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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.
| 201. |
A spring with one end attached to a mass and the other to a right support is stretched and releasedA. Magnitude of acceleration, when just released is maximumB. Magnitude of acceleration, when at equilibrium position, is maximumC. Speed is maximum when is at equilibrium positionD. Magnitude of displacement is always maximum whenever speed is minimum |
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Answer» Correct Answer - A::C When spring is stretched by x, restoring force will be `F = - kx` Potential energy of the stretched spring `= PE = 1/2 kx^(2)` The restoring force is central hence, when particles relased it will executes SHM about equilibrium position. Acceleration will be , `a = F/m = (-kx)/(m)` At equilibrium position, `x = 0 rArr a = 0` Hence, when just released , `x = x_(max)` Hence, acceleration is maximum. Thus options (a) is correct Hence, acceleration is maximum. Thus option (a) is correct. At equilibrium whole PE will be converted to KE hence, KE will be maximum and hence, speed will be maximum. |
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| 202. |
Depict the shown `v -x` graph `a - x` graph: A. B. C. D. |
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Answer» Correct Answer - A The given graph can be represented by the equation, `v = (-(v_(0))/(x_(0))) x + v_(0)` `:. A = v((dv)/(dx))` `a = [(-(v_(0))/(x_(0))) x + v_(0)] [- (v_(0))/(x_(0))]` `a = ((v_(0)^(2))/(x_(0)^(2))) x - [(v_(0)^(2))/(x_(0))]` which is equation of a straight line with positive slope and negative intercept. |
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| 203. |
What does `d|vec(v)|//dt` and `|d vec(v)//dt|` represent ? Can these be equal ? Can. (a) `d|vec(v)|//dt = 0` while `|d vec(v)//dt| != 0` (b) `d|vec(v)|//dt != 0` while `|d vec(v)//dt| = 0`? |
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Answer» `d |vec(v)|//dt` represents time rate of change of speed as `|vec(v)| = v`, while `|d vec(v)//dt|` represents magnitude of acceleration. If the motion of a particle is accelerated translatroy (without change in direction) as `vec(v) = |vec(v)| hat(n), (d vec(v))/(dt) = (d)/(dt) [|vec(v)| hat(n)]` or `(d vec(v))/(dt) = hat(n) (d)/(dt) |vec(v)|` [as `hat(n)` constant] or `|(d vec(v))/(dt)| = (d)/(dt) |vec(v)| (!= 0)` However, if the motion is uniform translartory, both these will still be equal but zero . (a) The given condition implies that: `|d vec(v) //dt| !=`, i.e., |acc.| `!=0` while `d|vec(v)|//dt = 0`, i.e., speed = constant This actually is the case of uniform circular motion. In case of uniform circular motion `|(d vec(v))/(dt)| = |vec(a)| = (v^(2))/(r) =` constt. `!= 0` while `|vec(v)| =` constt. i.e., `(d)/(dt) |vec(v)| = 0` (b) `|d vec(v)//dt| = 0` means `|vec(a)| = 0` means `|vec(a)| = 0`, i.e., `vec(a) = 0` or `(d vec(v)//dt) = 0` or `vec(v)=` constt. And when velocity `vec(v)` is constant speed will be constt., i.e., speed `= |vec(v)| =` constt. or `(d)/(dt) |vec(v)| = 0` So, it is not possible to have `|(d vec(v))/(dt)| = 0` while `(d)/(dt) |vec(v)| != 0` |
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| 204. |
A truck and a car are brought to a hault by application of same braking force. Which one will come to stop in a shorter distance if they are moving with same (a) velocity (b) kinetic energy and (c) momentum ? |
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Answer» By applying breakes the body is brought to rest, so `v = 0` and `a = (-F//m)` (as it is retardation). If s is the distance travelled in stopping (called stopping distance), from 3rd equation of motion, `v^(2) = u^(2) + 2as` we have `0 = u^(2) - 2 (F//m)s` i.e., `s = (mu^(2))/(2F)` But `KE = (1)/(2) mu^(2)` and also `KE = (p^(2))/(2m)` So, `s = (mu^(2))/(2F) = (KE)/(F) = (p^(2))/(2mF)` From this it is clear that: (a) If u is same: `S prop (m u^(2) //2F)`, i.e., `s prop m` Now as mass of car is less than that of truck, so car will stop in shorter distance. (b) If KE is same: `s prop (KE//F)` So, both will stop after travelling same distance. (c) If p is same: `s = (p^(2)//2mF)`, i.e., `s prop (1//m)` Now as mass of truck is more than that of car, so truck will stop in a shorter distance. |
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| 205. |
The velocity of the particle at any time t is given by `vu = 2t(3 - t) m s^(-1)`. At what time is its velocity maximum?A. 2 sB. 3 sC. 2/3 sD. 3/2 s |
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Answer» Correct Answer - D Given : `v = 2t(3 - t)` or v = `6t - 2t^(2)` `(dv)/(dt) = 6 - 4t`, For maximum velocity, `(dv)/(dt) = 0` `therefore 6 - 4t = 0 ` or `t = 3/2` s |
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| 206. |
The figure shows the velocity (v) of a particle plotted against time (t). A. The particle changes its direction of motion at some pointB. The acceleration of the particle remains constant.C. The displacement of the particle zeroD. The initial and final speeds of the particle are the same. |
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Answer» Correct Answer - A::B::C::D Particle changes direction of motion at `t=T`. Acceleration remains constant, because the velocity time graph is a straight line displacement is zero because net area is zero. Initial and final speed are equal. |
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| 207. |
Which of the following statements about distance are tire?A. It cannot be negativeB. It cannot be zeeroC. It can never be lesser than magnitude of displacementD. It can never decrease with time |
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Answer» Correct Answer - A::C::D Distance `ge` Displacement and it cannot be negative. |
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| 208. |
A ballon is moving vertically up with a velocity 4 m/s. When it is at a height `h`, a body is gently released from it. If it reaches ground in 4 sec, the height of balloon, when the body is released, is: (Take `g = 9.8 m//s^(2)`)A. 62.4 mB. 42.4 mC. 78.4 mD. 82.2 m |
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Answer» Correct Answer - A `s = ut + (1)/(2) at ^(2)` `h = -4 xx 4 + (1)/(2) xx 9.8 xx (4)^(2) = - 16 + 78.4 = 62.4 m` |
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| 209. |
A particle starts from point A moves along a straight line path with an acceleration given by a = p - qx where p, q are connected and x is distance from point A . The particle stops at point B. The maximum velocity of the particle isA. `p/q`B. `p/sqrt(q)`C. `q/p`D. `sqrt(q/p)` |
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Answer» Correct Answer - B Given : `a = p - qx` At maximum velocity, a = 0 `implies 0 = p - qx` or `x = p/q` `therefore` Velocity is maximum at `x = p/q` `since` Acceleration,` a = (dv)/(dt) = (dv)/(dx) (dx)/(dt) = v (dv)/(dx)` since` v = (dx)/(dt)` `therefore v (dv)/(dx) = a = p - qx` `vdv = (p - qx) dx` Integrating both sides of the above equation, we get `v^(2)/2 = px = (qx)^(2)/2` `v^(2) = 2px - qx^(2)` or v = `sqrt (2px - qx^(2)` At `x = p/q, v = v_max = sqrt (2p(p/q) - q(p/q)^(2))= p/sqrt q` |
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| 210. |
A particle starts from rest at t=0 and moves in a straight line with an acceleration as shown below. The velocity of the particle at t=3 s is A. `2 m//s`B. `3m//s`C. `4 m//s`D. `6 m//s` |
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Answer» Correct Answer - B The area of a-t graph gives change in velocity. Area of a-t graph between `t=0` and `t=3 s` is `(3xx2)+(-3xx1)=3` `v_(t=3)-v_(t=0)=`area of a-t graph `v_(t=3)-0=3impliesv_(t=3)=3 m//s` |
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| 211. |
A police party is chasing a dacoit in a jeep which is moving at a constant speed `v`. The dacoit is on a motor cycle. When he is at a distance `x` from the jeep, he accelerates from rest at a constant rate `alpha`. Which of the following relations is true, if the police is able to catch the dacoit ?A. `v^(2)leax`B. `v^(2)le2ax`C. `v^(2)ge2ax`D. `v^(2)geax` |
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Answer» Correct Answer - C If police is able to catch the dacoit after time t, then `vt=x+(1)/(2)alphat^(2)` this gives `(alpha)/(2)t^(2)-vt+x=0` or `t=(v+-sqrt(v^(2)-2alphax))/(alpha)` for t be real `v^(2)gealphax` |
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| 212. |
A police party is chasing a dacoit in a jeep which is moving at a constant speed `v`. The dacoit is on a motor cycle. When he is at a distance `x` from the jeep, he accelerates from rest at a constant rate `alpha`. Which of the following relations is true, if the police is able to catch the dacoit ?A. `V^(2) lt alpha X`B. `V^(2) lt 2alpha X`C. `V^(2) gt 2alphaX`D. `V^(2)=alphaX` |
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Answer» Correct Answer - C Distance travelled by the police party in `t` sec is vt. Distance travelled by thief `x+(1)/(2)alphat^(2)` `implies +(1)/(2)alphat^(2)le vt implies(alphat^(2))/(2)-vt+x=0` `alphat^(2)-2vt+2x=0 implies t=2y+-sqrt(4v^(2)-8alphax)` `4v^(2)gt8alphax,v^(2)gt2alphax` |
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| 213. |
Two particles start simultaneously from the same point and move along two straight lines. One with uniform velocity v and other with a uniform acceleration a. if `alpha ` is the angle between the lines of motion of two particles then the least value of relative velocity will be at time given byA. `(v)/(a)sinalpha`B. `(v)/(a)cosalpha`C. `(v)/(a)tanalpha`D. `(v)/(a)cotalpha` |
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Answer» Correct Answer - B At any time velocity of first car is V and that of second car is `v=v+at =0+at` `V_(rel)=sqrt(V^(2)+(at)^(2)-2vatcosalpha)` `V_(rel)` is minimum if `(d)/(dt)(v_(r)^(2))=0,t=(vcosalpha)/(a)` |
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| 214. |
A scooter acquires a velocity of ` 36 km//h` in `10` seconds just after the start . Calculate the acceleration of the scooter.A. `1 m//s^(2)`B. `2 m//s^(2)`C. `1//2 m//s^(2)`D. `3 m//s^(2)` |
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Answer» Correct Answer - A `v=u+at`, |
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| 215. |
A police party is chasing a dacoit in a jeep which is moving at a constant speed `v`. The dacoit is on a motor cycle. When he is at a distance `x` from the jeep, he accelerates from rest at a constant rate `alpha`. Which of the following relations is true, if the police is able to catch the dacoit ?A. `v^(2) lt ax`B. `v^(2) lt 2ax`C. `v^(2) ge 2ax`D. `v^(2) = ax` |
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Answer» Correct Answer - (c ) Let the police catch the thief after time t Then, `S_("Police"): = vt` and `S_"Thief" = 1/2 at^(2)` The police will catch the thief, `x + 1/2 at^(2) = vt` `1/2 at^(2) - vt + x = 0` `t= (v+- sqrt(v^(2)-2ax))/(a)` `v^(2)-2ax ge0` or `v^(2) ge 2ax` |
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| 216. |
A particle is moving with velocity `v=t^(3)-6t^(2)+4`, where v is in `m//s` and t is in seconds. At what time will the velocity be maximum//minimum and what is it equal to? |
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Answer» `v=t^(3)-6t^(2)+4` For v to be maximum or minimum, `a=(dv)/(dt)=3t^(2)-12t=3t(t-4)=0` `t=0,4 s` `(d^(2)v)/(dt^(2))=6t-12` At `t=0, (d^(2)v)/(dt^(2))=-12lt0`, v is maximum `v_(max)=4 m//s` At `t=4 s, (d^(2)v)/(dt^(2))=6(4)-12=12gt0`, v is minimum `v_(min)=(4)^(3)-6(4)^(2)+4=-28 m//s` |
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| 217. |
A point traversed half a circle of radius r during a time interval `t_(0)`, its mean speed and magnitude of mean velocity areA. `(pir)/t_(0),r/t_(0)`B. `(2pir)/t_(0),(2r)/t_(0)`C. `(pir)/t_(0),(2r)/t_(0)`D. `(2pir)/t_(0),r/t_(0)` |
| Answer» Correct Answer - C | |
| 218. |
The driver of a train moving at a speed ` v_(1)` sights another train at a disane ` d`, ahead of him moving in the same direction with a slower speed ` v_(2)`. He applies the brakes and gives a constant teradation ` a` to his train. Show that here will be no collision if ` d gt (v_(1) -v_(2))^(2) //2 a`.A. `d lt ((v_(1) + v_(2))^(2))/(a)`B. `d gt ((v_(1) - v_(2))^(2))/(2a)`C. `d gt ((v_(1) - v_(2))^(2))/(a)`D. `d lt ((v_(1) - v_(2))^(2))/(a)` |
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Answer» Correct Answer - B The velocity of train relative to good train `v_(1) - v_(2)` should become zero before the trains meet. `:. V_(r)^(2) = u_(r)^(2) + 2a_(r) s_(r)` `0 = (v_(1) - v_(2))^(2) - 2as` `:. s = ((v_(1) - v_(2))^(2))/(2a)` The trains will not collide if `d ge s, " i.e.," d ge ((v_(1) - v_(2))^(2))/(2a)` |
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| 219. |
Find the average velocity of a particle moving along a straight line such that its velocity changes with time as `v (m//s) = 4 sin. (pi)/(2) t`, over the time interval `t = 0` to `t = (2n - 1) 2` seconds. (n being any `+` ve integer). |
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Answer» Displacement over the interval `t = 0` to `t = 2 (2n - 1) s` `s = 4 underset(0)overset(2(2n - 1))int sin ((pi)/(2) t) dt` `= - (8)/(pi) [cos ((pi)/(2)t)]_(0)^(2(2n - 1)) = (16)/(pi) m` `:.` Average velocity `= (16)/(2(2n - 1)pi) = (8)/((2n - 1)pi) m//s` |
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| 220. |
(a) If the initial velocity of a particle is u and collinear acceleration at any time `t` is at, calculate the velocity of the particle after time `t`. (b) A particle moves along a straight line such that its displacement at any time t is given by `s = (t^(3) - 6 t^(2) + 3t + 4)m`. What is the velocity of the particle when its acceleration is zero? |
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Answer» By definition acceleration `= (dv//dt)` So, `(dv)/(dt) = at` (given) or `int_(u)^(v) dv = int_(0)^(t) at` or `v - u = (1)/(2) a t^(2)` or `v = u + (1)/(2) a t^(2)` (b) As according to given problem, `s = t^(3) - 6t^(2) + 3t + 4` instantaneous velocity `u (ds)/(dt) = 3t^(2) - 12 t + 3` and acceleration `a = (dv)/(dt) = (d^(2)s)/(dt^(2)) = 6 t - 12`....(i) So, acceleration will be zero when `6t - 12 = 0`, i.e., `t = 2` sec And so the velocity when acceleration is zero, i.e., at `t = 2` sec from Eqn. (i), will be `v = 3 xx 2^(2) - 12 xx 2 + 3 = - 9m//s` [Negative velocity means that body is moving towards the origin, i.e., as time increases displacement decreases.] |
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| 221. |
A ball is dropped from a bridge at a height of 176.4 m over a river. After 2s a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously ?A. 2.45 m/sB. 49 m/sC. 14.5 m/sD. 24.5 m/s |
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Answer» Correct Answer - D `F = v_("rel") ((dm)/(dt))` |
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| 222. |
A point moves in a straight line so its displacement `x` meter at time `t` second is given by `x^(2)=1+t^(2)`. Its acceleration in `ms^(-2)` at time `t` second is .A. `(1)/(x^(3//2))`B. `(-t)/(x^(3))`C. `(1)/(x)-(t^(2))/(x^(3))`D. `(1)/(x)-(1)/(x^(2))` |
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Answer» Correct Answer - C SOL: `x^(2)=1+t^(2)or x=(1+t^(2))^(1//2)` `(dx)/(dt)=(1)/(2)(1+t^(2))^(-1//2)2t=t(1+t^(2))^(-1/2)` `a=(d^(2)x)/(dt^(2))=(1+t^(2))^(-1//2)+t(-(1)/(2))(1+t^(2))^(-3//2)2t` `=(1)/(x)-(t^(2))/(x^(3))` |
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| 223. |
Two balls `A and B` of same masses are thrown from the top of the building `A`. Thrown upward with velocity `V and B`, thrown downward with velocity `V`, thenA. Velocity of A is more than B at the groundB. Velocity of B is more than A at the groundC. Both A and B strike the ground with same velocityD. None if these |
| Answer» Correct Answer - C | |
| 224. |
A juggler keeps n balls going with one hand, so that at any instant, (n - 1) balls are in air and one ball in the hand. If each ball rises to a height of x metres, the time for each ball to stay in his hand isA. `1/n -1((sqrt 2x)/g)`B. `2/n - 1((sqrt 2x)/g)`C. `2/n((sqrt 2x)/ g)`D. `1/n((sqrt 2x)/ g)` |
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Answer» Correct Answer - (b) Let u be the initial velocity of the ball while going upwards. The final velocity of the ball at height x is, v = 0. So u = `sqrt(2gx)` Time of flight, `T = (2u)/g = 2/g sqrt(2gx) = 2 sqrt(2x)/g` During time T, (n -1) balls will be in air and one ball will be in hand. So time for one ball in hand `T /(n - 1) = 2 sqrt(2x)/g(n - 1) = 2 / n - 1 ((sqrt 2x)/g)` |
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| 225. |
A juggler maintains four balls in vertically upwards motion. He attempts next ball after `(1)/(4)` seconds. For the show to go one, what should be the height for which he thrwos the ball ? `(g = 10 m//s^(2))`A. 1.25 mB. 5 mC. 2.5mD. 1.0 m |
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Answer» Correct Answer - A `H = (1)/(2) xx g xx ((1)/(2))^(2)` `H = 1.25 m` Time for one ball to reach at max. height is `(1)/(4) + (1)/(4) = (1)/(2)` sec |
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| 226. |
A particle exceutes the motion describes by `x(t)=x_(0)(1-e^(-gammat)),tge0,x_(0)0`. The maximum and minimum values of `v(t)` areA. `x_(0)` and `0`B. `x_(0)gamma` and 0C. 0 and `-x_(0)gamma^(2)`D. `x_(0)^(1+e^(-gamma))` |
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Answer» Correct Answer - B `x(t)=x_(0)(1-e^(-gammat))` `v(t)=(dx(t))/(dt)=x_(0)gammae^(-gammat)` `a(t)=(dv(t))/(dt)=-x_(0)gamma^(2)e^(-gammat)` |
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| 227. |
The displacement of a partical as a function of time t is given by `s=alpha+betat+gammat^(2)+deltat^(4)`, where `alpha`,`beta`,`gamma` and `delta` are constants. Find the ratio of the initial velocity to the initial acceleration. |
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Answer» First find the velocity and acceleration in terms of time t, then use t=0 to find the intial values. `s=alpha+betat+gammat^(2)+deltat^(4)` `v=(ds)/(dt)=0+beta.1+gamma.2t+delta.4t^(3)` `=beta+2gammat+4deltat^(3)` `a=(dv)/(dt)=0+2gamma.1+4delta.3t^(2)` `=2gamma+12deltat^(2)` At `t=0, v=beta, a=2gamma` `("Initial velocity")/("Initial acceleration")=beta/(2gamma)` |
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| 228. |
A partical is moving in a straight line under constant acceletation. If the motion starts from rest, Find the ratio of displacement in `n` second to that in the `n^(th)` second.A. `(2n-1)/n^(2)`B. `1/n`C. `n^(2)/(n-1)`D. `n^(2)/(2n-1)` |
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Answer» Correct Answer - D `s_(n)=1/2 a n^(2)` `s_(nth)=1/2 a(2n-1)` `s_(n)/s_(nth)=n^(2)/((2n-1))` |
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| 229. |
A partical is moving in a straight line under constant acceletation. If the motion starts from rest, Find the ratio of displacement in `n` second to that in the `n^(th)` second. |
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Answer» `mu=0, a=a, t=n` Displacement in n second `s=ut+1/2at^(2)=0+1/2an^(2)=1/2an^(2) ...(i)` Displacement in the `n^(th)` second `s_(t)=u+1/2a(2t-1)` `s_(n)=0+1/2a(2n-1) …(ii)` `s/s_(n)=(1/2an^(2))/(1/2a(2n-1))=n^(2)/(2n-1)` |
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| 230. |
A partical moves in a straight line as `s=alpha(t-2)^(3)+beta(2t-3)^(4)`, where `alpha` and `beta` are constants. Find velocity and acceleration as a function of time. |
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Answer» `s=alpha(t-2)^(3)+beta(2t-3)^(4)` `=alpha{(t-2)^(3)}+beta{(2t-3)^(4)}` `v=(ds)/(dt)=alpha.3(t-2)^(2).(1-0)+beta.4(2t-3)^(3).(2.1-0)` `=3alpha(t-2)^(2)+8beta(2t-3)^(3)` `=3alpha{(t-2)^(2)}+8beta{(2t-3)^(3)}` `a=(dv)/(dt)=3alpha.2(t-2).(1-0)+8beta.3(2t-3)^(2).(2.1-0)` `=6alpha(t-2)+48beta(2t-3)^(2)` Recall: First differential the outer layer, then the inner layer and then multiply the two differentials. |
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| 231. |
A stone is dropped into a well and the sound of inpact of stone on the water is heard after 2.056 sec of the release of stone form the top. If acc. Due to gravity is `980 cm//sec^(2)` and velocity of sound in air is `350 m//s`, calculate the depth of the well. |
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Answer» If the depth of well is `h` and time taken by stone to reach the bottom `t_(1)`, then `h = (1)/(2) g t_(1)^(2)` and time taken by sound to come to the surface `t_(2) = (h)/(350)` .....(ii) But according to given problem `t_(1) + t_(2) = 2.056`....(iii) Substituting `h` from Eqn. (i) in (ii) and then `t_(2)` from Eqn. (ii) in (iii), we get `t_(1) + (g t_(1)^(2))/(700) = 2.056` or `98 t_(1)^(2) + 7000 t_(1) - 14392 = 0` or `14 t_(1)^(2) + 1000 t_(1) - 2056 = 0` or `14 t_(1)^(2) - 28 t_(1) + 1028 t_(1) - 2056 = 0` or `14 t_(1) (t_(1) - 2) + 1028 (t_(1) - 2) = 0` or `(14 t_(1) + 1028) (t_(1) - 2) = 0` So `t_(1) = 2` or `t_(1) = -(1028//14)` Now as negative time is not physically acceptable, so `t_(1) = 2s` `:.` The depth of well `h = (1)/(2) xx 9.8 xx 2^(2) = 19.6 m` |
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| 232. |
A particle starts moving from the position of rest under a constant acc. If it travels a distance x in t sec, what distance will it travel in next t sec ? |
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Answer» As acc. is constt.,from 2nd equation of motion, i.e., `s = ut + (1)/(2) at^(2)` we have `x = (1)/(2) at^(2)` [as `u = 0`]....(i) Now if it travels a distance `y` in next t sec., the total distance travelled in `(t + t = 2t)` sec will be `x + y`, so `x + Y = (1)/(2) a (2t)^(2)` ....(ii) Dividing Eqn. (ii) by (i), `(x + y)/(x) = 4` or `y = 3x` Alternative solution: From 2nd equation of motion, we have `x = (1)/(2) at^(2)` The velocity of particle after time t from 1 st equation of motion will be `v = 0 + a t`, i.e., `v = at` Now for next `t` sec it will be the initital velocity, so again from 2nd Eqn. of motion, i.e., `s = u t + (1)/(2) at ^(2)` we have, `y = (at) t + (1)/(2) at^(2) = (3)/(2) at^(2)`.....(iv) Dividing eqn. (iv) by (iii), `(y//x) = 3`, i.e., `y = 3x` |
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| 233. |
The figure-1.125 shows the acceleration versus time graph of a train. If it starts from rest, the distance it travels before it comes to rest is : A. 30 mB. 26 mC. 13 mD. 40 m |
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Answer» Correct Answer - B Area below `v - t` curve will provide net displacement. |
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| 234. |
The velocity versus time graph of a body moving in a straight line is as follows. The distance travelled by the body is 5 sec is A. 2 mB. 3 mC. 4 mD. 5 m |
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Answer» Correct Answer - D Distance `= ((1)/(2) xx 3 xx 2) + ((1)/(2) xx 1 xx 2) + (1 xx 1) = 5 m` (In this problem displacement `= 3 m`) |
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| 235. |
The graph below shws the velocity versus time graph for a body Which of the following graph represents the corresponding acceleration v//s time graph?A. B. C. D. |
| Answer» Correct Answer - A | |
| 236. |
Velocity versus displacement graph of a particle moving in a straight line is shown in figure. Corresponding acceleration versus velocity graph will be. A. B. C. D. |
| Answer» Correct Answer - A | |
| 237. |
The acceleration-displacement graph of a particle moving in a straight line is as shown in figure, initial velocity of particle is zero. Find the velocity of the particle when displacement of the particle is `s=12 m`. |
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Answer» `v^(2)-u^(2)=2ax` `impliesax=(v^(2)-u^(2))/(2)=(v^(2))/(2)-(u^(2))/(2)` `because(u=0)` `impliesv=sqrt(2("area under a-x graph"))` Area under `a-x` graph Area of `DeltaOAE+` area of reactangle `ABEF` `+` area of trapenzium BFGC + Area of `DeltaCGD` Area `=(1)/(2)(2)(2)+6xx2+(1)/(2)(2+4)(2)+(1)/(2)xx2xx4=24` `impliesv=sqrt(2xx24)=4sqrt(3) m//s` |
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| 238. |
A particle moves in a straight line such that the displacement x at any time t is given by `x=6t^(2)-t^(3)-3t-4`. X is in m and t is in second calculate the maximum velocity (in `ms^(-1))` of the particle. |
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Answer» Correct Answer - 9 `x=6t^(2)-t^(3)-3t-4v=12t-3t^(2)-3` `V="max when" (dv)/(dt)=0impliest=2sec` `:. v(2)=9ms^(-1)` |
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| 239. |
The displacement `s` of a point moving in a straight line is given by: `s = 8 t^(2) + 3t - 5` `s` being in cm and `t` in s. The initial velocity of the particle is:A. 3 cm/sB. 16 cm/sC. 19 cm/sD. zero |
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Answer» Correct Answer - A `v = (ds)/(dt) = (d)/(dt) (8 t^(2) + 3t - 5) = 16 t + 3` Initial velocity (time, `t = 0`) is given by `u = 3 cm//s` |
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| 240. |
A partical is moving in a straight line such that `s=t^(3)-3t^(2)+2`, where `s` is the displacement in meters and `t` is in seconds. Find the (a) velocity at `t=2s`, (b) acceleration at `t=3s`, ( c) velocity when acceleration is zero and (d) acceleration when velocity is zero. |
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Answer» `s=t^(3)-3t^(2)+3t+2` `v=(ds)/(dt)=3t^(2)-6t+3` `a=(dv)/(dt)=6t-6` (a) At `t=2 s, v=3(2)^(2)-6(2)+3=3 m//s ` (b)At` t=1 s, a=6(3)-6=12 m//s^(2)` (c ) Acceleration `a=0implies6t-6=0impliest=1 s ` `At t=1 s, v=3(1)^(2)-6(1)+3=0` (d) When velocity is zero, `v=3t^(2)-6t+3=3(t^(2)=2t+1)=0` `(t-1)^(2)=0impliest=1 s ` At `t=1 s, a=6(1)-6=0` |
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| 241. |
The displacement of a particle moving in straight line is given as function of time as `s = ((t^(3))/(3) - (3t^(2))/(2) + 2t), s` is in m and t is in sec. The particle comes to momentary rest n times Find the value of `n` |
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Answer» Correct Answer - 2 `s = (t^(3))/(3) - (3t^(2))/(2) + 2t` `:. V = (ds)/(dt) = t^(2) - 3 t + 2` Since the particle comes to rest, therefore, `t^(2) - 3 t + 2 = 0` `(t - 1) (t - 2) = 0` ` t = 1 s " and " 2s` Hence, the particle comes to rest twice. `:. n = 2` |
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| 242. |
if `a=3t^(2)+2t`, initial velocity is `5 m//s`. Find the velocity at t=4s. The motion is in straight line, a is acceleration in `m//s^(2)` and t is time in seconds. |
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Answer» `a=(dv)/(dt)=3t^(2)+2t` `int_(5)^(v) dv=int_(0)^(4)(3t^(2)+2t)dt` `|v|_(5)^(v)=3t^(3)/3+2t^(2)/2=|t^(3)+t^(2)|_(0)^(4)` `v-5={(4)^(3)+(4)^(2)}-{0}` `v=85 m//s` |
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| 243. |
The displacement of a particla moving in straight line is given by `s=t^(4)+2t^(3)+3t^(2)+4`, where s is in meters and t is in seconds. Find the (a) velocity at `t=1 s`, (b) acceleration at `t=2 s`, (c ) average velocity during time interval `t=0` to `t=2 s` and (d) average acceleration during time interval `t=0` to `t=1 s`. |
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Answer» `s=t^(4)+2t^(3)+3t^(2)+4` `v=(ds)/(dt)=4t^(3)+6t^(2)+6t` `a=(dv)/(dt)=12t^(2)+12t+6` (a) `At t=1 s, v=4(1)^(3)+6(1)^(2)+6(1)=16 m//s` (b) `At t=2 s, a =12(2)^(2)+12(2)+6=78 m//s^(2)` (c ) `At t_(1)=0, s_(1)=4 m` At `t_(2)=2 s, s_(2)=(2)^(4)+2(2)^(3)+3(2)^(2)+4=48 m` `bar(v)=(Deltas)/(Deltat)=(s_(2)-s_(1))/(t_(2)-t_(1))=(48-4)/(2-0)=22 m//s` `At t_(1)=0, v_(1)=0` At `t_(2)=1 s, v_(2)=4(1)^(3)+6(1)^(2)+6(1)=16 m//s` `bar(a)=(Deltav)/(Deltat)=(v_(2)-v_(1))/(t_(2)-t_(1))=(16-0)/(1-0)=16 m//s` |
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| 244. |
A travelling wave in a stretched string is described by the equation `y = A sin (kx - omegat)` the maximum particle velocity isA. `A omega`B. `omega//k`C. `d omega//dk`D. `x//t` |
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Answer» Correct Answer - A `y = 3 cm//s` `:. V = (dy)/(dt) = - A omega cos (kx - omega t)` The maximum particle velocity is `|v_("max")| = A omega` |
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| 245. |
From the top of a tower of height 200 m, a ball A is projected up with speed `10ms^(-1)` and 2 s later, another ball B is projected vertically down with the same speed. ThenA. Both A and B will reach the ground simultaneouslyB. Ball A will hit the ground 2 s later than B hitting the ground.C. Both the balls will hit the ground with the same velocityD. Both the balls will hit the ground with the different velocity. |
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Answer» Correct Answer - A::C::D Ball A will return to the top of tower after `T=(2u)/(g)=(2xx10)/(10)=2s` With speed of `10ms^(-1)` downward. And this time B is also projected downwards with `10ms^(-1)`. So both reach ground simultaneously also they will hit the ground with the same speed. |
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| 246. |
A stone falls freely rest. The distance covered by it in the last second is equal to the distance covered by it in the first 2 s. The time taken by the stone to reach the ground isA. `2.5 s`B. `3.5 s`C. `4 s`D. `5 s` |
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Answer» Correct Answer - A Let t: time of journey `1/2g(2t-1)=1/2g(2)^(2)` `t=2.5 s` |
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| 247. |
A stone falls freely from rest from aheight `h` and it travels a distance `9h//25` in the last second. The value of `h` isA. 145 mB. 100 mC. 125 mD. 200 m |
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Answer» Correct Answer - C `h=1/2g t^(2) …(i)` `(9h)/25=1/2g(2t-1) …(ii)` Solving, `t=5 s, h=125 m` |
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| 248. |
A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of `18 km //h` while the other has the speed of ` 27 km //h`. The bird starts moving from first car towards the other and is moving with the speed of ` 36 km //h` and when the two cars were separated by ` 36 km`. What is the total distance covered by the bird ? What is the total displacement of the bird ? |
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Answer» Given, Speed of first car `= 18 km//h` Speed of second car `= 27 km//h` `:.` Relative speed of each car w.r.t each other `= 18 + 27 = 45 km//h` Distance between the cars `= 36 km` `:.` time of meeting the cars`(t) = ("Distance between the cars")/("Relative speed of cars") = 36/45 = 4/5 h = 0.8 h` Speed of th bird `(v_(b)) = 36 km//h` `:.` Distance covered by the bird `= v_(b) xx t = 36 xx 0.8 = 28.8 km` |
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| 249. |
Two town A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man leaving in either direction every in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. The period T of the bus service isA. 4.5 minB. 9 minC. 1.2 minD. 24 min |
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Answer» Correct Answer - B Let `v "km"h^(-1)` be the constant speed with which the bus travel ply between the towns A and B. Relative velocity of the bus from A to B with respect to the cyclist = `(v - 20) km h^(-1)` Relative velocity of the bus from B to A with respect to the cyclist `= (v + 20) km h^(-1)` Distance travelled by the bus in time T (minutes) = vT As per question `(vT)/v - 20 = 18` or `vT = 18v - 18 xx 20` and `(vT)/v + 20 = 6` ...(i) or `vT = 6v + 20 xx 6` ....(ii) Equaitons (i) and (ii), we get `18v - 18 xx 20 = 6v + 20 xx 6` or `12 v = 20 xx 6 + 18 xx 20 = 480` or v = `40 km h^(-1)` Putting this value of v in (i) , we get `40T = 18 xx 40 - 18 xx 20 = 18 xx 20` or T = `18 xx 20/40` = 9 min |
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| 250. |
On a two lane road , car (A) is travelling with a speed of `36 km h^(-1)`. Tho car ` B and C` approach car (A) in opposite directions with a speed of ` 54 km h^(-1)` each . At a certain instant , when the distance (AB) is equal to (AC), both being ` 1 km, (B) decides to overtake ` A before C does , What minimum accelration of car (B) is required to avoid and accident. |
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Answer» Velocity of a car A, `V_(A)=36 km//h=10 m//s` Velocity of car B, `V_(B)=54 km//h=15 m//s` Velocity of car C, `V_(C)=54 km//h=15 m//s` Relative velocity of car B with respect to car A `V_(BA)=V_(B)-V_(A)=15-10=5 m//s` Relative velocity of car C with respect to car A `V_(CA)=V_(C)+V_(A)=15+10=25 m//s` At a certain instance, both cars B and C are at the same distance from car A i.e., `s=1km=1000m` Time taken (t) by car C to cover `1000m` is `t=(1000)/(25)=40s` The acceleration rpduced by car B is `1000=5xx40+(1)/(2)at^(2)a=(1600)/(1600)=1 m//s^(2)` |
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