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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.
| 151. |
A drunkard waking in a barrow lane takes ` 5` steps forward and `3` steps backward, followed again `5` steps forward and `3` steps backward, and so on. Each step is ` 1 m`long and requires 1 s. Determine how long the drunkard takes to fall in a pit ` 13 m` away from start. |
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Answer» Distance of the pit from the start `=13-5=8m` Time taken to move first `5m=5sec` 5 steps (i.e., 5m) forward and 3 steps (i.e., 3m) backward means that net distance moved `=5-3=2m` and time taken during this process `=5+3=8sec` `:.` Time taken in moving `8m=(8xx8)/(2)=32sec` `:.` Total time to fall in the pit`=32+5=37sec` |
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| 152. |
A ball ` A` is dropped from a building of height ` 45 m`. Simultaneously another ball ` B` is thrown up with a speed ` 40 m//s`. Calculate the relative speed of the balls as a function of time.A. `20ms^(-1)`B. `40ms^(-1)`C. `30ms^(-1)`D. `0ms^(-1)` |
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Answer» Correct Answer - D Relative acceleration is zero as g is downwards for the both the bodies. |
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| 153. |
At the maximum height of a body thrown vertically upA. velocity is not zero but acceleration is zeroB. acceleration is not zero but velocity is zero.C. both acceleration and velocity are zero.D. both acceleration and velocity are not zero. |
| Answer» Correct Answer - B | |
| 154. |
A ball is dropped freely while another is thrown vertically downward with an initial velocity v from the same point simultaneously. After `t` second they are separated by a distance ofA. `(vt)/(2)`B. `(1)/(2)g t^(2)`C. `vt`D. `vt+(1)/(2)g t^(2)` |
| Answer» Correct Answer - C | |
| 155. |
In the question number 97, the speed of the child running opposite to the direction of motion of the belt isA. `4 km h^(-1)`B. `5 km h^(-1)`C. `9 km h^(-1)`D. `13 km h^(-1)` |
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Answer» Correct Answer - B In this case, `v_(BG) = 4 km h^(-1), v_(CB) = -9 km h^(-1)` `therefore` For an observer on a stationary platform, speed of child running opposite to the direction of motion of the belt is `v_(CG) = v_(CB) + v_(BG) = -9 km h^(-1) + 4 km h^(-1) = -5 km h^(_1)` -ve sign shows that the child appears to be moving from right to left with a speed of `5 km h^(-1)`. |
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| 156. |
In Q.111, if A lies to the left of B, time taken by the insect to travel from B to A will be:A. 12 secB. 15 secC. 18 secD. 21 sec |
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Answer» Correct Answer - B `t = (60)/(4) = 15 sec` |
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| 157. |
A bus starts from rest with a constant acceleration of `5 m//s^(2)` at the same time a car travelling with a constant velocity `50 m//s` over takes and passes the bus. How fast is the bus travelling when they are side by side?A. `10 m//s`B. `50 m//s`C. `100 m//s`D. `150 m//s` |
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Answer» Correct Answer - C `S=(1)/(2)at^(2), s=vt, v=at` |
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| 158. |
The area of the shaded portion of the graph represents : A. the average accelerationB. the maximum KEC. the momentumD. the displacement |
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Answer» Correct Answer - B Area blow `v - t` curve provides displacement |
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| 159. |
The velocity time curve of a moving point is shown in Fig. Find the retardation of the particle for the portion `CD`. .A. `1 cm//sec^(2)`B. `2 cm//sec^(2)`C. `3 cm//sec^(2)`D. `4 cm//sec^(2)` |
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Answer» Correct Answer - B `a = (v_(2) - v_(1))/(t_(2) - t_(1)) = ( 0 - 60)/(70 - 40) = - 2 m//s^(2)` |
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| 160. |
Two trans A and B 100 m and 60 m long are moving in opposite direction on parallel tracks. The velocity of the shorter train is 3 times that of the longer one. If the trains take 4s to cross each other, the velocities of the trains areA. `V_(A)=10ms^(-1),V_(B)=30ms^(-1)`B. `V_(A)=2.5ms^(=1),V_(B)=7.5ms^(-1)`C. `V_(A)=20ms^(-1),V_(B)=60ms^(-1)`D. `V_(A)=5ms^(-1),V_(B)=15ms^(-1)` |
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Answer» Correct Answer - A `{:(2V_(A)=V_(B)," "V_("rel")=V_(A)+V_(B)),(t=(I_(A)+I_(B))/(V_("rel"))," "S_(1)+S_(2)=(V_(A)+V_(B))t):}` |
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| 161. |
A player throws a ball vertically upwards with velocity u. At highest point,A. both the velocity and acceleration of the ball are zero.B. the velocity of the ball is u but its accelerationC. the velocity of the ball is zero but its acceleration is g.D. the velocity of the ball is u but its acceleration is g. |
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Answer» Correct Answer - C At the highest point velocity of the ball becomes zero, but its acceleration is equal to g. |
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| 162. |
A body covers 20 m, 22 m, 24 m, in `8^(th)`, `9 ^(th)` and `10^(th)` seconds respectively. The body startsA. from rest and moves with uniform velocity.B. from rest and moves with uniform acceleration.C. with an initial velocity and moves with uniform acceleration.D. with an initial velocity and moves with uniform velocity. |
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Answer» Correct Answer - C As distance travelled in successive seconds differ by 2 m each, therefore acceleration is constant = `2 m s^(-2)` From `D_n` = u + a/2 (2n - 1) `therefore` `D_8 = u + 2/2 (2 xx 8 - 1)` 20 = u + 15 or u = `5 m s^(-1)` |
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| 163. |
If a body starts from rest, the time in which it covers a particular displacement with uniform acceleration is :A. inversely proportional to the square root of the displacement.B. inversely proportional to the displacementC. directly proportional to the displacementD. directly proportional to the square root of the displacement. |
| Answer» Correct Answer - D | |
| 164. |
The displacement by the bolt during its free fall time w.r.t. ground frameA. 0.3 mB. 0.7 mC. 0.9 mD. 1 m |
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Answer» Correct Answer - B `vec(S)_(b)=vec(S)_(be)+vec(S)_(e)S_(b)=-2.7+1.97S_(b)approx0.7m` |
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| 165. |
The velocity-time graph of a body is shown in figure. The ratio of magnitude of average acceleration during the intervals OA and AB is A. `1`B. `(1)/(2)`C. `(1)/(3)`D. `3` |
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Answer» Correct Answer - C During OA, acceleration `=tan30^(@)=(1)/(sqrt(3))ms^(-2)` During AB, acceleration `=-tan60^(@)=-sqrt(3)ms^(-2)` requried ratio `= (1//sqrt(3))/(sqrt(3))=(1)/(3)` |
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| 166. |
The `v-s` and `v^(2)-s` graph are given for two particles. Find the accelerations of the particles at `s=0`. . |
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Answer» `a_(1) = (vdv)/(ds)`, where `v = 20` (at `s = 0`) and `(dv)/(ds) = - (2)/(5)` Then `a_(1) = -8 m//s^(2)` `a_(2) = (1)/(2) (d(v^(2)))/(ds)`, where `(d(v^(2)))/(ds) = - (2)/(5)` then , `a_(2) = - 0.2 m//s^(2)` |
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| 167. |
A body moving along a straight line travels one third of the total distance with a speed of `3.0 m s^(-1)`. The remaining distance is covered with a speed of `4.0 m s^(-1)` for half the time and `5.0 m s^(-1)` for the other half of the time. The average speed during the motion isA. `4.0 m s^(-1)`B. `6.0 m s^(-1)`C. `3.8 m s^(-1)`D. `2.4 m s^(-1)` |
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Answer» Correct Answer - C Let S be the total distance travelled by the body. Let `t_1` be the time taken to cover the first one third of the distance. Then `t_1 = (S/3) / 3 = S/9` Let `t_2` be the time taken for each of the remaining two journeys. Then `S-S/3 = 4t_2 + 5t_2` or `(2S)/3 = 9t_2` or `t_2 = (2S)/27` Average speed = Total distance travelled / Total time taken `= S/t_1 + 2t_2` `= (S)/(S/9) + 2xx(2S)/27 = 3.8 m s^(-1)` |
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| 168. |
Which of the following graphs represents the position time-graph of a particle moving with negative velocity?A. B. C. D. |
| Answer» Correct Answer - B | |
| 169. |
Velocity of a body moving with uniform acceleration of `3 m//s^(2)` is changed through 30 m/s in certain time. Average velocity of body during this time is 30 m/s. Distance covered by it during this time isA. 300 mB. 200 mC. 400 mD. 250 m |
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Answer» Correct Answer - A `a=(v-u)/(t),s=v_(avg)xxt` |
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| 170. |
A particle moves in a straight line with a constant acceleration. It changes its velocity from `10 ms^-1` to `20 ms^-1` while passing through a distance `135 m` in `t` seconds. The value of `t` is.A. `12`B. `9`C. `10`D. `1.8` |
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Answer» Correct Answer - B (2) `(20)^(2)=(10)^(2)-2axx135` `a=300/270=10/9 m//s^(2)` `20=10+a t=10+10/9 t` `t=9 s` OR `s=((u+v)/2)t` `135=((10+20)/2)timpliest=9s` |
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| 171. |
Two cars A and B are at rest at the origin O. If A starts with a uniform velocity of` 20 m//`s and `B` starts in the same direction with a constant acceleration of `2 m//s^(2)`, then the cars will meet after timeA. `10 s`B. `20 s`C. `30 s`D. `40 s` |
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Answer» Correct Answer - B Let the cars meet after time t. A (uniform motion): `s=20 t …(i)` B(accelerated motion): `s=(1)/(2).2.t^(2) …(ii)` Equating (i) and (ii), we get `t^(2) =20timpliest=20 s` |
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| 172. |
A particle is moving such that the velocity is given by `v=sqrt(2gammas)`, where `gamma` is constant and s is diplacement. Find the acceleration. |
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Answer» `v=sqrt(2lambdas)=sqrt(2lambda)s^(1//2)` `(dv)/(ds)=sqrt(2lambda)1/2.s^(-1//2)=1/2sqrt((2lambda)/s)` `a=v(dv)/(ds)=sqrt(2lambda) s^(1//2).1/2sqrt((2lambda)/s)=lambda` |
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| 173. |
The displacement of a particle is moving by `x = (t - 2)^2` where `x` is in metres and `t` in second. The distance covered by the particle in first `4` seconds is.A. 4mB. 8mC. 12 mD. 16 m |
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Answer» Correct Answer - B Given `x=(t-2)^(2)` velocity, `v=(dx)/(dt)=(d)/(dt)(t-2)^(2)=2(t-2)m//s` Acceleration `a=(dv)/(dt)=(d)/(dt)[2(t-2)]` `=2[1-0]=2 m//s^(2)` |
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| 174. |
The displacement of a particle is moving by `x = (t - 2)^2` where `x` is in metres and `t` in second. The distance covered by the particle in first `4` seconds is.A. 4 mB. 8 mC. 12 mD. 16 m |
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Answer» Correct Answer - (b) Given : x = `(t - 2)^(2)` At t = 0, x = `x_0 = (0 - 2)^(2)` = 4 m t = 1 s, x = `x_1 = (1 - 2)^(2)` = 1 m t = 2 s, x = `x_2 = (2 - 2)^(2)` = 0 m t = 3 s, x = `x_3 = (3 - 2)^(2)` = 1 m t = 4 s, x = `x_4 = (4 - 2)^(2)` = 4 m The distance covered bt the particle in `1^(st)` second is `D_1 = x_0 - x_1` = 3 m The distance covered by the particle in `II^(nd)` second is `D_2 = x_1 - x_2` = 1 m Similarly, `D_3` = 1 m, `D_4` = 3 m The distance covered by the particle in first 4 seconds is D = `D_1 + D_2 + D_3 + D_4` = 3 m + 1 m + 1 m + 3 m = 8 m |
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| 175. |
If a car covers `(2)/(5)^(th)` of the total distance with `v_1` speed and `(3)/(5)^(th)` distance with `v_2`. Then average speed isA. `1/2sqrt(v_(1)v_(2))`B. `(v_(1)+v_(2))/2`C. `(2v_(1)v_(2))/(v_(1)+v_(2))`D. `(5v_(1)v_(2))/(3v_(1)+2v_(2))` |
| Answer» Correct Answer - D | |
| 176. |
A vehicle travels half the distance (L) with speed ` V_1` and the other half with speed ` V_2`, then its average speed is .A. `v_1 + v_2/2`B. `2v_1 + v_2/v_1 + v_2`C. `2v_(1)v_2/v_1 + v_2`D. `L(v_1 + v_2)/v_(1)v_(2)` |
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Answer» Correct Answer - (c ) Time taken to travel first half distance, `t_1 = (L/2)/v_1 = L/(2v_1)` Time taken to travel second half distance, `t_2 = (L/2)/v_2 = L/(2v_2)` Total time taken = `t_1 + t_2 = L/(2v_1) + L/(2v_2)` Average speed `= ("Total distance travelled")/("Total time taken")` `= L/(L/(2v_1) + L/(2v_2)) = 1/(1/2[1/v_1 + 1/v_2]) = (2v_(1)v_(2))/(v_(2) + v_(1))` |
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| 177. |
If a car covers `(2)/(5)^(th)` of the total distance with `v_1` speed and `(3)/(5)^(th)` distance with `v_2`. Then average speed isA. `(1)/(2)sqrt(v_(1)v_(2))`B. `(v_(1)+v_(2))/(2)`C. `(2v_(1)v_(2))/(v_(1)+v_(2))`D. `(5v_(1)v_(2))/(3v_(1)+2v_(2))` |
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Answer» Correct Answer - D avg speed`=("total distance")/("total time")=(S_(1)+S_(2))/((S_(1))/(V_(1))+(S_(2))/(V_(2)))` |
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| 178. |
The position-time graph of an object moving in a straight line is shown below. The object has zero velocity at A. OB. CC. DD. F |
| Answer» Correct Answer - C | |
| 179. |
Assertion : The position-time graph of a body moving uniformly is a straight line parallel to position-axis. Reason : The slope of position-time graph in a uniform motion gives the velocity of an object.A. If both assertion and reason are trure and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanations of assertion.C. If assertion is true but reason is false.D. If both assertion and reason is false |
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Answer» Correct Answer - (d) If the position-time graph of a body moving uniformly in a straight line is parallel to position axis, it means that the position of body is changing at constant time, which is absurd and shows that the velocity of the body is infinite. |
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| 180. |
A body moving with a uniform acceleration crosses a distance of 65 m in the 5 th second and 105 m in 9th second. How far will it go in 20 s?A. 2040 mB. 240 mC. 2400 mD. 2004 m |
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Answer» Correct Answer - C `65 = u + (1)/(2) a (2 xx 5 -1)` `105 = u + (1)/(2) a (2 xx 9 - 1)` `5 = u (20) + (1)/(2) a (20)^(2)` |
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| 181. |
A body starts from rest and moves with constant acceleration. The ratio of distance covered by the body in `nth` second to that covered in `n` second is.A. `((2)/(n)-(1)/(n^(2)))`B. `((1)/(n^(2))-(1)/(n))`C. `((2)/(n^(2))-(1)/(n))`D. `(2)/(n)+(1)/(n^(2))` |
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Answer» Correct Answer - A `S_(n)=(a)/(2)(2n-1), S =(a)/(2)n^(2),(S_(n))/(S)=(2n-1)/(n^(2))` |
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| 182. |
If a light and a heavy body are released from same height:A. heavier body hits the ground with greater velocityB. lighter body hits the ground with greater velocityC. both hits the ground with same velocityD. which one will hit the ground with greater velocity depends on their shape |
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Answer» Correct Answer - C Motion under gravity is independent of mass of body. |
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| 183. |
Which of the following option is correct for the object having a staright line motion represented by the following graph?A. The object moves with constantly increasing velocity from O to A then it moves with constant velocity.B. velocity of the object increases uniformlyC. Average velocity is zero.D. The graph shown is impossible. |
| Answer» Correct Answer - C | |
| 184. |
If the position of a particle along Y axis is represented as a funtion of time t by the equation `y(t)=t^(3)` then find displacement of the particle during the period t to `t+Deltat` |
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Answer» Position at time t is `y(t)=t^(3)` Position at time `t+Delta t` is `y(t+Deltat)=(t+Deltat)^(3)` `:.` displacement of the particle from to `t+Delta t` is `y(t+Delta)-y(t)=(t+Deltat)^(3)-t^(3)` `=t^(3)+3t^(2)Deltat+3t(Deltat)^(2)+(Deltat)^(3)-t^(3)` `=3t^(2)Deltat+3t(Deltat)^(2)+(Deltat)^(3)` |
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| 185. |
The displacement of a particle as a funtion of time is shown in the figure. The figure shows thatA. the particle starts with certain velocity but the motion is retarded and finally the particle stopsB. the velocity of the particle is constant through out.C. the acceleration of the particle is constant throughoutD. the particle starts with constant velocity, the motion is accelerated and finally the particle move with another constant velocity. |
| Answer» Correct Answer - A | |
| 186. |
A uniform moving cricket ball is turned back by hitting it with a bat for a very shart time interval. Show the variation of its acceleration with time. (Take acceleration in the back ward direction as positive).A. B. C. D. |
| Answer» Correct Answer - A | |
| 187. |
The speed of a car was `50 km//hr` for the first 900s, then 40 km/hr for the 50 km and then the car decelerated uniformly at `10 km//hr^(2)` till it came to rest. The average speed of the car was :A. 50 km/hrB. 7.2 m/sC. 30 km/hrD. 9.0 m/s |
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Answer» Correct Answer - B Av. Speed `= ("Total distance")/("Total time")` `= ((v_(1)t_(1) + v_(2) t_(2) + (v_(2)^(2))/(2a_(2)))/(t_(1) + t_(2) + (v_(2))/(a_(2))))` |
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| 188. |
Which of the following statements is not correct regarding the motion of a particle in a straight line?A. x-t graph is a parabola, if motion is uniformly accelerated.B. v-t is a straight line inclined to the time-axis, if motion is uniformly accelerated.C. x-t graph is a straight line inclined to the time axis if motion is uniform and acceleration is zero.D. v-t graph is a parabola if motion is uniform and acceleration is zero. |
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Answer» Correct Answer - D For uniform motion with zero acceleration, v-t graph is a straight line parallel to the time axis. |
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| 189. |
The displacement of a body is proporticonal to the cube of time elapsed. What is the nature of the acceleration of the body ?A. increasing with timeB. decreasing with timeC. constant but not zeroD. zero |
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Answer» Correct Answer - A As `x propto t^(3)` Velocity, `v propto 3t^(2)` Acceleration, `a propto 6t` |
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| 190. |
Assertion : For a particle, acceleration-time graph gives the velocity. Reason : Rate of change of velocity is acceleration.A. If both assertion and reason are trure and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanations of assertion.C. If assertion is true but reason is false.D. If both assertion and reason is false |
| Answer» Correct Answer - (b) | |
| 191. |
Match the Column I with Column II. A. A - p, B - q, C - s, D - rB. A - q, B - p, C - r, D - sC. A - s, B - r, C - q, D - pD. A - r, B - q, C - s, D - p |
| Answer» Correct Answer - D | |
| 192. |
Match Column I with Column II. A. A - p, B - q, C - r, D - sB. A - r, B - s, C - p, D - qC. A - s, B - p, C- q, D - rD. A - q, B - r, C - s, D - p |
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Answer» Correct Answer - C Instaneouus velocity, `v=lim_(Deltato0) (Deltax)/(Deltat), A-s` Average Velocity, `vecv=(Deltax)/(Deltat), C-q` Average acceleration, `veca= (Deltav)/(Deltat), D-r` |
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| 193. |
The area under acceleration-time graph givesA. intial velocityB. final velocityC. change in velocityD. distance travelled |
| Answer» Correct Answer - C | |
| 194. |
A particle is moving in sand under acceleration `a=g-lambday^(2)`, where lambda is a positive constant and `y` is the distance traveled on the y-axis. If the initial velocity is `v_(0)` and after traveling a distance `y_(m)`, the particle stops, find `lambda` (the motion is along the y-axis). |
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Answer» `a=g-lambday^(2)` `v(dv)/(dy)=g-lambday^(2)` `int_(v_(0))^(0) v dv=int_(0)^(y_(m))(g-lambday^(2))dy` `|v^(2)/2|_(v_(0))^(0)=|gy-(lambday^(3))/3|_(0)^(y_(m))` `0-v_(0)^(2)/2=gy_(m)-(lambday_(m)^(3))/3` `(lambday_(m)^(3))/3=gy_(m)+v_(0)^(2)/2=(2g y_(m)+v_(0)^(2))/2` `lambda=(6 g y_(m)+3v_(0)^(2))/(2 y_(m)^(3))` |
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| 195. |
The displacement `x` of a particle varies with time `t` as `x = ae^(-alpha t) + be^(beta t)`. Where `a,b, alpha` and `beta` positive constant. The velocity of the particle will.A. go on decreasing with timeB. be independent of `alpha` and `beta`C. drop to zero when `alpha=beta`D. go on increasing with time |
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Answer» Correct Answer - D `x=a e^(-alphat)+b e^(betat)` `v=(dx)/(dt)=ae^(-alphat)(-alpha)+b e^(betat)(beta)` `=(-aalpha)/e^(alphat)+b beta e^(betat)` As t increases, `e^(alphat), e^(betat)` increases `(aalpha)/e^(alphat)` decreases hence `v` uncreases. |
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| 196. |
The velocity of a particle moving in the `x` direction varies as `V = alpha sqrt(x)` where `alpha` is a constant. Assuming that at the moment `t = 0` the particle was located at the point `x = 0`. Find the acceleration. |
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Answer» `v=alphasqrt(x)=alphax^(1//2) [Here v=fn(x)]` `(dv)/(dx)=alpha.(1)/2x^(-1//2)` Acceleration `a=v(dv)/(dx)=alphax^(1//2) alpha/2x^(-1//2)=alpha^(2)/2` Acceleration is constant. Using equation of motion, `v=u+at=0+alpha^(2)/2t` `=(alpha^(2)t)/2` |
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| 197. |
Refer to Example 30, if the time of journey is minimum, find the value of v in terms of given quantities and also the minimum time. |
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Answer» `t=d/v+v/2(1/alpha+1/beta)=dv^(-1)+v/2(1/alpha+1/beta)` for t to be minimum `(dt)/(dv)=d(-1)v^(-2)+1/2(1/alpha+1/beta)=0` `d/v^(2)=(alpha+beta)/(2alphabeta)` `v=sqrt((2alphabetad)/(alpha+beta))` `t_(min)=d/sqrt((2alphabetad)/(alpha+beta))+1/2sqrt((2alphabetad)/(alpha+beta)).((alpha+beta)/(2alphabeta))` `=sqrt(((alpha+beta)d)/ (2alphabeta))+=sqrt(((alpha+beta)d)/ (2alphabeta))` `sqrt((2(alpha+beta)d)/(alphabeta))` |
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| 198. |
The velocity of a particle is given by `v=v_(0) sin omegat`, where `v_(0)` is constant and `omega=2pi//T`. Find the average velocity in time interval `t=0 `to `t=T//2.` |
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Answer» `v=v_(0) sin omega t` `barv=(overset(T//2)underset(0)intvdt sin(omegat)dt)/(overset(T//2)underset(0)int dt)` `(v_(0)(|-cos(omegat)|_(0)^(T//2))/omega)/(|t|_(0)^(T//2)=(T/2-0))` `=(2v_(0))/(omegaT)[{-cos((omegaT)/2)}-{-cos(0)}]` `=(2v_(0))/(2pi)(-cospi+1)` `=v_(0)/pi{-(-1)+1} (cos pi=cos 180^(@)=-1)` `=2/piv_(0)` |
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| 199. |
The displacement-time graph for two particles A and B are straight lines inclined at angles of `30^(@)` and `60^(@)` with the time axis. The ratio of velocities of `V_(A) :V_(B)` isA. `1 : 2`B. `1 : sqrt3`C. `sqrt3 : 1`D. `1 : 3` |
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Answer» Correct Answer - D `(v_(A))/(v_(B)) = (tan 30^(@))/(tan 60^(@))` |
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| 200. |
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) gt t_(2)`B. `t_(1) = t_(2)`C. `t_(1) lt t_(2)`D. `t_(1)` can be greater or smaller depending upon the initial velocity of the body |
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Answer» Correct Answer - C In upward motion, resistance of air acts downward while in downward motion it acts upwards, always opposite to velocity. |
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