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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.
| 251. |
Velocity-time graph of a particle moving in a straight line is shown in figure. In the time interval from `t = 0` to `t = 14 s,` find (a) average velocity and (b) average speed of the particle. |
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Answer» Correct Answer - A::B Displacement s=net area of v-t graph `=40+40+40-20=100m` Distance `d=|"Total area"|` `=40+40+40+20=140m` (a) Average velocity=`s/t=100/14=50/7 m//s` (b) Average speed=`d/t=140/14=10 m//s` |
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| 252. |
An elevator without a ceiling is ascending up with an acceleration of `5 ms^-2.` A boy on the elevator shoots a ball in vertical upward direction from a height of 2 m above the floor of elevator. At this instant the elevator is moving up with a velocity of `10 ms^-1` and floor of the elevator is at a height of 50 m from the ground. The initial speed of the ball is `15 ms^-1` with respect to the elevator. Consider the duration for which the ball strikes the floor of elevator in answering following questions. (`g=10 ms^-2`) 3. Displacement of ball with respect to ground during its night would beA. 16.25 mB. 8.76 mC. 20.24 mD. 30.56 m |
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Answer» Correct Answer - D `S=25xx2/13-5xx(2.13)^2` `=30.56m` |
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| 253. |
A particle moves along a straight line and its velocity depends on time as `v = 3t - t^2.` Here, v is in `m//s` and t in second. Find (a) average velocity and (b) average speed for first five seconds. |
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Answer» Correct Answer - A::B::C (a) Average velocity= Displacement /Time =`(int_0^5 vdt)/5` `=(int_0^5(3t-t^2)dt)/5` `=-0.833 m//s` (b) Velocity of particle=0 at `t=3s` i.e, at 3s, particle changes its direction of motion. Average speed=(Total distance)/(Total time) =((Distance from 0 to 3s)+(Distance from 3s to 5s))/Time `d_(0-3)=int_0^3(3t-t^2)dt=4.5m` `d_(3-5)=int_3^5(t^2-3t)dt=8.67m` `:.` Average speed =`(4.5+8.67)/5` `=2.63 m//s` |
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| 254. |
A passenger arriving in a new town wishes to go from the station to a hotel located ` 10 km` away on a straight road from the station. A dishownest cabman takes him along a circuitons path ` 23 km` long and reaches the hotel in ` 28 minmtes` . What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ? |
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Answer» Here, actual path length travelled, ` S= 23 km , Displacement= 10 km , Time taken , ` t= 28 min = 28 // 60 h` (a) Average speed of the taxi ` (actual path length)/(tune taken ) = ( 23)/( 28 // 60 ) = 21 .4 km //h` The aveage speed is bot equal to the magnitude of aveage veloctiy. The two ar equal fromthe motion of taxis along a straight parh in one direction. |
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| 255. |
A force is inclened at `60 ^(@)` to the horizontal . If its horizontal coponent in the horizonal direction is `60N` find themagitude of theforce and its vertical componet. |
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Answer» Let (F) be the force applied, `theta =60^(00`. Horizontal component of force, `F_(x) =F cos theta` :. 60 =F cos 60^(0) =F xx 1//2` or `F=60 xx 2 =120 N` Verical component of force, `F sin theta =120 sin `60^(0)` `=120 xx sqrt 3 // 2` `=103 .92 N` . |
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| 256. |
A target situated on a hill can be seen from the foot at angle `alpha` above the horizontal .The distance of the target along the horizonal is R. if a shell is fired from the foot at an angle `beta` with the horizontal, find the velocity required for it to strike the target. |
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Answer» Correct Answer - `sqrt((R g cos alpha)/(sin (beta-alpha) cos beta))` |
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| 257. |
A bead moves along a straight horizontal wire of length L, starting from the left end with velocity v0. Its retardation is proportional to the distance that remains to the right end of the wire. Find the initial retardation (at left end of the wire) if the bead reaches the right end of the wire with a velocity `(v_(0))/(2)`. |
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Answer» Correct Answer - `(3v_(0)^(2))/(4L)` |
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| 258. |
A ball is dropped from the roof of a tower of height (h). The total distanc coverd by it in the last second of its motion is equal to the distance covered by in first three seconds. What will be the velocity at the end of `9 secind ` ? |
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Answer» Let the ball remain in air for (t) seconds. Then , ` D_1 = u + a/ 2 ( 2t- 1) =0 + ( 10) /2 ( 2t-t)` ` = 10 t-5` …(i) Distance covered in first three secnds. ` S= 1/2 gt^2 = 1/2 xx 10 3^2 = 45` …(ii) As per question ` 10 t- 5 = 45` or ` t= 5 s` :. ` h = 1/2 at^2 = 1/2 xx 10 xx 10 5^2 = 125 m` . |
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| 259. |
An object is moving with uniform acceleration. Its velocity after ` 4s is 20 ms^(-1)` and after 7 second is 29 ms"^(-1)` . Find the distanc etravelled by the object in 10 th second. |
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Answer» As, ` v= u + at` ltbRgt Fro case (i), ` 20 = u + a xx 4` ..,(i) for case (ii) ` 29 =u + a xx 7` …(ii) Solving (i) and (ii) we get ` a= 3 ^(-2), u=8 ms^(-1)` ` S_(10) = u + a/2 92 xx 10 - 1 ] = 8 + 3/ [ 19] = 36.5`. |
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| 260. |
Two identical discs of same radius `R` are rotating about their axes in opposite directions with the same constant angular speed `omega` . The discs are in the same horizontal plane. At time `t = 0` , the points `P` and `Q` are facing each other as shown in the figure. The relative speed between the two points `P` and `Q` is `v_(r)`. In one time period `(T) ` of rotation of the discs , `v_(r)` as a function of time is best represented by A. B. C. D. |
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Answer» Correct Answer - A AT an instatnt, speed of ` P=v`, going in clockwese direction Speed of ` Q = v`, going in anticlokwise direction Relative andgular velocity of ` P w.r.t. Q = omega - (-omega)` `=2 omega` Relative angular separation of ` P` and ` Q` in time (t) ` theta = 2 omega t`. Relative speed between the points ` P` and ` Q` at time (t)` ` | vec _r | = sqrt (v^2+vec v^2 - 2 vv cos (2 omega t))` ` =2 v is omega t` Since, ` | vec_r|` will not have any negative value so the lower part of the sine wave will come upper siede. Hence option (a) is true. |
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| 261. |
Two particles A and B projected simultaneously from a point situated on a horizontal place. The particle A is projected vertically up with a velcity `v_(A)` while the particle B is projected up at an angle `30^(@)` with horizontal with velocity `v_(B)`. After 5s the particles were observed moving mutually perpendicular to each other. The velocity of projection of the particle `v_(A) and v_(B)` respectively are:A. `50ms^(-1),100ms^(-1)`B. `100ms^(-1),50ms^(-1)`C. any value greater than`25ms^(-1),100ms^(-1)`D. none of the above |
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Answer» Correct Answer - C |
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| 262. |
Velocity-time graph of a particle moving in a straight line is shown in figure. At time `t= 0. s = 20 m.` Plot a-t and s-t graphs of the particle. |
| Answer» Correct Answer - A::C::D | |
| 263. |
A body is projected up with a speed `v_0` along the line of greatest slope of an inclined plane of angle of inclination `beta`. If the body collides elastically perpendicular to the inclined plane, find the time after which the body passes through its point of projection. |
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Answer» Correct Answer - `[(6V)/(g sqrt((1 + 8 sin^(2) alpha)))]` |
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| 264. |
A ball is projected vertically up with an initial speed of `20m//s` on a planet where acceleration due to gravity is `10m//s^(2)` (a) How long does it takes to reach the highest point? (b) How high does it rise above the point of projection? (c) How long wil it take for the ball to reach a point `10m` above the point of projection? |
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Answer» As here motion is vertically upwards, `a =- g` and `v = 0` (a) From `1st` equaiton of motion, i.e., `v = u +at`, `0 = 20-10t rArr t = 2sec.` (b) Using `v^(2) =u^(2) +2as` `0 = (20)^(2) -2x 10x h rArr h = 20m`. (c ) Using `s = ut +(1)/(2)at^(2), rArr` `10 =20t (-(1)/(2)) x10xt^(2) rArr t^(2)-4t+2 =0` `rArr t = 2+- sqrt(2), rArr t = 0.59 sec.` or `3.41sec.` i.e., there are two times, at which the ball passes thorugh `h = 10m`, once while going up and then coming down. |
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| 265. |
A small ball is projected up a smooth inclined plane with an initial speed of `10m//s` at `30^(@)` from the bottom edge of the slope. It returns to the edge after `2s`. The ball is in contact with the inclined plane throughout the process. The inclination angle (in degree) of the plane is `N`. The value of `N//6` is |
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Answer» Correct Answer - 5 |
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| 266. |
A ball is projected directly upward with an initial speed `v_(0)`. Bounces elastically from a roof inclined at an angle `45^(@)` as shown in figure and then it strikes a table at a horizontal distance 2D from its starting point. Find `v_(0)`. |
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Answer» Correct Answer - `[sqrt(6gD)]` |
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| 267. |
A particle of mass `m` is projected up from the bottom of an inclined plane with initial velocity `v_(0)` at angle `45^(@)` with an inclined plane of inclineation `30^(@)` as shown in figure. At the same time a small block of same mass `m` is released from rest at a height `h`. The particle hits the block at some point on inclined plane. If the particle sticks to the block after collision, the velocity of block parallel to the inclined plane just after collision will be (take `g=10m//s^(2)`)A. `(v_(0))/(sqrt(2))`B. `(v_(0))/2`C. `(v_(0))/(2sqrt(2))(1-4/(sqrt(3)))`D. `(v_(0))/(2sqrt(2))(1+4/(sqrt(3)))` |
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Answer» Correct Answer - C |
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| 268. |
Find a unit vector parallel to the resultant of vectors ` vec A = 3 hat I + 3 hat j - 2 hat k` and ` vec B= hat i- 5 hat j + hat k`. |
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Answer» The resultant of ` vec A ` and `vec B` is ` vec R= vec A + vec B = ) 3 hat i+ 3 hat j- 2 hat j - 2 hat k) + ( hat i+ 5 hat j + hat k)` ` ( 4 hat i- 2 hat j- hat k)` ` R= sqrt (4^2 + (- 2)^2 + (-1)^2 ) = sqrt 21` ` hat R = ( hat R)/R = ( 4 hat i- 2 jat j - aht k)/( sqrt 21)`. |
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| 269. |
The resultant of two forces is `20 N`. When one of the force is `10 sqrt2` and angle between two torces is `30^@`, then what is the value of the second force? |
| Answer» Correct Answer - (b) | |
| 270. |
The resultant ot two forces acting at an angle of ` 15^@` ii ` 10 N` and is peroendicular to one of the forces. The other force is .A. `20 // sqrt 2 N`B. `10 // sqrt 3 N`C. ` 20 N`D. ` 20 sqrt 3 N` |
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Answer» Correct Answer - C ` 10^@ = A^2 + b^2 +2 AB cos 150^@` ` = A^2 +b^2 - 2 AB ( sqrt 3 //2)` …(i) ` tan 90^@ = ( B sin 150^@)/( A+ B cos ` 15^@) =0` or ` A+B cos 1 50^@ =0` or ` A= B xx sqrt 3//3` …(ii) ` From (i), ` 100 = 3/ 4 B^2 + B^@ - ( 3 B^2)/2 = B^2/4` ` or ` B = 20 N`. |
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| 271. |
Two persons (P) and (Q) are standing ` 54 m` apart on a long moving belt. Person (P) rolls a round staone towards person (Q) with a speed of ` 9 ms^(-1)` with respect to belt . If the belt is moving with a speed of `4 ms^(-1)` in the direction from (P) to (Q) (a) What will be the speed of the stone with respect to an obsever on a stationary platform if person (Q) rolls the stone with a velocity of ` 9 ms^(-1)` with respect to the belt towards person (P) and the time taken by the stoine to travel from (Q) to (P) ? |
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Answer» (a) Let the direction from (P) to (Q) be positive. Given , speed of the belt, ` v_b =+ 4 ms^(-1)` , speed of the stone w.r.t. belt, ` v_s = + 9 ms^(-1) speed of stpne w.r.t. a stationary observe ` = v_S + v_b = 9 + 4 = 13 ms^(-1)` (b) Since, persons ` P and Q` are located on belt, the speed of stone w.r.t. `P` or ` Q` will be ` 9 ms^(-1)` . As distance between ` (P) and ` Q= 54 m`, so time taken `= 54//9 = 6 s. (c ) Here , ` v_b = + 4 ms^(-1)` , v_s =-9 ms^(-1)` :. Speed of the stone w.r.t. observe ` v_s + v_b =- 9 + 4 =- 5 ms^(-1)` Here, negative singn shows that the stone is moving in a direction opposite to the direction in which the belt is moving. In this case also , the speed of stone w.r.t. ` P or ` Q` is still ` 9 ms^(-1)`. So, time taken ` = ( 534) /9 = 6 s`. |
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| 272. |
the resultant of two forces is ` 10 sqrt 13 N`, when one of the force is ` 10 sqrt 3 N` and angle between two forces is 1 30^@` then what is the value of second force ?A. ` 10 N`B. ` 20 N`C. ` 20 sqrt 23 N`D. ` 10 sqrt 3 N` |
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Answer» Correct Answer - B Here, ` R= 10 sqrt 13 N, A= 10 sqrt 3 N, theta = 30^@, B= ?` ` R^2 =B^2 + 2 = 2 zz 10 sqrt 3 xx B cos 30^@` ` (10 sqrt 13)^2 = ( 10 sqrt 3 )^2 + B^2 + 2 xx 10 sqrt 3 xx B cos 30^2` ` 1300 = + B^^@ = 30 B ` or B^2 + 30 B- 1000 =0` ` (B+ 50 ) (B- 20 ) = 0, B= 20 N ` or ` 50 N` , |
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| 273. |
A bullet loses ` 1//20` of its velocity in passing through a plank. What is the least number of plank required to stop the bullet .A. (a) ` 8`B. (b) ` `7`C. (c ) ` 11`D. (d) ` 14` |
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Answer» Correct Answer - C Let (u) be the initial velocity of bullet and (a) be the retardation of bullet while passign through a plank. Velocity of bullet after passing theough plank ` v= u - u/(20) = (19 u)/(20)` Taking motion of bullet through one plank and using the relation, ` v^2 =u^2 +2 as`, we hve (19)/(20) u)^@ = u^2 -2 aS` or ` 2 aS =u^2 - ((10)/(20)u)^2` [ :. motion is retarted ]` ` 2 a S=u^2 - ( 361)/(400)u^2 = (39u^2)/(400)` ltbRgt Taking motion bullet through (n) plaks ` ` v=0, S=n S` ` 0 u^2 -2 a nS` or ` n=u^2/(2 aS0 = u^2 /( 39u^2 //400) = (400)/(39) = 10.25 ~~ 11`. |
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| 274. |
A bullet loses ` 1//2th1 of its velocity is passing through a plank. What is the least number of planks required to stop the bullet ? |
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Answer» Let (u) the initial velocity of butter and (a) be the retardation of the bullet while passing through a plank. Velcoty of bullet after passing through a plank is ` u =u - u//20 = 19 u //20` Taking motion of bullet through one pland and using the relation, `v^2 =u^2 = 2 a S, we have ` (19 u //20 )^2 -2 aS1 or ` `2 aS= u^2 - (36 1//400) u^2 = 39 u^2 //400` Let (n) number of plands be required to stop the bullet. Taking motion of bullert through (n) planks, we have ` ` u =u, v= 0, S= nS. a=-a, then` ` O^2 =u^2 -2 a Sn` or ` n= u^2 / (2 a S) = u^2 /((39u^2//400) = (4000/(39) ~~ 11`. |
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| 275. |
A man is standing on top of a building 100 m high. He throws two ball vertically, one at `t=0` and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. At `t=2`, both the balls reach to their and second ball is `+15m`. Q. The speed of first ball is |
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Answer» Let the first ball be thrown with vleocity (u) and second ball with a time interv al (x) Taking vertical motion of ball first for time (t = x), we have S_1 =u (t+ x) - 1/2 g (t + x)^2` …(i) ` Taking upaward motion of second ball for time (t), we have ` S_2 = u/2 t 1/2 gt^2` ....(ii) :. S_1 - s_2 = u (tt+ x - t/2 g (t^2 + 2 xt = x^2 - t^2)` `= u (t/2 + x) - 1/2 g ( 2 xt + x^2)` As per question, `S_1 -S_2 = 15 m , t=2 s and x = 1 s [ as x lt 2 s]` Then, ` 15 =u (2/2 + 1) - 1/2 xx 10 (2xx 1 xx 2 1^2)` or ` 15 =u xx 2 - 25` or ` u = 40 //2 = 20 m//s` The velocity other ball ` = u/ 2 = (20)/2 = 10 m//s`. Time interval ` x = 1 s` |
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| 276. |
A particle with a velcoity (u) so that its horizontal ange is twice the greatest height attained. Find the horizontal range of it. |
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Answer» Max. height , `H= (u^2 sin^2 theta)/ (2g)` Horizontal range, ` R= ( u^2 sin 2 theta ) /g = ( 2 u^2 sin theta cos theta)/( 2g)` As per question, ` R= 2 H` ` ( 2u^2 sin theta cos theta)/g = ( 2 u^2 sin ^2 theta)/(2g)` ltbRgt Solving we get, ` tan theta =2 , therfore ` sin theta= ( 2/ sqrt2)` ltbRgt and ` cos theta = (1 /sqrt 4)` ltbRgt :. Horizontal range, ` R= (2 u^2)/g xx 2/(sqrt5) xx 1/(sqrt 5) = (4u^@)/(5g)`. |
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| 277. |
If ` | vec A | =2 and | vec B| =4`, then match the relations in column I with the ange ` theta` between ` vec A and vec B` in column II. Column I, Column II (a) ` |vec A xx vec B| =0` , (i) ` theta =30^@` (b) `| vec A xx vec B|=0` , (ii) ` theta =45^@` (c ) ` vec A xx vec B| =4` , (iii) ` theta =90^@` (d) ` | vec A xxx vec B| = 4 sqrt 2` , (iv) ` theta =0^@`. |
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Answer» Here ` A= 2 and B=4`. `(a) | vec A xx vec B| =0 = AB sin theta =2 xx 4 sin theta or =0 and theta =0^@`. It matches with option (iv). `(b) | vec A xx vec B| =8 = AB sin theta 2 xx 4 sin theta or sin theta =1 and theta =90^@`. It matches with option (iii). `(c ) | vec A xx vec B| =4 =2 xx 4sin theta or sin theta =1/2 or theta =30^@`. It matches with option (i). `(d) | vec A xx vec B| = 4 sqrt 2=2 xx 4 or sin theta =1 /(sqrt 2) or theta =45^@`. It matches with option (ii). |
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| 278. |
It is common observation that rain clouds can be at about a kilometre altitude above the grond. (a) If a rain drop falls from such a height freely under gravity, what will be its speed ? Also calculate in km //h. (g=10 m//s^2) (b) A typical rain drop is about ` 4 mm` diameter. Vomentum is mass xxspeed in vagnitude . Estimate its momentum when it hits ground. (c ) Estimate the time required to flatten the drop. (d) Rete of change of momentum id force. Fstmate how much forece such a drop would exert you. (e) Estiate the order of magitude force of magnitude force on umbrella.Typical lateral separation between tao rain drops is ` 5 cm`. (Assume that umbrella is circular and has a diameter of `1 m` and cloth is not prierced through it) |
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Answer» Here, ` h= 1000 m, g=10 m//s^2` (a) Velcoity of rain drop after fallin height (h) si ` v = sqrt (2 gh) = sqrt (2 xx 10 xx 1000) = 100 sqrt 2 m//s = ( 100 sqer 2 xx 60 xx 60)/(1000) km// h = 36 0 sqrt 2~~ 510 km //h` (b) Diameter of drop, ` d = 2 =4 mm or r= 2 mm = 2 xxx 10^(-3)m` Mass of rain drop, ` m= 4/3/ pi r^3 roh = 4/3 xx (220/7 xx ( 2 xx 10^3)^3 xx 10^(-3)^3 xx 10^3 ~~ 3.4 xx 10 ^(-5) kg` Mumentum of rain drop when it hits the ground, ` P = mu = ( 3.4 xx 10 ^(-5) xx 100 sqrt 2 ~~ 4.7 xx 10^(-3) kg m//s` (c ) Time required to flatten the frop is ` Delta t = d/v = (4 xx 10^(-3))/(100 sqrt 2) = 0/028 xx 10^(-3) = 28 xx 10^(-6) s` (d) Force, ` F= (change in momentum )/(time) = ( P-0)/(Delta t) = ( 4. 7 xx 10^(-3))/(28 xx 10^(-6)) ~~ 16 8 N` (e) Given , rakies of umbrella, ` R = 1//2 m` Area of umbrella ` =pi R^2 = 922) /7 9 1/2 )^2 = (22)/(28) = (11)/(14) ~~ 0.8 m^@` with average separaion of ` 5 cm (= 5 xx 10^(-2) m), number of drops those will hit nubrella simultanceously ` = (0.8)/((5 xx 10^(-2))^2` Net force on umbrella`= 320 xx 168 = 53760 N ~~ 54000 N`. |
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| 279. |
a rain is falling vertically with a speed of ` 24 ms^(-1)`. A woman rides a bicycle with a speed of `12 ms^(-10` in each to west direction. The direction which woman should hold her umbrellat to proterct from rain is. |
| Answer» Correct Answer - (d) | |
| 280. |
Given that ` vec A + vec B= vec R and A^(20 + B^(2) -R^(2)`, find the angle between ` vec A and vec B` . |
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Answer» ` cos theta = (R^2 -A^2 -B^2)/(2 AB) = (R^2 -R^2)/(2 AB) =0` or ` theta = 90^@`. |
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| 281. |
A car of mass `m` starta from rest and accelerates so that the instyantaneous power delivered to the car has a constant magnitude `P_(0)`. The instaneous velocity of this car is proportional toA. (a) ` t^2 P_0`B. (b) ` t^(1/2)`C. (c )` - 1/2)` (d) t/(sqertm)`D. |
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Answer» Correct Answer - B Constant power of car, ` P_0 = F xx v` ` = ma xx v = m (dv)/(dt) xx v` or ` P_0 dt = mv dv` ltbr. Integratignboth sides, we get ` int_0^t P_o dt = int _0^v mv d v = ( mv^2 /2 , i.e. , P_0 t =( mv^2)/2` ` v= sqrt ( 2 P_0 t)/m` or ` v prop sqrt t` . |
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| 282. |
A player initially at rest throws a ball with an initial speed `u = 19.5 m//s` at an angle `theta = sin^(-1) ((12)/(13))` to the horizontal. Immediately after throwing the ball he starts running to catch it. He runs with constant acceleration (a) for first 2 s and thereafter runs with constant velocity. He just manages to catch the ball at exactly the same height at which he threw the ball. Find ‘a’. Take `g = 10 m//s^(2)`. Do you think anybody can run at a speed at which the player ran? |
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Answer» Correct Answer - `a = 5.19 m//s^(2)` |
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| 283. |
A 100 m long train is travelling from North to South at a speed of 30 ms-1. A bird is flying from South to North at a speed of 10-1. How long will the bird take to, cross the train?(a) 3 s (b) 2.5 s(c) 10 s (d) 5 s |
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Answer» (b) Length of train = 100 m Relative velocity = 30+ 10 = 40 ms-1 Time taken to cross the train (t) = \(\frac{distnace}{R.velocity}\) = \(\frac{100}{40}\) =2.5 s |
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| 284. |
An object is moving in a straight line with uniform acceleration a, the velocity-time relation is –(a) u = v + at(b) v = u + at(c) v2 = u2 + a2t2(d) v2 – u2 = at |
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Answer» Correct answer is (b) v = u + at |
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| 285. |
The displacement x of a particle varies with time as x = 4t2 – 15t + 25. Find the position, velocity and acceleration of the particle at t = O. |
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Answer» Position, x = 25 m Velocity = \(\frac{dx}{dt}\) 8t – 15 t = 0, v = 0 – 15 = -15 m/s acceleration, a = \(\frac{dx}{dt}\)= 8 ms-2 |
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| 286. |
A swimmer can swim in still water at of 10 ms-1 While crossing a river his average speed is 6 ms-1 . If he crosses the river in the shortest possible time, what is the speed of flow of water?(a) 16 ms-1 (b) 4 ms-1 (c) 60 ms-1 (d) 8 ms-1 |
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Answer» (d) The resultant velocity of swimmer must be perpendicular to speed of water to cross the river in a shortest time ∴ \( v^2_S = v^2 +v^2_w\) \(v^2_w = v^2_s -v^2\) =100-36 = 64 ∴ V = 8 m/s-1 |
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| 287. |
What is non uniform circular motion? |
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Answer» If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration. |
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| 288. |
What is average acceleration? |
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Answer» The average acceleration is defined as the ratio of change in velocity over the time interval aavg= \(\frac{Δ\vec V}{Δt}\) it is a vector quantity. |
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| 289. |
Define acceleration. |
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Answer» Acceleration of a particle is defined as the rate of change of velocity or it can also be defined as the ratio of change in velocity to the given interval of time. |
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| 290. |
The Moon is orbiting the Earth approximately once in 27 days, what is the angle transformed by the Moon per day? |
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Answer» Given, period of moon = 27 days i.e. in 27 days moon covers 360° In one day angle traversed by moon = 360°/2H = 13.3° |
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| 291. |
The radius of the earth was measured by – (a) Newton (b) Eratosthenes(c) Galileo (d) Ptolemy |
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Answer» (b) Eratosthenes |
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| 292. |
In relation between linear and angular velocity is –(a) ω = vr(b) ω = \(\frac{v}{r}\)(c) ω = \(\frac{r}{v}\)(d) ω = \(\frac{r}{ω}\) |
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Answer» Correct answer is (b) ω = \(\frac{v}{r}\) |
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| 293. |
An object is moving in a straight line with uniform acceleration, the displacement-time relation is –(a) S = ut2 + \(\frac{1}{2}\)at2(b) S = ut - \(\frac{1}{2}\) at2(c) S = ut +\(\frac{1}{2}\) at2(d) S = ut - at2 |
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Answer» (c) S = ut + \(\frac{1}{2} at^2\) |
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| 294. |
Define angular displacement and angular velocity |
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Answer» 1. Angular displacement: The angle described by the particle about the axis of rotation in a given time is called angular displacement. 2. Angular velocity: The rate of change of angular displacement is called angular velocity. |
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| 295. |
When a projectile is at its maximum height, the direction of its velocity and acceleration are – (a) parallel to each other (b) perpendicular to each other (c) anti – parallel to each other (d) depends on its speed |
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Answer» (b) perpendicular to each other |
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| 296. |
One radian is equal to – (a) \(\frac{π}{180}\) degree(b) 60° (c)57.295° (d) 53.925° |
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Answer» Correct answer is (c) 57.295° |
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| 297. |
Define a radian? |
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Answer» One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle. 1 rad = 57.295° |
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| 298. |
From this velocity – time graph, which of the following is correct?(a) Constant acceleration (b) Variable acceleration (c) Constant velocity (d) Variable acceleration |
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Answer» (b) Variable acceleration |
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| 299. |
An object is executing uniform circular motion with an angular speed \(\frac{π}{12}\) of radian per second. At t = 0 the object starts at angle θ = 0. What is the angular displacement of the particle after 4 s? |
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Answer» Given: Angular speed = \(\frac{π}{12}\)rad/ sec Angular speed = \(\frac{angular\, displacement}{time \, taken}\) Angular displacement = \(\frac{π}{12}\) x 4 = \(\frac{π}{12}\) = 60° |
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| 300. |
A bullet is dropped from some height, when another bullet is fired horizontally from the same height. They will hit the ground –(a) depends upon mass of bullet (b) depends upon the observer (c) one after another (d) simultaneously |
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Answer» (d) simultaneously |
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