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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 301. |
If a particle executes uniform circular motion in the xy plane in clockwise direction, then the angular velocity is in (a) +y direction (b) +z direction (c) -z direction (d) -x direction |
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Answer» Correct answer is (c) -z direction |
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| 302. |
A particle is in clockwise uniform circular motion the direction of its acceleration is radially inward. If sense of rotation or particle is anticlockwise then what is the direction of its acceleration? |
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Answer» Radial in ward. |
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| 303. |
What is the average value of acceleration vector in uniform circular motion over one cycle? |
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Answer» Null vector. |
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| 304. |
Is the acceleration of a particle in circular motion not always towards the center. Explain. |
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Answer» No acceleration is towards the center only in case of uniform circular motion. |
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| 305. |
Is the acceleration of a particle in circular motion not always towards the centre. Explain. |
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Answer» No acceleration is towards the centre only in case of uniform circular motion. |
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| 306. |
STATEMENT -1 : For an observer looking out through the window of a fast moving train , the nearby objects appear to move in the opposite direction to the train , while the distant objects appear to be stationary . STATEMENT - 2 : If the observer and the object are moving at velocities ` vec v_(1)` and `vec v_(2)` respecttively with refrence to a laboratory frame , the velocity of the object with respect to a laboratory frame , the velocity of the object with respect to the observer is `vecv_(2) - vecv(1) ` . (a) Statement -1 is True, statement -2 is true , statement -2 is a correct explanation for statement -1 (b) Statement 1 is True , Statement -2 is True , statement -2 is NOT a correct explanation for statement -1 (c) Statement - 1 is True , Statement -2 is False (d) Statement -1 is False, Statement -2 is TrueA. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-2C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - B | |
| 307. |
A person sitting in a train moving at constant velocity throws a ball vertically upwards. How will the ball appear to move to an observer. (i) Sitting inside the train (ii) Standing outside the train |
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Answer» (i) Vertical straight line motion (ii) Parabolic path. |
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| 308. |
A ball thrown vertically upwards with a speed of 19.6 ms-1 from the top of a tower returns to the earth in 6s. Find the height of the tower, (g = 9.8 m/s) |
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Answer» s = ut + \(\frac{1}{2}\)at2 -h = 19. 6 x 6 + x \(\frac{1}{2}\)(-9.8) x (6)2 h = 58.8 m. |
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| 309. |
A person sitting in a train moving at constant velocity throws a ball vertically upwards. How will the ball appear to move to an observer. Sitting inside the train Standing outside the train |
Answer»
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| 310. |
A rifle shoots a bullet with a muzzle velocity of `400 m s^-1` at a small target `400 m` away. The height above the target at which the bullet must be aimed to hit the target is `(g = 10 m s^-2)`.A. 1 mB. 5 mC. 10 mD. 0.5 m |
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Answer» Correct Answer - B (b) Time taken by bullet to reach the target is (Distance)/(Velocity) = (Distance)/(u cos theta) [As theta is very small, `cos theta = 1`] Time = `(Distance)/(u) = (400)/(400) = 1 s` Vertical deflection of bullet is =`(1)/(2) "gt"^2 = (1)/(2) xx 10 xx (1)^2 = 5 m`. |
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| 311. |
A bulley with muzzle velocity `100 m s^-1` is to be shot at a target `30 m` away in the same horizontal line. How high above the target must the rifle be aimed so that the bullet will hit the target ? . |
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Answer» Horizontal range of bullet is `30 m`. Using range formula, `R = (u^2 sin 2 theta)/(g) = 30` or `sin 2 theta = (30 xx 10)/((100)^2) or sin 2 theta = 0.03` For small `theta, sin theta ~~ theta ~= tan theta, i.e., 2 theta = 0.03` Therefore, `theta = 0.015 rad` In `Delta OAP, tan theta = (AP)/(OP) rArr AP = OP tan theta` The rifle must be aimed at an angle `theta = 0.015` above horizontal. Height to be aimed `= 30 tan theta ~~ 30(theta) = 30 xx 0.015 = 45 cm`. |
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| 312. |
A man is riding on a flat car travelling with a constant speed of `10m//s`.He wishes to throw a ball through a stationary hoop `15 m` above the height of his hands in such a manner that the ball will move horizontally as it passes through the hoop. He throws the ball with a speed of `12.5 m//s` w.r.t himself. At what horizontal distance in front of the hoop must he release the ball? |
| Answer» You can attempt the problem in a reference frame fixed to the car or in the reference frame fixed to the ground. Fundamentally, these two reference frames are equivalent in all respects, but in the reference frame fixed to the car, the calculations are slightly easier. | |
| 313. |
The greatest height to which a man can throw a stone is h. What will be the greatest distance upto which he can throw the stone? |
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Answer» Maximum height: H = \(\frac{u^2sin^2θ}{g}\) ⇒ Hmax \(\frac{u^2}{2g}\) = h(at θ = 90) Maximum range Rmax = \(\frac{u^2}{2g}\) = 2h |
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| 314. |
A boy can throw a stone up to a maximum height of `10 m`. The maximum horizontal distance that the boy can throw the same stone up to will be :A. 20 mB. `20sqrt(2) m`C. `10 m`D. `10 sqrt(2) m` |
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Answer» Correct Answer - 1 `H_("max")=u^(2)/(2g)=10 m` and `R_("max")=u^(2)/g=20 m` |
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| 315. |
A man can throw a stone with initial speed of `10m//s`. Find the maximum horizontal distance to which he can throw the stone in a room of height h m for: (i) `h = 2m` and (ii) `h = 4m`. |
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Answer» Correct Answer - (i) `4sqrt(6)` (ii) `10m` for maximum horizontal rnage `theta = 45^(@)` for `theta = 45^(@)` max. height achieved `= (u^(2)sin^(2) theta)/(2g) = (10^(2)sin6(2)45^(@))/(2xx10) = 2.5m` (i) Since height of room `= 2m` which is less than height found above `h = (u^(2)sin^(2)theta)/(2g) rArr 2 = (10^(2)sin^(2) theta)/(2xx10)` `rArr sin theta = sqrt((2)/(3)) & cos theta = sqrt((3)/(5))` `R = (u^(2)sin 2 theta)/(g) rArr R = (u^(2)2sin theta cos theta)/(g) = 4sqrt(6)m` (ii) `theta = 45^(@)` so `R = (u^(2)sin 2 theta)/(g)` `rArr R = (10^(2) xx sin 90^(@))/(g) rArr R = 10m` |
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| 316. |
The current velocity of river grows in proportion to the distance from its bank and reaches the maximum value `v_0` in the middle. Near the banks the velocity is zero. A boat is moving along the river in such a manner that the boatman rows his boat always perpendicular to the current. The speed of the boat in still water is u. Find the distance through which the boat crossing the river will be carried away by the current, if the width of the river is c. Also determine the trajectory of the boat. |
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Answer» Correct Answer - `[(cv)/(4u), y^(2) = (u c x)/(v_(0))]` |
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| 317. |
The velocity of a particle moving in the positive direction of `x`-axis varies as `v = alpha sqrt(x)` where `alpha` is positive constant. Assuming that at the moment `t = 0`, the particle was located at `x = 0`, find (i) the time dependance of the velocity and the acceleration of the particle and (ii) the mean velocity of the particle averaged over the time that the particle takes to cover first `s` meters of the path. |
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Answer» `a=(dv)/(dt) = alpha (d)/(dt) sqrt(x) = alpha. (1)/(2)x^(-1//2) (dx)/(dt)` `= alpha.(1)/(2sqrt(x)).alphasqrt(x) rArr a = (alpha^(2))/(2)` |
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| 318. |
The velocity of a particle moving in the positive direction of the x axis varies as `v=alphasqrtx`, where `alpha` is a positive constant. Assuming that at the moment `t=0` the particle was located at the point `x=0`, find: (a) the time dependence of the velocity and the acceleration of the particle, (b) the mean velocity of the particle averaged over the time that the particle takes to cover the first s metres of the path. |
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Answer» (i) Given that `v - alpha sqrt(x)` or `(dx)/(dt) = alpha sqrt(x)` `:. (dx)/(sqrt(x)) = alpha dt` or `int_(0)^(x) (dx)/(sqrt(x)) = int_(0)^(t) alpha dt` Hence `2sqrt(x) = alpha t`or `x = (alpha^(2)t^(2)//4)` Velocity `(dx)/(dt) = (1)/(2)alpha^(2)t` and Acceleration `(d^(2)x)/(dt^(2)) = (1)/(2)alpha^(2)` (ii) Time taken to cover first `s` meters `s = (alpha^(2)t^(2))/(4) rArr t^(2) = (4s)/(alpha^(2)) rArr t = (2sqrt(s))/(alpha)` `barv = ("total distance")/("total time") = (sa)/(2sqrt(s)) rArr barv = (1)/(2) sqrt(s) alpha` |
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| 319. |
A person, intending to cross a river by the shortest path, starts at an angle `prop` with the downstream. If the speed of the person be less than that of water current, show that `prop` must be obtuse. |
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Answer» Let `v and u` be the velocities of the person and the water current, and `d` the width of the river. If `t` be the time taken by the person to cross the river, then `t = (d)/(v sin prop)` and distance carried downstream in this time =`(d)/(v sin prop) [u + v cos theta] = x (say)` Now, eliminating `prop`, we get `t^2 (u^2 - v^2) - 2 xut + (x^2 + d^2) = 0` Clearly `t` will have real solutions, if `"discriminant ge 0"` `rArr 4 x^2 u^2 ge 4 (u^2 - v^2)(x^2 + d^2)` `rArr x ge (d)/(v) (sqrt(u^2 - v^2))` Distance swayed away by the water current (x) has to be minimum, for the shortest path. Again `t = (d)/(v sin prop) and x = (d)/(v) sqrt(u^2 - v^2)=(d)/(v sin prop) (u + v cos prop)` `rArr (u^2 - v^2) sin^2 prop = (u + v cos prop)^2` `rArr u^2 (1 - sin^2 prop) + 2 u v cos prop + v^2 = 0` `rArr u^2 cos^2 prop + 2 uv cos prop + v^2 = 0` `rArr (u cos prop + v)^2 = 0 rArr cos prop = -(v)/(u)`. |
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| 320. |
A launch travels across a river from a point `A` to a point `B` of the opposite bank along the line `AB` forming angle `prop` with the bank. The flag on the mast of the launch makes an angle `beta` with its direction of motion. Determine the speed of the launch w.r.t. the bank. The velocity of wind is `u` perpendicular to the stream. . |
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Answer» Correct Answer - `[u sin(alpha+beta - pi//2)sin beta]` |
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| 321. |
A particle moves for total time T sec in a straight line in three consecutive parts such that its acceleration during the first, second and third parts is in the ratio 1:2:7. The distances covered in the first and the third parts are a and b meters while the time taken for each of the is tseconds. Find the average velocity of the particle during the second part. |
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Answer» Correct Answer - `[(a+b)/(2t) + 3((a-b)/(4t))]` |
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| 322. |
A point mass starts moving in a straight line with a constant acceleration a. At a time `t_(1)` a fter the beginning of motion, the acceleration changes sign, remaining the same in magnitude. Determine the time t from the beginning of motion in which the point mass returns to the initial position. |
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Answer» Correct Answer - `[t_(1)(2+sqrt(2))]` |
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| 323. |
Which of the following velocity-time graphs shows a realistic situation for a body in motion ?A. B. C. D. |
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Answer» Correct Answer - B Two value of velocity (at the same instant) is not possible. |
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| 324. |
A body is thrown vertically upwards from `A`. The top of a tower . It reaches the fround in time `t_(1)`. It it is thrown verically downwards from `A` with the same speed it reaches the ground in time `t_(2)`, If it is allowed to fall freely from `A`. then the time it takes to reach the ground. . |
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Answer» Suppose the body be projected vertically upwards from `A` with a speed `u_(0)`. Using equation `s=ut+((1)/(2))a t^(2)` For first case `-h=u_(0)t_(1)-((1)/(2))g t_(1)^(2)` (i) For second case, `-h=-u_(0)t_(2)-((1)/(2))g t_(2)^(2)` (ii) From (i)-(ii) `rArr 0=u_(0)(t_(2)+t_(1))+((1)/(2))g (t_(2)^(2)-t_(12)^(2))` or `u_(0)=((1)/(2)) g (t_(1)-t_(2))` (iii) Put the value of `u_(0)` in (ii), we get `-h=-((1)/(2))g (t_(1)-t_(2))t_(2)-((1)/(2))g t_(2)^(2)` `rArr h=(1)/(2) g t_(1)t_(2)` For third case, `u=0, t=`? `-h=0 xxt-((1)/(2))g t^(2)` or `h=((1)/(2))g t^(2)` Combining (iv) and (v), we get `(1)/(2) g t^(2)=(1)/(2) g t_(1)t_(2)` or `t=sqrtt_(1)t_(2)`. |
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| 325. |
A body is thrown vertically upwards from `A`. The top of a tower . It reaches the ground in time `t_(1)`. It it is thrown vertically downwards from `A` with the same speed it reaches the ground in time `t_(2)`, If it is allowed to fall freely from `A`. then the time it takes to reach the ground. .A. `t=(t_(1)+(t_(2))/(2)`B. `t=(t_(1)t_(2))/(2)`C. `t=sqrt(t_(1)t_(2)`D. `t=sqrt((1_(1)/(t_(2))` |
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Answer» Correct Answer - C Suppose the body be projected vertically upwards from `A` with s speed `u_(0)`, `Uisng equation `s=ut + ((1)/(2)) at^(2)`, we get For first case: `-h=u_(0)t_(10 - ((1)/(2)) g t_(1)^(2)` For second acse:`-h =- u_(0)t_(2) -((1)/(2)) g t_(2)^(2)` (i) -(ii) `rArr 0=u_(0) (t_(2) +t_(1)) + ((1)/(2)) g (t_(2)^(2)-t_(1)^(2))` `rArr u_(0) =((1)/(2))g (t_(1)-t_(2))` Putting the value of `u_(0)` in (ii), we get `-h=- ((1)/(2)) g (t_(1)-t_(2)) -((1)/(2)) g t_(2)^(2)` `rArr h=(1)/(2) g t_(1)t_(2)` (iv) For thir case: `u =0, t=`? `-h=0xx t-((1)/(2 0 g t^(2)` or `h=((1)/(2)) g t^(2)` (v) Combining Eq. (iv) and Eq. (v), we get `(1)/(2) g t^(2) =(1)/(2) g t_(1))t_(2)` or `tsqrt t_(1) t_(2)`. |
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| 326. |
Two persons pull each other through a massless rope in `tug of war` game. Who will win?A. exerts greater force on the ropeB. exerts greater force on the groundC. exerts a force on the rope which is greater than the tension in the ropeD. makes a smaller angle with the vertical |
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Answer» Correct Answer - B |
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| 327. |
Ball `1` is released from the top of a smooth inclined plane, the at the same instant ball `2` is projected from the foot of the plance with such a velocity that they meet halfway up the incline. Determine: a.the velocity with which balls are projected and b. the velocity of each ball when they meet. . |
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Answer» For ball 1: `u_(1)=0, a=g sun theta=g ((h)/(l))` and `s=l//2` Using `v^(2)=u^(2)+2as rArr v^(2)=0+2g ((h)/(l)).((1)/(2))` `rArr v=sqrtgh` Again using `v=u+at rArr v=0+g ((h)/(l)).t` `sqrtgh=g ((h)/(l))t rArr t=(l)/(sqrtgh)` For nall 2: `u=u_(2), v=v_(2)` and `a=-g sin theta=-((h)/(l))` Using `v^(2)=u^(2)2as rArr v_(2)^(2)=u_(2)^(2)+2(-g(h)/(l)).((l)/(2))` ...(iii) Asing using `v=u+at rArr v_(2)=u_(2)-(g(h)/(l)).(l)/(sqrtgh)` ...(iv) Form (iii) and (iv) `u_(2)=sqrtgh and v_(2)=0` Hence velocity of ball `2` at the of throwing will be and when it meets ball `1` its velocity becomes zero. The velocity of ball `1` when it meets ball `2` is `sqrtgh`. |
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| 328. |
Difference between Uniform acceleration and Non-uniform acceleration. |
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Answer» Uniform acceleration : A body is said to have uniform acceleration if magnitude and direction of the acceleration remains constant during particle motion. • If a particle is moving with uniform acceleration, this does not necessarily imply that particles is moving in straight line, e.g., Projectile motion. Non-uniform acceleration : A body is said to have non-uniform acceleration, if magnitude or direction or both, change during motion. |
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| 329. |
Define linear kinematics, distance and displacement, speed and velocity, acceleration. |
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Answer» Kinematics ⇒ A subset of kinematics that is particular to motion in a straight line is called linear kinematics. Distance ⇒ Distance refer to a physical length. When body moves from one location to another, the length of the path that the body follow is the distance, e.g. when a runner completes 2 laps around a 400 m track, the distance that the runner has covered is equal to (400 + 400) 800 m. Displacement ⇒ It is change in position displacement is the shortest distance from the initial position to the final position of a point. It is measured in a straight line e.g. in the completion of a lap around the 400 m track, the displacement is zero because the starting and finishing positions are the same. Speed ⇒ Speed is the rate at which an object covers distance. It is defined as the distance covered, divided by the time taken to cover it. \(\bar s=\frac{1}{\triangle t}\) l = length of path t = time Δ = delta (or) change in time \(\bar s\) = average speed Velocity ⇒ Velocity is the change in position or the displacement that occurs during a given period of time. \(v=\frac{d}{\triangle t}\) v = velocity d = displacement t = time Δ = delta (or) change in time Acceleration ⇒ The rate of change of velocity is called acceleration. May be positive may be negative. a = v/t or a d/t2 = m/sec2 unit Positive acceleration : the final velocity (v2) will be greater than initial velocity (v1) [deacceleration] Unit are meters/second2 |
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| 330. |
What are the Kinematics Variables? |
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Answer» These are the various relations between u, v, a, t and s for the moving particle where the notations are used as : u = Initial velocity of the particle at time t = 0 sec v = Final velocity at time t sec a = Acceleration of the particle s = Distance travelled in time t sec sn = Distance travelled by the body in nth sec |
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| 331. |
A ball is projected at an angle of `30^(@)` above with the horizontal from the top of a tower ans strikes the ground in `5sec` at angle of `45^(@)` with the horizontal. Find the hieght of the lower and the speede with which it was projected. |
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Answer» Correct Answer - `[50 (sqrt(3)-1)m//s, 125(2-sqrt(3))m]` |
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| 332. |
A parojectile of mass `1kg` is projected with a velcoity of `sqrt(20)m//s` such that it strikes on the same level as the point of projection at a distance of `sqrt(3)m`. Which of the following options are incorrect:A. The maximum height reached by the projectile can be `0.25m`.B. the minimum velocity during its motion can be `sqrt(15)m//s`C. the time taken for the flight can be `sqrt((3)/(5))sec`.D. maximum potential enegry during its motion can be `6J`. |
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Answer» Correct Answer - D `R = (u^(2)sin 2 theta)/(g) rArr sqrt(3) = ((sqrt(20))^(2)sin 2 theta)/(10)` `sin 2 theta = (sqrt(3))/(2) rArr 2 theta = 60^(@), 120^(@)` `rArr theta = 30^(@), 60^(@)` (A) For `theta = 30^(@)` `H = (v^(2)sin^(2) theta)/(2g) rArr H = ((sqrt(20))^(2)(1//2)^(2))/(2xx10) = (1)/(4) = 0.75m` (B) At heghiest point of motion `u` is minimum `u_(min) = u cos theta = sqrt(20) xx cos 30^(@) = sqrt(15) m//s` (C ) `T = (2u sin theta)/(g)` For `theta = 30^(@), T = (2sqrt(20)xx(1)/(2))/(10) = (1)/(sqrt(15))` For `theta = 60^(@) , T = (2xxsqrt(20))/(10) xx (sqrt(3))/(2) = sqrt((3)/(5))sec`. (D) `U_(max) = mgH = 1 xx 10 xx 0.25 = 2.5 J` |
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| 333. |
Velocity of a particle varies with time as `v=athati+2ht^(2)hatj`. If the particle starts from point `(0,c)`, the trajectory of the particle isA. `y=(bx^(3//2))/a+c`B. `y=(4sqrt(2)b)/3(x/a)^(3//2)+c`C. `y=(4sqrt(2)b)/3(x/a)^(3//2)-c`D. `y=(bx^(3//2))/a-c` |
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Answer» Correct Answer - A |
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| 334. |
Two bodies start moving in the same straight line at the same instant of time from the same origin. The first body moves with a constant velocity of `40ms^-1`, and the second starts from rest with a constant acceleration of `4ms^-2`.Find the time that elapses before the second catches the first body. Find the also the greatest distance between them prior to it and time at which this occurs. |
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Answer» Correct Answer - [20 s, 200 mts] |
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| 335. |
Two particles start moving from the same point along the same straight line. The first moves with constant velocity `v` and the second with constant acceleration `a`. During the time that elapses before the sound catches the first, the greatest distance between the particle is.A. `(v^(2))/a`B. `(v^(2))/(2a)`C. `(2v^(2))/a`D. `(v^(2))/(4a)` |
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Answer» Correct Answer - B |
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| 336. |
A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The moves horizontally with a speed of `2m//s`. At what angle `alpha` with the vertical should the wind screen placed so that the rain drops falling vertically downwards with velcoity `6m//s` strike the wind screen perpendicularly?A. `tan^(-1)(1//3)`B. `tan^(-1)(3)`C. `cos^(-1)(3)`D. `sin^(-1)(1//3)` |
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Answer» Correct Answer - B |
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| 337. |
A sailor in a boat, which is going due east with a speed of `8m//s` observes that a submarine is heading towards north at a speed of `12m//s` and sinking at a rate of `2 m//s`.The commander of submarine observes a helicopter ascending at a rate of `5 m//s` and heading towards west with `4 m//s`.Find the speed of the helicopter with respect to boat. |
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Answer» Taking east, north and vertically upward directions along `+x`, `+y` and `+z` axes respectively. For boat `vecV_(B)=8hati` Relative velocity of submarine w.r.t. boat `vecV_(S//B)=12hatj-2hatk` `:. vecV_(S)-vecV_(B)=12hatj-2hatk rArr vecV_(S)=8hati+12hatj-2hatk` Relative velocity of helicopter w.r.t. submarine `vecV_(H//S)=vecV_(H)-vecV_(S)` `5hatk-4hati=vecV_(H)-(8hati+12hatj-2hatk)` `vecV_(H)=+4hati+12hatj+3hatk` `|vecV_(H)|=sqrt(4^(2)+12^(2)+3^(2))=13m//s` `vecV_(H//B)=vecV_(H)-vecV_(B)=(4hati+12hatj+3hatk)-8hati= -4hati+12hatj+3hatk` `|vecV_(H//B)|=13m//s` |
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| 338. |
A particle moving with a speed v changes direction by an angle `theta`, without change in speed.A. The change in the magnitude of its velocity is zeroB. The change in the magnitude of its velocity is `2v sin (theta//2)`C. The magnitude of the change in its velocity is `2v sin(theta//2)`D. The magnitude of the change in its velocity is `v (1-cos theta)` |
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Answer» Correct Answer - A::C |
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| 339. |
A particle is thrown with a speed u at an angle θ to the horizontal. When the particle makes an angle ϕ with the horizontal, its speed changes to v.(a) v = ucos θ(b) v = ucos θ . cos ϕ(c) v = ucos θ . sec ϕ (d) v = usec θ . cos ϕ |
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Answer» Correct answer is: (c) v = ucos θ . sec ϕ The horizontal component of the velocity remains constant. Hence, ucos θ . vcos ϕ. |
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| 340. |
A particle moving with a speed v changes direction by an angle θ, without change in speed. (a) The change in the magnitude of its velocity is zero. (b) The change in the magnitude of its velocity is 2vsin (θ/ 2). (c) The magnitude of the change in its velocity is 2vsin (θ/ 2). (d) The magnitude of the change in its velocity is v(1 - cos θ). |
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Answer» Correct Answer is: (a, c) |
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| 341. |
A particle moving with velocity V changes its direction of motion by an angle `theta` without change in speed. Which of the following Statement is not correct?A. The magnitude of the change in its velocity is 2V sin`((theta)/(2))`.B. The change in the magnitude of its velocity is zero.C. The change in its velocity makes an angle `(pi)/(2)+(theta)/(2)` with its initial direction of motion.D. The change in velocity is equal to the negative of the resultant of the initial and final velocities. |
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Answer» Correct Answer - D |
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| 342. |
A block rests on a rough floor. A horizontal force which increases linearly with time (t), begins to act on the block at `t=0`. Its velocity (v) is plotted against t. Which of the given graphs is correct?A. B. C. D. |
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Answer» Correct Answer - D |
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| 343. |
A block rests on a rough floor. A horizontal force which increases linearly with time (t), begins to act on the block at t = 0. Its velocity (v) is plotted against t. Which of the given graphs is correct? |
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Answer» Correct Answer is: (d) The block will begin to move only when the external force becomes equal to the limiting friction at the floor, i.e., at t > 0. Also, as the external force is variable, its acceleration will be variable. The plot of v against t will be a curve. |
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| 344. |
A block of mass of mass 1 kg moves under the influence of external forces on a rough horizontal surface. At some instant it has a speed of 1m/s due east and an acceleration of `1 m//s^(2)` due north . The force of friction acting on it is F.A. F acts due west.B. F acts due south.C. F acts in the south-west direction.D. The magnitude of F cannot be found from the given data. |
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Answer» Correct Answer - A::D |
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| 345. |
A ball falls vertically onto a floor with momentum `p`, and then bounces repeatedly. If the coefficient of restitution is `e`, then the total momentum imparted by the ball on the floor till the ball comes to rest isA. `p(1+e)`B. `(p)/(1-e)`C. `p(1+(1)/(e))`D. `p((1+e)/(1-e))` |
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Answer» Correct Answer - D |
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| 346. |
A ball falls vertically onto a floor, with momentum p, and then bounces repeatedly. The coefficient of restitution is e. The total momentum imparted by the ball to the floor is(a) p(1 + e)(b) p/1-e(c) p (1 + 1/e)(d) p (1 + e / 1- e) |
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Answer» Correct answer is: (d) p (1 + e / 1 - e) 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. Thus, in this case, the change in momentum for the first impact is ep - (-p) = p(1 + e), for the second impact it is e(ep) - (-ep) = ep(1 + e) and so on. The total change in momentum is p(1 + e)[1 + e + e2 + ...]. |
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| 347. |
A ball falls from rest from a height h onto a floor, and rebounds to a height h/4. The coefficient of restitution between the ball and the floor is(a) 1/√2(b) 1/2(c) 1/4(d) 3/4 |
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Answer» Correct answer is: (b) 1/2 |
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| 348. |
A ball falls from rest from a height h onto a floor, and rebounds to a height `h//4`. The coefficient of restitution between the ball and the floor isA. `(1)/(sqrt(2))`B. `(1)/(2)`C. `(1)/(4)`D. `(3)/(4)` |
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Answer» Correct Answer - B |
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| 349. |
A particle strikes a horizontal frictionless floor with a speed u, at an angle θ to the vertical, and rebounds with a speed v, at an angle ϕ to the vertical. The coefficient of restitution between the particle and the floor is e. The magnitude of v is .The magnitude of v is(a) θ (b) tan -1 [etan θ] (c) tan -1 [ 1/e tan θ ](d) (1 + e) θ |
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Answer» Correct answer is: (c) tan -1 [ 1/e tan θ ] |
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| 350. |
In a gravity free space, man of mass `M` standing at a height `h` above the floor, throws a ball of mass `m` straight down with a speed `u`. When the ball reaches the floor, the distance of the man above the floor will be.A. `h(1+(m)/(M))`B. `h(2-(m)/(M))`C. 2hD. a function of `m,M,h` and u |
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Answer» Correct Answer - A |
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