1496 + Interview Questions in KINEMATICS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	From the top of at all building (height27.3m), a boy throws an apple upward, which strikes ground after 16 s. Take `g = 9.8 m//s^(2)`, find the speed of apple with which it was thrown and the maxium height reached by it.
Answer» Correct Answer - [76.8 m/s, 327.45 m]

Discussion

2.	A car travels along a straight line for first half time with speed `40 km//h` and the second half time with speed `60 km//h`. Find the average speed of the car.
Answer» Correct Answer - (b)

Discussion

3.	A vechicle travels along a straight path between two places `alpha and beta`. It travels first half of the distance with a velocity of 72 km `h^(-1)` and the remaining distance with a velocity of 36 km `h^(-1)` then the average velocity of the vehicle is ______ m`s^(-1)` A. 11.3B. 13.3C. 15D. 14
Answer» Correct Answer - B A-d A body covering 5 m in every subsequent second moves with constant speed. B-b if a body covers distance in unequal intervals of time, then is said to move with non- uniform speed. C-a A lunar month is based on the periodic motion of the moon around the earth. D-c The age a person is expressed in years.

Discussion

4.	In the case of a simple pendulum, a graph is drawn between the displacement and time taken for its oscillations as shown in the figure . Then From the given figure, the number of oscillation performed by the bob at the end of 21 s is ________ A. 3B. 4C. 2D. `3^(1//2)`
Answer» Correct Answer - D

Discussion

5.	`{:("ColumnA","ColumnB"),(A. "Periodic motion",a."maximum displacement of vibrating particle from its mean position"),(B."Vibrtory motion" ,b."The pistion of a motor car engine at uniform car engine at uniform speed"),(C."Body at rest" ,c."Non- uniform motion"),(D."Amplitude",d.18//5 kmh^(-1)),(E."Variable speed" ,e."OScillating objects undergo change in shape of size"),(F. 1ms^(-1) , f. ("Total displacement")/("Total time")),(G."Average velocity" ,g. "Zero velocity"):}`
Answer» Correct Answer - A::B::C::D

Discussion

6.	`{:( "Column A"," ColumnB"),(A. "Non- uniform motion","a. To measure speed of the vehicle"),(B."Oscillatory motion","b. Uniform motion"),(C."Mean position" , "c. A criket ball that rolls over ground"),(D."Speedometer","d.The needle of sewing machine that moves up and down"),(E."Constant speed","e. Freely suspended pendulum at rest"),(F. "Acceleration","f".("Total change in velocity")/("Total time taken for change")):}`
Answer» Correct Answer - A::B::C::D A-c A cricket ball rolls over the ground , its speed decreases. B-d The needle of sewing machine that moves uo and down is an example of oscillatory motoin. C-e Mean position of a freely suspended pendulum is at that position where it stays at rest . D-a Speedometer is used to measure the speed of the vehicle. E-d A body moving with constant speed is said to be in uniform motion. F-f The rate of change in velocity is called acceleration.

Discussion

7.	`{:( "Column A"," ColumnB"),(A."Speed",a."Uniform motion"),(B."Time",b."Non-uniform motion"),(C."Average speed",c."km h"^(-1)),(D."Uniform speed in a straight line",d."second"),("E. Freely falling body","e. Change in time period"),("F. Change in length of a pendulum","f".("Total distance")/("Total time")):}`
Answer» Correct Answer - A::B::C::D A-c km `h^(-1)` is a unit of speed . B-d Unit of time is second. C-f,c average speed = total distance coverd/ total time km `h^(-1)` is a unit of average speed. D-a An object moving along a straight line with aconstant speed is said to be in uniform motion . E-b Freely falling body is an example of non-uniform motion. F-e Change in length of a simple pendulum results in change in its time period.

Discussion

8.	`{:("Column A" ,"ColumnB"),((A)"A body covers 5 m in every subsequent second",(a)"Lunar month"),((B)"Equal distances in unequal intervals of time",(b)"Non-uniform speed"),((C) "Periodic motion of the moon around the earth",(c)"Years"),((D)"Approprite unit to express the age of a person",(d) "Constant speed"):}`A. `A to b, B to c, C to d,D to a `B. `A to d, B to b, C to a, D to c`C. `A to d, B to a, C to b, D to c`D. ` A to a, B to b, C to c, D to d`
Answer» Correct Answer - B

Discussion

9.	A ball thrown by one player reaches the other in `2 s`. The maximum height attained by the ball above the point of projection will be about.A. 2.5 mB. 5 mC. 7.5 mD. 10 m
Answer» Correct Answer - B (b) `t = (2 u sin theta)/(g)` or `2 = (2 u sin theta)/(g)` or `u sin theta = g` `h_m = (u^2 sin^2 theta)/(2 g) = (g^2)/(2 g) = (g)/(2) = 5 m`.

Discussion

10.	A thief in a stolen car passes through a police check post at his top speed of `90 kmh^-1.` A motorcycle cop, reacting after 2 s, accelerates from rest at `5 ms^-2.` His top speed being `108 kmh^-1.` Find the maximum separation between policemen and thief.A. `112.5 m`B. `115 m` (c) 116.5 m (1) None of theseC. `116.5 m`D. None of these
Answer» Correct Answer - A `90 km//h=90xx5/18=25 m//s` `180 km//h=108xx5/18=30 m//s` At maximum separation their velocities are same. `:.` Velocity of motorcycle=`25 m//s` or `at=25` or `t=5s` But thief has travelled up to 7s. `s_1` =displacement of thief `=v_1 t_1=25xx7=175m` `s_2`=displacement of motorcycle `=1/2xxa_2t_2^2` `=1/2xx5xx(5)^2` `=62.5m` `:.` Maximum separation `=s_1-s_2=112.5 m`

Discussion

11.	In above problem, if the police man is behind the thief and the initial distacen between them is 10 m, how much time will it take of the policmane to catch the thief?
Answer» Let the distacne travelled by the thief by d, when the polceman catches him. The distace travelled by the policeman `=(0.1d)` km= (100). Let the time taken be t `:.2=(d)/(t)` `d=2t` `d=(d+0.1)/(t)` `5=(2t+0.1)/(t)` `5t=2t+0.13t=0.1` `t=(0.1)/(3)h` `h=0.1xx60` minutes `0.1xx20` minutes =2 mintues

Discussion

12.	A policeman is running with a uniform velocity of `5 km h^(-1)` towards east and a thief is running with a uniform velocity of `2 km h^(-1)` towards east. Is it possible for the policemna to catch the theif? Explain
Answer» As the velcity of policeman is greer than the velocity of the thief and both are moving in the same direction, policeman can catch the thief

Discussion

13.	Theree girls skating on a circular ice ground of radius ` 200 m` start from a point (P) on the edge of the ground and reach a point ` Q` diametrically opposite to (P) following different paths as shown in Fig. (NCT) . 17. What is the magnitude of the displacemnt vector for each ? which girl is thsi equal to the actual length of path skated ? .
Answer» Deplacement for each gilr ` = vec (PQ)` :. Magnitude of the displacement for each girl `= PQ = diameter of circular ice ground = 2 xx 200 = 400 m`. For girl (B), the magnitude of displacemnet is equal to the actual length of path skated.

Discussion

14.	Angular position `theta` of a particle moving on a curvilinear path varies according to the equation `theta=t^(3)-3t^(2)+4t-2`, where `theta` is in radians and time t is in seconds. What is its average angular acceleration in the time interval `t=2s` to `t=4s`?
Answer» Like average linear acceleration, the average angular acceleration `alpha_(av)` equals to ratio of change in angular velocity `Deltaomega` to the concerned time interval `Deltat`. `alpha_(av)=(Deltaomega)/(Deltat)=(omega_("final")-omega_("initial"))/(t_("final")-t_("initial"))` ...(i) The angular velocity `omega` being rate of charge in angular position can be obtained by equation `omega=(d theta)/(dt)` Substituting the given expression of the angular position `theta`, we have `omega=3t^(2)-6t+4` ...(ii) From the above eq. (ii), angular velocities `omeha_(2)` and `omega_(4)` at the given instants `t=2s` and `4s` are `omega_(4)=4 rad//s` and `omega_(4)=28 rad//s` Substituting the above values in eq. (1), we have `alpha_(av)=12 rad//s^(2)`

Discussion

15.	Assertion : Average velocity can’t be zero in case of uniform acceleration. Reason : For average velocity to be zero, a non zero velocity should not remain constant.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.
Answer» Correct Answer - D If a particle is projected upwards then `s=0,` when it returns back to its acceleration is constant=g

Discussion

16.	Assertion : Displacement-time equation of two particles moving in a straight line are, `s_1 = 2t - 4t^2` and `s_2 = -2t + 4t^2.` Relative velocity between the two will go on increasing. Reason : If velocity and acceleration are of same sign then speed will increase.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.
Answer» Correct Answer - D `v_1=(ds_1)/(dt)=(2-8t)` `v_2=(ds_2)/(dt)=(-2+8t)` `:. v_(12)=v_1-v_2=(4-16t)` `v_(12)` does not keep on incresing.

Discussion

17.	A helicopter takes off along the vertical with an acceleration `a = 3m//s^(2)` and zero initial velocity. In a certain time the pilot switches off the engine. At the point of take off, the sound dies away in a time `t_(2) = 30 sec`. Determine the velocity of the helicopter at the moment when its engine is switched off assuming that velocity of sound is 320 m/s.
Answer» Correct Answer - `80 ms^(-1)` Let t= time of accelerated motion of the helicopter. Distance travelled by helicopter + Distanace travelled by sound `rArr (1)/(2) xx 3 xx t^(2) =320 (30-t)rArr t=(80)/(3)sec` Final velocity of helicopter `v=u+at =0 +3 xx (80)/(3)= 80 m//s`.

Discussion

18.	A stone is released from an elevator going up with an acceleration of `g/2.` What is the acceleration of the stone just after release ?
Answer» Correct Answer - A::D Stone comes under gravity `:. F=mg` or `a=F/m=g`

Discussion

19.	A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is(a) a upward(b) (g — a) upward(c) (g— a) downward(d) g downward.
Answer» (d) g downward Explanation: Gravity is the only force acting on the stone when it is released. And, we know that gravity is always in the downward direction.

19.

A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is(a) a upward(b) (g — a) upward(c) (g— a) downward(d) g downward.

Answer»

(d) g downward

Explanation:

Gravity is the only force acting on the stone when it is released. And, we know that gravity is always in the downward direction.

Discussion

20.	(a) A car moves around a circular arc subtending an angle of `60°` at the centre. The car moves at a constant speed `u_(0)` and magnitude of its instantaneous acceleration is `a_(0)`. Find the average acceleration of the car over the `60°` arc. (b) The speed of an object undergoing uniform circular motion is `4 m//s`. The magnitude of the change in the velocity during 0.5 sec is also `4 m//s`. Find the minimum possible centripetal acceleration (in `m//s^(2))` of the object.
Answer» Correct Answer - (a) `lt ltagt gt = (3)/(pi) a_(0)` (b) `8.37 m//s^(2)`

Discussion

21.	A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball in `T/3` second?A. ` h//9 m`B. ` 7h//9 m`C. ` 8h//9m`D. ` 17 h//18 m`
Answer» Correct Answer - C We have `h=(1)/(2)g T^(2)` In `T//3` second, distance fallen `=(1)/(2) g((T)/(3))^(2) =(h)/(9)` So position of the ball from the ground is `h-(h)/(9)=(8h)/(9)m`.

Discussion

22.	A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball in `T/3` second?A. `h/9`metre from the groundB. `(7h//9)` metre from the groundC. `(8h//9)` metre from the groundD. `(17h//18)` metre from the ground
Answer» Correct Answer - C `h=1/2 gT^2` At `T/3` second, distance fallen `d=1/2 g(T/3)=h/9` `:.` Height from ground =`h-d=(8h)/9`

Discussion

23.	Water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap, the instant the first drop touches the ground. How far above the ground is the second drop at that instant. `(g = 10 ms^-2)`A. (a) ` 1 . 25 m`B. (b) ` 2. 50 m`C. (c ) ` 3. 75 m`D. (d) ` 6 sec`
Answer» Correct Answer - C Time taken by one drop to fall through ` 5 m` is ` sqrt 9 2S)/g ` = sqrt (2 xx 50/(10) = 1 sec` :. Time interval for successive two drops is 91//20 sceond. The distance travelled by falling drop ` in (1//2) second is , ` S = 1/2 xx 10 x 91/2)^2 = 1.25 m` :. Height of second drop from ground ` -= 5 - 1.25 m`.

Discussion

24.	A body released from the top of a tower falls through half the height of tower in `3` seconds. If will reach the ground after nearly .A. (a) ` 3 . 5 sec `B. (b) ` ` 4 . 24 sec`C. (c )` 4. 71 sec`D. (d)` 6 sec`
Answer» Correct Answer - B Let (h) be the total height of tower , tower , taking fall of the body in ` 3 s, u=0` ` S =h/2 , = g = 9.8 m//s^2 , t=3 sec` ,bRgt or ` S =ut + 1/2 at^2 :. h/2 =0 xx 3 = 1/2 xx (9.8) (3)^2` or ` h=9.8 xx 9 m` ltbRgt Now, if body takes total tiem (t) to fall therough height (h) , then ` h= 1/2 xx 9.8 xx t^2 ` or t= =sqrt (2 h)/(9.8)` or ` t= sqrt ((2 xx 9.0 xx 9 )/(9.8) = 3 sqrt 2 sec = 4 . 24 sexs`.

Discussion

25.	Two trains ` A and B` of length `400 m` each are moving on two parallel tracks with a uniform speed of ` 72 km h^(-1)` in the same direction with ` A ahead of B` . The driver of ` (A) , what was the originaldistance between them ?
Answer» For train ` A, u= 72 h^(-1) = ( 72 xx 1000)/( 60 xx 60) =20 mS^(-10 , t = 50 s, t= 50 , a =0, S=S_A` As. ` S= ut = 1/2 at^2 :. `S_A` = `20 xx 50 + 1/2 xx 0 xx 50^2 = 1000 m` For train ` B` , u= 72 km s^(-10 = 20 ms^(-1) , a= 1 m//s^2 , t = 50 s, S= S_B` As, ` S =ut = 1/2 at^2` :. S_B = 20 xx 50 = 1/2 xx 1 xx 50^2 = 22 50 = S-B` Taking the guard of the (B) in the last copartment of the train (B), it follows that original distance between the two trains = length of train ` A + length of train ` B= S_B - S_A` or orginal isanc ebetween the two trains ` + 400 + 400 = 2250 - 1000 = 1250` or original distance between the two trains `= 1250 =- 800 = 450 m`.

Discussion

26.	A body travels a distance of `2 m` in `2` seconds and `2.2m` next `4 seconds`. What will be the velocity of the body at the end of `7` the second from the start?
Answer» Here, case 9i) `S=2 m, `t=2 s` Case (ii) `S=2 +22.2 =42 , t=2=6 s` Let (u) and (a) be the initial velocity and unitorm acceleration of the body. we know that, `S=ut + 1/2 at^(2)` Case 9i), `2=u xx 2 + 1/2 a xx 2^(2)` or `1=u+a` .....(i) Case (ii) , `4.2 =u xx 6 +1/2 axx 6^(2)` or `0.7 =u +3a` ......(ii) Subtracting (ii) form(i), we ger `0.2 =0-2a =- 2a` or `a =- 0- 0.3//2 =- 0.15 ms^(-20` From (i0, `u=1 -a =1 +0.15 = 1.15 ms^(-1)` For the velcity of body at the end of `7 thsecond, we have `u=1.15 ms^(-1)` , `a=- 0. 15 ms^(-2) , v-? t=7 s` As, v=u +at ` :. `v=1.15 +(-0.15) xx 7 =0.1 ms^(-1)` .

Discussion

27.	The distance moved by a freely falling body (startibg from rest) during ` st, 2nd, 3nd,……nth` second of its motion are propotional to .A. Even numbersB. Odd numbersC. All integral numbersD. Squares of integral numbere
Answer» Correct Answer - D The required ratio is 1:3:5:…. So on .

Discussion

28.	Prove that the distances traversed during equal intervals of time by a body falling from rest, stand to one another in the same fatio as the odd mumbers beginning with unity [namely 1: 3: 5: …………….].
Answer» Here, `u=0, a=g, Distance travelled I nth second is given by `D_(n) =u +a/2 (2n -1)` Distance travelled in 1st` second` [i.e. when `n=1] will be Distance travelled in 2th second [i.e. when `n=2 ] will be `D_(2) 0+ g/2 92 xx1) =(3g)/2` Distance travelled in 3th second [i.e. when `n3 ] will be `D_(3) 0+ g/2 (2 xx5-1) =(5g)/2` Distance travelled in 4th second [i.e. when 4n ] will be `D_(4) =0+ g/2 (2 xx 4-4 -1) =(7g)/2` `:. D_(1) : D_(2) : D_(3): D_(4) 1: 3: 5: 7 (Proved)`.

Discussion

29.	A force of ` 6 hat i+ 7 hat j` newton makes a body move on a rough loane with a velocity of (4 hat j = 3 hat k) ms^(-1)` . Calculate the power in watt.
Answer» Here, ` vec F = 6 hat I + 7 hat j , vec v= 34 hat j + 3 hat k` Power, ` P= Fvec F. vec v= ( 6 hat I + 7 hat j).(4 hat + 3 hat k)` ` = 24 ( hat i.hat j) + 18 ( hat . Hat k)+ 28 ( hat j. hhat j) + 21 (hat j. hat k)` ltbRgt `= 24 (0) + 18 (0) + 28 (1) + 21 (0)` ` =28 W`.

Discussion

30.	A juggler throws ball into air. He throus one whenever the previus one is at its highest point. How high do the balls rise if he throus (n) balls each second. Acceleration the to gravity=g`.
Answer» Since the juggler is throwing (n) balls each second and he throws second ball when the first ball is at the highest point, so time taken by each ball to reach the highest point is ` t= 1//n`. Taking vertical upward motion ofball up to the highest point,we have, ` v=0., a=- g, t=1/n, u=?` As, ` v=u +at,` so `0= u+ (-g) 1//n` or `u= g//n` ...(i) 25 s`, Also, ` v^(2) =u^(2) + 2 as, so ` 0=u^(2) -2gh`, e.l. ` h=(u^(2) //2 g) =g (2 n^(2)`.

Discussion

31.	A juggler throws ball into air. He throus one whenever the previus one is at its highest point. How high do the balls rise if he throus (n) balls each second. Acceleration the to gravity=g`.A. ` 5m`B. ` 3,75 m`C. ` 2.50 m`D. ` 1.25 m`
Answer» Correct Answer - A Time of ascent=`1 rArr (u)/(g) =rArr u=10 msd^(-1)` Maximum height attained `=u^(2)/(2g)=(10^(2))/(2xx10) =5 `.

Discussion

32.	A particle rotates in a circle with angular speed `omega_(0)`. A retarding force decelerates it such that angular deceleration is always proportional to square root of angular velocity. Find the mean angular velocity of the particle averaged over the whole time of rotation.
Answer» Correct Answer - `(omega_(0))/(3)`

Discussion

33.	A particle is thrown from the origin,at an angle `theta(0lttheta lt 90^(@))` such that it just crosses a wall of height `9m`. Wall is at `x=12m`. Speed of projection is `nsqrt(3)m/s` and particle strikes the ground at `x=48m`. Value of `n` is `(g=10m//s^(2))`
Answer» Correct Answer - 4

Discussion

34.	A man throws ball into the air one after the other. Throwing one when other is at the highest point. How high the balls rise if he throws twice a second.A. (a) ` 2. 45 m`B. (b) ` 1. 22 5 m`C. (c ) ` 19 .6 m`D. (d) ` 4. 9 m`
Answer» Correct Answer - B The time taken by each ball to go from starting point to highest point , `t = 1//2 sec`, which is equal to tiem taken by each ball to fall back to starting point (= 1 //2 sec) ` :. S= 1/2 xx 9.8 xx 1/2 xx (1/2)^@ = (9.8)/8 m 1.225 m`.

Discussion

35.	A particle is moving along positive X direction and is retarding uniformly. The particle crosses the origin at time `t = 0` and crosses the point `x = 4.0 m` at `t = 2 s`. (a) Find the maximum speed that the particle can possess at `x = 0`. (b) Find the maximum value of retardation that the particle can have.
Answer» Correct Answer - (a) `4m//s` (b) `2m//s^(2)`

Discussion

36.	The velocity time graph for two particles (1 and 2) moving along X axis is shown in fig. At time `t = 0`, both were at origin. (a) During first 4 second of motion what is maximum separation between the particles? At what time the separation is maximum? (b) Draw position (x) vs time (t) graph for the particles for the given interval.
Answer» Correct Answer - (a) `X_(max) = 4m; t = 2s`

Discussion

37.	A ball is dropped from the top of a building. The ball takes `0.5s` to fall the `3m` length of a window some distance from the to of the building. If the speed of the ball at the top and at the bottom of the window are `v_(T)` and `v_(T)` respectively, then `(g=9.8m//s^(2))`A. `v_(T)+v_(B)=12ms^(-1)`B. `v_(T)-v_(B)=4.9ms^(-1)`C. `v_(B)+v(T)=1ms^(-1)`D. `(v_(B))/(v_(T))=2`
Answer» Correct Answer - A

Discussion

38.	A steel ball is dropped from the roof of a building.An observer standing in front of a window 1.2 m high notes that the ball takes `1/8` s to fall from the top to the bottom of the window.The ball continues to fall, makes a completely elastic collision with a horizontal sidewalk and reappears at the bottom of the window 2 s after passing it on the way down. How tall is the building ?
Answer» Correct Answer - 20.43 m

Discussion

39.	A ball is dropped vertically from `a` height `d` above the ground . It hits the ground and bounces up vertically to a height ` (d)//(2). Neglecting subsequent motion and air resistance , its velocity `v` varies with the height `h` above the ground asA. B. C. D.
Answer» Correct Answer - A `v^(2)=2gh` [it is parabola] and direction of speed (velocity) charges.

Discussion

40.	A ball is projected vertically up from ground. Boy A standing at the window of first floor of a nearby building observes that the time interval between the ball crossing him while going up and the ball crossing him while going down is `t_(1)`. Another boy B standing on the second floor notices that time interval between the ball passing him twice (during up motion and down motion) is `t_(2)`. (a) Calculate the height difference (h) between the boy B and A. (b) Assume that the height of boy A from the point of projection of the ball is also equal to h and calculate the speed with which the ball was projected.
Answer» Correct Answer - (a) `h = (g(t_(1)^(2) - t_(2)^(2)))/(8)` (b) `u = (g)/(2) sqrt)2t_(1)^(2) - t_(2)^(2))`

Discussion

41.	A piece of stone is allowed to fall from a balloon floating in air. It goes down past a tower 30 m in height in the last quarter of the last second of its flight.Find the height of the stone at the instant when it was dropped and its velocity as it reaches the ground. `(g=9.8 ms^(-2))`
Answer» Correct Answer - 749.54 m , `121.21 ms^(-1)`

Discussion

42.	A moving car possesses average velocities of `5 m s^(-1), 10 m s^(-1)`, and `15 m s^(-1)`, in the first, second, and third seconds, respecticely. What is the total destance coverd by the car in these `3 s`.?A. `15 m`B. `30`C. `55 m`D. `None of these
Answer» Correct Answer - B Distance coverd `=S=v_(av) xx` time For first second: `s_(1)=5 xx 1=5 m` For second second: `s_(2)=10 xx 1=10 m` For thirss second: `S_(3)=15 xx 1=15 m` Total distance travelled `S=S_(1)+S_(2)S_(3) =5+ 10 +15 =30 m` .

Discussion

43.	When the speed of the car is v, the minimum distance over which it can be stopped is x. If the speed becomes nv, what will be the minimum distance over which it can be stopped during same time:A. `x//n`B. `nx`C. `x//n^(2)`D. `n^(2)x`
Answer» Correct Answer - B

Discussion

44.	When the speed of a car is `u`, the mimimum distance over which it canbe stopped is `a`, If speed becomes `nu`, what will be the mimimum distance over which it can be stopped during the same time?A. ` s//h`B. ` ns`C. ` s//n^(2)`D. `n^(2) s`
Answer» Correct Answer - D `v^(2)-u^(2)=2as` , `v=0` `prop u^(2)`, When the initial velocity is made `n` times, the distance over which it can be stopped becomes `n^(2)` times.

Discussion

45.	A ball is dropped into a well in which the water level is at a depth `h` below the top. If the speed of sound id `C`, then the time after which the splash is heard will be give by.A. `h[sqrt((2)/(gh))+(1)/(c)]`B. `h[sqrt((2)/(gh))+(1)/(c)]`C. `h[(2)/(g)+(1)/(c)]`D. `h[(2)/(g)+(1)/(c)]`
Answer» Correct Answer - A Time of fall `=sqrt((2h)/(g))` Time taken by the sound to come out =`(h)/(c)` `Total time `=sqrt((2h)/(g)) +(h)/(c) =h [ sqrt((2)/(gh)) +(1)/(2)]`.

Discussion

46.	A particle is dropped from rest from a large height Assume `g` to be constant throughout the motion. The time taken by it to fall through successive distance of `1 m` each will be :A. All equal, being equal to `sqrt(2//g) second `B. In the ratio of the square roots roots of the integers `1, 2, 3,.............C. In the ratio of the disfference in the square roots of the integers, i.e., sqrt1, (sqrt2, -sqrt1),(sqrt3-sqrt2), (sqrt4-sqrt3)`......D. In the ratio of the rectiprocals of the square roots of the integers, i.e., (1)/(sqrt1), (1)/(sqrt2), (1)/(sqrt(3)`,........
Answer» Correct Answer - C Time taken to cover first `n` meter is givenby `n=(1)/(2) g t_(n)^(2)` or `t_(n) =sqrt((2n)/(g))` Time taken to cover (n+1 )th `m` is givenby `t _(n +t)= sqrt ((2 ( n+1))/(g)` So time taken to cover (n+1 )th `m` is given by ` t_(n)+1-t_(n)=sqrt((2(n+1))/(g))-sqrt((2n)/(g))=sqrt((2)/(g))[sqrt(n+1)-sqrtn]` This gives the required ratio as `sqrt1, sqrt2 -sqrt1), sqrt3- sqrt2),...,etc.., ("starting from n=0")`.

Discussion

47.	A body is thrown up with a speed `49 m//s`. It travels 5m in the last second of its upward journey. If the same body is thrown up with a velocity `98 m//s`, how much distance (in m) will it travel in the last second. `(g=10 m//s^(2))`
Answer» Correct Answer - 5 In last second of upward journey, all bodies travel same distance `(=g//2=5m)`

Discussion

48.	a particle is moving in a circle of radius R in such a way that at any instant the normal and the tangential component of its acceleration are equal. If its speed at `t=0` is `v_(0)` then time it takes to complete the first revolution is `R/(alphav_(0))(1-e^(-betapi))`. Find the value of `(alpha+beta)`.
Answer» Correct Answer - 3 `(dv)/(dt)=v^(2)/RrArr underset(v_(0))overset(v)(int)(dv)/v^(2)=1/Runderset(0)overset(t)(int)dt rArr (-1/v)_(v_(0))^(v)=t/RrArr v=v_(0)/(1-v_(0)/Rt)rArr (ds)/(dt)=v_(0)/(1-v_(0)/Rt)rArr underset(0)overset(2piR)(int)ds=underset(0)overset(t)(int)v_(0)/((1-v_(0)/Rt))dt` `rArr 2pi R=-R[ln(1-v_(0)/Rt)]_(0)^(t)rArr 2pi=-ln(1-v_(0)/Rt)rArr 1-v_(0)/Rt=e^(-2pi)rArr t=R/v_(0)(1-e^(-2pi))` `rArr alpha=1, beta=2 rArr (alpha+beta)=(1+2)=3`

Discussion

49.	What is the property of two vectorsx if. ` vec A + vec B =vec A- vec B` ?
Answer» Given ` vec A + vec B =vec A -vec B or 2 vec B=0 or vec B =0` i.e. `vec B` is a null vector.

Discussion

50.	Acceleration versus velocity graph of a particle moving in a straight line starting form rest is as shown in figure. The corresponding velocity-time graph would be : A. B. C. D.
Answer» Correct Answer - D

Discussion

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A body is thrown up with a speed `49 m//s`. It travels 5m in the last second of its upward journey. If the same body is thrown up with a velocity `98 m//s`, how much distance (in m) will it travel in the last second. `(g=10 m//s^(2))`

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