1582 + Interview Questions in WORK IN GENERAL AWARENESS Page 7 InterviewSolution

301.	An ideal massless spring `S` can compressed `1.0` m in equilibrium by a force of `1000 N`. This same spring is placed at the bottom of a friction less inclined plane which makes an angle `theta =30^@` with the horizontal. A (10 kg) mass (m) is released from rest at the top of the incline and and is brought to rest momentarily after compressing the spring by `2 m. if `g =10 ms^(-1)`, what is the speed of just before it touches the spring? .A. (a) `sqrt20 ms^(-1)`B. (b) `sqrt30 ms^(-1)`C. (c) `sqrt10 ms^(-1)`D. (d) `sqrt40 ms^(-1)`
Answer» Correct Answer - A `F=kx` `:. k=(F)/(x) =(100)/(1)` `=100 N//m` `E_(i)=E_(f)` `:. 1/2 xx 10 xx v^(2) =1/2 xx 100 xx (2)^(2)-(10)(10)(2 sin 30^@)` Solving we get, `v=sqrt20 m//s`

Discussion

302.	A small disc A slides down with initial velocity equal to zero from the top of a smooth hill of height H having a horizontal portion. What must be the height of the horizontal portion h to ensure the maximum distance s covered by the disc? What is it equal to?
Answer» Correct Answer - B In order to obtain the velocity point (B), we apply the law of conservation of energy. So, Loss in `PE`=Gain in `KE` `m g (H-h) =1/2mv^(2)` `:. v=(sqrt[2g(H-h)])` Further `h=1/2gt^(2)` `:. t=sqrt((2h)/g)` Now, `s=v xx t =sqrt[[2g(H-h)]] xx sqrt((2 h//g))` or `s=sqrt[[4h(H-h)]]` .....(i) For maximum value of s, `(ds)/(dh)=0` `:. 1/(2sqrt[[4h(H-h)]]xx4(H))=0` or `h=H/2` Substituting `h=H//2` in E.q. `(1)`, we get `s=sqrt[[4(h//2)(H-//2)]]=sqrt(H^(2))=H`

Discussion

303.	A ball suspended by a thread swings ia a vertical plane so that its acceleration in the extreme position and lowest position are equal. The angle `theta` of thread deflection in the extreme position will beA. `tan^(-1)(2)`B. `tan^(-1)(sqrt2)`C. `tan^(1)((1)/(2))`D. `2 tan^(-1)((1)/(2))`
Answer» Correct Answer - D

Discussion

304.	Two ball of same temperature collide which is conserved? A) Temperature B) Velocity C) Kinetic energy D) Momentum
Answer» D) Momentum Explanations: • Due to the impact or the impulsive force velocities of balls change and hence total K.E. , being a scalar quantity, also changes. • While momentum being vector is conserved in collision. The loss of K.E. in impact is responsible for change in temperature.

304.

Two ball of same temperature collide which is conserved? A) Temperature B) Velocity C) Kinetic energy D) Momentum

Answer»

D) Momentum

Explanations:

• Due to the impact or the impulsive force velocities of balls change and hence total K.E. , being a scalar quantity, also changes.

• While momentum being vector is conserved in collision.

The loss of K.E. in impact is responsible for change in temperature.

Discussion

305.	The potential energy for a force filed `vecF` is given by `U(x,y)=cos(x+y)`. The force acting on a particle at position given by coordinates `(0, pi//4)` isA. (a) `-1/sqrt2(hati+hatj)`B. (b) `1/sqrt2(hati+hatj)`C. (c) `(1/2hati+sqrt3/2hatj)`D. (d) `(1/2hati-sqrt3/2hatj)`
Answer» Correct Answer - B `F_x=-(delU)/(delX)=sin(x+y)=1/sqrt2` `F_y=-(delU)/(delY)=sin(x+y)=1/sqrt2` `vecF=1/sqrt2[hati+hatj]`

Discussion

306.	A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to (i) `t^(1//2)` (ii) t (iii) `t^(3//2)` (iv) `t^(2)`
Answer» From `v=u+at` `v=0+at=at` As power, `P=Fxxv ` `:.P=(ma)xxat=ma^(2)t` As m and a are constants, therefor, `P prop t`

Discussion

307.	Define work done by a varying force.
Answer» It can be defined as the definite integral of the varying force over some displacement. It is given by the area under a F(x) – x graph, i.e., variable force and displacement graph.

Discussion

308.	What is the dimensional formula of work?
Answer» The dimensional formula of work is [ML²T^-2]

Discussion

309.	Consider a one-dimensional motion of a particle with total energy E. there are four regions A, B, C and D in which the relation between potential energy V, kinetic energy (k) and total energy E is as given below :Region A : V > ERegion B : V < ERegion C : K > ERegion D : V > KState with reason in each case whether a particle can be found in the given region or not.
Answer» Region A : No, V > E ⇒ E = V + K, K = E – V ⇒ V > E, so K < 0 as KE will become negative. Region B : Yes, V < E, K = E – V, K > 0, total energy can be greater than PE for non zero K.E. Region C : Yes, K > 0, V = E – K, V < 0, KE can be greater than total energy if its PE is negative. Region D : Yes, V > K, K = E – V as PE can be greater than KE.

309.

Consider a one-dimensional motion of a particle with total energy E. there are four regions A, B, C and D in which the relation between potential energy V, kinetic energy (k) and total energy E is as given below :Region A : V > ERegion B : V < ERegion C : K > ERegion D : V > KState with reason in each case whether a particle can be found in the given region or not.

Answer»

Region A : No, V > E ⇒ E = V + K, K = E – V ⇒ V > E, so K < 0 as KE will become negative.

Region B : Yes, V < E, K = E – V, K > 0, total energy can be greater than PE for non zero K.E.

Region C : Yes, K > 0, V = E – K, V < 0, KE can be greater than total energy if its PE is negative.

Region D : Yes, V > K, K = E – V as PE can be greater than KE.

Discussion

310.	200 kg of water is heated from `20^(@)C` to `100^(@)C`. If specifice heat of water is `4.2xx10^(3)Jkg^(-1)0C^(-1)`, calculate the increase in the mass of water.
Answer» Here, `m=200kg,` `DeltaT=(100-20)^(@)C=80^(@)C` `s=4.2xx10^(3)Jkg^(-1).^(@)C^(-1)` Heat Energy absorbes, `E=smDeltaT` `=(4.2xx10^(13))xx200xx80J` `E=6.72xx10^(7)J` Increase in mass of water, `m=(E)/(C^(2))=(6.72xx10^(7))/((3xx10^(8))^(2))` `m=7.46xx10^(-10)kg`

Discussion

311.	An engine pumps up `100 kg` water through a height of `10 m` in `5 s`. If efficiency of the engine is `60%`. What is the power of the engine? `Take g = 10 ms^(2)`.A. (a) `33kW`B. (b) `3.3kW`C. (c) `0.33kW`D. (d) `0.033kW`
Answer» Correct Answer - B Power used to pump the water`=(mgh)/(t)=(100xx10xx10)/(5)` Power of engine`=2000xx100/60=3.3kW`

Discussion

312.	If 2000kg of water is heated from `0^(@)`C to `100^(@)C` , what will be the corresponding increase in mass of water? Take specific heat of water `=1cal. g^(-1).^(@)C^(-1)` .
Answer» Here, `m=2000 kg = 2xx10^(6)g, Delta T = 100-0=100^(@)C, c = 1 cal. g^(-1).^(@)C^(-1)`. `:.` Heat energy supplied `//`gained by water, `DeltaW=E=2xx10^(8)xx4.2J=8.4xx10^(8)J.` Hence, increase in mass , `Deltam=(E)/(c^(2))=(8.4xx10^(8))/((3xx10^(8)))=9.3xx10^(-9)kg`

Discussion

313.	A rubber ball falls on a floor from a height of `19*6m. ` Calculate the velocity with which it strikes the ground. To what height will the ball rebounce if it loses `25%` of its energy on striking the groun?
Answer» Here, `h_(1)=196m, v_(1)=?` Loss of energy `=25% h_(2)=?` On striking th egroun, K.E. `=` P.E `(1)/(2)mv_(1)^(2)=mgh_(1)` `v_(1)=sqrt(2gh_(1))=sqrt(2xx98xx196)=196m//s` As the body loses `25%` of its energy, final energy `E_(2)=(75)/(100)` of initial energy i.e., `mgh_(2)=(75)/(100)(mgh_(1))` `(h_(2)=(3)/(4)h_(1)=(3)/(4)xx196=147m`

Discussion

314.	An engine pumps up `100 kg` water through a height of `10 m` in `5 s`. If efficiency of the engine is `60%`. What is the power of the engine? `Take g = 10 ms^(2)`.
Answer» Here, `m=100kg, h=10m, t=5s` `g=10m//s^(2),eta=60%` output power, `P_(0)=(mgh)/(l)=(100xx10xx10)/(5)` Input power `P_(i)=?` As `eta=(P_(o))/(P_(i)) :.P_(i)=(P_(o))/(eta)=(2)/(60//100)=3.33kW`

Discussion

315.	A lift is designed to carry a load of 4000 kg through 10 floors of a building, avarage 6m per floor, in 10s. Calculate the power of the lift.
Answer» Here, `m=4000kg, h=10xx6m,t=10s,` `P=(W)/t0=(mgh)/(t)=(4000xx9.8xx60)/(10)wat t` Horse power of lift `=(4000xx9.8xx60)/(10xx746)=315.3`

Discussion

316.	A man pushes a wall and fails to displace it.He doesA. negative workB. positive but not maximum workC. maximum positive workD. no work at all
Answer» Correct Answer - D (d) As, work done (W) `=F.s =F s costheta` When a man pushes against a wall but fails to move it, then displacement of wall, s=0 `:. "Work done", W=Fxx0xx cos theta =0` Therefore, man does no work at all.

Discussion

317.	A body of mass `M` moves with velocity `v` and collides elasticity with another body of mass `m(M gt gt m)` at rest , then the velocity of the body of mass `m` isA. vB. 2vC. `v//2`D. zero
Answer» Correct Answer - B

Discussion

318.	A man pushes a wall and fails to displace it.He doesA. Negative workB. positive but not maximum workC. No work at allD. Maximum work
Answer» Correct Answer - C

Discussion

319.	A body of mass 5 kg moving with a velocity 10 m/s collides with another body of the mass 20 kg at, rest and comes to rest. The velocity of the second body due to collision isA. 2.5 m/sB. 5m/sC. 7.5 m/sD. 10 m/s
Answer» Correct Answer - A

Discussion

320.	If a skater of weight 3 kg has initial speed 32 m / s and second one of weight 4 kg has 5 m / s . After collision, they have speed (couple) 5 m / s . Then the loss in K.E. isA. 48 JB. 96 JC. zeroD. none of these
Answer» Correct Answer - D

Discussion

321.	If the heart pushes 1 cc of blood in one second under pressure 20000 N/m 2 the power of heart isA. 0.02 WB. 400 WC. `5xx10` WD. 0.2 W
Answer» Correct Answer - A

Discussion

322.	If the heart pushes 1 cc of blood in one second under pressure 20000 N/m 2 the power of heart isA. 0.02 WB. 400 WC. `5xx10^(-10)`WD. 0.2 W
Answer» Correct Answer - A (a) Power `=("Workdone")/("Time")=("Pressure"xx"changein volume")/("Time")` `=(20000xx1xx10^(-6))/(1)=2xx10^(-2)=0.02" W"`

Discussion

323.	A ball moving with velocity `2 ms^(-1)` collides head on with another stationary ball of double the mass. If the coefficient of restitution is `0.5`, then their velocities (in `ms^(-1)`) after collision will beA. `0,2`B. `0,1`C. `1,1`D. `1,0.5`
Answer» Correct Answer - B Here, `u_(1)=2 m//s, u_(2)=0, e=0.5,` `m_(1)=m, m_(2)=2m` As momentum is conserved `:. m_(1)u_(1)+m_(2)u_(2)=m_(1)upsilon_(1)+m_(2)upsilon_(2)` ` mxx2+2 mxx0=m upsilon_(1)+2 m upsilon_(2)` or `upsilon_(1)+2 upsilon_(2)=2 ....(1) ` As coefficient of restitution `e=(upsilon_(2)-upsilon_(1))/(u_(1)-u_(2))` `0.5=(upsilon_(2)-upsilon_(1))/(2-0)` or `upsilon_(2)-upsilon_(1) ......(ii)` Adding `(i)` and `(ii)` `3 upsilon_(2)=3` or `upsilon_(2)=1 m//s` `upsilon_(1)=0 m//s`

Discussion

324.	Particle A makes a perfectly elastic collision with anther particle B at rest. They fly apart in opposite direction with equal speeds. If the masses are `m_(A)&m_(B)` respectively, thenA. `2m_(A)=m_(B)`B. `3m_(A)=m_(B)`C. `4m_(A)=m_(B)`D. `sqrt3m_(A)=m_(B)`
Answer» Correct Answer - B

Discussion

325.	A ball moving with velocity `2 ms^(-1)` collides head on with another stationary ball of double the mass. If the coefficient of restitution is `0.5`, then their velocities (in `ms^(-1)`) after collision will be
Answer» Here, `u_(1)=2m//s, u_(2)=0,m_(1)=m, m_(2)=2m` `e=0.5, upsilon_(1)=?upsilon_(2)=?` As `e=(upsilon_(2)-upsilon_(1))/(u_(1)-u_(2))=0.5` `:. upsilon_(2)-upsilon_(1)=0.5(u_(1)-u_(2))=0.5(2-0)=1` …(i) According to the law of conservation of linear momentum, `m_(1)upsilon_(1)+m_(2)upsilon_(2)=m_(1)upsilon_(1)+m_(2)upsilon_(2)` `m upsilon_(1)+2m upsilon_(2)=mxx2+0` `upsilon_(1)+2upsilon_(2)=2` ..(ii) Add (i) and (ii) `:3upsilon_(2)=3,upsilon_(2)=1m//s` From (i), `upsilon_(1)=upsilon_(2)-1=1-1=0`

Discussion

326.	A block of mass 1.2 kg moving at a speed of 20 cm /s collides head on with a similar block kept at rest. The coefficient of restitution is 3/5. find the loss of kinetic energy during the collision.
Answer» Here, `m_(1)=1.2kg, u_(1)=20cm//s, m_(2)=1.2kg, u_(2)=0` If `upsilon_(1)` and `upsilon_(2)` are velocities of the two blocks after collision, then accordin to the principle of conservation of momentum, `m_(1)u_(1)+m_(2)u_(2)=m_(1)upsilon_(1)+m_(2)upsilon_(2)` `1.2xx20+0=1.2upsilon_(1)+1.2upsilon_(2) ` `:. upsilon_(1)+upsilon_(2)=20(cm//s)` ...(i) velocity of approach `=u_(1)-u_(2)=20cm//s,` velocity of separation `=upsilon_(2)-upsilon_(1)` By definition, `e=(upsilon_(2)-upsilon_(1))/(u_(1)-u_(2))` `(3)/(5)=(upsilon_(2)-upsilon_(1))/(20) :. upsilon_(2)-upsilon_(1)=(20xx3)/(5)=12` ...(ii) From (i) and (ii), `upsilon_(1)=4cm//s, upsilon_(2)=16cm//s` Loss in K.E. `=(1)/(2)m_(1)u_(1)^(2)-(1)/(2)(m_(1))upsilon_(1)^(2)-(1)/(2)m_(2)upsilon_(2)^(2)=(1)/(2)xx1.2((20)/(100))^(2)-(1)/(2)(1.2)((4)/(100))^(2)-(1)/(2)xx1.2((16)/(100))^(2)` `=2.4xx10^(-2)-0.096xx10^(-2)-1.536xx10^(-2)`

Discussion

327.	A ball of mass m moving with velocity v collides head-on which the second ball of mass m at rest. I the coefficient of restitution is e and velocity of first ball after collision is `v_(1)` and velocity of second ball after collision is `v_(2)` thenA. `v_(1)=((1-e)u)/(2),v_(2)=((1+e)u)/(2)`B. `v_(1)=((1+e)u)/(2),v_(2)=((1-e)u)/(2)`C. `v_(1)=u/2,v_(2)=-u/2`D. `v_(1)=(1+e)u,v_(2)=(1-e)u`
Answer» Correct Answer - A

Discussion

328.	A ball of mass m moving with a speed `2v_0` collides head-on with an identical ball at rest. If e is the coefficient of restitution, then what will be the ratio of velocity of two balls after collision?A. `(1-e)/(1+e)`B. `(1+e)/(1-e)`C. `(e-1)/(e+1)`D. `(e+1)/(e-1)`
Answer» Correct Answer - A According to law of conservation of linear momentum , we get `m(2v_0)+mxx0=mv_1 + mv_2` where `v_1` and `v_2` are the velocities of the balls after collision. `v_1+v_2 =2v_0`..(i) By definition of coefficient of restitution `e=(v_2-v_1)/(u_1-u_2)=(v_2-v_1)/(2v_0)" " (because u_1=2v_0, u_2=0)` `v_2-v_1=2ev_0`...(ii) Adding (i) and (ii), we get `2v_2-2v_0+2ev_0 , v_2=(1+e)v_0` Subtract (ii) from (i), we get `2v_1 =2v_0 - 2ev_0 , v_1=v_0 (1-e)` Their corresponding ratio is `v_1/v_2=(1-e)/(1+e)`

Discussion

329.	A small ball is dropped from rest from height 10 m on a horizontal floor. If coefficient of restitution between ground and body is `0.5` then find the maximum height it can rise after collision.
Answer» The ballis dropped from height `h_(1)` so velocity just before hitting the ground= velocity of approach `=V_("approach"=sqrt(2gh_(1)))` The ball rises to the height `h_(2),` so velocity when just leaves the ground = velocity of sparation `=sqrt(2gh_(2))=V_("separation")` Coefficient of restitution, `e=(V_("separation"))/(V_("approach"))` `impliese=sqrt((2gh_(2))/(2gh_(1)))=sqrt((h_(2))/(h_(1)))` `impliesh_(2)=e^(2)h_(1)` `=1/4xx10=2.5m`

Discussion

330.	A ball falls vertically onto a floor with momentum `p` and then bounces repeatedly. If coefficient of restitution is `e`, then the total momentum imparted by the ball to the floor isA. `p(1+e)`B. `(p)/(1-e)`C. `p((1+e))/((1-e))`D. `p(1-(1)/(e))`
Answer» Correct Answer - C When a particle undergoes normal collision with a floor or a wall, with coefficient of restitution `e`, the speed after collision is `e` times the speed before collision. Therefore, change in momentum after first impact `= ep-(-p) =p(1+e)` After the second impact, change in momentum would be `e(ep)-(-ep)=ep(1+e)` and so on. ltbr. Therefore, total change in momentum of ball `=` momentum imparted to floor `=p(1+e) [1+e+e^(2)+......]` `[p(1+e)[(1)/(1-e)]=(p(1+e))/((1-e)) `

Discussion

331.	A man of `50 kg` mass is standing in a gravity free space at a height of `10m` above the floor. He throws a stone of `0.5 kg` mass downwards with a speed `2m//s`. When the stone reaches the floor, the distance of the man above the floor will beA. `9.9 m`B. `10.1 m`C. `1.0 m`D. `20 m`
Answer» Correct Answer - B Here, `m_(1)=50 kg, m_(2)=0.5kg` `u_(1)=?,u_(2)=2m//s` By momentum conservation, `m_(1)u_(1)+m_(2)u_(2)=0` `=u_(1)=(-m_(2)u_(2))/(m_(1))=(-0.5xx2)/(50)=(1)/(50)m//s` Negative sign is for upward motion of man. In gravity free space, time taken by stone to reach the floor, `t=(s)/(m_(1))=(10)/(2)=5s` Upward distance moved by the man `=u_(1)xxt` `=(1)/(50)xx5=0.1m.` `:.` Distance of man above the floor `=(10+0.1)m=10.1m`

Discussion

332.	A metal ball falls from a height of 32 metre on a steel plate. If the coefficient of restitution is 0.5, to what height will the ball rise after second bounceA. 2 mB. 4 mC. 8 mD. 16 m
Answer» Correct Answer - A

Discussion

333.	A body falls on a surface of coefficient of restitution 0.6 from a height of 1 m . Then the body rebounds to a height ofA. 0.6 mB. 0.4 mC. 1 mD. 0.36 m
Answer» Correct Answer - D

Discussion

334.	A sphere of mass `m` moving with constant velocity `u`, collides with another stationary sphere of same mass. If `e` is the coefficient of restitution, the ratio of the final velocities of the first and second sphere isA. `(1-e)/(1+e)`B. `(1+e)/(1-e)`C. `(e+1)/(e-1)`D. `(e-1)/(e+1)t^(2)`
Answer» Correct Answer - A

Discussion

335.	A body of mass `m_(1)` is moving with a velocity V. it collides with another stationary body of mass `m_(2)`. They get embedded. At the point of collision, the velocity of the system.A. increasesB. decreases but does not become zeroC. remains sameD. become zero
Answer» Correct Answer - B

Discussion

336.	An elevator which can carry a maximum load of `1800kg` is moving with a constant speed of `2ms^(-1)`. The frictional force opposing the motino is `5000N`. Calculate the minimum power delivered by the motor to the elevator in watt and hp ? Take `g=10m//s^(2)`
Answer» Here, `m=1800kg, upsilon=2ms^(-1),` `f=5000N,P=?` Minimum force required to be applied, `F=mg+f=1800xx10+5000=23000N` `P=Fxxupsilon=23000xx2=46000W=46kW` `=(46000)/(746)hp=61.66hp`

Discussion

337.	A particle of mass m moving with speed u collides perfectly inelastically with another particle of mass 3 m at rest. Loss of KE of system in the collision isA. `3/4mu^(2)`B. `3/8mu^(2)`C. `1/4mu^(2)`D. `1/8mu^(2)`
Answer» Correct Answer - B

Discussion

338.	An object of mass 80 kg moving with velocity `2ms^(-1)` hit by collides with another object of mass 20 kg moving with velocity `4 ms ^(-1)` Find the loss of energy assuming a perfectly, inelastic collisionA. `12J`B. `24J`C. `30J`D. `32J`
Answer» Correct Answer - D

Discussion

339.	Asseration : In circular motion work done by all the forces acting on the body is zero. Reason : Centripetal force and veloity are mutually perpendicular.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true and the Reason is not the correct explanation of the Assertion.C. If Assertion is true, but the Reason is falseD. If Assertion is false but the Reason is true.
Answer» Correct Answer - D In non-uniform circular motion (when speed constant) work done by all the forces is zero.

Discussion

340.	An object of mass `40kg` and having velocity `4 m//s` collides with another object of mass `60 kg` having velocity `2 m//s` . The loss of energy when the collision is perfectly inelastic isA. `392J`B. `440J`C. `48J`D. `110J`
Answer» Correct Answer - C Here, `m_(1)=40 kg, u_(1)=4 m//s ` `m_(2)=60 kg, u_(2)= 2 m//s` Final vel. After collision, `upsilon=(m_(1)u_(1)+m_(2)u_(2))/((m_(1)+m_(2))) ` `upsilon=(40xx4+60xx2)/((40+60))=2.8 m//s` Loss of energy `=(1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2)-(1)/(2)(m_(1)+m_(2))upsilon^(2)` `=(1)/(2)xx40xx4^(2)+(1)/(2)xx60xx2^(2)-(1)/(2)(40+60)xx(2.8)^(2)` `=320+120-392=48` joule

Discussion

341.	A horizontal force F pulls a 20 kg box at a constant speed along a rough horizotal floor. The coefficient of friction between the box and the floor is 0.25. The work done by force F on the block in displacing it by 2 m isA. 49 JB. 98 JC. 147 JD. 196 J
Answer» Correct Answer - B (b) Frictional force acting on the box, `F=mu mg=0.25xx20xx9.8=49 N` This force must be same as applied force F so that the box moves with constant speed. Work done by the applied force, F `W=Fs=49xx2=98 J`

Discussion

342.	Two sphere `A and B` of masses `m_(1) and m_(2)` respectivelly colides. A is at rest initally and `B` is moving with velocity `v` along x-axis. After collision `B` has a velocity `(v)/(2)` in a direction perpendicular to the original direction. The mass `A` moves after collision in the direction.A. same as that of `B`B. opposite to that of `B`C. `theta=tan^(-1)(1//2)`to the `x-`axisD. `theta=tan^(-1)(-1//2)`to the `x-`axis
Answer» Correct Answer - D Here, mass of `A=m_(1), ` mass of `B=m_(2)` Initial velocity of `A,U_(1)=0` Initial velocity of `B,u_(2)=upsilon, `along `x-` axis. ltbr. After collision, let final velocity of `A` along `x-` axis be `v_(x)` and final velocity `A` along `y-` axis be `upsilon_(y)`. Final velocity of `B=(upsilon)/( 2)` along `Y-` axis. Applying principle of conservation of linear momentum along `x-` axis. `m_(1)xx0+ m_(2) upsilon=m_(1) upsilon_(x)+ m_(2)xx0` `upsilon_(x)=( m_(2)upsilon)/(m_(1)) ....(i)` Applying principle of conservation of linear momentum along `Y-axis`, `m_(1)xx0 + m_(2)xx0=m_(1)upsilon_(y)+m_(2)((upsilon)/(2))` or `upsilon^(y)=-(m_(2)upsilon)/(2m_(1)) ...(ii)` If mass `A` moves at angle `theta` with the `X-`axis, then `tan theta=(upsilon_(y))/( upsilon_(x))=-(m_(2)upsilon)/(2m_(1))xx(m_(1))/(m_(2)upsilon)=-(1)/(2)` `:. theta= tan^(-1)((1)/(2)) ` to the `x-`axis

Discussion

343.	A Force `F` acting on an object varies with distance `x` as shown in the here . The force is in newton and `x` in metre. The work done by the force in moving the object from `x = 0` to `x = 6m` is A. `4.5 J`B. `13.5 J`C. `9.0 J`D. `18.0 J`
Answer» Work done = Area enclosed by `F - x` graph ` = (1)/(2) xx (3 +6) xx 3 = 13.5 J`

Discussion

344.	A Force `F` acting on an object varies with distance `x` as shown in the here . The force is in newton and `x` in metre. The work done by the force in moving the object from `x = 0` to `x = 6m` is A. 4.5JB. 13.5JC. 9.0JD. 18.0J
Answer» Correct Answer - b

Discussion

345.	A body of mass `3 kg` is under a constant force which causes a displacement `s` metre in it, given by the relation `s=(1)/(3)t^(2)`, where `t` is in seconds. Work done by the force in 2 seconds isA. `(19)/(5)J`B. `(5)/(19)J`C. `(3)/(8)J`D. `(8)/(3)J`
Answer» Correct Answer - D Here, `s=(1)/(3)t^(2)` `upsilon=(ds)/(dt)(2t)/(3)` `a=(dupsilon)/(dt)=(2)/(3)` `F=ma=3xx(2)/(3)=2N` When `t=2s, s=(t^(2))/(3)=(4)/(3)` As work done `=Fxxs :. W=2xx(4)/(3)=(8)/(3)J`

Discussion

346.	A vertical spring with force constant `k` is fixed on a table. A ball of mass `m` at a height `h` above the free upper end of the spring falls vertically on the spring , so that the spring is compressed by a distance `d`. The net work done in the process isA. `mg(h+d)+(1)/(2)kd^2`B. `mg(h+d)-(1)/(2)kd^2`C. `mg(h-d)-(1)/(2)kd^2`D. `mg(h-d)+(1)/(2)kd^2`
Answer» Correct Answer - B Loss in P.E. `=` gain in the spring energy `mg(h+d)=(1)/(2)kd^2` net work done `=mg(h+d)-(1)/(2)kd^2`

Discussion

347.	A space station 960 m in diameter rotates fast enough that the artificial gravity at the outer edge is 1.5 g (a) What is the frequency of rotation ? (b) What is its period ? (c) At what distance from the center will the artificial gravity be 0.75 g ? Take `g=10m//s^(2)`
Answer» Correct Answer - `[0.0281 sec^(-1),35.52 sec, 240 m]`

Discussion

348.	Calculate the angle of banking required for a curve of 200 m radius so that a car rounding the curve at 80 kph would have no tendency to skid outward or inward. Assume the surface is frictionless. Take `g=10m//s^(2)`
Answer» Correct Answer - `[tan^(-1) (0.247) approx 13.87^(@)]`

Discussion

349.	In figure, block A is released from rest, when spring is at its natural unstretched length. For block B of mass M to leave contact with the ground at some stage, the minimum mass of A must be A. 2 MB. MC. M/2D. A function of M and the force constant of the sping.
Answer» Correct Answer - C

Discussion

350.	Two unequal masses are ties together with a cord with a compressed spring in between. When the cord is burnt with a match releasing the spring, the two masses fly apart with equalA. (a) Kinetic energyB. (b) SpeedC. (c) MomentumD. (d) Acceleration
Answer» Correct Answer - C As `F_(ext)=0`, according to the law of conservation of momentum, `vecP_1+vecP_2=const ant`. but initially both the blocks were at rest, so `vecP_1+vecP_2=0`, i.e., `vecP_2=-vecP_1` i.e., at any instant, the two blocks will have momentum equal in magnitude but opposite in direction (through they may have different values of momentum at different positions).

Discussion

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