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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: A body is thrown vertically upwards with an initial speed 25 m/s from a position 1. It falls back to position 1 after some time. During this time duration, total change of velocity of the body is zero. R : Average acceleration of the body during this time is zeroA. If both A and B are true and R is the correct explanation of AB. If both A and R are true but R is not correct explantion of AC. If A is true but R is falseD. If both A and R are false |
| Answer» Correct Answer - D | |
| 102. |
A ball thrown vertically upwards with an initial velocity of 1.4 m/s returns in 2s. The total dispalcement of the ball isA. 22.4 cmB. zeroC. 44.8 mD. 33.6 m |
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Answer» Correct Answer - B Since the ball return back to its initial position, the displacement is zero. |
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| 103. |
A body starting with a velocity v returns to its initial position after t second with the same speed, along the same line. Acceleration of the particle isA. `(-2v)/(t)`B. zeroC. `(v)/(2t)`D. `(t)/(2v)` |
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Answer» Correct Answer - A `a=(v-u)/(t)` |
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| 104. |
A body starting from rest moving with uniform acceleration has a displacement of `16 m` in first `4` seconds and `9 m` in first `3` seconds. The acceleration of the body is :A. `1ms^(-2)`B. `2ms^(-2)`C. `ms^(-2)`D. `4ms^(-2)` |
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Answer» Correct Answer - B `S_(n)=a(n-(1)/(2)), S=(1)/(2)an^(2)` |
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| 105. |
Velocity-time graph for a moving object is shown in the figure. Total displacement of the object during the time interval when there is non-zero acceleration and retardation is. A. 60 mB. 50 mC. 30 mD. 40 m |
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Answer» Correct Answer - B area under trapezium gives displacement. |
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| 106. |
Velocity-time graph for a moving object is shown in the figure. Total displacement of the object during the time interval when there is non-zero acceleration and retardation is. A. 60 mB. 50 mC. 30 mD. 40 m |
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Answer» Correct Answer - B Area under trapezium gives displacement. |
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| 107. |
Given a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of `v` and `a` in the three intervals. What are the accelerations at the points ` A, B , C` and ` D` ? . |
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Answer» Average acceleration is greatest in interval 2 Average speed is greatest in interval 3 v is positive in intervals 1, 2, and 3 a is positive in intervals 1 and 3 and negative in interval 2 a = 0 at A, B, C, D Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time. Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval. Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, average speed of the particle is the greatest in interval 3. In interval 1: The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval. In interval 2: The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity. In interval 3: The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval. Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero. |
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| 108. |
Which of the following statements is incorrect? (i) Average velocity is path length divided by time interval. (ii) In general, speed is greater than the magnitude of the velocity. (iii) A particle moving in a given direction with a non-zero velocity can have zero speed. (iv) The magnitude of average velocity is the average speed.A. (ii) and (iii)B. (ii) and (iv)C. (i), (iii) and (iv)D. All four |
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Answer» Correct Answer - C (c ) : Average velocity = Displacement/Time interval A particle moving in a given direction with non-zero velocity cannot have zero period. In general, average speed is not equal to magnitude of average velocity. However it can be so if the motion is aolong a straight line without change in direction . |
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| 109. |
The three initial and final positions of a man on the x-axis are given as (i) (-3 m, 7 m) (ii) (7 m, -3 m) (iii) (-7 m, 3 m)A. (i)B. (ii)C. (iii)D. i) and iii) |
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Answer» Correct Answer - B (b) : Displacement = Final position - Initial position `delta x = x_2 - x_1` Then, (i) `delta x` = 7 m - (-3 m) = 10 m (ii) `delta x` = - 3 m - 7 m = - 10 m `delta x` = - 3 m - (- 7 m) = 10 m |
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| 110. |
An elevator car whose floor to ceiling distance is equal to `2.7 m` starts ascendiung with constant acceleration `1.2(m)/(s^(2))`, 2 sec after the start a bolt begins falling from the ceiling of the car. Answer the following question `g=9.8 m//s^(2)` Distance covered by the bolt during the free fall time w.r.t. ground frame.A. 0.7 mB. 0.9 mC. 1.1 mD. 1.3 m |
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Answer» Correct Answer - D Distance `s|Deltaty|+(2u^(2))/(2g)=0.7+((2.4)^(2))/(9.8)=1.3m` |
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| 111. |
In Q.94, time for which the car decelerates is:A. `(alpha)/(alpha + beta)t`B. `(beta)/(alpha + beta) t`C. `(alpha)/(beta) t`D. `(beta)/(alpha)t` |
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Answer» Correct Answer - A `t_(1) = (v_(m))/(alpha) = ((beta)/(alpha + beta))t` |
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| 112. |
An elevator car whose floor to ceiling distance is equal to `2.7 m` starts ascendiung with constant acceleration `1.2(m)/(s^(2))`, 2 sec after the start a bolt begins falling from the ceiling of the car. Answer the following question `g = 9.8 m//s^(2)` Distance moved by elevator car w.r.t. ground frame during the free fall time of the bolt.A. 1.44 mB. 1.63mC. 1.68mD. 1.97m |
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Answer» Correct Answer - D `y=ut+(1)/(2)at^(2)=(2.4)(0.7)+(1)/(2)(1.2)(0.7)^(2)` |
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| 113. |
In a detective story a body is found 5 m away from the base of a building and beneath an open window `25 m` above. Would you guess the death to be accidental or not ? Explain your answer. `(g = 10 m//s^(2))` |
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Answer» Correct Answer - Accidental As `t = sqrt((2h//g)) = sqrt((2 xx 25)/(10)) = (sqrt5) s`, the horizontal velocity of body would have been `u = (x)/(t) = 5//sqrt5 = sqrt5 m//s ~= 223 m//s` (about 20 % of world class sprint speed of 10 m//s). As this horizontal launch speed cannot be achieved by a standing person himself without a push from behind or by throwing him, so the case is not accidental |
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| 114. |
Check up only the correct statement in the following.A. A body has a constant velocity and still it can have a varying speedB. A body has a constant speed but it can have a varying velocityC. A body having constant speed cannot have any accelerationD. A body in motion under a force acting upon it must |
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Answer» Correct Answer - B Just like on circular motion where speed is constant but velocity is variable. |
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| 115. |
Starting from rest a particle moves in a straight line with acceleration `a = (25 - t^(2))^(1//2) m//s^(2) " for " 0 le t le 5s` `a = (3pi)/(8) m//s^(2) " for " t gt 5s` The velocity of particle at `t = 7 s` is:A. 11 m/sB. 22 m/sC. 33 m/sD. 44 m/s |
| Answer» Correct Answer - B | |
| 116. |
A car starts from rest and moves with uniform acceleration `a`, At the same instant from the same point a bike crosses with a uniform velocity `u`. When and there will they meet ? What is the velocity of car with respect to the bike at the time of meeting? |
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Answer» `underset("car")(s)=(1)/(2)at^(2)to(1),underset("bike")(s)=ut to(2)` if they meet at the same point `underset("car")(s)=underset("bike")(s)` `(1)/(2)at^(2)=ut impliest=(2u)/(3)` ltbr `S_("bike")=ut=u(2u)/(a)=(2u^(2))/(a)` `V_("car")=at=2u` `V_("car")` w.r.t. bike at the time of meeting, `|vec V_(cb)|=|vec(V)_("car")-vec(V)_("bike")|=2u-u=u` |
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| 117. |
A ball is dropped on the floor from a height of 10m. It rebounds to a height of 2.5 m if the ball is in contact with floor for 0.01 s then the average acceleration during contact is nearlyA. `500sqrt(2) m//s^(2)` upwardsB. `1800sqrt(2) m//s^(2)` downwardsC. `1500sqrt(2) m//s^(2)` upwardsD. `1500sqrt(2) m//s^(2)` downwards |
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Answer» Correct Answer - C `a=(sqrt(2gh_(2))-(-sqrt(2gh_(1))))/(Delta t)` |
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| 118. |
A body starting from rest moves along a straight line with a constant acceleration. The variation of speed (v) with distance (s) is represented by the graph:A. B. C. D. |
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Answer» Correct Answer - C `a = (vdv)/(ds)`, for constant acceleration. `int v dv = int a dv` `v^(2) = 2as` Graph is parabolic |
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| 119. |
A body is moving with uniform velocity of `8 m s^(-1)`. When the body just crosses another body, the second one starts and moves with uniform acceleration of `4 m s^(-2)`. The distance covered by the second body when they meet is .A. 8 mB. 16 mC. 24 mD. 32 m |
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Answer» Correct Answer - D `s=8t=8xx4=32m` |
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| 120. |
A body moves from rest with a constant acceleration. Which one of the following graphs represents the variation of its kinetic energy K with the distance travelled x?A. B. C. D. |
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Answer» Correct Answer - C `E = (1)/(2) mv^(2)` `E = (1)/(2) m xx 2a xx s` `E prop s` |
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| 121. |
If `S_(n)=2+0.4n` find initial velocity and accelerationA. `2.2` units. `0.4` unitsB. 2.1 units, 0.3 unitsC. 1.2 units, 0.4 unitsD. 2.2 units, 0.3 units |
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Answer» Correct Answer - A `s_(n)=u+an-(1)/(2)a=(u-(1)/(2)a)+an` ..(1) `s_(n)=2+0.4n` …(2) From (1) and (2) `0.4n=an,u-(1)/(2)a=2` |
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| 122. |
A body falls freely from a height of 125 m `(g=10 m//s^(2))`. After 2 sec gravity ceases to act find time taken by it to reach the ground? |
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Answer» Distance covered in 2s under gravity `S_(1)=(1)/(2)g t^(2)=(1)/(2)(10)(2)^(2)=20m` Velocity at the end of 2s `V=g t=10 xx2=20 m//s` Now at this instant gravity ceases to act, there after velocity becomes constant. The remaining distance which is `125-20=105m` is covered by body with constant velocity `20 m//s` time taken to cover 105 m with constant velocity is given by `t_(1)=(S)/(V)impliest_(1)=(105)/(20)=5.25s` Total time taken `=2+5.25=7.25s` |
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| 123. |
It is a common observation that rain clouds can be at about a kilometer altitude above the ground . (a) If a rain drop falls from such a height freely under gravity, what will be its speed ? Also calcualte in `km//h (g = 10 m//s^(2))`. (b) A typical rain drop is about 4 mm diameter. Momentum is mass `xx` speed in magnitude. Estiamate its momentum when its hits ground. (c) Estimate the time required to flatten the drop. (d) Rate of change of momentum is force. Estiamate how much force such a drop would exert on you. (e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drop is 5 cm. (Assume that umbrealla is circular and has a diameter of 1 m and cloth is not peicreced through.) |
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Answer» Given, height `(h) = 1 km = 1000 m` `g = 10 m//s` (a) velocity attained by the rain drop in freely falling through a height h. `v = sqrt(2gh) = sqrt(2 xx 10 xx 1000) = 100sqrt(2) m//s` `= 100 sqrt(2) xx (60 xx 60)/(1000) km//h` `= 360sqrt(2) km//h ~~ 510 km//h` (b) Diameter of the drop `(d) = 2r = 4 mm` `:.` Radius of the drop `(r) = 2 mm = 2 xx 10^(-3) m` Mass of rain drop `(m) = V xx rho` `= 4/3 pir^(3)rho` `= 4/3 xx 22/7 xx (2 xx 10^(-3))^(2) xx 10^(3)` , (`:.` Density of water `= 10^(3) kg//m^(3)`) `= 3.4 xx 10^(-5) kg` Momentum of the rain drop `(p) = mv` `= 3.4 xx 10^(-5) xx 100sqrt(2)` `= 4.7 xx 10^(-3) kg-m//s` `= 5 xx 10^(-3) kg-m//s` (c ) Time required to flatten the drop `=` time taken by the drop to travel the distance equal to the diameter of the drop near the ground `t = (d)/(v) = (4 xx 10^(-3))/(100sqrt(2)) = 0.028 xx 10^(-3) s` `= 2.8 xx 10^(-5) s = 30s` (d) Force exerted by a rain drop `F = ("Change in momentum")/("Time") = (p - 0)/(t)` `= (4.7 xx 10^(-3))/(2.8 xx 10^(-5)) ~~ 168 N` (e) Radius of the umbrella `(R) = 1/2 m` `:.` Area of the umbrella `(A) = piR^(2) = (22)/(7) xx ((1)/(2))^(2) = (22)/(28) = 11/14 ~~ 0.8 m^(2)` Number of drops striking the umbrella simultaneously with average separation of 5 cm `= 5 xx 10^(-2) m` `= (0.8)/((5 xx 10^(-2))^(2)) = 320` `:.` Net force exerted on umbrella `= 320 xx 168 = 53760 N = 54000 N` |
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| 124. |
A stone projected upwards with a velocity u reaches two points P and Q separated by a distance h with velocities `u//2` and `u//3`. The maximum height reached by it isA. `(9h)/(5)`B. `(18h)/(5)`C. `(36h)/(5)`D. `(72h)/(5)` |
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Answer» Correct Answer - C `v^(2)-u^(2)=2as,H=(u^(2))/(2g)` |
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| 125. |
A body projected vertically upwards with a velocity of `19.6 m//s` reaches a height of 19.8 m on earth. If it is projected vertically up with the same velocity on moon, then the maximum height reached by it isA. 19.18 mB. 3.3 mC. 9.9 mD. 118.8 m |
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Answer» Correct Answer - D `h=(u^(2))/(2g),halpha(1)/(g)implies(h_(E))/(h_(M))=(g_(M))/(g_(E))` |
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| 126. |
A train of `150 m` length is going toward north direction at a speed of `10 ms^-1`. A parrot flies at a speed of `5 ms^-1` toward south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to.A. `30 s`B. `15 s`C. `8 s`D. `10 s` |
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Answer» Correct Answer - D `t = (150)/(10 + 5)` |
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| 127. |
A freely falling body travels-- of total distance in 5th secondA. `8%`B. `12%`C. `25%`D. `36%` |
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Answer» Correct Answer - D `(S_(n))/(h)=(2n-1)/(n^(2))` |
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| 128. |
A stone projected vertically up from the ground reaches a height `y` in its path at `t_(1)` seconds and after further `t_(2)` seconds reaches the ground. The height y is equal toA. `(1)/(2)g(t_(1)+t_(2))`B. `(1)/(2)g(t_(1)+t_(2))^(2)`C. `(1)/(2)g t_(1)t_(2)`D. `g t_(1)t_(2)` |
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Answer» Correct Answer - C `h=y=(1)/(2)g t_(1)t_(2)` |
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| 129. |
A stone thrown vertically up from the ground reaches a maximum height of 50 m in 10s. Time taken by the stone to reach the ground from maximum height isA. 5 sB. 10 sC. 20 sD. 25 s |
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Answer» Correct Answer - B `t_(a)=t_(d)` |
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| 130. |
A man walks on a straight road from his home to a market ` 2.5 km` away with a speed of `5 km //h`. Finding the market closed, he instantly turns and walks back with a speed of `7.5 km//h`. What is the (a) magnitude of average velocity and (b) average speed of the man, over the interval of time (i) `0` to 30 min `. (ii) 0 to 50 min (iii) 0 to 40 min ?A. 0,4B. 0,6C. 0,8D. 0,12 |
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Answer» Correct Answer - B (a) `overline(V)_("avg")=(overline(S)_(1)+overline(S)_(2))/(t_(1)+t_(2))`, (b) `V_("avg")=(S_(1)+S_(2))/(t_(1)+t_(2))` |
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| 131. |
A particles is dropped from certain height. The tame taken by it to fall through successive distance of 1 m each will beA. all equal being equal to `sqrt(2//g)` secondB. in the ratio of the square roots of the integers 1,2,3…C. in the ratio of the difference in the square roots of the integers i.e., `sqrt(1),(sqrt(2)-sqrt(1)),(sqrt(3)-sqrt(2)),(sqrt(4)-sqrt(3))`,….D. In the ratio of the reciprocals of the square roots of the integers, i.e., `(1)/(sqrt(1)),(1)/(sqrt(2)),(1)/(sqrt(3))` ,,… |
| Answer» Correct Answer - C | |
| 132. |
Two bodies of different masses `m_(a)` and `m_(b)` are dropped from two different heights, viz, a and b. The ratio of time taken by the two to drop through these distance is:A. `a : b`B. `m_(a)//m_(b) : b//a`C. `sqrta : sqrtb`D. `a^(2) : b^(2)` |
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Answer» Correct Answer - C `a = (1)/(2) g t_(1)^(2) rArr t prop sqrth` |
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| 133. |
A ball of mass `m_(1)` and another ball of mass `m_(2)` are dropped from equal height. If the time taken by the balls are `t_(1)` and `t_(2)`, respectively, thenA. `t_(1)=t_(2)`B. `t_(1)=2t_(2)`C. `t_(1)/t_(2)=m_(1)/m_(2)`D. `t_(1)/t_(2)=m_(2)/m_(1)` |
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Answer» Correct Answer - A Time is independent of mass. `h=1/2g t_(1)^(2)=1/2g t_(2)^(2)impliest_(1)=t_(2)` |
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| 134. |
Two bodies of different masses are dropped simultaneously from the top of a tower. If air resistance is proportional to the mass of the body then,A. The heavier body reaches the ground earlier.B. the lighter body reaches the ground earlier.C. both the bodies reach the ground simultaneously.D. cannot be decided |
| Answer» Correct Answer - C | |
| 135. |
At a metro station, a girl walks up a stationary escalator in time `t_1` If she remains stationary on the escalator, then the escalator take her up in time `t_2`. The time taken by her to walk up the moving escalator will be.A. `(t_(1)+t_(2))//2`B. `t_(1)t_(2)//(t_(2)-t_(1))`C. `t_(1)t_(2)//(t_(2)+t_(1))`D. `t_(1)-t_(2)` |
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Answer» Correct Answer - C velocity of girl `v_(g)=(L)/(t_(1))` Velocity of escalator `v_(e)=(L)/(t_(2))` Net velocity of the girl`=v_(g)+v_(e)=(L)/(t_(1))+(L)/(t_(2))` `(L)/(t)+(L)/(t_(1))+(L)/(t_(2))impliest=(t_(1)t_(2))/(t_(1)+t_(2))` |
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| 136. |
A car , moving with a speed of `50 km//hr` , can be stopped by brakes after at least ` 6m `. If the same car is moving at a speed of `100 km//hr`, the minimum stopping distance isA. 12 mB. 18 mC. 24 mD. 6 m |
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Answer» Correct Answer - C `v^(2)-u^(2)=2as,v=0,u^(2)props` |
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| 137. |
A car moving with a speed of `40 km//h` can be stopped by applying the brakes after at least 2 m. If the same car is moving with a speed of `80 km//h`, what is the minimum stopping distance?A. `8 m`B. `2 m`C. `4 m`D. `6 m` |
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Answer» Correct Answer - A `v^(2)=u^(2)-2as` `0=(40)^(2)-2axx2 …(i)` `0=(80)^(2)-2axxd …(ii)` `d=8 m` |
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| 138. |
A person standing on the roof of a building throws a ball vertically upward at an instant `t = 0`. The ball leaves his hand with an upward speed 20 m/s and it is then in free fall. The ball rises to a certain height and then moves down. On its way down, the ball just misses to hit the roof of the building and keeps falling towards the earth. the ball hits earth at `t = 5`sec. considering that (i) the vertically upward direction is the positive Y-direction (ii) the position of ball at `t = 0` is the origin (iii) the ball does not rebound and comes to rest at the same place where it hits earth and (iv) air resistance is negligible. (Take `g = 10 m//s^(2)`) Position -time graph for the given motion of the ball is :A. B. C. D. |
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Answer» Correct Answer - A `- H = ut - (1)/(2) g t^(2)` Position 1 st increases then decreases. Velocity 1 st decrease then increases. Acceleration remains constant. |
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| 139. |
A person standing on the roof of a building throws a ball vertically upward at an instant `t = 0`. The ball leaves his hand with an upward speed 20 m/s and it is then in free fall. The ball rises to a certain height and then moves down. On its way down, the ball just misses to hit the roof of the building and keeps falling towards the earth. the ball hits earth at `t = 5`sec. considering that (i) the vertically upward direction is the positive Y-direction (ii) the position of ball at `t = 0` is the origin (iii) the ball does not rebound and comes to rest at the same place where it hits earth and (iv) air resistance is negligible. (Take `g = 10 m//s^(2)`) Maximum displacement of the ball from the initial position is :A. `45 hat(j) m`B. `-45 hat(j)m`C. `25 hat(j) m`D. `-25 hat(j)` |
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Answer» Correct Answer - D `- H = ut - (1)/(2) g t^(2)` Position 1 st increases then decreases. Velocity 1 st decrease then increases. Acceleration remains constant. |
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| 140. |
A person standing on the roof of a building throws a ball vertically upward at an instant `t = 0`. The ball leaves his hand with an upward speed 20 m/s and it is then in free fall. The ball rises to a certain height and then moves down. On its way down, the ball just misses to hit the roof of the building and keeps falling towards the earth. the ball hits earth at `t = 5`sec. considering that (i) the vertically upward direction is the positive Y-direction (ii) the position of ball at `t = 0` is the origin (iii) the ball does not rebound and comes to rest at the same place where it hits earth and (iv) air resistance is negligible. (Take `g = 10 m//s^(2)`) Average velocity of the ball from `t = 0` to `t = 5` secA. `10 hat(j) m//s`B. `-5 hat(j) m//s`C. `-8 hat(j) m//s`D. `-9 hat(j) m//s` |
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Answer» Correct Answer - B `- H = ut - (1)/(2) g t^(2)` Position 1 st increases then decreases. Velocity 1 st decrease then increases. Acceleration remains constant. |
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| 141. |
Consider a rubber ball freely falling from a height ` h = 4.9 m` onto a horizontally elastic plate. Assume that the duration of collision is negligible and the collisions with the plate is totally elastic . Then the velocity as a function of time and the height as a function of time will be :A. B. C. D. |
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Answer» Correct Answer - B Consider downward direction negative and upward positive. for downward motion, `v = - g t`. For upward motion, `v = + g t`. As the collisions are elastic, the magnitude of velocity will remain the same. The direction will change after each collision, starting with negative direction. Also, `v prop t`, hence the graph will be straight line. Also, `h = 4.9 - (1)/(2) g t^(2)` (before collision) `h = (1)/(2) g t^(2)` (after collision) As `h prop t^(2)`, hence graph will be parabolic with `h = 4.9 m` at `t = 0`. |
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| 142. |
A person standing near the edge of the top of a building throws two balls A and B. The ball A is thrown vertically upward and B is thrown vertically downward with the same speed. The ball A hits the ground with speed `v_A` and the ball B hits the ground wiht a speed `v_B`. We haveA. `V_(A) lt V_(B)`B. `V_(A) lt V_(B)`C. `V_(A)=V_(B)`D. the relation between `V_(A)` and `V_(B)` depends on height of the building above the ground. |
| Answer» Correct Answer - C | |
| 143. |
Two balls are projected simultaneously with the same velocity u from the top of a tower, one vertically upwards and the other vertically downwards. Their respective times of the journeys are `t_(1)` and `t_(2)` At the time of reaching the ground, the ratio of their final velocities isA. `1:1`B. `1:2`C. `2:3`D. `2:1` |
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Answer» Correct Answer - A `v=sqrt(u^(2)+2gh)` is same for both the bodies |
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| 144. |
A stone is allowed to fall from the top of a tower 300 m height and at the same time another stone is projected vertically up from the ground with a velocity `100 m//s`. Find when and where the two stones meet? |
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Answer» Height of the tower `h=300m` Suppose the two stones meet at a height x from Z ground after t second `t=(S_(r))/(u_(r)),u_(r)=u+0=u,S_(r)=h` `t=(h)/(u)=(300)/(100)=3sec` Height of the stone from the point of projection is `X=ut-(1)/(2 g t^(2))=100xx3-(1)/(2)xx9.8xx9=255.9m` |
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| 145. |
A ball is projected vertically upwards with a velocity of `25ms^(-1)` from the bottom of a tower. A boy who is standing at the top of a tower is unable to catch the ball when it passes him in the upward direction. But the ball again reaches him after 3 sec when it is falling. Now the boy catches it then the height of the tower is `(g=10ms^(-2))`A. 5mB. 10 mC. 15 mD. 20 m |
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Answer» Correct Answer - D `t=(2v)/(g),v^(2)u^(2)=-2gh` |
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| 146. |
A ball is thrown vertically upwards with a speed of `10 m//s` from the top of a tower 200 m height and another is thrown vertically downwards with the same speed simultaneously. The time difference between them on reaching the ground is `(g=10 m//s^(2))`A. 12 sB. 6 sC. 2 sD. 1 s |
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Answer» Correct Answer - C `Deltat=(2u)/(g)` |
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| 147. |
A stone is dropped from the top of a `400 m` high tower. At the same time another stone is projected vertically upwards from the ground with a speed of `50 m//s`. The two stones will cross each other after a timeA. 2 sB. 4 sC. 6 sD. 8 s |
| Answer» Correct Answer - D | |
| 148. |
A ball is dropped from the top of an 80 m high tower After 2 s another ball is thrown downwards from the tower. Both the balls reach the ground simultaneously. The initial speed of the second ball isA. `10 m//s`B. `20 m//s`C. `30 m//s`D. `40 m//s` |
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Answer» Correct Answer - C Let the first ball reaches the ground after time t, the second ball will take time `(t-2)`. `80=1/2g t^(2)impliest=4 s` Let u: initial velocity of the second ball `80=u(t-2)+1/2g (t-2)^(2)` `=u(4-2)=1/2xx10(4-2)^(2)` `=2u+20` `u=30 m//s` |
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| 149. |
A stone is thrown vertically upwards with an initial speed `u` from the top of a tower, reaches the ground with a speed `3 u`. The height of the tower is :A. `(v^(2))/(g)`B. `(2v^(2))/(g)`C. `(4v^(2))/(g)`D. `(8v^(2))/(g)` |
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Answer» Correct Answer - C According to the third equation of motion `v^(2)-u^(2)=2as` given `v=3v,u=v and a=g` or `(3v)^(2)-v^(2)=2gs or s=(4v^(2))/(g)` |
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| 150. |
A stone is dropped from a height h simultaneously another stone is thrown up from the ground with such a velocity so that it can reach a height of 4h. The time when two stones cross each other is `sqrt(((h)/(kg)))` where `k = "_____"` |
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Answer» Correct Answer - 8 `S_(1)=(1)/(2)g t^(2), S_(2)=ut-(1)/(2)g t^(2)` `S_(1)+S_(2)=h,4h=(u^(2))/(2g)impliesu=sqrt(8gh)` `impliesut=h` `impliessqrt(8gh)t=h` `impliest=sqrt((h)/(8g))` |
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