Correct Answer - A::B

Discussion

263.	A bead of mass `1/2 kg` starts from rest from A to move in a vertical place along a smooth fixed quarter ring of radius `5 m`, under the action of a constant horizontal force `f=5 N` as shown. The speed of bead as it reaches the point (B) is [Take `g=10 ms^(-2)`] A. `14.14 ms^(-1)`B. `7.07 ms^(-1)`C. `5 ms^(-1)`D. `25 ms^(-1)`
Answer» Correct Answer - A (a) From work-energy theorrem `W_(F)+W_(mg)=(1)/(2)mv^(2)` `rArr " " F.R+mgR=(1)/(2)mv^(2)` `rArr " " 5xx5+(1)/(2)xx10xx5=(1)/(2)xx(1)/(2)xxv^(2)` `rArr " " v=sqrt(200)=14.14" ms"^(-1)`

Discussion

264.	A bead of mass `1/2 kg` starts from rest from A to move in a vertical place along a smooth fixed quarter ring of radius `5 m`, under the action of a constant horizontal force `f=5 N` as shown. The speed of bead as it reaches the point (B) is [Take `g=10 ms^(-2)`] A. (a) `14.14ms^-1`B. (b) `7.07ms^-1`C. (c) `5ms^-1`D. (d) `25ms^-1`
Answer» Correct Answer - A Applying the work-energy theorem, we get `1/2xxmv^2-0=FxxR+mgxxR` `1/2xx1/2xxv^2=5xx5+1/2xx10xx5=50` `v=sqrt(200)=14.14ms^-1`

Discussion

265.	A particle slides along a track with elevated ends and a flat central part as shown in figure. The flat part has a length `l=3m`. The curved portions of the track are frictionless. For the flat part, the coefficient of kinetic friction is `mu_k=0.2`. The particle is released at point A which is at height `h=1.5m` above the flat part of the track. Where does the particle finally come to rest?
Answer» Correct Answer - [Mid point]

Discussion

266.	A block of mass 1 kg slides down a vertical curved track that is one quadrant of a circle of radius 1m. Its speed ast the the bottom si `2 m//s`. The work done by frictional force is : A. 8 JB. `-8 J`C. 4 JD. `-4 J`
Answer» Correct Answer - B

Discussion

267.	A body of mass m=3.90 kg slides on a horizontal frictionless table with a speed of v=120 `ms^(-1)`. It is brought to rest in compressing a spring in its path. How much does spring is compressed, if its force constant k is 135 `Nm^(-1)`?A. 0.204 mB. 0.408 mC. 0.804 mD. 4.04 m
Answer» Correct Answer - A (a) Given, m=3.9 kg, v=1.20 `ms^(-1)` and k for spring=135 `Nm^(-1)` From law of conservation of energy, `(1)/(2)mv^(2)=(1)/(2)kx^(2)rArrx=sqrt((mv^(2))/(k))=sqrt((3.9xx1.2xx1.2)/(135))=0.204 m`

Discussion

268.	A block of mass 5 kg slides down a rough inclined surface. The angle of inclination is `45^(@)`. The coefficient of sliding friction is 0.20. When the block slides 10 cm, the work done on the block by force of friction isA. `-(1)/(sqrt2)J`B. 1JC. `-sqrt2J`D. `-1J`
Answer» Correct Answer - A (a) Work done by frictional force, `W_(f)=fs cos 180^(@)` `=(mu mg cos theta)(-1)(s)` `=-0.2xx5xx10xx(1)/(sqrt2)xx0.1=-(1)/(sqrt2)J`

Discussion

269.	A body of mass m moving with velocity v makes a head-on collision with another body of mass 2 m which is initially at rest. The loss of kinetic energy of the colliding body (mass m ) isA. `1:1`B. `2:1`C. `4:1`D. `9:1`
Answer» Correct Answer - D

Discussion

270.	A body of mass m moving with velocity v makes a head-on collision with another body of mass 2 m which is initially at rest. The loss of kinetic energy of the colliding body (mass m ) isA. `(1)/(2)` of its intial kinetic energyB. `(1)/(9)` of its initial kinetic energyC. `(8)/(9)` of its initial kinetic energyD. `(1)/(4)` of its initial kinetic energy
Answer» Correct Answer - C

Discussion

271.	A body of mass m having an initial velocity v , makes head on collision with a stationary body of mass M . After the collision, the body of mass m comes to rest and only the body having mass M moves. This will happen only whenA. `m gt gt M`B. `m lt lt M`C. `m=M`D. `m=(1)/(2)M`
Answer» Correct Answer - C

Discussion

272.	Figrue shows the F-x graph. Where F is the force applied x is the distance covered by the body along a straight line path. Given that F is in newton and x in metre, what is the work done?A. 10 JB. 20 JC. 30 JD. 40 J
Answer» Correct Answer - A

Discussion

273.	A car of mass m 1600 kg, while moving on any road, experience resistance to its motion given by `(m+nV^(2))` newton, where m and n are positive constants. On a horizontal road the car moved at a constant speed of 40 m/s when the engine developed a power of 53 KW. When the engine developed an output of 2 KW the car was able to travel on a horizontal road at a constant speed of 10 m/s. (a) Find the power that the engine must deliver for the car to travel at a constant speed of 40 m/s on a horizontal road. (b) The car is able to climb a hill at a constant speed of 40 m/s with its engine working at a constant rate of 69 KW. Calculate the inclination of the hill (in degree)
Answer» Correct Answer - (a) 53 KW (b) `theta=1.43^(@)`

Discussion

274.	The pointer reading vs load graph for a spring balance is as given in the figure. The spring constant isA. 0.1 kg/cmB. 5 kg/cmC. 0.3 kg/cmD. 1 kg/cm
Answer» Correct Answer - A

Discussion

275.	Instentenous power delivered by engine of a car of mass 18 kg moving on +x-axis is given as `p=(2x+5)` watt, where x is (in meter) position of car. Car starts from origin from rest (choose the correct statement(s).A. Power increases with time.B. Power decreases with time.C. At x = 1m, speed of car is v = `1 m//s`D. At x = 1m, speed of car is v = `2 m//s`
Answer» Correct Answer - A::C

Discussion

276.	A spring of force constant 10N / m has an initial stretch 0.20 m . In changing the stretch to 0.25 m , the increase in potential energy is aboutA. 0.1 jouleB. 0.2 jouleC. 0.3 jouleD. 0.5 joule
Answer» Correct Answer - A

Discussion

277.	A particle free to move along the (x - axis) hsd potential energy given by `U(x)= k[1 - exp(-x^2)] for -o o le x le + o o`, where (k) is a positive constant of appropriate dimensions. Then.A. At point away from the origin, the particle is in unstable equilibriumB. For any finite non-zero value of x , there is a force directed away from the originC. If its total mechanical energy is k /2, it has its minimum kinetic energy at the originD. For small displacements from x = 0, the motion is simple harmonic
Answer» Correct Answer - D

Discussion

278.	A force `F` acting on a body depends on its displacement `S` as `FpropS^(-1//3)` . The power delivered by `F` will depend on displacement asA. `s^(2//3)`B. `s^(-5//3)`C. `s^(1//2)`D. `s^(0)`
Answer» Correct Answer - D

Discussion

279.	A spring 40 mm long is stretched by the application of a force. If 10 N force required to stretch the spring through 1mm, then the work done in stretching the spring through 40 mm isA. 84 JB. 68 JC. 23 JD. 8 J
Answer» Correct Answer - D `k=10(N)//(mm)` `W_(1rarr2)=(1)/(2)kx^2-0=(1)/(2)xx10xx10^3(40xx10^-3)^2` `=8J`

Discussion

280.	The force required to stretch a spring varies with the distance a shown in the figure. If the experiment is performed with the above spring of half length, the line OA willA. Shift towards F-axisB. Shift towards X-axisC. Remain as it isD. Become double in length
Answer» Correct Answer - A

Discussion

281.	The force required to stretch a spring varies with the distance a shown in the figure. If the experiment is performed with the above spring of half length, the line OA willA. shift towards F axisB. shift towardsC. X-axisD. remain as it is
Answer» Correct Answer - A

Discussion

282.	The force required to stretch a spring varies with the distance a shown in the figure. If the experiment is performed with the above spring of half length, the line OA willA. shift towards F-axisB. shift towards X-axisC. remain as it isD. become double in length
Answer» Correct Answer - A (a) `K=(F)/(x)`=slope of F-x graph. `K prop(1)/(l)` Length is reduced to half. Therefore, K will become two times. Slope will increase.

Discussion

283.	A force `F=50N` is applied at one end of a string, the other end of which is tied to a block of mass `10kg`. The block is free to move on a frictionless horizontal surface. Take initial instant as `theta=30^@` and final instant as `theta=37^@`. For the time between these two instants, answer teh following questions? Find the ratio of initial acceleration to final acceleration of the blockA. (a) `(3sqrt5)/(8)`B. (b) `(8sqrt3)/(5)`C. (c) `(8sqrt5)/(3)`D. (d) `(5sqrt3)/(8)`
Answer» Correct Answer - D Initial acceleration `=(F cos 30^@)/(m)` Final acceleration `=(F cos 37^@)/(m)` Ratio `=(cos 30^@)/(cos 37^@)=(sqrt3xx5)/(2xx4)=(5sqrt3)/(8)`

Discussion

284.	A chain of length l and mass m lies of the surface of a smooth hemisphere of radius `Rgtl` with one end tied to the top of the hemisphere. Taking base of the hemisphere as reference line, find the gravitational potential energy of the chain.
Answer» The mass is distributed in chain uniformly along its length. Choose a small element of chain of width `dtheta` at an angle `theta` from the vertical. The mass of the element, `dm=(m/lRd theta)` The gravitational potential energy of the element `dU=(dm)gy` Thus, the gravitational potential energy of whole chain `U=int(dm)gy` `=underset0overset((l//R))int(m/lRd theta)g(Rcostheta)` `=(mR^2g)/(l)underset0overset((l/R))intcosthetad theta` `=(mgR^2)/(l)\|sintheta\|_0^(l//R)=(mgR^2)/(l)sin(l/R)`

Discussion

285.	One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacment, the work done by the spring is `+(1/2)kx^(2)`. The possible cases are.A. The spring was initially compessed by a distance x and was finally in its natural length .B. It was initially stretched by a distance x and finally was in its natural length.C. It was initially natural length and finally in a compressed position.D. It was initially in its natural length and finally in a stretched position.
Answer» Correct Answer - A::B Spring force is always towareds mean position If displacement is also towards mean position, F and S will be of same sign and work done will be posititve.

Discussion

286.	A smooth narrow tube in the form of an arc AB of a circle of centre O and radius r is fixed so that A is vertically above O and OB is horizontal. Particles P of mass m and Q of mass 2 m with a light inextensible string of length (pi r//2) connecting them are placed inside the rube with P at A and Q at B and released from rest. Assuming the string remains taut during motion, find the speed of particles when P reaches B. .
Answer» Correct Answer - A::B::C All surfaces are smooth. Therefore, mechanical energy of the system will remain conserved. `:.` Decrease in PE of both the blocks =increase in KE of both blocks `:. (mgr)+(2mg)(pir/2)=1/2(m+2m)v^(2)` or `v =sqrt((2)/(3)(1+pi)gr)`.

Discussion

287.	A car of mass `m` starta from rest and accelerates so that the instyantaneous power delivered to the car has a constant magnitude `P_(0)`. The instaneous velocity of this car is proportional toA. `t^(2) P_(0)`B. `t^(1//2)`C. `t^(-1//2)`D. `t//sqrtm`
Answer» Power `p_(0) = Fv = (m(dv)/(dt)) v = mv (dv)/(dt)` Integrating both sides `int p_(0) dt = int mv dv implies p_(0)t = (mv^(2))/(2)` `v^(2) =(2 p_(0) t)/(m) implies v prioop t^(1//2)`

Discussion

288.	A particale moves under the effect of a force `F = Cs` from `x = 0` to `x = x_(1)`. The work down in the process isA. `Cx_(1)^(2)`B. `(1)/(2)Cx_(1)^(2)`C. `Cx_(1)`D. zero
Answer» Correct Answer - B

Discussion

289.	When an automobile moves with constant speed down a highway , most of the power developed by the engain is used to compensate on the car by the air and the road, friction forces exerted on the car by the air and the road. If the power developed by an engine is `175 hp`, estimate the total friction force (apporox) acting on the car when it is moving at a speed of `25 m//s`. One horsepower equals `746 W`.A. `360 kN`B. `373 kN`C. `250 kN`D. `500 kN`
Answer» Correct Answer - D When the car moves at constant speed on a level roadway , the power used to veer come the total friction force equal the power input from the engine . or `P_(output) =f_("total")v=P_(imput)`. This gives `f_("total")=(P_(imput))/(v) =((175 hp)/(1 hp))` `5.22xx10^(5) N` or about `5xx10^(5) N` or `500 kN` (approx.).

Discussion

290.	A particale moves under the effect of a force `F = Cs` from `s = 0` to `s = s_(1)`. The work down in the process isA. `Cx_(1)^(2)`B. `(1)/(2)Cx_(1)^(2)`C. `Cx_(1)`D. Zero
Answer» Correct Answer - B `W underset(0)overset(x_(t))intF.dx =underset(0)overset(x_(t))intCx dx =C [(x^(2))/(2)]_(0)^(x_(t))=(1)/(2) C_(x_(1)^(2)`.

Discussion

291.	A ball is projected vertically down with an initial velocity from a height of `20 m` onto a horizontal floor. During the impact it loses `50%` of its energy and rebounds to the same height. The initial velocity of its projection isA. `20ms^(-1)`B. `15ms^(-1)`C. `10ms^(-1)`D. `5ms^(-1)`
Answer» Correct Answer - A

Discussion

292.	A body of mass `6kg` is under a force which causes displacement in it given by `S= (t^(2))/(4)` maters where `t` is time . The work done by the force in `2` sec isA. `12 J`B. `9 J`C. `6J`D. `3 J`
Answer» Correct Answer - C `s = (t^(2))/(4) :. Dx = (1)/(2) dt` `F = ma = (md^(2)s)/(dt^(2)) = (6d^(2))/(dt^(2)) [(t^(2))/(4)] = 3N` Now `W = int _(0)^(2) F ds = int_(0)^(2) 3 (t)/(2)dt = (3)/(2) [(t^(2))/(2)]_(0)^(2) = (3)/(4) [(2)^(2) - (0)^(2)] = 3 J`

Discussion

293.	A body of mass `6kg` is under a force which causes displacement in it given by `S= (t^(2))/(4)` maters where `t` is time . The work done by the force in `2` sec isA. `12 J`B. `9 J`C. `6 J`D. `3 J`
Answer» Correct Answer - D `s = (t^(2))/(4) :. Dx = (1)/(2) dt` `F = ma = (md^(2)s)/(dt^(2)) = (6d^(2))/(dt^(2)) [(t^(2))/(4)] = 3N` Now W=underset(0)overset(2)intF.dS=underset(0)overset(2)intma.dS` `underset(0)overset(2)int3xx2t xxt^(2)dt=underset(0)overset(2)int6t^(3)dt=(3)/(2)[t^(4)]_(0)^(2)=24J`.

Discussion

294.	An inelastic ball falls from a height of `100` metres. It loses `20%` of its total energy due to impact. The ball will now rise to a height ofA. 80 mB. 40 mC. 60 mD. 20 m
Answer» Correct Answer - A 20% energy is lost that means 80% still remains so by that 80% mgh the mass is not given assume to be 1 kg 80/100mgh = 0 therefore h =100mg/80 Correct option (A) 80 m Percentage loss of total energy = \(\frac{h_1-h_2}{h_1}\times100\) 20 = \((\frac{100-h_2}{100})\times100\) 20 = (100 - h₂) h₂ = 100 - 20 h₂ = 80 m

294.

An inelastic ball falls from a height of `100` metres. It loses `20%` of its total energy due to impact. The ball will now rise to a height ofA. 80 mB. 40 mC. 60 mD. 20 m

Answer» Correct Answer - A

20% energy is lost that means 80% still remains so by that

80% mgh

the mass is not given assume to be 1 kg

80/100mgh = 0

therefore

h =100mg/80

Correct option (A) 80 m

Percentage loss of total energy = \(\frac{h_1-h_2}{h_1}\times100\)

20 = \((\frac{100-h_2}{100})\times100\)

20 = (100 - h₂)

h₂ = 100 - 20

h₂ = 80 m

Discussion

295.	A body of mass `6kg` is under a force which causes displacement in it given by `S= (t^(2))/(4)` maters where `t` is time . The work done by the force in `2` sec isA. 12 JB. 9 JC. 6 JD. 3 J
Answer» Correct Answer - D

Discussion

296.	If the stone is thrown up vertically and return to ground, its potential energy is maximumA. During the upward journeyB. At the maximum heightC. During the return journeyD. At the bottom
Answer» Correct Answer - B

Discussion

297.	A body of mass `3 kg` is under a force , which causes a displacement in it is given by `S = (t^(3))/(3)` (in metres). Find the work done by the force in first `2` seconds.A. 2 JB. 3.8 JC. 5.2 JD. 24 J
Answer» Correct Answer - D

Discussion

298.	A body of mass `2 kg` is thrown up vertically with kinetic energy of `490 J`. If `g = 9.8m//s^(2)`, the height at which the kinetic energy of the body becomes half of the original value, isA. 50 mB. 12.5 mC. 25 mD. 10 m
Answer» Correct Answer - B

Discussion

299.	A body of mass 10 kg is dropped to the ground from a height of 10 metres. The work done by the gravitational force is `(g=9.8m//sec^(2))`A. `-490` joulesB. `+490` joulesC. `-980` joulesD. `+980` joules
Answer» Correct Answer - D

Discussion

300.	If the water falls from a dam into a turbine wheel 19.6m below, then the velocity of water at the turbine is `(g=9.8m//s^(2))`A. 9.8 m/sB. 19.6 m/sC. 39.2 m/sD. 98.0 m/s
Answer» Correct Answer - B

Discussion

Explore topic-wise InterviewSolutions in .

A ball of mass `m` moves with speed `v` and stricks a wall having infinite mass and it returns with same speed then the work done by the ball on the wall isA. zeroB. mv JC. m/v. JD. v/m J

A particle of mass `0.5kg` travels in a straight line with velocity `v=ax^(3//2)` where `a=5m^(-1//2)s^-1`. What is the work done by the net force during its displacement from `x=0` to `x=2m`?

A force `F=(10+0.50x)` acts on a particle in the x direction, where F is in newton and x in meter./find the work done by this force during a displacement form x=0 tox=2.0m

A particle of mass `0.5kg` travels in a straight line with a velocity `v=(5x^(5//2))m//s`. How much work is done by the net force during the displacement from `x=0` to `x=2m` ?

A force `F=(10+0.50x)` acts on a particle in the x direction, where F is in newton and x in meter./find the work done by this force during a displacement form x=0 tox=2.0mA. 31.5 JB. 63 JC. 21 JD. 42 J

A raindrop falling from a height `h` above ground, attains a near terminal velocity when it has fallen through a height `(3//4)h`. Which of the diagrams shown in figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground ?A. B. C. D.

A raindrop falling from a height `h` above ground, attains a near terminal velocity when it has fallen through a height `(3//4)h`. Which of the diagrams shown in figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground ?A. B. C. D.

A mass of `5kg` is moving along a circular path or radius `1m`. If the mass moves with 300 revolutions per minute, its kinetic energy would beA. `250pi^2 J`B. `100 pi^2 J`C. `5pi^2 J`D. 0 J

A mass of `5kg` is moving along a circular path or radius `1m`. If the mass moves with 300 revolutions per minute, its kinetic energy would beA. `250pi^(2)`B. `100pi^(2)`C. `5pi^(2)`D. 0

A block of mass 1 kg slides down a vertical curved track that is one quadrant of a circle of radius 1m. Its speed ast the the bottom si `2 m//s`. The work done by frictional force is : A. 8 JB. `-8 J`C. 4 JD. `-4 J`

A body of mass m=3.90 kg slides on a horizontal frictionless table with a speed of v=120 `ms^(-1)`. It is brought to rest in compressing a spring in its path. How much does spring is compressed, if its force constant k is 135 `Nm^(-1)`?A. 0.204 mB. 0.408 mC. 0.804 mD. 4.04 m

A block of mass 5 kg slides down a rough inclined surface. The angle of inclination is `45^(@)`. The coefficient of sliding friction is 0.20. When the block slides 10 cm, the work done on the block by force of friction isA. `-(1)/(sqrt2)J`B. 1JC. `-sqrt2J`D. `-1J`

A body of mass m moving with velocity v makes a head-on collision with another body of mass 2 m which is initially at rest. The loss of kinetic energy of the colliding body (mass m ) isA. `1:1`B. `2:1`C. `4:1`D. `9:1`

A body of mass m having an initial velocity v , makes head on collision with a stationary body of mass M . After the collision, the body of mass m comes to rest and only the body having mass M moves. This will happen only whenA. `m gt gt M`B. `m lt lt M`C. `m=M`D. `m=(1)/(2)M`

Figrue shows the F-x graph. Where F is the force applied x is the distance covered by the body along a straight line path. Given that F is in newton and x in metre, what is the work done?A. 10 JB. 20 JC. 30 JD. 40 J

The pointer reading vs load graph for a spring balance is as given in the figure. The spring constant isA. 0.1 kg/cmB. 5 kg/cmC. 0.3 kg/cmD. 1 kg/cm

A spring of force constant 10N / m has an initial stretch 0.20 m . In changing the stretch to 0.25 m , the increase in potential energy is aboutA. 0.1 jouleB. 0.2 jouleC. 0.3 jouleD. 0.5 joule

A force `F` acting on a body depends on its displacement `S` as `FpropS^(-1//3)` . The power delivered by `F` will depend on displacement asA. `s^(2//3)`B. `s^(-5//3)`C. `s^(1//2)`D. `s^(0)`

A spring 40 mm long is stretched by the application of a force. If 10 N force required to stretch the spring through 1mm, then the work done in stretching the spring through 40 mm isA. 84 JB. 68 JC. 23 JD. 8 J

The force required to stretch a spring varies with the distance a shown in the figure. If the experiment is performed with the above spring of half length, the line OA willA. Shift towards F-axisB. Shift towards X-axisC. Remain as it isD. Become double in length

The force required to stretch a spring varies with the distance a shown in the figure. If the experiment is performed with the above spring of half length, the line OA willA. shift towards F axisB. shift towardsC. X-axisD. remain as it is

The force required to stretch a spring varies with the distance a shown in the figure. If the experiment is performed with the above spring of half length, the line OA willA. shift towards F-axisB. shift towards X-axisC. remain as it isD. become double in length

A chain of length l and mass m lies of the surface of a smooth hemisphere of radius `Rgtl` with one end tied to the top of the hemisphere. Taking base of the hemisphere as reference line, find the gravitational potential energy of the chain.

A car of mass `m` starta from rest and accelerates so that the instyantaneous power delivered to the car has a constant magnitude `P_(0)`. The instaneous velocity of this car is proportional toA. `t^(2) P_(0)`B. `t^(1//2)`C. `t^(-1//2)`D. `t//sqrtm`

A particale moves under the effect of a force `F = Cs` from `x = 0` to `x = x_(1)`. The work down in the process isA. `Cx_(1)^(2)`B. `(1)/(2)Cx_(1)^(2)`C. `Cx_(1)`D. zero

A particale moves under the effect of a force `F = Cs` from `s = 0` to `s = s_(1)`. The work down in the process isA. `Cx_(1)^(2)`B. `(1)/(2)Cx_(1)^(2)`C. `Cx_(1)`D. Zero

A ball is projected vertically down with an initial velocity from a height of `20 m` onto a horizontal floor. During the impact it loses `50%` of its energy and rebounds to the same height. The initial velocity of its projection isA. `20ms^(-1)`B. `15ms^(-1)`C. `10ms^(-1)`D. `5ms^(-1)`

A body of mass `6kg` is under a force which causes displacement in it given by `S= (t^(2))/(4)` maters where `t` is time . The work done by the force in `2` sec isA. `12 J`B. `9 J`C. `6J`D. `3 J`

A body of mass `6kg` is under a force which causes displacement in it given by `S= (t^(2))/(4)` maters where `t` is time . The work done by the force in `2` sec isA. `12 J`B. `9 J`C. `6 J`D. `3 J`

An inelastic ball falls from a height of `100` metres. It loses `20%` of its total energy due to impact. The ball will now rise to a height ofA. 80 mB. 40 mC. 60 mD. 20 m

A body of mass `6kg` is under a force which causes displacement in it given by `S= (t^(2))/(4)` maters where `t` is time . The work done by the force in `2` sec isA. 12 JB. 9 JC. 6 JD. 3 J

If the stone is thrown up vertically and return to ground, its potential energy is maximumA. During the upward journeyB. At the maximum heightC. During the return journeyD. At the bottom

A body of mass `3 kg` is under a force , which causes a displacement in it is given by `S = (t^(3))/(3)` (in metres). Find the work done by the force in first `2` seconds.A. 2 JB. 3.8 JC. 5.2 JD. 24 J

A body of mass `2 kg` is thrown up vertically with kinetic energy of `490 J`. If `g = 9.8m//s^(2)`, the height at which the kinetic energy of the body becomes half of the original value, isA. 50 mB. 12.5 mC. 25 mD. 10 m

A body of mass 10 kg is dropped to the ground from a height of 10 metres. The work done by the gravitational force is `(g=9.8m//sec^(2))`A. `-490` joulesB. `+490` joulesC. `-980` joulesD. `+980` joules

If the water falls from a dam into a turbine wheel 19.6m below, then the velocity of water at the turbine is `(g=9.8m//s^(2))`A. 9.8 m/sB. 19.6 m/sC. 39.2 m/sD. 98.0 m/s