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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.
| 101. |
A rod or length 2m starts to roll on a horizontal surface as shown in figure. At the same time an ant starts from one end of the rod to other and when the rod rools through a distance 10m the ant dreaches the other end (B). Calcualte the distance and displacement of the ant (a) with respect to the rod (b) with respect to the ground. |
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Answer» Distance travelled by the ant with respect to the rod =2m Displacement with respect to the rod =2m Distance with respect to the ground =10+2=12m Displacement with respect to the ground `=sqrt(2^(2)+10^(2))=sqrt(4+100)=sqrt(104)m` |
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| 102. |
Two trains `A` and `B` `, 100km` apart are travelling towards each other on different tracks with same starting speed of `50km//h`. The train `A` accelerates at `20km//h^(2)` and the train `B` retards at the rate `20km//h^(2)`. The distance covered by the train `A` when they cross each other isA. `70km`B. `55km`C. `65km`D. `60km` |
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Answer» Correct Answer - D |
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| 103. |
To cross the river in shortest distance, a swimmer should swimming an angle `theta with the upsteram. What is the ratio of the time taken to swim across in the shortest time to that in swimming across over shortest distance. [Asume that the speed of swimmer in still water is greater than the speed of river flow]A. `cos theta`B. `sin theta`C. `tan theta`D. `cot theta` |
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Answer» Correct Answer - B |
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| 104. |
A person walks up a stationary escalator in time `t_(1)`. If he remains stationary on the escalator, then it can take him up in time `t_(2)`. How much time would it take for him to walk up the moving escalator?A. `(t_(1)+t_(2))/2`B. `sqrt(t_(1)t_(2)`C. `(t_(1)t_(2))/(t_(1)+t_(2))`D. `t_(1)+t_(2)` |
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Answer» Correct Answer - C |
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| 105. |
Raindrops are falling vertically with a velocity `10 m//s`. To a cyclist moving on a straight road the rain drops appear to be coming with a velocity of `20m//s`. The velocity of cyclist is :-A. `10m//s`B. `10sqrt(3)m//s`C. `20m//s`D. `20sqrt(3)m//s` |
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Answer» Correct Answer - B |
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| 106. |
A ball is dropped from an elevator at an altitude of `200 m` (Fig.4. 39). How much time will the ball take to reach the ground if the elevatior is . a. Stationary? b. Ascending with velocity `10 m s^(-1)` c. Descending with velocity `10 m s^(-1)? |
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Answer» a. `s=ut+(1)/(2)at^(2) rArr -200=0 xx+(1)/(2) (-10)t^(2)` `rArr t=sqrt40 s` b. `-200=10t-(1)/(2) t^(2)` `rArr t^(2)-2t-40=0 rArr t=1+sqrt41 s` c. `-200=-10t-(1)/(2) 10t^(2) rArr t^(2)+2t-40=0` `rArr t=-t+ sqrt41 s`. |
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| 107. |
A stone is dropped from a balloon at an altitude of ` 300 m`. How ling will the stone take to reach the ground of (a) the balloon is ascending with a velocity of ` 5 ,s^(-1). (b) the ballon is descending with a velocity of ` 5 ms^(-1)` (c ) the balloon is shationary ? |
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Answer» (a) Dere, ` u =- u ms^(-1) , a= 9.8 ms^(-1), ` S = 300 m, t= ?` As,` S= ut + 1/2 at^2` :. ` 300 =- 5 xx t + 1/2 xx 9.8 xx t^2` Sove it for (t). (b) Hder, u = 5 ms^(-3)` , a = 9.8 ms(-2)` S= 300 m`, ` t= ? Use , S= ut + 1/2 at^2` and sove it for (t) . ltBrgt (c ) Here, ` u =0 a -= 9.8 ms^(-2)` , S= 300 m, t= ?` Use, ` S= ut + 1/2 at^2` and sove it for (t). |
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| 108. |
If a stone is to hit at a point which is at a distance `d` away and at a height `h` (Fig. 5.200) above the point from where the stone starts, then what is the value of initial speed `u` if the stone is launched at an angle `theta` ? .A. `(g)/(cos theta) sqrt((d)/(2(d tan theta - h)))`B. `(d)/(cos theta) sqrt((g)/(2(d tan theta - h)))`C. `sqrt((g d^2)/(h cos^2 theta))`D. `sqrt((g d^2)/(d - h))` |
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Answer» Correct Answer - B (b) `h = (u sin theta) t - (1)/(2) "gt"^2` `d = (u cos theta) t` or `t = (d)/(u cos theta)` `h = u sin theta (d)/(u cos theta) -(1)/(2) (g)(d^2)/(u^2 cos^2 theta)` `u = (d)/(cos theta) sqrt((g)/(2(d tan theta - h)))`. |
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| 109. |
Assertion: A balloon ascends from the surface of earth with constant speed. When it was at a height `50 m` above the ground, a packet is dropped from it. To an observer on the balloon, the displacement of the packet, from the moment it is dropped to the moment it reaches the surface of earth, is `50 m` Reason: Displacement (vector) depends upon the reference frame used to measure it.A. Statement -1 is True, Statement-2 is Ture, Statement-2 is a correct explanantion for Statement-1.B. Statement-1 is Ture, Statement-2 is Ture, Statement-2 is Not a correct explanantion for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - D To an observer on the ballon, the ground is moving. |
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| 110. |
A stone projected at angle `53^(@)` attains maximum height 25m during its motion in air. Then its distance from the point of projection where it will fall isA. 75mB. `400/3 m`C. `50 m`D. `60 m` |
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Answer» Correct Answer - A velocity of projectile `=(50sqrt(5))/4 m//s` (by appling equation `v^(2)=u^(2)+2as` in vertical direction ) time of flight `=(8v)/(5g)` distance travelled `=v cos 53^(@) xxT=3/5 v xx8/5 v/g=75 m` (by putting the value of v ) |
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| 111. |
A storne is dropped from the `25` th storey of a multistored building and it reaches the ground in `5 s`. In the first second, it passes through how many storey of the buliding?A. `1`B. `2`C. `3`D. none of ther |
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Answer» Correct Answer - A Suppose `h` be the height of each storedy. Then `25 h= 0+ (1)/(2) xx10 xxt^(2) =(1)/(2) xx 10 xx5^(2)` or `h=5 m` In first second, let the stone passes through `n` storey. So `nxx5 (1)/(2) xx10 xx(1) ^(2)` or `n=1`. |
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| 112. |
A ballon starts risintg from ground from rest at some constant acceleration. After some time, a storne is dropped from it. If the stone reaches the ground in the same time in which balloon reached the dropping poing from ground, find the acceleration of the balloon. |
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Answer» Correct Answer - `(g)/(3)` For balloon, `v=at` (i) `S=(1)/(2)at^(2)` (ii) For stone, -`S=vt-(1)/(2)g t^(2)` (iii) From equation(i), (ii) and (ii) `-(1)/(2)at^(2)=(at) t-(1)/(2)at^(2) rArr a=(g)/(3)`. |
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| 113. |
A ballon moves up vertically such that if a stone is thrown from it with a horizontal velocity `v_(0)` relative to it the stone always hits the ground at a fixed point `2v_(0)^(2)//g` horizontally away from it. Find the height of the balloon as a function of time. |
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Answer» Range `=v_(0)xx"time of flight" (t)` `rArrt=(2v_(0)^(2))/(gv_(0))=(2v_(0))/(g)` If `y` is the height of balloon at any instant `t` and `(dy)/(dt)` its velocity then `y= -((dy)/(dt))((2v_(0))/(g))+(1)/(2)g((2v_(0))/(g))^(2)` ` rArr (dy)/(dt)+(g)/(2v_(0))y-v_(0)=0` `(dy)/(v_(0)-(g)/(2v_(0))y)=dt` ` rArr -(2v_(0))/(g)ln(v_(0)-(g)/(2v_(0))y)=t+c` At `t=0`, `y=0rArrc= -(2v_(0))/(g)lnv_(0)` Simplfying we get `y=(2v_(0)^(2))/(g)[1-e^(-"gt"//2v_(0))]` |
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| 114. |
A stone is projected with speed `20 m//s` at angle `37^(@)` with the horizontal and it hits the ground with speed `12m//s` due to air resistance. Assume the effect of air resistance to reduce only horizontal component of velocity. Then the time of flight will be-A. greater than `2.4 sec`B. less than `2.4 sec`C. `2.4 sec`D. depends on other data |
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Answer» Correct Answer - C Time of flight only depends on the vertical component of the velocity. Hence, `T=(2 u sin theta)/g=2.4 s` |
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| 115. |
A projectile si thrown with velocity of `50m//s` towards an inclined plane from ground such that is strikes the inclined plane perpendiclularly. The angle of projection of the projectile is `53^(@)` with the horizontal and the inclined plane is inclined at an angle of `45^(@)` to the horizonta.. (a) Find the time of flight. (b) Find the distance between the point of projection and the foot of inclined plane. |
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Answer» Correct Answer - (a) `t = 7 sec`, (b) `175 m` `v_(x) = u_(x) = 50 cos 53^(@) = 30m//s` at striking point `v_(y) =- u_(x) =- 30 m//s` (a) form time of flight `v_(y) = u_(y) +a_(y) t rArr -30 = 40 +(-10)t rArr t = 7 sec` ltbrltgt (b) for distance distance `= OQ - MQ = u_(x)t - (u_(y)t +(1)/(2)a_(y)t^(2))` distance `= 30 xx 7- [40 xx 7 +(1)/(2) xx (-10) xx 7^(2)]` distance `= 210 - (280 - 245) rArr distance `= 175 m` |
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| 116. |
A motor boat is to reach to a point `30^(@)` upstream on other side of a river flowing with velocity `5 m//s`.Velocity of motor boat with respect to water is `5sqrt3m//sec`.The driver should steer the boat at an angle ofA. `30^(@)` up w.r.t. the line of destination from the starting pointB. `60^(@)` up w.r.t. normal to the bankC. `120^(@)` w.r.t. stream directionD. None of these |
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Answer» Correct Answer - A, B, C |
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| 117. |
A stone is projected from ground. Its path is a shown in figure. At which point its speed is decreasing at fastest rate? A. AB. BC. CD. D |
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Answer» Correct Answer - A |
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| 118. |
A projectile strikes the inclined plane perpendicularly as shwon in the figure. Which of the following statemetns is correct? A. Just before striking the inclined plane, component of final velocity of particle parallel to the inclined plane is non-zeroB. Initial and final velocities are perpendicular to each other.C. Component of acceleration of particle parallel to the inclined plane before striking the plane is zero.D. None of the above |
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Answer» Correct Answer - D |
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| 119. |
A body slides down a smooth inclind placne when released form the tip, whille another body fall freely the same point . Which on will strike the ground earlier ? |
| Answer» A bod falling freely will reach the ground earlier because its accelerationis g (i.e. acceleration due to gravity) which is greater than the acclceration of ther body `=g sin theta , there theta` is the inclination of the plane with the horizontal. | |
| 120. |
A man who swims at a speed of `5km//h` wants to cross a `500m` wide stream flowing at `4km//h` and reach the point which is directly opposite to his starting point. If he reaches a point somewhere else he has to walk back to desitination, his walking speed being `2km//h`. Find the minimum time in which he can reah his destination. A. 5 minB. 10 minC. 15 minD. 20 min |
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Answer» Correct Answer - B |
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| 121. |
The string in fig. is passing over small smooth pulley rigidly attached to trolley A. If the speed of trolley is constant and qual to `v_(A)` towards right, speed and magnitude of acceleration of block B at the instant shown in figure are A. `v_(A)`B. `4/5v_(A)`C. `3/4 v_(A)`D. `3/5 v_(A)` |
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Answer» Correct Answer - D |
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| 122. |
The string in fig. is passing over small smooth pulley rigidly attached to trolley A. If the speed of trolley is constant and qual to `v_(A)` towards right, speed and magnitude of acceleration of block B at the instant shown in figure are A. `V_(B)=V_(A), a_(B)=0`B. `a_(B)=0`C. `a_(B)=3/5 v_(A)`D. `a_(B)=(16 v_(A)^(2))/(125)` |
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Answer» Correct Answer - C, D |
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| 123. |
On a frictionless horizontal surface , assumed to be the ` x-y` plane , a small trolley `A` is moving along a straight line parallel to the `y-axis `( see figure) with a constant velocity of `(sqrt(3)-1) m//s ` . At a particular instant , when the line `OA` makes an angle of `45(@)` with the `x - axis ` , a ball is thrown along the surface from the origin `O`. Its velocity makes an angle `phi` with the `x -axis and it hits the trolley . (a) The motion of the ball is observed from the frame of the trolley . Calculate the angle `theta` made by the velocity vector of the ball with the ` x-axis in this frame . (b) Find the speed of the ball with respect to the surface , if ` phi = (4 theta )//(4)`. |
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Answer» Correct Answer - (a) `45^(@)` (b) `2m//s` |
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| 124. |
Water drops fall at regular intervals from a roof. At an instant when a drop is about to leave the roof, the separations between 3 successive drops below the roof are in the ratioA. `1 : 2 : 3`B. `1 : 4 : 9`C. `1 : 3 : 5`D. `1 : 5 : 13` |
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Answer» Correct Answer - C |
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| 125. |
A particle is travelling on a curved path. In an interval `Deltat` its speed changed from v to 2v. However, the change in magnitude of its velocity was found to be `|vec(DeltaV)| =sqrt(5)v`. What can you say about the direction of velocity at the beginning and at the end of the interval `(Delta t)`? |
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Answer» Correct Answer - The two velocities are perpendicular |
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| 126. |
Can a body have a. Zero instantaneous velocity and yet be accelerating? b. Zero average speed but non-zero average velocity? c. Negative acceleration and yet be speeding up? d. Magnitude of average velocity be equal to average speed ? |
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Answer» a. Yes for example, a freely falling particle at its highest point has zero velocity but acceclerating downward. b. No , if distance is zero then displacement is also zero, c. Yes if velocity is also in the negative direction. d. Yes if motion takes place continuously in one dirction. |
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| 127. |
(a) Is it possible to be accelerating if you are travelling at constant speed? (b) Is it possible to move on a curved path with zero acceleration, constant acceleration, variable acceleration |
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Answer» Correct Answer - A::B::C (a) On a curvilinear path (a path which is not straight line), even if speed is constant, velocity will change due to change in direction. `:. a!=0` (b) (i) On a curved path velocity will definitely change (at least due to change in direction). `:. a!=0` (ii) In projectile motion, path is curved, but acceleration is constant (=g). (iii) Variable acceleration on curved path is definitely possible. |
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| 128. |
Find the speed of the block B when the wedges A and C are moving toward each other with speed v and the strings connected to block make and angle `theta` with the vertical, as shown in figure |
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Answer» Correct Answer - `[v_(B)=(v(1-sin theta))/(cos theta)]` |
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| 129. |
If the wedge A shown in figure-1.93, is moving toward left with acceleration `3 m/s^(2)`, find the net acceleration of block B which is constrainedtoslide along the wedge surface. `(theta=30^(@))` |
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Answer» Correct Answer - `[(3sqrt(5-2sqrt(3)))m//s^(2)]` |
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| 130. |
A person can throw a ball to a maximum horizontal distance of ` 90.m` Calculat the maximum vertcal heitht to which he can through the ball. Fiven ` g= 10 ms^(-2)`. |
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Answer» Max. Horizontal distace . ` R_(max) = u^2/g = 90` or ` u^2 = 90 g= 90 xx 10 = 900 so `u=30 m//s` ` Max. vertical height, ` H = u^2 /(2g) = (900)/(2 xx 10) = 45 m` . |
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| 131. |
A ball is released from the top of a tower of height `H m`. After `2 s` is stopped and then instantaneously released. What will be its heitht after next `2 s`?.A. `(H-5) m`B. ` (H-10) m`C. ` (H-20) m`D. `(H-40) m` |
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Answer» Correct Answer - D Distance covered by by the object in first `2 s` `h_(1)=(1)/(2) g t^(2) =(1)/(2) xx10xx2^(2) =20 m` Similarly , destance covered by the object invext `2 s` will also be `20 m`, hence the required height `=H-20-20=H-40 m`. |
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| 132. |
A student is standing on the open platform of a moving trains at a speed of `10m//s`. The student throws a ball into the air along a path that the, he judges to make an initial angle of `60^(@)` with the horizotal and to be in line with the track. The professor, who is standing on the ground nearby, observes the ball to rise vertically. Find the height reached by the ball. |
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Answer» Correct Answer - [15m] |
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| 133. |
Trajectories of two projectiles are shown in figure.Let `T_(1)` and `T_(2)` be the time periods and `u_(1)` and `u_(2)` their speeds of projection.Then A. `T_(2)gtT_(1)`B. `T_(1)=T_(2)`C. `u_(1)gtu_(2)`D. `u_(1)ltu_(2)` |
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Answer» Correct Answer - B::D |
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| 134. |
Two balls projected at different times have trajectories `A` and `B` as shown below. The balls do no collie. Which statement(s) is/are correct? A. velocity of ball `B` must be greater than velocity of ball `A`B. Ball `A` is in the air for a longer time than ball `B`C. Ball `B` is in the air for a longer time than ball `A`D. Ball `B` has same acceleration as that of ball `A` |
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Answer» Correct Answer - C::D |
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| 135. |
A boy throws a ball up ward with a speed of 12m/s The wind imparts a horizontal acceleration of `0.4 m//s^(2)`. At what angle `theta` to the vertical, the ball must be thrown so that it returns to the point of release. |
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Answer» Correct Answer - `[tan^(-1)((1)/(25))]` |
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| 136. |
The figure shows a velocity-time graph of a particle moving along a straight line If the particles starts form the position `x_(0) =- 15m`, then its position at `t = 2s` will beA. `-5`B. `5m`C. `10m`D. `15m` |
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Answer» Correct Answer - A displacement `=` Area convered by `v-t` graph on time axis `x - x_(0) = (1)/(2) xx 2xx 10` `x - (-15) = 10 rArr x =- 5m` |
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| 137. |
A particle of mass `2kg` starts to move at position `x=0` and time `t=0` under the action of force `F=(10+4x)N` along the `x- ` axis on a frictionless horizontal track . Find the power delivered by the force in watts at the instant the particle has moved by distance `5m`.A. `20`B. `5`C. `30`D. `40` |
| Answer» Correct Answer - C | |
| 138. |
A gun is set up in such a wat that the muzzle is at around level as in figure. The hoop A is located at a horizontal distance 40m from the muzzle and is 50m above the ground level. Shell is fired with initial horizontal component of velocity as `40 m//s`. Which of the following is `//`are correct? A. The vertical component of velocity of the shell just after it is fired is `55 m//s`, if the shell has to pass through the hoop a.B. The shell will pass through both the hoops if `x=40 m`C. The shell will pass through both the hoops if `x=20 m`D. The vertical component of velocity of the shell just after it is fired is `45 m//s`, if the shell is to pass through both the hoops |
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Answer» Correct Answer - A::B As given horizontal velocity `=40 m//s` `u cos theta xxt=40, t=1` sec at `t=1`, height `=50 m` `:. 50=u sin theta xx1-1//2xxgxx1rArr u sin theta=55` `:.` Initial vertical component `=u sin theta=55 m//s` As hoop is on same height of the trajectory. So by symmetry x will be 40 m. |
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| 139. |
If time taken by the projectile to reach `Q` is `T`, than `PQ =` A. `Tv sin theta`B. `Tv cos theta`C. `Tv sec theta`D. `Tv tan theta` |
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Answer» Correct Answer - D `T = (2v)/(g cos theta), R = (1)/(2) g sin thetaT^(2)`, `R = (1)/(2)g sin theta T.T = (1)/(2) g sin theta (2v)/(g cos theta) T = TV tan theta` |
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| 140. |
STATEMENT-1: When a particle moves in a circle with a uniform speed, it velocity and acceleration both changes. STATEMENT-2: The centripetal acceleration in circular motion is depdendent on angular velocity of the body.A. Statement -1 is True, Statement-2 is Ture, Statement-2 is a correct explanantion for Statement-1.B. Statement-1 is Ture, Statement-2 is Ture, Statement-2 is Not a correct explanantion for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - B In uniform circular motion, the magnitude of velocity and acceleration remains same, but due to change in direction of motion, the direction of velcotiy and acceleration changes. Also the centripetal acceleration is given by `a = omega^(2) r`. |
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| 141. |
For motion of an object along the x-axis the velocity `v` dipends on the displacement `x` an `v=3x^(2)`, then what is the acceleration at `x=2 m`.A. `48 ms^(-2)`B. `80 ms^(-2) m`.C. `18 ms^(-2)`D. `10 ms^(-2)` |
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Answer» Correct Answer - B Given `v=3x^(2)-2x`, differentiating `v`, we get `(dv)/(dt) =(6x -2) (dx)/(dt) =(6x -2) v` ` rArr a =(6xx 2 -2 ) (3x^(2)-2x)` Now put `x=2 m` `rArr a =(6 xx 2-2) (3(2)^(2)-2xx2)=80 ms^(-2)`. |
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| 142. |
A particle moves along a straight line its velocity dipends on time as `v=4t-t^(2)`. Then for first `5 s`:A. Average velcotu is `25//3 m s^(-1)`B. Average speed is `10 m s^(-1)`.C. Average velcotu is `5//3 m s^(-1)`D. Acceleration is `4 m s^(-2)` at `t=o` |
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Answer» Correct Answer - C::D Average velocity. `vec v = (int _(0)^(5)v dt)/(int_(0)^(5) dt) = (int_(0)^(5) (4t-t^(2)) dt)/(int_(0)^(5) dt)` `=[2t^(2) -t^(3)/3]_(0)^(3)/5 =(50 -(125)/3 )/5 =(25)/(3xx5) =5/3` For average speed, let us put `v=0`, which gives `t=0 and `t =4` s` For average speed, let us put `v=0`, which gives` t=0 and t=4 s` becuase Average speed `=(|underset(0)overset(4)(int)v dt|+|underset(0)overset(5)(int)v dt|)/(underset(0)overset(5)(int)dt)=(|underset(0)overset(4)(int)4t-t^(3)dt|+|underset(4)overset(5)(int)vdt|)/(5)` `=([2t^(2)-(t)^(3)/(3)]_(0)^(4)+[2t^(2)-(t^(3))/(3)]_(4)^(5))/(5)` `=(|[2t^(2)-(t^(3))/(3)]_(0)^(4)|+|[2t^(2)-(t^(3))/(3)]_(4)^(5)|)/(5)=(13)/(5) ms^(-s)` For aceleration: `a=(dv)/(dt)=(d)/(dt)(4t-t^(2))=4-2t`. At `t=0, a=4 ms^(-2)` Therefore, optione (c) and (d) are correct, and options (a) and (b) and wrong. |
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| 143. |
A stone is dropped form certain height which can reach the ground in `5s`. If the stone is stopped after `3s` of its fall and then allowed to fall again. Find the time taken (in second) by the stone to reach the ground for the remaining distance. |
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Answer» Correct Answer - 4 |
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| 144. |
An object , moving with a speed of ` 6.25 m//s `, is decelerated at a rate given by : `(dv)/(dt) = - 2.5 sqrt (v)` where `v` is the instantaneous speed . The time taken by the object , to come to rest , would be : |
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Answer» Given, (dv)/(dt) =- 2.5 sqrt v` or ` (dv)/9sqrtv) =- 2.4 dt` Let the object will come to resta after time (t), then velocity of object will be changing in time (t) from `0 625 ms^(-1)` rozero. Integrating the above relation within the conditions of motion, we get ` int _(6.25) ^(0) (du)/(sqrtv) = int _(0)^(t) -2.5 dt or [2 sqrt 2]_(6.25)^(0) =- 2.5 [t] _(0)^(t)` or ` 0- 2 sqrt 6.25 =- 2.5 t or -2 xx 2.5 =- 2.5 t` or t= 2 s`. |
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| 145. |
Starting from rest a particle is first accelerated for time `t_1` with constant acceleration `a_1` and then stops in time `t_2` with constant retardation `a_2.` Let `v_1` be the average velocity in this case and `s_1` the total displacement. In the second case it is accelerating for the same time `t_1` with constant acceleration `2a_1` and come to rest with constant retardation `a_2` in time `t_3.` If `v_2` is the average velocity in this case and `s_2` the total displacement, thenA. `v_(2)=2v_(1)`B. `2v_(1)ltv_(2)lt4v_(1)`C. `s_(2)=2s_(1)`D. `2s_(1)lts_(2)lt4s_(1)` |
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Answer» Correct Answer - A::D |
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| 146. |
Let `v` and `a` be the instantaneous velocity and acceleration of a particle moving in a plane. Then rate of change of speed `(dv)/(dt)` of the particle is equal toA. `|a|`B. `(v.a)/v`C. the component of `a` parallel to `v`D. the component of `a` perpendicular to `v` |
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Answer» Correct Answer - B::C |
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| 147. |
Determine the sine of the angle between the vectors . `2 hat I + 3 hat j -4 hat k` and ( hat i-2 hat k). |
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Answer» Here `vec A =( 2 hat I + 3 hat j -4 hat k)` and `vec B =(hat I -2 hat j + 2 hat k)` ` (vec A xx vec B) =|(hat I, hat j, hat k), (2, 3, -4),(1, -2, 2)|` `=hat I (6-8 ) + hat j (-4 -4 ) + hat k (-4 -3)` `=- 2 hat I -8 hat j-7 hat k` `| vec A xx vec B| = sqrt ((-2)^(2) + (-8)^(2) + (-7) ^(2)) = sqrt (29),` `| vec B | = sqrt (1^(2) + (-2)^(2) + 2^(2)) =sqrt 9` `sin theta = (|vec A xx vec B|)/(|vec A | | vec B|) =(sqrt117)/(sqrt 29 sqrt9) =(sqrt 23)/(29)` ` or `theta = sin^(-1) sqrt (13)/(29)`. |
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| 148. |
Assertion : We know the relation `a = v. (dv)/ds` . Therefore, if velocity of a particle is zero, then acceleration is also zero. Reason : In the above equation, a is the instantaneous acceleration.A. If the both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D It is not necessary that if `v=0` then acceleration is also zero. If a particle is thrown upwards, then at highest point `v=0`but `!=0` |
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| 149. |
A particle moves in XY plane such that its position, velocity and acceleration are given by `vec(r)=xhat(i)+yhat(j), " "vec(v)=v_(x)hat(i)+v_(y)hat(j), " "vec(a)=a_(x)hat(i)+a_(y)hat(j)` Which of the following condition is correct if the particle is speeding down?A. `xv_(x)+yv_(y) lt 0`B. `xv_(x)+yv_(y) gt 0`C. `a_(x)v_(x)+a_(y)v_(y) lt 0`D. `a_(x)v_(x)+a_(y)v_(y) gt 0` |
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Answer» Correct Answer - C For speeding down `vec(a).vec(v) lt 0 rArr a_(x)v_(x)+a_(y)v_(y) lt 0` |
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| 150. |
The trajectory of a projectile is given by `y=x tantheta-(1)/(2)(gx^(2))/(u^(2)cos^(2)theta)`. This equation can be used for calculating various phenomen such as finding the minimum velocity required to make a stone reach a certain point maximum range for a given projection velocity and the angle of projection required for maximum range. The range of a particle thrown from a tower is define as the distance the root of the tower and the point of landing. A tower is at a distance of `5m` from a man who can throw a stone with a maximum speed of `10m//s`. What is the maximum height that the man can hit on this tower.A. `4.5m`B. `4m`C. `5m`D. `3.75m` |
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Answer» `tantheta=(u^(2))/(Rg)` `H=R tantheta-(1)/(2)g.(R^(2))/(u^(2))(1+tan^(2)theta)` `=(u^(2))/(g)-(1)/(2)g. (R^(2))/(u^(2))-(1)/(2)(u^(2))/(g)=10-1.25-5=3.75m` |
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