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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.
| 51. |
A projectile is fired horizontally with an initial speed of 20 m/s/ its horzontal speed 3 sec later is:A. 20m/sB. 6.67m/sC. 60m/sD. 29.4m/s |
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Answer» Correct Answer - a Horizontal speed remains same. |
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| 52. |
A man can swim in still water at a speed of `6 kmph` and he has to cross the river and reach just opposite point on the other bank.If the river is flowing at a speed of `3 kmph`,and the width of the river is `2km`, the time taken to cross the river is (in hours)A. `(2)/(27)`B. `(2)/sqrt(27)`C. `(2)/(3)`D. `(2)/sqrt(45)` |
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Answer» Correct Answer - b `v_(m)=6` kmph, `v_(w)=3` kmph, `t=(d)/(sqrt(v_(m)^(2)-v_(w)^(2)))` |
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| 53. |
From a point `A` on bank of a channel with still water a person must get to a point `B` on the opposite bank.All the distances are shown in figure.The person uses a boat to travel across the channel and then walks along the bank of point `B`.The velocity of the boat is `v_(1)` and the velocity of the walking person is `v_(2)`.Prove that the fastest way for the person to get from `A` to `B` is to select the angles `alpha_(1)` and `alpha_(2)` in such a manner that A. `(sin alpha_(1))/(sin alpha_(2))=(u_(2))/(v_(1))`B. `(sin alpha_(1))/(sin alpha_(2))=(u_(1))/(v_(2))`C. `(cosalpha_(1))/(cos alpha_(2))=(v_(2))/(v_(1))`D. `(cos alpha_(2))/(cos alpha_(1))=(v_(1))/(v_(2))` |
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Answer» Correct Answer - a `S_(1)=(x)/(sin alpha_(1)),S_(2)=(d-x)/(sin alpha_(2))` `t_(1)=(S_(1))/(v_(1))=(x)/(v_(1)sin alpha_(1))` and `t_(2)=(S_(2))/(v_(2))=(d-x)/(v_(2)sinalpha_(2))` `r=x[(1)/(v_(1)sinalpha_(1))-(1)/(v_(2)sin alpha_(2))]+(d)/(v_(2)sin alpha_(2))` For t be minimum `(dt)/(dx)=0` or `(1)/(v_(1)sin alpha_(1))=(1)/(v_(2)sinalpha_(2)),(v_(1))/(v_(2))=(sin alpha_(2))/(sinalpha_(1))` |
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| 54. |
An insect trapped in a circular groove of radius `12 cm` moves along the groove steadily and complete `7` revolutions in `100` seconds.The linear speed of the motion in `cm//s`A. 5.3B. 4C. 3D. 5 |
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Answer» Correct Answer - a `v=(s)/(t)=(2pirN)/(t)` |
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| 55. |
A particle of mass m moves in a circle of radius R in such a way that its speed `(v)` varies with distance (s) as `v = a sqrt(s)` where a is a constant. Calcualte the acceleration and force on the particle.A. `a^(2)sqrt((1)/(4)-(S^(2))/(R^(2)))`B. `a^(2)sqrt((1)/(4)+(S^(2))/(R^(2)))`C. `asqrt((1)/(2)+(S^(2))/(R^(2)))`D. `a^(2)sqrt((1)/(2)+(S^(2))/(R^(2)))` |
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Answer» Correct Answer - b `a=sqrt(a_(1)^(2)+a_(1)^(2))` `=sqrt(((dv)/(dt))^(2)+((v^(2))/(R))^(2))` |
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| 56. |
At a given instant of time the position vector of a particle moving in a circle with a velocity `3hati-4hatj+5hatk is hati+9hatj-8hatk`.Its anglular velocity at that time is:A. `((13hati-29hatj-31hatk))/(sqrt(146)`B. `((13hati-29hatj-31hatk))/(146)`C. `((13hati+29hatj-31hatk))/(sqrt(146))`D. `((13hati+29hatj+31hatk))/(146)` |
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Answer» Correct Answer - b `vecomega=(vecrxxvecv)/(|vecr|^(2))` |
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| 57. |
A boatman finds that he can save `6 s` in crossing a river by the quickest path than by the shortest path. If the velocity of the boat and the river be, respectively, `17 ms^-1 and 8 ms^-1`, find the river width.A. 675mB. 765mC. 567mD. 657m |
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Answer» Correct Answer - b `R=(u^(2)sin 2 theta)/(g)` |
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| 58. |
A small block is connected to one end of a massless spring of un - stretched length `4.9 m`. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by `0.2 m` and released from rest at `t = 0`. It then executes simple harmonic motion with angular frequency `(omega) = (pi//3) rad//s`. Simultaneously at `t = 0`, a small pebble is projected with speed (v) from point (P) at an angle of `45^@` as shown in the figure. Point (P) is at a horizontal distance of `10 m from O`. If the pebble hits the block at `t = 1 s`, the value of (v) is `(take g = 10 m//s^2)`. .A. `sqrt(50)m//s`B. `sqrt(51)m//s`C. `sqrt(52)m//s`D. `sqrt(53)m//s` |
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Answer» Correct Answer - a For a projectile, the vertical displacement is zero after t=1s. `therefore t=(2u sin 45^(@))/(g)` `1=(2u)/(10sqrt(2))` `u=5sqrt(2)=sqrt(50)m//s` |
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| 59. |
Two particles A and B are projected from the same point with the same velocity of projection but at different angles `alpha `and `beta` of projection, such that the maximum height of A is two-third of the horizontal range of B. then which of the following relations are true?A. `3(1-cos 2alpha)=8sin 2beta`B. Range of A=maximum height of BC. Maximum vlaue of `beta` is `(1)/(2)sin^(-1)(3)/(4)`D. Maximum horizontal range of A `=(u^(2))/(g)` and this occurs when `beta=(1)/(2)sin^(-1)(3)/(8)` |
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Answer» Correct Answer - a,c,d `(u^(2)sin^(2)alpha)/(2g)=(2)/(3)(u^(2)sin 2beta)/(g)` `sin^(2)alpha=(4)/(3)xx2sin beta.cosbeta` `((1-cos2alpha))/(2)=(4)/(3) sin 2beta` `beta_("max")=sin^(-1)((3)/(4))` |
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| 60. |
A particle is dropped from point P at time t = 0. At the same time another particle is thrown from point O as shown in the figure and it collides with the particle P. Acceleration due to gravity is along the negative y-axis. If the two particles collide 2 s after they start, find the initial velocity `v_0` of the particle which was projected from O. Point O is not necessarily on ground. A. `sqrt(6)m//s^(-1),theta=tan^(-1)(1)` with x-axisB. `sqrt(26)m//s^(-10theta=tan^(-1)(5)` with x-axisC. `sqrt(2)m//s^(-1),theta=tan^(-1)(2)` with x-axisD. `sqrt(13)m//s^(-1),theta=tan^(-1)(4)` with axis |
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Answer» Correct Answer - b `x=(u cos theta)t=2, u cos theta=1` `(u sin theta)t-(1)/(2)g t^(2)=-10, u sin theta=5` `v=sqrt((u cos theta)^(2)+(u sin theta)^(2))=sqrt(26), tan theta=5` |
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| 61. |
If a body A of mass M is thrown with velocity V at an angle of `30^(@)` to the horizontal and another body B of the same mass is thrown with the same speed at an angle of `60^(@)` to the horizontal. The ratio of horizontal range of A to B will beA. `1:sqrt(3)`B. `sqrt(3):1`C. `1:3`D. `1:1` |
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Answer» Correct Answer - d R`=sin 2 theta` |
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| 62. |
A ball of mass `m` is thrown vertically upwards. Another ball of same mass is thrown at an angle `theta` to the horizontal. If the time if flights for both is same, the ratio of maximum height attained by themA. `1:2`B. `2:1`C. `1:1`D. `1:cos theta` |
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Answer» Correct Answer - c `H=(sin^(2)theta)`. Hence height will remain same. |
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| 63. |
A train is moving along a straight line with a constant acceleration a. A body standing in the train throws a ball forward with a speed of `10ms^(-1)`, at an angle of `60^(@)` to the horizontal . The body has to move forward by 1.15 m inside the train to cathc the ball back to the initial height. the acceleration of the train. in `ms^(-2)` , is: |
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Answer» Correct Answer - 5 `T=(2u sin theta)/(g)=(2xx10xxsqrt(3))/(10xx2)=sqrt(3)s` `R=u cos thetaT-(1)/(2)aT^(2)` `1.15=10x(1)/(2)sqrt(3)-(1)/(3)a(sqrt(3))^(2)` `(3)/(2)a=5sqrt(3)-1.15=8.65-1.15=7.5` `a=7.5xx(2)/(3)=6ms^(-2)` |
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| 64. |
A projectile is thrown in the upward direction making an angle of `60^(@)` with the horizontal direction with a velocity of 147 `ms^(-1)`. Then the time after which its inclination with the horizontal is `45^(@)`, is:A. 15sB. 10.98sC. 5.49sD. 2.745s. |
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Answer» Correct Answer - c Let the velocity of the projectile is v. `v cos 45^(@)=u cos 60^(@) =u=147m//s` Now, `v sin 45^(@)=u sin 60^(@)-gt` |
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| 65. |
A body projected vertically upwords from the top of a tower reaches the ground in `t_(1)` second . If it projected vertically downwards from the some top with same velocity ,it reaches the ground in `t_(2)`seconds . If it is just dropped from the top it reaches the ground in t second .prove that `t=sqrt(t_(1)t_(2))` |
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Answer» Taking downward direction as postive and applying equation of motion `s=ut+((1)/(2))at^(2)`, we get `h=-ut_(1)+(1)/(2)gt_(1)^(2)`............(i) and `h=ut_(2)+(1)/(2)gt_(2)^(2)` ..............(ii) Mu ltiplying Eqn. (i) by `t_(20` and Eqn (ii) by `t_(1)`and adding the two, we get `h(t_(1)+t_(2))=(1)/(2)gt_(1)t_(2)(t(1)+t_(2))` or `h=(1)/(2)gt_(1)t_(2) ["as"t_(1)+t_(2)cancel=0]` or `(1)/(2)g t^(2)=(1)/(2)gt_(1)t_(@) ["as"h=(1)/(2)g t^(2)]` or `t=sqrt(t_(1)t_(2))` |
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| 66. |
A particle is projected at an angle of `45^(@)` with a velocity of `9.8 ms^(-1)`. The horizontal range will be (Take, `g = 9.8 ms^(-2))`A. 9.8mB. 4.9mC. 9.8/ `sqrt(2)m`D. `9.8sqrt(2)m` |
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Answer» Correct Answer - a `R=(u^(2)sin2theta)/(g)` |
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| 67. |
When a particle is projected at an angle to the horizontal, it has range R and time of flight `t_(1)`. If the same projectile is projected with same speed at another angle to have the saem range, time of flight is `t_(2)`. Show that: `t_(1)t_(2)=(2R//g)` |
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Answer» `t_(1)=(2u sin theta)/(g)` and `t_(2)=(2u sin (90^(@)-theta))/(g)=(2u cos theta)/(g)` `therefore t_(1)t_(2)=(2u sin theta)/(g)xx(2u cos theta)/(g)=(2)/(g)((u^(2)sin 2theta)/(g))=(2R)/(g)` |
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| 68. |
Two particles were projected simu ltaneously in horizontal plane with same velocity u perpendicu lar to each other. The time after which their velocities makes angle `60^(@)` with each other is `k(u)/(g)`. Find the value of k |
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Answer» Correct Answer - 1 Velocity of 1st particle after time t `vecv_(1)=uhati-gt hatk` For 2nd particle `vecv_(2)=uhatj-gt hatk` Now, `vecv_(1).vecv_(2)=v_(1)v_(2) cos theta` `g^(2)t^(2)=(u^(2)+g^(2))cos theta` `cos theta=(g^(2)t^(2))/((u^(2)+g^(2)t^(2)))` For `theta=60^(@)` `g^(2)t^(2)=u^(2)` ltb u=gt |
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| 69. |
A body is projected with a speed (u) at an angle to the horizontal to have maximum range. What is its velocity at the highest point ?A. zeroB. uC. `u//sqrt(2)`D. `usqrt(2)` |
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Answer» Correct Answer - c At highest point `u cos theta=u cos 45^(@)` |
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| 70. |
A particle is projected from surface of the inclined plane with speed u and at an angle `theta` with the horizontal. After some time the particle collides elastically with the smooth fixed inclined plane for the first time and subsequently moves in vertical direction. Starting from projection, find the time taken by the particle to reach maximum height. (Neglect time of collision).A. `(2u cos theta)/(g)`B. `(2u sin tehta)/(2)`C. `(u (sin theta+cos theta))/(g)`D. `(2u)/(g)` |
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Answer» Correct Answer - c Particle will strike the point B if velocity of particle with respect to platform is along AB or component of its relative velocity along AD is zero i.e., `u cos theta=v` or `theta=cos^(-1)((v)/(u))` |
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| 71. |
A particle is projected with speed `u` at angle `theta` to the horizontal. Find the radius of curvature at highest point of its trajectoryA. `(u^(@)cos^(2)theta)/(2g)`B. `(sqrt(3)u^(2)cos^(2)theta)/(2g)`C. `(u^(2)cos^(2)theta)/(g)`D. `(sqrt(3)u^(2)cos^(2)theta)/(g)` |
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Answer» Correct Answer - c `r=(u^(2)cos^(2)theta)/(g)` |
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| 72. |
An aeroplane is flying horizontally at a height of `980 m` with velocity `100 ms^(-1)` drops a food packet.A person on the ground is `414 m` ahead horizontally from the dropping point.At what velocity should he move so that he can catch the food packet.A. `50sqrt(2)ms^(-1)`B. `(50)/sqrt(2)ms^(-1)`C. `100ms^(-1)`D. `200ms^(-1)` |
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Answer» Correct Answer - a `T=sqrt((2h)/(g))` is independent of velocity u of the projectile released horizontally. |
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