1496 + Interview Questions in KINEMATICS IN GENERAL AWARENESS Page 8 InterviewSolution

351.	The position-time (x-t) graphs for two children `A` and `B` returning from their school `O` to their homes `P` and `Q`, respectively, are shown in . Choose the . a.` (A//B)` Iives closer to school than `(B//A)`. b. `(A//B)` starts from the school earlier than `(B//A)`. c. `(A//B)` walks faster than `(B//A)`. d. `A` and `B` reach home at the (same//differnt) time. e. (A//B) overtakes on the road (once//twice).
Answer» a. It is clear from the graph that `OP lt OQ`. A loves closer to the school than `B`. b. As starts from `t=0` will `B` sstarts little later. So `A` starts from the school earlier than `B`. c. The slope of `x-t` for motion of `B` g t slpoe of `x-t` A. therce `B` walks fasster than `A` . d. The value of `t` corresponding to positions `P` and `Q` of the there homes is samd, So `A` and `B` reach home at the same time. e. It is clear form the gragh `B` overtakes `A` once on the road.

Discussion

352.	A man of mass m stands on a frame of mass M. He pulls on a light rope, which passes over a pulley. The other end of the rope is attached to the frame. For the system to be in equilibrium, what force must the man exert on the rope? A. `(1)/(2)(M+m)g`B. `(M+m)g`C. `(M-m)g`D. `(M+2m)g`
Answer» Correct Answer - A

Discussion

353.	Two balls `A and B` are thrown with speeds `u and u//2`, respectively. Both the balls cover the same horizontal distance before returning to the plane of projection. If the angle of projection of ball `B is 15^@` with the horizontal, then the angle of projection of `A` is.A. `sin^-1 ((1)/(8))`B. `(1)/(2) sin^-1 ((1)/(8))`C. `(1)/(3) sin^-1 ((1)/(8))`D. `(1)/(4) sin^-1((1)/(8))`
Answer» Correct Answer - B (b) `(u^2 sin 2 theta)/(g) = ((u//2)^2 sin 30^@)/(g) = (u^2)/(8 g)` :. `sin 2 theta = (1)/(8)` or `theta = (1)/(2) sin^-1 ((1)/(8))`.

Discussion

354.	A man pulls a block heavier than himself with a lioght rope. The coefficient if friction is the same between the man and the ground, and between the block and the ground.A. The block will not move unless the man also moves.B. The man can move even when the block is stationary.C. If both move, the acceleration of the man is greater than the acceleration of the block.D. None of the above assertions is correct.
Answer» Correct Answer - A::B::C

Discussion

355.	Two projectiles are thrown simultaneously in the same plane from the same point. If their velocities are `v_(1)` and `v_(2)` at angles `theta_(1)` and `theta_(2)` respectively from the horizontal, then ansewer the following questions If `v_(1)costheta_(1) = v_(2)cos theta_(2)`, then choose the incrorrect statementA. one particle will remain exactly below or above the other particleB. the trajectory of one with respect to other will be a vertical straight lineC. both will have the same rangeD. none of these
Answer» Correct Answer - C `y = 0 xx x` `theta = 90^(@)` for same range `U_(x1) U_(y1) = V_(x1) V_(y1)`

Discussion

356.	Two projectiles are thrown simultaneously in the same plane from the same point. If their velocities are `v_(1)` and `v_(2)` at angles `theta_(1)` and `theta_(2)` respectively from the horizontal, then ansewer the following questions The trajectory of particle `1` with respect to particle `2` wil beA. a parabolaB. a straight lineC. a vertical striaght lineD. a horizontal straight line
Answer» Correct Answer - B `V_(12(x)) = v_(1) cos theta_(1) - v_(2) cos theta_(2)` `a_(12(x) = 0 rArr V_(12(y)) = V_(1) sin_(1)^(theta) - V_(2) sin_(2)^(theta)` `rArr a_(12(y)) = 0` `x = v_(12(x)) t …(1)` `y = v_(12(y)) t ….(2)` From (1)& (2) `Y = (V_(12(y)))/(V_(12(y)))`

Discussion

357.	Find the speed of the box-3, if box-1 and box-2 are moving with speeds `v_(1)` and `v_(2)` as shown in figure when the string makes an angle `theta_(1)` and `theta_(2)` with the horizontal at its left and right end.
Answer» Correct Answer - `[((v_(1) + v_(2))cos theta_(1))/(cos theta_(2))-v_(2)]`

Discussion

358.	An insect of mass m is initially at one end of a stick of length L and mass M, which rests on a smooth floor. The coefficient of friction between the insect and the stick is k. The minimum time in which the insect can reach the other end of the stick is t. Then t2 is equal to(a) 2L/kg (b) 2Lm/kg (M + m) (c) 2LM/kg (M + m) (d) 2Lm/kgM
Answer» Correct Answer is: (c) 2LM/kg (M + m) Maximum force of friction = kmg. ∴ maximum acceleration of insect = a₁ = kmg / m = kg and maximum acceleration of stick = a₂= kmg / M . ∴ acceleration of insect with respect to stick = a = a₁ - (-a₂) = kg (1 + m/M). ∴ L = 1/2 at² or t² = 2L/a = 2ML / kg(M +m).

358.

An insect of mass m is initially at one end of a stick of length L and mass M, which rests on a smooth floor. The coefficient of friction between the insect and the stick is k. The minimum time in which the insect can reach the other end of the stick is t. Then t2 is equal to(a) 2L/kg (b) 2Lm/kg (M + m) (c) 2LM/kg (M + m) (d) 2Lm/kgM

Answer»

Correct Answer is: (c) 2LM/kg (M + m)

Maximum force of friction = kmg.

∴ maximum acceleration of insect = a₁ = kmg / m = kg

and maximum acceleration of stick = a₂= kmg / M .

∴ acceleration of insect with respect to stick

= a = a₁ - (-a₂) = kg (1 + m/M).

∴ L = 1/2 at² or t² = 2L/a = 2ML / kg(M +m).

Discussion

359.	Trajectory of two particles projected from origin with speeds `v_(1)` and `v_(2)` at angles `theta_(1)` and `theta_(2)` with positive `x`- axis respectively are as shown in the figure. Given that `(g=-10m//s^(2)(hatj))`. Choose the correct option related to diagram. A. `v_(1)-v_(2)=2v_(1)`B. `theta_(2)-theta_(1)=2theta_(1)`C. `3(v_(1)-v_(2))=v_(1)`D. `3(theta_(2)-theta_(1))=theta_(1)`
Answer» Correct Answer - D

Discussion

360.	Two stones are projected from level ground. Trajectory of two stones are shown in figure. Both stones have same maximum heights above level ground as shown. Let `T_(1)` and `T_(2)` be their time of flights and `u_(1)` and `u_(2)` be their speeds of projection respectively (neglect air resistance). then A. `T_(2)gtT_(1)`B. `T_(1)=T_(2)`C. `u_(1)gtu_(2)`D. `u_(1)=u_(2)`
Answer» Correct Answer - B Since maximum heights are same, their time of flight should be same `:. T_(1)=T_(2)` Aslo, vertical components of initial velocity are same. `:.` Horizontal component of velocity of 2 gt horizontal component of velocity of 1. Hence `u_(2) gt u_(1)`

Discussion

361.	The trajectory of a projectile in a vertical plane is `y=sqrt(3)x-2x^(2)`. `[g=10 m//s^(2)]` Angle of projection `theta` is :-A. `30^(@)`B. `60^(@)`C. `45^(@)`D. `sqrt(3)` rad
Answer» Correct Answer - B `y=sqrt(3)x-2x^(2)` Trajectory equation is `y=x tan theta-(gx^(2))/(2u^(2) cos^(2) theta)` `tan theta=sqrt(3)rArr theta=60^(@)` & `(g)/(2u^(2)cos^(2) theta)=2` `rArr u=5/(2xx1/4)=sqrt(10)`

Discussion

362.	A particle is projected with initial velocity of `hati+2hatj`. The eqaution of trajectory is `(take g=10ms^(-2))`A. `y=2x-5x^(2)`B. `y=x-5x^(2)`C. `4y=2x-5x^(2)`D. `y=2x-25x^(2)`
Answer» Correct Answer - A

Discussion

363.	A heavy particle is projected with a velocity at an angle with the horizontal into the uniform gravitational field. The slope of the trajectory of the particle varies asA. .B. .C. .D. .
Answer» Correct Answer - B::C::D (b.,c.,d.) If the particle is projected with velocity `u` at an angle `theta`, then equation of its trajectory will be `y = x tan theta -(gx^2)/(2 u^2 cos^2 theta)` We know slope is given by `m = (dy)/(dx)` Therefore, slope `m = tan theta - (gx)/(u^2 cos^2 theta)` It implies that the graph between slope and `x` will be straight line having negative slope and a non - zero positive intercept on y - axis. But `x` is directly proportional to the time `t`, therefore, the shape of graph between slope and time will be same as that of the graph between slope and `x`. Hence, only option (a) is correct, i.e., options (b),(c)and (d) are incorrect.

Discussion

364.	The trajectory of a projectile in a vertical plane is `y = ax - bx^2`, where `a and b` are constant and `x and y` are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.A. `(b^2)/(2 a), tan^-1 (b)`B. `(a^2)/(b), tan^-1 (2b)`C. `(a^2)/(4 b), tan^-1 (a)`D. `(2 a^2)/(b), tan^-1 (a)`
Answer» Correct Answer - C ( c) `y = ax - bx^2,` for height of `y` to be maximum : `(d y)/(d x) = 0` or `a - 2 bx = 0` or `x = (a)/(2 b)` (i) `y_(max) = a((a)/(2 b)) - b ((a)/(2 b))^2 = (a^2)/(4 b)` (ii) `((d y)/(d x))_(x = 0) = a = tan theta_0`, where `theta_0` = angle of projection `theta_0 = tan^-1 (a)`.

Discussion

365.	The trajectory of a projectile in a vertical plane is `y = ax - bx^2`, where `a and b` are constant and `x and y` are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.A. `(b^(2))/(2a),tan^(-1)(b)`B. `(a^(2))/b, tan^(-1)(2a)`C. `(a^(2))/(4b),tan^(-1)(a)`D. `(2a^(2))/b,tan^(-1)(a)`
Answer» Correct Answer - C

Discussion

366.	An object is projected with a velocitiy of `20m//s` making an angle of `45^(@)` with horizontal. The equation for trajectory is `h=Ax-Bx^(2)` where `h` is height `x` is horizontal distance. `A` and `B` are constants. The ratio `A.B` is `x/0.1`. Find value of `x.(g=10m//s^(2))`
Answer» Correct Answer - 4

Discussion

367.	At a metro station, a girl walks up a stationary escalator in time `t_1` If she remains stationary on the escalator, then the escalator take her up in time `t_2`. The time taken by her to walk up the moving escalator will be.A. (a) ` (t-1+t_2)//2`B. (b) ` T_1 T_2 //(t_2- T_1)`C. (c ) ` t_1 t_2 //(t_2+t_1)`D. (d) ` ` T_1 -t_2`
Answer» Correct Answer - C Velocity fo girl, ` v_g = L/t_1 , velocity of escalatro , ` v_e = L/ t_1` Effective velocity of girl on escalatio `= v_g + v_e L/t_1 + L/T_2` If (t) is the time taken, then ` L/t = L/ t-1 = L/t_2 or ` t t_1 t_2` gt/t_1+ t_2`.

Discussion

368.	A car , moving with a speed of `50 km//hr` , can be stopped by brakes after at least ` 6m `. If the same car is moving at a speed of `100 km//hr`, the minimum stopping distance isA. `12 m`B. `18 m`C. `24 m`D. `6 m`
Answer» Correct Answer - C `v^(2) = u^(2) +2as` `0 = (50 xx 5//18)^(2) +2a xx 6` `{:(a=-16m//s^(2),,,(a=retardation),):}` {:(`:."Again",,,v^(2)u^(2)+2as,):}` `0 = (100 xx 5//18^(2)) - 16 xx 2xx s` `S = (100 xx 5)^(2)` `S = ((100xx5)^(2))/(18 xx 18 xx 32) = 24.1 = 24 m`

Discussion

369.	A particle starts moving with initial velocity `3m//s` along `x`- axis from origin. Its acceleration is varying with `x` co-ordinate in parobalic nature as shown in the figure. At `x=1m `, tangent to the graph makes an angle `45^(@)` with positive `x`- aixs. Then at `x=3m`, A. velocity of the particle is `3sqrt(2)m//s`B. velocity of the particle is `3sqrt(2)m//s`C. acceleration of the particle is `4.5m//s^(2)`D. acceleration of the particle is `9m//s^(2)`
Answer» Correct Answer - A::C

Discussion

370.	A particle starts from rest with a time varying acceleration `a=(2t-4)`. Here `t` is in second and `a` in `m//s^(2)` Maximum velocity of particle in negative direction is at `t=`…………… secondA. 3B. 4C. 2D. 1
Answer» Correct Answer - C

Discussion

371.	If a particle is moving along a straight line and following is the graph showing acceleration varying with time then choose correct statement(s). At `t=0, x=0` and `v_(0)=7 ms^(-1)` A. Its displacement can never become zeroB. Its velocity can never become zeroC. Its displacement can become zeroD. Its velocity can become zero
Answer» Correct Answer - A, D

Discussion

372.	A car starts from rest to cover a distance s. the coefficient of friction between the road and the tyres is `mu`. The minimum time in which the car can cover the distance is proportional toA. `mu`B. `sqrt(mu)`C. `(1)/(mu)`D. `(1)/(sqrt(mu))`
Answer» Correct Answer - D

Discussion

373.	Block A is placed on block B. There is friction between the blocks, while the ground is smooth. A horizontal force P, increasing linearly with time, begins to act on A. The accelerations `a_(1) & a_(2)` of A and B respectively. Are plotted against time (t). Time correct graph is : A. B. C. D.
Answer» Correct Answer - C

Discussion

374.	STATEMENT-1: A particle si projected at an angle `theta ( lt 90)` to horizontal, with a velcotiy `u`. When particle strikes the ground its speed is again `u`. STATEMENT-2: Velocity along horizontal direction remains same but velocity along vertical direction is changed. When particle strikes the ground, magnitude of final vertical velocity is equal to magnitude of initla vertical velocity.A. Statement -1 is True, Statement-2 is Ture, Statement-2 is a correct explanantion for Statement-1.B. Statement-1 is Ture, Statement-2 is Ture, Statement-2 is Not a correct explanantion for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - A `{:(U_(x) =0 a_(x) =0,,rArr,V_(x)=U_(x) +a_(x)T = U_(x),),(U_(y)=usin theta, a_(y) =- g,,rArr,V_(y) = u sin theta - g t,):}` `= u sin theta - g.(2u sin theta)/(g), V_(y) =- u sin theta. `

Discussion

375.	A person travelling on a straight line moves with a uniform velocity v1 for some time and with uniform velocity v2 for the next equal time. The average velocity is v given by (a) v =(v1+v2)/2(b) v=√v1v2(c) 2/v=1/v1+1/v2(d) 1/v=1/v1+1/v2
Answer» (a) v = (v₁ + v₂)/2. Explanation: Let t be the time related to each velocity. Total distance travelled =v₁t+v₂t. Total time taken=2t. Hence average velocity = v₁t+v₂t/2t =(v₁+v₂)/2.

375.

A person travelling on a straight line moves with a uniform velocity v1 for some time and with uniform velocity v2 for the next equal time. The average velocity is v given by (a) v =(v1+v2)/2(b) v=√v1v2(c) 2/v=1/v1+1/v2(d) 1/v=1/v1+1/v2

Answer»

(a) v = (v₁ + v₂)/2.

Explanation:

Let t be the time related to each velocity. Total distance travelled =v₁t+v₂t. Total time taken=2t. Hence average velocity = v₁t+v₂t/2t =(v₁+v₂)/2.

Discussion

376.	Two graphs of the same projectile motion (in the x - y plane) projected from origin are shown in (Fig. 5 .10). X - axis is along horozontal direction and Y - axis is vertically upwards. Take `g = 10 m s^2` Find (i) The y component of initial velocity and (ii) the `X` component of initial velocity. .
Answer» From garph (i) : `v_y = 0 at t = (1)/(2) s`, i.,e., time taken to reach maximum height `H` is `t = (u_y)/(g) = (1)/(2) rArr u_y = 5 m s^-1` From graph (2) , `v_y = 0 at x = 2 m`, i.,e., when the particle is at maximum height, its displacement along horizontal, `x = 2 m`. `x = u_x xx t rArr 2 = u_x xx (1)/(2) rArr u_ x = 4 m s^-1`.

Discussion

377.	A jet airplance travelling at the speed of ` 500 km ^(-1)` ejects its products of combustion at the speed of ` 1500 km h^(-1)` relative to the jet plane. What is the speed of the later with respect to observer on the ground ?
Answer» Let (v -p)` be the velocity of the products w.r.t. ground, Let us consiedr the direction of moton of airplane to be positive direction of X-axis. Here, speed of jet plane, ` v_A =500 km h^(-1)` Relative speed of products of combustion w.r.t to jet plane , v_(pA0 = - 1500 =- 1000 km h^(-10` Relative velocity of the projucts w.r.t jet plane is ` v_(PA) = v_(p) =- v_A =- 1500 or v_p =v _A- 1500 =500 - 1500 =- 1000 km h^()-1)` Here- ve sign shows that the direction of products of combustion os opposite to tht of the airplane. Thus the magnitude of relative velocity is ` 1000 km h^(-1)` .

Discussion

378.	An airplane flies northward from town `A` and `B` and then back again. There is a steady wind blowing towards the north so that for the first state of the trip, the airplane is flying in the same direction as the wind and for the return trip of the journey, the airplane is flying opposite of the wind. The total trip time `T_(omega)` as compared to the total trip time in the absence of any winds `T_(0)` is:A. `T_(w)=T_(0)`B. `T_(w)gtT_(0)`C. `T_(w)ltT_(0)`D. Data is insufficient
Answer» Correct Answer - A

Discussion

379.	A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity `20 ms^-1`. The second stone will overtake the first after travelling a distance of `(g=10 ms^-2)`A. `13 m`B. `15 m`C. `11.25 m`D. `19.5 m`
Answer» Correct Answer - C `1/2g t^2=20(t-1)+1/2g(t-1)^2` Solving this equation we get, `:.t=1.5s` Now, `d=20(t-1)+1/2g(t-1)^2` `=11.25 cm`

Discussion

380.	A stone is dropped from a certain height which can reach the ground in `5 s`. It is stopped after `3 s` of its fall and then it is again released. The total time taken by the stone to reach the ground will be .A. (a) `2 s`B. (b) 3 s`C. ( ` 4s`D. (d)` none of these`.
Answer» Correct Answer - C Height from where the stoneis dropped = 1/2 gt^2` ` = 1/2 xx 9.8 8 xx 5%2 = 4.9 xx 9 m` Vertoca, dpwmward dostamce covered by stone in `3` seconds ` 1/2 xx 9.0 xx 3^2 = 4.9 xx 9 m` Velocity attained `= 9.8 xx 3 ms^(-1)` Remaining distance ` 4.9 xx 25-4.9 xx 9` `= 4.9 xx 16 m` If ` t_1` is the tiem taken to cover the remaining distance when stone is dropped again (u=0)` ` 4.9 xx 16 1/2 xx 9.8 t^2` ` t= sqrty16 = 4 s`.

Discussion

381.	The length of a minure hand of clock is 3 cm. Find the aerage velocity of the tip of the minute hand when it moves during a time interval form `4:00` p.m. to `4:15` p.m.
Answer» (i) Conditions at which constant velocity becomes equal to velocity of speed (ii) `0.0047 cm s^(-1)`

Discussion

382.	A table clock has its minute hand ` 5.0 cm` long. Find the average velocity of the tip of the miute hand (a) between ` 6.000 a.m. ` to 6. 15 a.m. and (b) between ` 6.00 a.m. to 6.30. p.m.`
Answer» (a) AT ` 6. 00 a.m.` the tip of the miute hand is at ` 12` mark and at ` 6. 15 a.m` it is ` 19^@` away. Thus, displacement `= sqrt(R^2 + R^2 ) = sqrt 2 R = sqrt xx 5 cm` Time taken from ` 6.00 a.m.` to ` 6.15 a.m. ` is `15 minutes = 15 xx 60 s`. The average velocity ` = ( displacement)/( tiem taken ) = ( sqrt 2 xx 5 cm)/( 15 xx 60 s)` = 7. 86 xx 10^# cm s^-1` Direction of average velocity is ` 45^@` with vertical. (b) AT ` 6.00 a.m.` the minute hand is at ` 12 ` mard and at ` 6. 30 p.m is ` 180^@` away. Thus, displacement ` =R+ R = 2 R = 2 xx 5 = 10 cm`. Time taken from ` 6.00 a.m. to ` 6. 30 p.m. is `12 ` hours and ` 30 minutes ` 45000 s`. The averge velocity, ` v_(av0 = ( desplacement ) / ( tiem) = (10) /( 45 000)` `= 2.2.xx 10^(-4) cm//s` Direction of average velocity is `12 ` mark to the ` 6 ` mark on clock panel i.e. at ` 0^@` angle with vertical.

Discussion

383.	The sum of the magnitudes of two forces acting at a point is 18 and the magnitude of their resultant is 12. If the resultant is at `90^(@)` with the force of smaller magnitude, What are the magnitudes of forces?
Answer» Let (A) and (B) be two forces acting at a point and ` theta` be the angle between them. Then `A+B =18 ` …(i) and ` A^2 + B^2 +2 AB cos theta = 12^2 = 144` ….(ii) If (A) si smaller force thea as per question ` tan 90^2 = (B sin theta)/(A+B cos theta)` or ` prop = (B sin theta)/(A+B cos theta)` or ` A+B cos theta =0 ` or ` cos theta =- A//B` ` From (ii) , ` A^2 -B^2 + 2 AB (- A//B) = 144` or ` B^2 - A^2 = 144` or` (B-A(B+A) = 144` or `B- A = (144)/(B+A) = (144)/(18) = 8 `...(iii) Sloving (i) and (iii), we shall get , ` A= 5 N`.

Discussion

384.	A drunkard waking in a borrow lane takes ` 6 ` steps forward and `4` steps backward, following againg ` 6` steps forward and ` 4` steps backward and so on Each step is ` 1 `m long and requires ` 1s`. Determine how long the drunkard takes to fall in a pit ` 15 m` away from the start.
Answer» Efferctive distance travelled by drunkard in ` 10 ` steps ` =- 4 = 2 m`. Therefore he takes ` 50 ` steps to moves `10 m`. Now he will hve to civer ` 5 m` more to reach the pot. For this he has to take only `5` forward steps. Thus total steps taken to fall in pit ` = 50 + 5 = 55 ` As each step is taken in ` 1 s`. so the time taken to fall in pit ` = 55 s`.

Discussion

385.	A drunkard is walking along a stsraight road. He takes five steps forward and three steps backward and so on. Each step is `1 m` long and takes `1 s`. There is a pit on the road `11 m`, away from the starting point. The drunkard will fall into the pit after.A. ` 29 s`B. ` 21 s`C. ` 37 s`D. ` 31 s`
Answer» Correct Answer - A Since the last five steps covering `5 m` land the drunkard fell into the pit, displacement prior to theis is 11-5 `m=6 m` . Time taken for edight steps (displacement in first eight steps `=5-3=2 m`) `=8 s`. Then time taken to cover first `6 m` of journey `=(6)/(2)xx8 =24 s` Time taken to cover last `5 m=5 s` Total time `=24 +5=29 s` .

Discussion

386.	The figure shows the velocity (v) of a particle ploted against time (t). A. The particle changes its direction of motion at some pointB. The acceleration of the particle remains constantC. The displacement of the particle is zeroD. The initial and final speeds of the particle are the same.
Answer» Correct Answer - A::B::C::D

Discussion

387.	A uniform chain of mass m hangs from a light pulley, with unequal lengths of the chain hanging from the two sides of the pulley. The force exerted by the moving chain on the pulley isA. mgB. `gt mg`C. `lt mg`D. either (b) or (c) depending of the chain
Answer» Correct Answer - C

Discussion

388.	A unifrom heavy chain is placed on a table with a part of it hanging over the edge. It just begins to slide when this part is one-third of its length. The coefficient of friction between the table and the chain isA. `(1)/(2)`B. `(1)/(3)`C. `(2)/(3)`D. `(3)/(4)`
Answer» Correct Answer - A

Discussion

389.	A unifrom chain of mass m hannges from a light pulley ,with unequal lenghts hanging from the two sides of the pulley .The force exerted by the moving chain on the pulley isA. `mg`B. `gtmg`C. `ltmg`D. May be any these depending on the time elapsed
Answer» Correct Answer - C

Discussion

390.	A ball ` A` is dropped from a building of height ` 45 m`. Simultaneously another ball ` B` is thrown up with a speed ` 40 m//s`. Calculate the relative speed of the balls as a function of time.
Answer» Speed of falling ball (A) after time ` t, v_(1) = gt (downwards ) Speed of projected upward ball (B) after time ` t, v_1 = (40 -gt) (upwards ) =- (40- gt) (downwards) Relative velocity of ball (A) w.r.t. another velocity of ball ` B = v_1 -v_2 = gt -[-(40-gt0] =40 m//s` At time ` t=0`, the relative velocity of ball ` A w.r.t ball (B), u_(AB) =0 - (-40) = ms^-1). Since ` v_(AB)=u_(AB)`, therefore, there is no acceleration of ball ` A w.r.t. ball B`. Hence the relative speed of ball (A) at any instant of time remains constant .

Discussion

391.	When a rifle is fired at a distant target, the barrel is not lined up exactly on the target. Why ?
Answer» As soon as a bullet is fired from a gun whose barrel is lined up exactly on the target, it starts falling downwards on acconunt of acceleration due to gravity. Due to it, the bullet will hit below the target just to avoid it, the barrel of the gun is lined up little above the target, so that the bullet, after travelling in parabolic path hits the distant target.

Discussion

392.	Is the rocket in flight is an illustration of projectile ?
Answer» No, because it is propelled by combustion of fuel and does not move under the effect of gravity alone.

Discussion

393.	Rocket in flight is not an illustration of projectle .Rocket takes flight due to combustion of fuet and does not move undr the graveity effect alone.A. (a) Statement-1 is true , Statement-2 is true , Statvement -2 is correct explanation of Statement-1 .B. (b) Statement-1 is true , statement -2 is true , statement -2 is not coerrecrt explanation of Statement-1.C. (C ) Statement-1 is true , Statement-2 is false.D. (d) Statement-1 is false , Statement-2 is true.
Answer» Correct Answer - B A projectile is the name given to a body which is once projected coan move uner gravity in two dimensions without consuming any fuel or propelled by any enging. Hence both Assertion and Reason are correct.

Discussion

394.	Can the speed of a body be begative ?A. positiveB. negativeC. zeroD. Both (a) and (b)
Answer» Correct Answer - B The speed of a body cannot be negative.

Discussion

395.	If `T = 2pi sqrt(l/g)` is the time period of a simple pendulu, then the unit of `4pi^(2) l/T^(2)` in the SI system is ________ .A. ` m s^(-1)`B. `s^(-2)`C. `m s^(-2)`D. `s^(-1)`
Answer» Correct Answer - C ` g = 4 pi^(2) l//T^(2)` it unit is m/s^(2)` that is ` m s ^(1)`

Discussion

396.	A cat, on seeing a rat at a distance `d=5 m`, starts velocity `u=5 m s^(-1)` and moves with acceleration `alpha =2.5 m s^(-2)` in order to catch it, while the rate with acceleration `beta` starts from rest. For what value of `beta` will the overtake the rat?. (in `m s^(-2)`).
Answer» Correct Answer - 5 For rat `S=(1)/(2) betat^(2)` (i) For cat `S =d =ut +(1)/(2) alpha t^(2) 0` Putting the value os `S` from Eq. (i) in Eq. (ii), `(a-b) t^(2) + 2 ut -2d =0` ` t=(2u+-sqrt( 4u^(2)-8 d (beta-alpha)))/(2(beta -alpha)` For `t` to be real, `(u^(2))/(2d) le (beta-alpha) becuase vbeta =alpha (u^(2))/(2d)` Substituting , a, d and `u` we get `beta =2.5 +(5^(2^(.)))/(2xx5) =2.5 +2 .5 =5 m s^(-2)`.

Discussion

397.	Choose the corrct statements (s)A. speed and velocity both have same units.B. if a body has a speed of `50m s^(-1)` in a striaght line path , then its velocity is 180 km h^(-1)`C. speed ofa vehicle is measured by a device called speedometer.D. All the above
Answer» Correct Answer - D

Discussion

398.	Three stones are projected simultaneously with same speed u from the top of a tower. Stone 1 is projected horizontally and stone 2 and stone 3 are projected making an angle q with the horizontal as shown in fig. Before stone 3 hits the ground, the distance between 1 and 2 was found to increase at a constant rate u. (a) Find `theta` (b) Find the rate at which the distance between 2 and 3 increases.
Answer» Correct Answer - (a) `theta = 60^(@)` (b) `sqrt(3) u`

Discussion

399.	One stone is dropped from a tower from rest and simultaneously another stone is projected vertically upwards from the tower with some initial velocity. The graph of distance (s) between the two stones varies with time (t) as (before either stone hits the ground).A. B. C. D.
Answer» Correct Answer - A

Discussion

400.	A stone is projected vertically up from a point on the ground, with a speed of `20 m//s`. Plot the variation of followings with time during the entire course of flight –A. VelocityB. SpeedC. Height above the groundD. distance travelled
Answer» Correct Answer - A::B::C

Discussion

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A drunkard is walking along a stsraight road. He takes five steps forward and three steps backward and so on. Each step is `1 m` long and takes `1 s`. There is a pit on the road `11 m`, away from the starting point. The drunkard will fall into the pit after.A. ` 29 s`B. ` 21 s`C. ` 37 s`D. ` 31 s`

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When a rifle is fired at a distant target, the barrel is not lined up exactly on the target. Why ?

Is the rocket in flight is an illustration of projectile ?

Can the speed of a body be begative ?A. positiveB. negativeC. zeroD. Both (a) and (b)

If `T = 2pi sqrt(l/g)` is the time period of a simple pendulu, then the unit of `4pi^(2) l/T^(2)` in the SI system is ________ .A. ` m s^(-1)`B. `s^(-2)`C. `m s^(-2)`D. `s^(-1)`

A cat, on seeing a rat at a distance `d=5 m`, starts velocity `u=5 m s^(-1)` and moves with acceleration `alpha =2.5 m s^(-2)` in order to catch it, while the rate with acceleration `beta` starts from rest. For what value of `beta` will the overtake the rat?. (in `m s^(-2)`).

Choose the corrct statements (s)A. speed and velocity both have same units.B. if a body has a speed of `50m s^(-1)` in a striaght line path , then its velocity is 180 km h^(-1)`C. speed ofa vehicle is measured by a device called speedometer.D. All the above

One stone is dropped from a tower from rest and simultaneously another stone is projected vertically upwards from the tower with some initial velocity. The graph of distance (s) between the two stones varies with time (t) as (before either stone hits the ground).A. B. C. D.

A stone is projected vertically up from a point on the ground, with a speed of `20 m//s`. Plot the variation of followings with time during the entire course of flight –A. VelocityB. SpeedC. Height above the groundD. distance travelled