1496 + Interview Questions in KINEMATICS IN GENERAL AWARENESS Page 15 InterviewSolution

701.	A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of `3 m s^-2` for `0.5 min`. If the maximum height reached by it is `80 m`, then the angle of projection is `(g = 10 ms^-2)`.A. `tan^-1 3`B. `tan^-1(3//2)`C. `tan^-1 (4//9)`D. `sin^-1(4//9)`
Answer» Correct Answer - C ( c) `H = (u^2 sin^2 theta)/(2 g)` or `80 = (u^2 sin^2 theta)/(2 xx 10)` or `u^2 sin^2 theta = 1600` or `u sin theta = 40 ms^-1` Horizontal velocity `= u cos theta = 3 xx 30 = 90 ms^-1` `(u sin theta)/(u cos theta) = (40)/(90)` or `tan theta = (4)/(9)` or `theta = tan^-1 ((4)/(9))`.

Discussion

702.	a particle moving along a straight line with uniform acceleration has velocities `7 m//s` at A and `17 m//s` at C. B is the mid point of AC. Then :-A. The velocity at B is `12 m//s`B. The average velocity between A and B is 10 m/sC. The ratio of the time to go from A to B to that from B to C is `3 : 2`D. The average velocity between B and C is 15m/s
Answer» Correct Answer - B, C, D

Discussion

703.	When a body is thrown up in a lift with a velocity `u` relative to the lift, the time of flight is found to be `t`. The acceleration with which the lift is moving up isA. `(u-"gt")/(t)`B. `(u+"gt")/(t)`C. `(2u-"gt")/(t)`D. `(u)/(t)-2g`
Answer» `(2u)/(g+a)=t rArr a =(2u)/(t)-g`

Discussion

704.	A balloon starts rising from the ground with an acceleration of 1.25 m/s 2. After 8 s, a stone is released from the balloon. The stone will(a) cover a distance of 40 m (b) have a displacement of 50 m (c) reach the ground in 4 s (d) begin to move down after being released
Answer» Correct answer is: (c) reach the ground in 4 s At the instant of release, the balloon is 40 m above the ground and has an upward velocity of 10 m/s. For the motion of the stone from the balloon to the ground, u = 10 m/s, s = -40 m, a = -10 m/s ²(g).

704.

A balloon starts rising from the ground with an acceleration of 1.25 m/s 2. After 8 s, a stone is released from the balloon. The stone will(a) cover a distance of 40 m (b) have a displacement of 50 m (c) reach the ground in 4 s (d) begin to move down after being released

Answer»

Correct answer is: (c) reach the ground in 4 s

At the instant of release, the balloon is 40 m above the ground and has an upward velocity of 10 m/s. For the motion of the stone from the balloon to the ground, u = 10 m/s, s = -40 m, a = -10 m/s ²(g).

Discussion

705.	Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. They hit the ground with speeds vA, vB and vC respectively.(a) vA = vB = vC(b) vB > vC > vA (c) vA = vB > vC (d) vA > vB = vC
Answer» Correct answer is: (a) v_A = v_B = v_C The increase in kinetic energy = the loss in potential energy. This is the same for all three.

705.

Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. They hit the ground with speeds vA, vB and vC respectively.(a) vA = vB = vC(b) vB > vC > vA (c) vA = vB > vC (d) vA > vB = vC

Answer»

Correct answer is: (a) v_A = v_B = v_C

The increase in kinetic energy = the loss in potential energy. This is the same for all three.

Discussion

706.	From the top of a tower, a stone is thrown up and it reaches the ground in time t1. A second stone is thrown down with the same speed and it reaches the ground in time t2. A third stone is released from rest and it reaches the ground in time t3.(a) t3 = 1/2 (t1 + t2)(b) t3 = √ t1t2(c) 1/t3 = 1/t2 - 1/t1(d) t32 = t12 - t22
Answer» Correct answer is: (b) t₃ = √ t₁t₂

Discussion

707.	The displacement of a body is gives to be proporticonal to the cube of time elapsed. What is the nature of the acceleration of the body ?
Answer» Given, ` A prop t^(3) or S =kt^(3)` v elocity, ` v=(ds)/(dt) =3 kt^(2)` Acceleration, ` a = (dv)/(dt) 6 kt i.e. a prop t` Thus acceleration increases with time.

Discussion

708.	The ceiling of a long hall is 25 m high.What is the maximum horizontal distance that a ball thrown with a speed of `40 ms^(-1)` can cover without hitting the ceiling of the hall ? At what angle and from must it be thrown ?
Answer» Correct Answer - 11 s

Discussion

709.	Which of the following statements are true for a moving body?A. If its speed changes, its velocity must change and it must have some accelerationB. If its velocity changes, its speeds must change and it must have some accelerationC. If its velocity changes, its speed may or may not change, and it must have some accelerationD. If its speed changes but direction of motion does not change, its velocity may remain constant
Answer» Correct Answer - A::C `vec(v)=\|vec(v)\|hat(v), " "[\|vec(v)\| rarr "speed"]` Velocity may charge by changing either speed or direction and by both.

Discussion

710.	In a tennis match Maria Sharapova returns an incoming ball at an angle that is `4°` below the horizontal at a speed of `15 m//s`. The ball was hit at a height of 1.6 m above the ground. The opponent, Sania Mirza, reacts 0.2 s after the ball is hit and runs to the ball and manages to return it just before it hits the ground. Sania runs at a speed of `7.5 m//s` and she had to reach 0.8 m forward, from where she stands, to hit the ball. (a) At what distance Sania was standing from Maria at the time the ball was returned by Maria? Assume that Maria returned the ball directly towards Sania. (b) With what speed did the ball hit the racket of Sania? `[g = 9.8 m//s^(2)]`
Answer» Correct Answer - (a) `12.13 m` (b) `16 m//s`

Discussion

711.	Path of a particle moving in `x-y` plane is `y=3x+4`. At some instant suppose `x`- component of velocity is `1m//s` and it is increasing at a constant rate of `1m//s^(2)`. Then at this instant.A. speed of particle is `sqrt(10)m//s`B. acceleration of particle is `sqrt(10)m//s`C. velocity time graph is parabolaD. acceleration time graph is parabola
Answer» Correct Answer - A::B

Discussion

712.	An object while moving may not have .A. variable speed but constant velocityB. variable velocity but constant speedC. non-zero acceleration but constant speedD. non - zero acceleration but constant velocity
Answer» Correct Answer - A::D A body while movig on a crculara may have veriabl velocity but constant speed or non-zero acceleratio but constant speed.

Discussion

713.	A particle moves along the X-axis as `x=u(t-2s)=at(t-2)^2`.A. the initial velocity of the particle is `u`B. the acceleration of the parabola is `u`C. the acceleration of the particle is `2a`D. at `t=2s` particle is at the origin
Answer» Correct Answer - C::D

Discussion

714.	A particle moves along the X-axis as `x=u(t-2s)=at(t-2)^2`.A. The initial velocity of the particle is uB. The acceleration of the particle is aC. The acceleration of the particle is 2aD. At `t=2s` particle is at the origin
Answer» Correct Answer - C::D `x=u(t-2)+a(t-2)^(2)`…(i) `v=(dx)/(dt)=u+2a(t-2)` Therefore `v(0)=u-4a` `a=(d^(2)x)/(dt^(2))=2a" "` Hence [C] `x(2)=0`[From (i)]. Hence [D]

Discussion

715.	A particle is projected from ground with velocity `40sqrt(2)m//s` at `45^(@)`. At time `t=2s`A. displacement of particle is `100m`B. vertical component of velocity is `20m//s`C. velocity makes an angle `tan^(-1)(2)` with verticalD. particle is at height of `60m` from ground
Answer» Correct Answer - A::B::C::D

Discussion

716.	A particle moves along the X-axis as `x=u(t-2s)=at(t-2)^2`.A. the acceleration of particle is `a`B. the initial velocity of particle is `u`C. at ` t=2 s`, the particle is at originD. the acceleration of particles is `2` `a`
Answer» Correct Answer - B::C::D When ` t= 2 s`, then ` S=0`, so the particle is at the origin. Compairing , the giveen equation with the equation of uniformly accelerated motion of a body in one dimensons , ` S = ut + 1/2 At^2 `, we have Initial velocity =`u`, acceleration ` A = 2 a`.

Discussion

717.	Two particles are projected from ground with same intial velocities at angles `60^(@)` and `30^(@)` (with horizontal). Let `R_(1)` and `R_(2)` be their horizontal ranges, `H_(1)` and `H_(2)` their maximum heights and `T_(1)` and `T_(2)` are the time of flights.ThenA. `(H_(1))/(R_(1))gt(H_(2))/(R_(2))`B. `(H_(1))/(R_(1))lt(H_(2))/(R_(2))`C. `(H_(1))/(T_(1))gt(H_(2))/(T_(2))`D. `(H_(1))/(T_(1))gt(H_(2))/(T_(2))`
Answer» Correct Answer - A::C

Discussion

718.	If T is the total time of flight, h is the maximum height and R is the range for horizontal motion, the x and y coordinates of projectile motion and time t are related asA. `y=4h(t/T)(1-t/T)`B. `y=4h(x/R)(1-x/R)`C. `y=4h(T/t)(1-T/t)`D. `y=4h(R/x)(1-R/x)`
Answer» Correct Answer - A::B::D

Discussion

719.	Find the magnitude of the centripetal acceleration of a particle on the tip of a blade, ` 0 .30 ` metre in diameter, rotating at `1200` revolution per minute.
Answer» Here, ` r=0.3//2 =0.15 m, ` v(12000/(60) =200 ros` Centripetal acceleration, `a_cromega^2 =r 4 pi^2 v^2` ` =0.15 xx 4xx (22)/7)^2 xx (20)^2 = 2370.6 ms^(-2)`.

Discussion

720.	The last soldier of an `80m` long marching troops runs from the end to its front, and then it runs back to the end with the same speed. During this, the marching troop covers a distance of `150m`. The distance covered by the soldier isA. `310m`B. `250m`C. `230m`D. `160m`
Answer» Correct Answer - B

Discussion

721.	A projectile is fired at an angle of `30^(@)` to the horizontal such that the vertical component of its initial velocity is `80m//s`. Its time of fight is `T`. Its velocity at `t=T//4` has a magnitude of nearly.A. `200m//s`B. `300m//s`C. `140m//s`D. `100m//s`
Answer» Correct Answer - C

Discussion

722.	From `v-t` graph shown in figure. We can draw the following conclusions A. between `t=1s` to `t=2s` speed of particle is decreasingB. between `t=2s` to `t=3s` speed of particle is increasingC. between `t=5s` to `t=6` acceleration of particle is negativeD. between `t=0` to `t=4s` particle changes its directioin of motion twice
Answer» Correct Answer - C::D

Discussion

723.	A particle `P` is projected upwards with `80m//s`. One second later another particle `Q` is projected with initial velocity `70m//s`. Before either of the particle srikes the ground `(g=10m//s^(2))`A. both particle are at rest with respect to each otherB. after `2s` distance between the particles is `75m`C. when particle `P` is at highest point, particle `Q` is moving downwardsD. when particle `P` is at highest point, particle `Q` is moving upwards
Answer» Correct Answer - A::B

Discussion

724.	Two balls of equal masses are thrown upwards, along the same vertical direction at an interval of 2 seconds, with the same initial velocity of `40m//s`. Then these collide at a height of (Take `g=10m//s^(2)`).A. `120m`B. `75m`C. `200m`D. `45m`
Answer» Correct Answer - B

Discussion

725.	Two balls of equal masses are thrown upwards, along the same vertical direction at an interval of 2 seconds, with the same initial velocity of `40m//s`. Then these collide at a height of (Take `g=10m//s^(2)`).A. 50 mB. 75 mC. 100 mD. 125 m
Answer» Correct Answer - D `s_1=s_2` `:. (40)t-1/2xx10xxt^2=(40)(t-2)` `-1/2xx10xx(t-2)^2` Solving this equation, we get `t=5s` Then `s_1=(40)(5)-1/2xx10xx(5)^2` `=75m`

Discussion

726.	A cart moves with a constant speed along a horizontal circular path. From the cart, a particle is thrown up vertically with respect to the cart. (a) The particle will land somewhere on the circular path. (b) The particle will land outside the circular path. (c) The particle will follow an elliptical path. (d) The particle will follow a parabolic path.
Answer» Correct Answer is: (b, d)

Discussion

727.	A cart moves with a constant speed along a horizontal circular path. From the cart, a particle is thrown up vertically with respect to the cart.A. The particle will land somewhere on the circular pathB.C. The particle will follow an elliptical pathD. The particle will follow a parabolic path
Answer» Correct Answer - B::D

Discussion

728.	Two eual forces act at a point. The squard of their resultant is ` 3 ` times their product, Find the angle between them.
Answer» Here, ` A = B=x` and ` R^2 = 3 AB= 3 x x = 3 x^2` ` cos theta (R^2 - A^2 - B^2 ) /( 2 AB) = ( 3 x^2 - x^2 - x^2)/( 2 xx x xx x)` `= 1/2 = cos 60^@` ` theta = 60^@`.

Discussion

729.	A projectile is thorwn with a velocity of `50ms^(-1)` at an angle of `53^(@)` with the horizontal Detemrine the instants at which the projectile is at the same heightA. `t = 1s` and `t = 7s`B. `t = 3s` and `t = 5s`C. `t = 2s` and `t = 6s`D. all the above
Answer» Correct Answer - D For same height `y = u_(y)t +(1)/(2) a_(y)t^(2) rArr y = u_(y)t - (1)/(2) "gt"^(2)` `"gt"^(2) - 2u_(y) +2h = 0 rArr t = (2u_(y)+-sqrt(4u_(y)^(2)-8gh))/(2g)` `t_(1) = `(u_(y)-sqrt(u_(y)^(2)-2gh))/(g), t_(2) =(u_(y)-sqrt(u_(y)^(2)-2gh))/(g)` `t_(1) +t_(2) =(2u_(y))/(g) = (2 xx 50 xx sin 53^(@))/(10) = 8sec`.

Discussion

730.	A particle starts with an initial velocity `2.5 m//s` along the posiive x-direction and it accelerates uniformly at the rate `0.50 m//s^2.` (a) Find the distance travelled by it in the first two seconds (b) How much time does it take to reach the velocity `7.5 m//s` ? (c) How much distance will it cover in reaching the velocity `7.5 m//s`?
Answer» Correct Answer - A::B::C (a) `S=ut+1/2 at^2` `=(2.5xx2)+1/2(0.5)(2)^2` `=6m=distance also` (b) `v=u+at` `7.5=2.5+0.5xxt rArr t=10s` (c) `v^2=u^2+2as` `:. (7.5)^2=(2.5)^2+(2)(0.5)s` `rArr s=50 cm`=distance also

Discussion

731.	find the values of ` T_1` and ` T_2` for the sustem shown in Fig. 2 (c ) . 76 , ` g= 10 ms^2`. .
Answer» ` T_2 sin 50^@ = mg = 60 xx 10 600 N` or 1 T-2 = 600/0.766 = 783 N` ` T-1 = T_2 cos 50^@ = 7 83 xx 0.643 = 503 N`.

Discussion

732.	A body is thrown at an angle `theta_0` with the horizontal such that it attains a speed equal to `(sqrt((2)/(3))` times the speed of projection when the body is at half of its maximum height. Find the angle `theta_0`.
Answer» At any height `y`, the speed of the projectile is `v = sqrt(v_0^2 - 2 g y))`, where `y = (y_(max))/(2)` Sunstituting `y_(max) = (v_0^2 sin^2 theta_0)/(2 g)`, we have `v = v_0 (sqrt(1 - (sin^2 theta_0)/(2)))` Since `v = sqrt((2)/(3)) v_0` (given), we have `1 -(sin^2 theta_0)/(2) = (2)/(3)` This given `sin theta_0 = sqrt((2)/(3))`. Hence, `theta_0 = sin^-1 (sqrt((2)/(3)))`.

Discussion

733.	A particle is moveint along the x-axis whose instantaneous speed is given by `v^(2)=108-9x^(2)`. The acceleration of the particle is.A. `-9x ms^(-2)`B. `-18x ms^(-2)`C. `(-9x)/(2) ms^(-2)`D. None of there
Answer» Correct Answer - A `v^(2)=18-9x^(2)` `a=(dv)/(dt)=(dv)/(dx).(dx)/(dt) =(d(sqrt108-9x^(2)))/(dx).(dx)/(dt)` `a=(1(-18x))/(2sqrt(108-9x^(2))).sqrt(108-9x^(2)) =-9x^(-2)` Alternative: Differentiating w.r.t. we get `2v(dv)/(dx)=-18x` `(vdx)/(dx)=-9x rArr a=-9x because (v(dv)/(dx)=a)`.

Discussion

734.	An airplane is moving with velocity `v_0` horizontally at a height h. If a projectile is fired at the instant when the plane is overhead, what must be the angle of projection and minimum velocity of projection in order that it may hit the airplane ?
Answer» Correct Answer - `u_("min")=sqrt(v_0^2+2gh)` and `alpha=cos^(-1) v_0/(sqrt(v_0^2+2gh))`

Discussion

735.	There are two hills A and B and a car is travelling towards hill A along the line joining the two hills. Car is travelling at a constant speed u. There is a wind blowing at speed u in the direction of motion of the car (i.e., from hill B to A). When the car is at a distance `x_(1)` from A and `x_(2)` from B it sounds horn (for very short interval). Driver hears the echo of horn from both the hills at the same time. Find the ratio `(x_(1))/(x_(2))` taking speed of sound in still air to ve V.
Answer» Correct Answer - `(x_(1))/(x_(2)) = (v+u)/(v-u)`

Discussion

736.	Find the angle between the vectors ` vec A = hat I + 2 hat j- hat k` and ` vec B =- hat I + hatj - 2 hat k`.
Answer» ` vec A= hat i+ 3 hat j- hat k` ` :. \| vec A\| =sqrt ((1)^2 + (2)^2+ (-1) ^2) = sqrt6` ` vec B=- hat I + hat j- 2 hat k` :. ` \| vec B\|= sqrt( (-1) ^2 + (1) ^2 + (-2)^2 ) = sqrt 6` ` vec A.vec B = ( hat i+ 2 hat j- hat k).( -hati+ hat j-2 jat k)` =- 1 + 2 + 2 =3` ` cos theta = (vec A. vec B)/(\| vec A\|\|vec B\|) = 3/( sqrt 6 sqrt 6) = 1/2 = cos 60^^@~ or ` theta 60^@` .

Discussion

737.	If the magnitude of two vectors are ` 3 ` and `6` and ther scalar product is `9`, find the angle between the two vectors.
Answer» Here, ` A = 3 , B= 6 , vec A. vec B=9` ` cos theta (vec A.vec B)/(AB) = 9/(3 xx 6) = 1/2 = cos 60^2 ` or theta = 60^@`.

Discussion

738.	The magnitude of the resultant of two vectors is ……………………………………..When they act in the same direction and is ……………………………………….. When they act in opposite direction.
Answer» Correct Answer - maximum , minimum

Discussion

739.	A body moves aling a circular track of radius (r ). It starts from one end of a disameter, moves along the circuar track anf completes one and a half revolutins. The ration of distance travelled by the body to tis displacement is.
Answer» Correct Answer - (d)

Discussion

740.	Those vectors which are having equal or unequal magnitudes and are acting aling the parallel straight lines are called ……………………………. .
Answer» Correct Answer - cllinear vectors

Discussion

741.	When a vector ` vec A` is multiplied by a scalar (S), it becomes a vector ………………………whose unit is ……………………from the unit of vector `vec A`.
Answer» ` S vec A` , different

Discussion

742.	Forces proportional to AB, BC & 2 CA act along the sides of triangle ABC in order, their resultant represented in magnitude and direction as:A. CAB. ACC. BCD. CB
Answer» Correct Answer - A

Discussion

743.	The distance travelled by a particle moving along a st. line is given by `x=4 t + 5t^(2) +6^(3)` mette. Find (i) the initial velcity of the particle (ii) the velcoty at the end of `4 s` and (iii) the acceleration of the particle at the end of `5 second.
Answer» (i) velocity, `v=(dx)/(dt) =d/(dt) (4t +5t^(2) +6t^(3))` `=4 +10 t + 18 (0) ^(2) =4 m// s` (ii) Velocity of the particle when `t=4 s` `v=4 +40 xx 4 + xx (4)^(2) =332 m//s` `v=4 + 10 xx 4 18 xx (40 ^(2) =332 m//s` (iii) Accelration, `a=9dt)/(dt) d/(dt) (4+10 t+18) `= 0 + 10 + 36 t =10 36 t` When `t=5 s`, a=10 + 36 xx 5=190 ms^(-2)` .

Discussion

744.	The velocity of a particle is given by `v=u_(0) + g t+ 1/2 at^(2)`. If its position is `x =0` at `t=0`, then what is its displacement after `t=1 s` ?
Answer» Here, `v=(dx) /(dt) v_(0) + gt +1/2 ft^(2)` or `dx =(v_(0) + gt +1/2 ft^(2) ) dt` Integration it the conditions of motion, i.e. as (t) changes 0 to `1 s`, (x) changes `o to x`, we get `int _(0)^(x) dx =int _(0)^(1) (v_(0) + gt + 1/2 ft^(2) ) dt` `x=v_(0) (t) _(0)^(1) +g (t^(2)/2)_(0)^(1) + 1/2 (t^(3)/3)_(0)^(1)` `=v_(0) + g/2 +f/6`.

Discussion

745.	A unti bvlurd og ` a and b` are ` 0.6 and 0.8` respectively find the value od (c ),
Answer» Fiven, ` vec A =a hat I +vec b hat j+ c hat k =0. 6 hat j + c hat k`. Hence, ` \| vec A\| =1 =sqrt( (0.6)^(2) + (0.8) ^(2) + c^(2))` or ` 1 =0.36 + 0 64 + c^(2) or c^(@) =c or c=0`.

Discussion

746.	A vector `vec A` is exparessed as ` vec A = A_(x) hat I + A_(y) hat j` where ` vec A and vec B` are its components along x-axis and y-axis respectively. If vector `vec A` makes an angle theat with x-axis, then theta is given by which expressinon ?
Answer» Hence, ` A_(x) = A cos theta A_(y) =A sin theta` :.` A_(y)/A_(x) =)A sin theta)/(A cos theta) = tan tehta or theta = tan ^(-1) (A_(y)/A_(x))`.

Discussion

747.	The system shown in figure starts from rest, and each block moves with a constant acceleration. If the relative acceleration of block C with respect to block B is `6m//s^(2)` upward and the relative acceleration of block D with respect to block C after 3s from starts.
Answer» Correct Answer - `[57m//s darr]`

Discussion

748.	Acceleration is defined as the rate of change of velocity. Suppose we call the rate of change of acceleration as ` SLAP ?. (i) What is the unit of SLAP` . (ii) How can we calculate instantaneous SLAP ?
Answer» Given, SLAP ` (change in acceleration )/(time taken ) (i) The (SI) unti of ` SLAP is ` (m//s^(2))/s =m s^(-3)` (ii) Instantaneous ` SLAP` is the value of ` SLAP` at the given instant, which is the first derivative of accelertion at the given instant and can be written as Instantaneous ` SLAP= (Limit)/(Delta t rarr0) (Delta vec a)/(Delta t)) = (d vec a)/d t).

Discussion

749.	The acceleration time graph of a particle is shown in the figure-1.140. What is the velocity of particle at t=8s, if initial velocity of particle is 3 m/s? (Assume motion is 1 dimension) : A. 4 m/sB. 5 m/sC. 6 m/sD. 7 m/s
Answer» Correct Answer - D

Discussion

750.	A ball is thrown vertically upwards from the ground. If `T_1` and `T_2` are the respective time taken in going up and coming down, and the air resistance is not ignored, thenA. ` t_1 =t_2`B. ` t_1lt ty_2`C. `t_1 gtt_2`D. `t_2-ltt_1`
Answer» Correct Answer - B Let (a) be the retarding acceleration acting on body due to air friction. Therefore, effective value of (g) for upward motion, ` g_1 = g+a` and for downward motion, ` g-2 = (g-a) `. We know that, ` t= sqrt ( 2h)/(g_(effective)` or ` t prop 1/(sqrt (g_(effective )1 :. ` t_1 /t_2 = sqrt g_2/g_1` = sqrt ((g-1)/(g+a))) lt 1 ` or ` t_1 lt t_2`.

Discussion

Explore topic-wise InterviewSolutions in .

When a body is thrown up in a lift with a velocity `u` relative to the lift, the time of flight is found to be `t`. The acceleration with which the lift is moving up isA. `(u-"gt")/(t)`B. `(u+"gt")/(t)`C. `(2u-"gt")/(t)`D. `(u)/(t)-2g`

A balloon starts rising from the ground with an acceleration of 1.25 m/s 2. After 8 s, a stone is released from the balloon. The stone will(a) cover a distance of 40 m (b) have a displacement of 50 m (c) reach the ground in 4 s (d) begin to move down after being released

The displacement of a body is gives to be proporticonal to the cube of time elapsed. What is the nature of the acceleration of the body ?

The ceiling of a long hall is 25 m high.What is the maximum horizontal distance that a ball thrown with a speed of `40 ms^(-1)` can cover without hitting the ceiling of the hall ? At what angle and from must it be thrown ?

An object while moving may not have .A. variable speed but constant velocityB. variable velocity but constant speedC. non-zero acceleration but constant speedD. non - zero acceleration but constant velocity

A particle moves along the X-axis as `x=u(t-2s)=at(t-2)^2`.A. the initial velocity of the particle is `u`B. the acceleration of the parabola is `u`C. the acceleration of the particle is `2a`D. at `t=2s` particle is at the origin

A particle moves along the X-axis as `x=u(t-2s)=at(t-2)^2`.A. The initial velocity of the particle is uB. The acceleration of the particle is aC. The acceleration of the particle is 2aD. At `t=2s` particle is at the origin

A particle is projected from ground with velocity `40sqrt(2)m//s` at `45^(@)`. At time `t=2s`A. displacement of particle is `100m`B. vertical component of velocity is `20m//s`C. velocity makes an angle `tan^(-1)(2)` with verticalD. particle is at height of `60m` from ground

A particle moves along the X-axis as `x=u(t-2s)=at(t-2)^2`.A. the acceleration of particle is `a`B. the initial velocity of particle is `u`C. at ` t=2 s`, the particle is at originD. the acceleration of particles is `2` `a`

If T is the total time of flight, h is the maximum height and R is the range for horizontal motion, the x and y coordinates of projectile motion and time t are related asA. `y=4h(t/T)(1-t/T)`B. `y=4h(x/R)(1-x/R)`C. `y=4h(T/t)(1-T/t)`D. `y=4h(R/x)(1-R/x)`

Find the magnitude of the centripetal acceleration of a particle on the tip of a blade, ` 0 .30 ` metre in diameter, rotating at `1200` revolution per minute.

The last soldier of an `80m` long marching troops runs from the end to its front, and then it runs back to the end with the same speed. During this, the marching troop covers a distance of `150m`. The distance covered by the soldier isA. `310m`B. `250m`C. `230m`D. `160m`

A projectile is fired at an angle of `30^(@)` to the horizontal such that the vertical component of its initial velocity is `80m//s`. Its time of fight is `T`. Its velocity at `t=T//4` has a magnitude of nearly.A. `200m//s`B. `300m//s`C. `140m//s`D. `100m//s`

Two balls of equal masses are thrown upwards, along the same vertical direction at an interval of 2 seconds, with the same initial velocity of `40m//s`. Then these collide at a height of (Take `g=10m//s^(2)`).A. `120m`B. `75m`C. `200m`D. `45m`

Two balls of equal masses are thrown upwards, along the same vertical direction at an interval of 2 seconds, with the same initial velocity of `40m//s`. Then these collide at a height of (Take `g=10m//s^(2)`).A. 50 mB. 75 mC. 100 mD. 125 m

A cart moves with a constant speed along a horizontal circular path. From the cart, a particle is thrown up vertically with respect to the cart.A. The particle will land somewhere on the circular pathB.C. The particle will follow an elliptical pathD. The particle will follow a parabolic path

Two eual forces act at a point. The squard of their resultant is ` 3 ` times their product, Find the angle between them.

A projectile is thorwn with a velocity of `50ms^(-1)` at an angle of `53^(@)` with the horizontal Detemrine the instants at which the projectile is at the same heightA. `t = 1s` and `t = 7s`B. `t = 3s` and `t = 5s`C. `t = 2s` and `t = 6s`D. all the above

find the values of ` T_1` and ` T_2` for the sustem shown in Fig. 2 (c ) . 76 , ` g= 10 ms^2`. .

A body is thrown at an angle `theta_0` with the horizontal such that it attains a speed equal to `(sqrt((2)/(3))` times the speed of projection when the body is at half of its maximum height. Find the angle `theta_0`.

A particle is moveint along the x-axis whose instantaneous speed is given by `v^(2)=108-9x^(2)`. The acceleration of the particle is.A. `-9x ms^(-2)`B. `-18x ms^(-2)`C. `(-9x)/(2) ms^(-2)`D. None of there

An airplane is moving with velocity `v_0` horizontally at a height h. If a projectile is fired at the instant when the plane is overhead, what must be the angle of projection and minimum velocity of projection in order that it may hit the airplane ?

Find the angle between the vectors ` vec A = hat I + 2 hat j- hat k` and ` vec B =- hat I + hatj - 2 hat k`.

If the magnitude of two vectors are ` 3 ` and `6` and ther scalar product is `9`, find the angle between the two vectors.

The magnitude of the resultant of two vectors is ……………………………………..When they act in the same direction and is ……………………………………….. When they act in opposite direction.

A body moves aling a circular track of radius (r ). It starts from one end of a disameter, moves along the circuar track anf completes one and a half revolutins. The ration of distance travelled by the body to tis displacement is.

Those vectors which are having equal or unequal magnitudes and are acting aling the parallel straight lines are called ……………………………. .

When a vector ` vec A` is multiplied by a scalar (S), it becomes a vector ………………………whose unit is ……………………from the unit of vector `vec A`.

Forces proportional to AB, BC & 2 CA act along the sides of triangle ABC in order, their resultant represented in magnitude and direction as:A. CAB. ACC. BCD. CB

The distance travelled by a particle moving along a st. line is given by `x=4 t + 5t^(2) +6^(3)` mette. Find (i) the initial velcity of the particle (ii) the velcoty at the end of `4 s` and (iii) the acceleration of the particle at the end of `5 second.

The velocity of a particle is given by `v=u_(0) + g t+ 1/2 at^(2)`. If its position is `x =0` at `t=0`, then what is its displacement after `t=1 s` ?

A unti bvlurd og ` a and b` are ` 0.6 and 0.8` respectively find the value od (c ),

A vector `vec A` is exparessed as ` vec A = A_(x) hat I + A_(y) hat j` where ` vec A and vec B` are its components along x-axis and y-axis respectively. If vector `vec A` makes an angle theat with x-axis, then theta is given by which expressinon ?

The system shown in figure starts from rest, and each block moves with a constant acceleration. If the relative acceleration of block C with respect to block B is `6m//s^(2)` upward and the relative acceleration of block D with respect to block C after 3s from starts.

Acceleration is defined as the rate of change of velocity. Suppose we call the rate of change of acceleration as ` SLAP ?. (i) What is the unit of SLAP` . (ii) How can we calculate instantaneous SLAP ?

The acceleration time graph of a particle is shown in the figure-1.140. What is the velocity of particle at t=8s, if initial velocity of particle is 3 m/s? (Assume motion is 1 dimension) : A. 4 m/sB. 5 m/sC. 6 m/sD. 7 m/s

A ball is thrown vertically upwards from the ground. If `T_1` and `T_2` are the respective time taken in going up and coming down, and the air resistance is not ignored, thenA. ` t_1 =t_2`B. ` t_1lt ty_2`C. `t_1 gtt_2`D. `t_2-ltt_1`