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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.
| 1. |
Two forces whose magnitudes are in the ratio `3:5` give a resultant of 28N. If the angle of their inclination is `60^@`, find the magnitude of each force. |
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Answer» Let `F_(1)` and `F_(2)` be the two forces. Then `F_(1)=3x,F_(2)=5x,R=28N` and `theta=60^(@)` `R=sqrt(F_(1)^(2)+F_(2)^(2)+2F_(1)F_(2)cos theta` ` rArr 28=sqrt((3x)+(5x)^(2)+2(3x)(5x)cos 60^(@)` `rArr 28=sqrt(9x^(2)+25x^(2)+15x^(2))=7x` `rArr x=28/7=4` `:. F_(1)=3xx4=12N,F_(2)=5xx4=20N`. |
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| 2. |
An arrow shots from a bow with velocity v at an angle ` theta` with the horizontal range R . Its range when it is projected at angle ` 2 theta` with the same velocity isA. 2RB. `R/2`C. `2R cos 2theta`D. `R/(2cos^(2) theta)` |
| Answer» Correct Answer - C | |
| 3. |
The initital speed of an arrow shot from s bow, at an elevation of `30^(@)` , is ` 15 m s^(-1)` , Find the velocity when it hits the ground back. |
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Answer» Here ` v_(0) 15 " m s"^(-1)`, angle of projection , ` theta_(0) = 30^(@)` Therefore, ` v_(0x) = v_(0_ cos theta_(0)` ` = 15 cos(30 ^(@))` ` = ( 15 sqrt3)/2 " m s"^(-1)` And ` v_(0y) = v_(0)sin theta_(0)` ` = 15/2 m s ^(-1)` Horizontal component of velocity remians constant throughout the fight i.e. ` v_(x) = v_(0x) = (15 sqrt3)/2 ms^(-1)` Vertical component of velocity is given by ` v_(y) = v_(0) sin theta_(0) - gt ` Put ` t= T_(f) = ( 2v_(0) sin theta_(0))/g ` ` Rightarrow v_(y) = v_(0) sin theta_(0) -g xx ( 2v_(0) sin theta_(0))/ g` ` v_(y) = - v_(0) sintheta_(0)` `1 v_(y) - 15/2 m s^(-1)` Thus, total velocity ` v = sqrt(v_(x)^(2) + v_(y)^(2))` ` sqrt( ( 15^(2)xx 3)/4 + 15^(2)/4))` `v = 15 m s ^(-1)` Let the final velocity make and angle ` theta` with the positive x - axis , then ` tetha = tan^(-1) (v_(y)/v_(x))` ` theta = tan^(-1) (((-15)/2)/(15 sqrt3/2))` ` = tan^(-1) ((-1)/sqrt3)` ` theta = - 30^(@)` |
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| 4. |
A particle move a distance `x` in time `t` according to equation `x = (t + 5)^-1`. The acceleration of particle is alphaortional to.A. `("velocity")^(3//2)`B. `("distance")^(2)`C. `("distance")^(-2)`D. `("velocity")^(2//3)` |
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Answer» Correct Answer - A Given, `x=(1)/(t+5)` Differentiating both sides w.r.t. t, we get `(dx)/(dt)=-(1)/((t-5)^(2))xx1` `therefore` Velocity, `upsilon =(dx)/(dt)=-(1)/((t+5)^(2))` ….(i) Again differentiating both sides w.r.t. t, we get `(d upsilon)/(dt)=(1xx2)/((t+5)^(3))(1)=(2)/((t+5)^(3))` ....(ii) Now, from Eqs. (i) and (ii), we get `(d upsilon)/(dt)=-2upsilonxx(1)/((t+5))` `=(-2upsilon)(sqrt(upsilon))=-2(upsilon)^(3//2)` `therefore` Acceleration, `a=(d upsilon)/(dt)=-2upsilon^(3//2)rArr a prop ("velocity")^(3//2)` |
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| 5. |
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. `2.50m`B. `3.75 m`C. `4.00 m`D. `1.25 m` |
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Answer» Correct Answer - B Time taken by first drop to reach ground `t = sqrt((2h)/(g)` `rArr t = sqrt((2xx5)/(10))=1s` As the water drops fall at regular intervals from a tap therefore time difference between any two drops `=(1)/(2)s`. In this given time, distance of second drop from the tap `=(1)/(2)g((1)/(2))^(2)=(10)/(8)=1.25 m` Its distance from the ground `= 5-1.25=3.75 m` |
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| 6. |
The ration of the distance traversed, in successive intervals of time by a body, falling from rest, areA. 1 : 3 : 5 : 7 : 9 : …B. 2 : 4 : 6 : 8 : 10 : …C. 1 : 4 : 7 : 10 : 13 : …D. None of the above |
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Answer» Correct Answer - A Here, u = 0, a = g Distance travelled in nth second is given by `D_(n)=u+(a)/(2)(2n-1)` `therefore D_(n) prop(2n-1)` `therefore D_(1):D_(2):D_(3):D_(4):D_(5)…=1:3:5:7:9:..` |
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| 7. |
A stone is just dropped from the window of a train moving along a horizontal straight track with uniform speed. The path of the stone isA. a parabola for an observer standing by the side of the trackB. a horizontal straight line for an ovserver inside the trainC. both (1)&(2) are trueD. (1) is true but (2) is false |
| Answer» Correct Answer - C | |
| 8. |
A particle goes uniformly in circular motion with an angualr speed `pi/8 rad s^(-1)` . What is its time period ? |
| Answer» Correct Answer - 16 s | |
| 9. |
If and object completes 49 revolutions in a minute around a circular path with a speed ` 7 ms ^(-1)` find the radius of the path. |
| Answer» Correct Answer - r = 1.36 m | |
| 10. |
A wheel completes 2000 revolutions to cover the 9.5 km. distance. then the diameter of the wheel is |
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Answer» Given, number of revolutions n = 200 Distance `x=9.5 km = 9.5xx10^(3)m=9500 m` `because` Distance covered in n reveloutions is equal to the circumference of the wheel. `therefore x=n . 2pi r rArr x=n . pi D` (`because` Diameter `= 2xx` radius) `9500=2000xxpi xx D` `rArr` Diameter, `D=(9500)/(2000xx3.14)=1.5m` |
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| 11. |
The position `x` of a particle with respect to time `t` along the x-axis is given by `x=9t^(2)-t^(3)` where `x` is in meter and `t` in second. What will be the position of this particle when it achieves maximum speed along the positive `x` directionA. 24 mB. 32 mC. 54 mD. 81 m |
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Answer» Correct Answer - C The position x of a particle w.r.t. to time t along X-axis `x=9t^(2)-t^(3)` ….(i) Differentiating Eq. (i) w.r.t. time, we get speed, i.e., `upsilon = (dx)/(dt)=(d)/(dt)(9t^(2)-t^(3)) rArr upsilon = 18t-3t^(2)` …(ii) Again differntiating Eq. (ii) w.r.t. time, we get acceleration, i.e., `a=(d upsilon)/(dt)=(d)/(dt)(18t-3t^(2)) rArr a=18 -6t` ...(iii) Now, when speed of particle is maximum, its acceleration is zero. `a= 0 rArr 18-6t=0 rArr t=3s` Putting in Eq. (i), we obtain position of particle at that time `x=9(3)^(2)-(3)^(3)=9(9)-27=81-27=54m` |
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| 12. |
In the previous problem, if `H_(1)` and `H_(2)` are the maximum heights in the two cases, thenA. `R = 2sqrt(H_(1)H_(2))`B. `R = 4sqrt(H_(1)H_(2))`C. `R = sqrt(H_(1)^(2) + H_(2)^(2))`D. `R = H_(1) - H_(2)` |
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Answer» Correct Answer - B `theta: H_(1) = (u^(2) sin^(2) theta)/(2g)` `(90 - theta): H_(2) = (u^(2) sin^(2) (90 - theta))/(2g) = (u^(2) cos^(2) theta)/(2g)` `R = (u^(2))/(g).2 sin theta cos theta` `sqrt(H_(1)H_(2)) = sqrt((u^(2) sin^(2) theta)/(2g).(u^(2) cos^(2) theta)/(2g))` `= (u^(2) sin theta cos theta)/(2 g) = (1)/(4) ((u^(2))/(g).2 sin theta cos theta)` `= (1)/(4)R` `R = 4sqrt(H_(1)H_(2))` |
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| 13. |
Assertion : In the s-t graph as shown in figure, velocity of particle is negative and acceleration is positive. Reason : Slope of s-t graph is negative and increasing in magnitude.A. If both Assertion and Reason are correct and Reason is the correct explanation of assertion.B. If both Assertion and Reason are correct but Reason in not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - D Slope is negative. Therefore velocity is negative. Slope (therefore velocity) is increasing in magnitude. Therefore acceleration is also negative. |
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| 14. |
Assertion : A body of mass 4 kg has an initial velocity `5hat(i)ms^(-1)`. It is subjected to a force of `4hat(j)N`. The displacement of body from origin after 4 s will be `21.5 m`. Reason : The equation `v=u+a t` can be applied to obtain v if a is constant.A. If both Assertion and Reason are correct and Reason is the correct explanation of assertion.B. If both Assertion and Reason are correct but Reason in not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - B `a=(F)/(m)=1 hat(j)N` `s= u t +(1)/(2)a t^(2)=20 hat(i)+8 hat(j)` `therefore `s=21.5m` |
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| 15. |
A ball is projected from the ground with velocity `u` such that its range is maximum ThenA. Its velocity at half the maximum height is `sqrt3/2u`B. Its velocity at the maximum height is `u`C. Change in its velocity when it returns to the ground is `u`D. all the above are true |
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Answer» Correct Answer - A At half of max height velocity `(V)=usqrt((1+cos^(2)theta)/(2))` |
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| 16. |
A body is projected obliquely from the ground such that its horizontal range is maximum.If the change in its maximum height to maximum height, is `P`, the change in its linear momentum as it travels from the point of projection to the landing point on the ground will beA. `P`B. `sqrtP`C. `2 P`D. `2sqrt2 P` |
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Answer» Correct Answer - D `p=(mv sin theta)/sqrt2,p^(1)=2mv sin theta` |
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| 17. |
A particle moves in the `x-y` plane under the action of a force `vecF` such that the value of its linear momentum `vecP` at any time `t is P_(x) = 2 cos t, P_(y) = 2 sin t`. The angle `theta` between vecF and vecP` at a given time `t` will be:A. `theta=0^(@)`B. `theta=30^(@)`C. `theta=90^(@)`D. `theta=180^(@)` |
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Answer» Correct Answer - C `vecp=2 cos thati+2sinthatj` `vecF=(dvecp)/(dt)=-2sint hati+2 cos hattj` `vecp.vecF=0 rArr vecp_|_ "to" vecF` |
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| 18. |
`BC` is divided into four equal parts by `P,Q` and `R`.The resultant of `vec(AB)` and `3vec(AC)` is A. `vec(AR)`B. `4vec(AR)`C. `4vec(AP)`D. `vec(PQ)` |
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Answer» Correct Answer - B `vec(AC)+vec(CB)=vec(AB),vec(AC)+vec(4CR)=vec(AB)` `vec(AC)+vec4((AR)-vec(AC)),vec4(AR)-3vec(AC)=vec(AB)` `:.vec(AB)+3vec(AC)=4vec(AR)` |
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| 19. |
A stone is projected with a velocity `20sqrt2 m//s` at an angle of `45^(@)` to the horizontal.The average velocity of stone during its motion from starting point to its maximum height isA. `10sqrt5 m//s`B. `20sqrt5 m//s`C. `5sqrt5 m//s`D. `20m//s` |
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Answer» Correct Answer - A `v=u/2sqrt(1+3 cos^(2)theta)` |
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| 20. |
A stone is projected with a velocity `20sqrt2 m//s` at an angle of `45^(@)` to the horizontal.The average velocity of stone during its motion from starting point to its maximum height isA. `10sqrt5 m//s`B. `20sqrt5 m//s`C. `5sqrt5 m//s`D. `20 m//s` |
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Answer» Correct Answer - A `V_(av)=sqrt((V_(av))_(x)^(2)+(V_(av))_(y)^(2)),v=u/2sqrt(1+3cos ^(2)theta)` |
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| 21. |
A particle is projected from ground at an angle `45^(@)` with initial velocity `20sqrt2 ms^(-1)`.The magnitude of average velocity in a timer interval from `t=0` to `t=3 s` in `ms^(-1)` isA. 20.62B. 10.31C. 41.14D. 5.15 |
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Answer» Correct Answer - A `u_(i)=u_(x)i+u_(y)j` `u_(f)=u_(x)i+(u_(y)-"gt")j,V_(avg)=(u_(i)+u_(f))/2` |
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| 22. |
A ball is projected with `20sqrt2 m//s` at angle `45^(@)` with horizontal.The angular velocity of the particle at highest point of its journey about point of projection isA. `0.1 rad//s`B. `1 rad//s`C. `0.3 rad//s`D. `0.4 rad//s` |
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Answer» Correct Answer - B `omega=v/r` |
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| 23. |
A body, rpojected horizontally with a speed `u` from the top of a tower of height `h`, reaches the ground at a horizontally distance `R` from the tower. Another body, projected horizontally from the top of a tower of height `4h`, reaches the ground at horizontal distance `2R` from the tower. The initial speed of the second body isA. `u`B. `2u`C. `3u`D. `4u` |
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Answer» Correct Answer - A `y = (g x^(2))/(2 u^(2))` `h = (g R^(2))/(2 u^(2)), 4h = (g(2 R)^(2))/(2 u_(1)^(2))` `u_(1) = u` |
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| 24. |
The motion of a particle along a straight line is described by equation : `x = 8 + 12 t - t^3` where `x` is in metre and `t` in second. The retardation of the particle when its velocity becomes zero is.A. `24 ms^(-2)`B. zeroC. `6 ms^(-2)`D. `12 ms^(-2)` |
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Answer» Correct Answer - D Given, `x=8+12t-t^(3)` We know, `upsilon=(dx)/(dt)` and `a=(d upsilon)/(dt)` So, `upsilon=12-3t^(2)` and `a=-6t` At `t = 2 s` `upsilon = 0` and `a = - 6t` `rArr a=-12 ms(-2)` So, ratardation of the particle `= 12 ms^(-2)`. |
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| 25. |
A person walks up a stationary escalator in time `t_(1)`. If he remains stationary on the escalator, then it can take him up in time `t_(2)`. How much time would it take for him to walk up the moving escalator? |
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Answer» Let `L` be the length of escalator. Speed of man w.r.t. escalator is `V_(ME)=L/t_(1)` Speed of escalator `v_(E)=L/t_(2)` Speed of man with respect to ground would be `V_(M)=V_(ME)+V_(E)=L(1/t_(1)+1/t_(2))` `:.` The desired time is `t=L/v_(M)=(t_(1)t_(2))/(t_(1)+t_(2))` |
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| 26. |
Raindrops are falling with velocity `10 sqrt2` m/s making and angle `45^(@)` with the vertical. The drops appear to be falling vertically to a man running with constant velocity . The velocity of rain drops change such that the rain drops now appear to be falling vertically with `sqrt3` times the velcoity it appeared earlier to the same person running with same velocity . The magnitude of velocity of man w.r.t ground isA. `10sqrt2 m//s`B. ` 10sqrt3 m//s`C. 10 m/sD. 20 m/s |
| Answer» Correct Answer - C | |
| 27. |
If a stone is to hit at a point which is at a distance `d` away and at a height `h` (see fig) above the point from where the stone starts,then what is the value of initial speed `u` if the stone is launched at an angle `theta`? A. `d/sinthetasqrt(g/(2(d tan theta-h))`B. `d/costhetasqrt(g/(2(d tan theta-h))`C. `sqrt((gd)/(h cos ^(2)theta)`D. `sqrt((gd)/((d-h))` |
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Answer» Correct Answer - B `h=(u sin theta)t-1/2"gt"^(2),d=(u cos theta)t` |
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| 28. |
A stone is projected from the ground in such a direction so as to hit a bird on the top of a telegraph post of height h and attains the maximum height of 2h above the ground. If at the insatant of projection, the bird were to fly away horizontally with a uniform speed, find the ratio between the horizontal velocity of bird and the horizontal component of velocity of stone, if the stone hits the bird while descending.A. `2/(sqrt2+1)`B. `1/(sqrt2+1)`C. `2/(sqrt2-1)`D. `1/(sqrt2-1)` |
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Answer» Correct Answer - A Let the stone is projected with a velocity `u` at an angle `theta` with the horizontal, we have `(2h)=((u sin theta)^(2))/(2g) or u sin theta=2sqrtgh` Suppose `t` is the time taken by the stone to reach the height `h` above the ground.Then `h=u sin theta t-1/2g t^(2) or (g t^(2))/2-u sin thetat+h=0` As we have `u sin theta =2sqrt(gh)` `:. (g t)^(2)/2-2sqrt(gh)t+h=0` Solving above equation for `t`, we get `t=(2sqrt(gh)+-sqrt((2sqrt(gh))^(2)-4xxg/hh))/(2xxg/2)&t_(2)sqrt((2h)/g)(sqrt2+1)` Where `t_(1)` and `t_(2)` correspond to `P` and `Q` in the figure.Suppose `v` is the horizontal velocity of the bird. Then `PQ=vt_(2)`. |
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| 29. |
Read each statement below carefully and state with reasons, with it is true or false : (a) The magnitude of vector is always a scalar. (b) Each component of a vector is always a scalar. (c) The total path length is always equal to the magnitude of the displacement vector of a particle. (d) The average speed of a particle (defined as total path length divided by the time taken to cover the path) is greater or equal to the magnitude of average velocity of the particle over the same interval of time. (e) three vectors not lying in a plane can never add up to give a null vector. |
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Answer» a) True, magnitude of the veloicty of a body moving in a straight line may be equal to the speed of the body. b)False, each component of a vector is always a vector, not scalar c)False, total path length can be also be more than the magnitude of displacement vector of a particle. d) True, because the total path length is either greater than or equal to the magntitude of the displacement vector. e) True, this is because the resultant of two vectors will not be in the plane of third vector and hence, cannot cancel its effect to give null vector. |
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| 30. |
Which of the following is not a scalar quantity ?A. TemperatureB. Coefficient of frictionC. ChargeD. Impulse |
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Answer» Correct Answer - D Among the given physical quantities impulse is a vector quantity whereas all other are scalar quantities. |
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| 31. |
Read each statement below carefully and state with reason and examples, if it is true or false. A scalar quantity is one that (a) is conserved in a process(b) can never take negative values (c) must be dimensionless (d) does not vary from one point to another in space (e) has the same value for observers with different orientations of axes. |
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Answer» a) False, because kinetic energy is a scalar but does not remain conserved in an inelastic collision. b)False, because potential energy in a gravitational field may have negative values. c) False, because mass, length, time, speed, work etc. all have dimensions. False, because speed, energy etc., vary from point to point in space. e) True, because a scalar quantity will have the same value for observes with different orientations of axes since a scalar has no direction of its own. |
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| 32. |
A cannon is fixed with a smooth massive trolley car at an angle `theta` as shown in the figure.The trolley car slides from rest down the inclined plane of angle of inclination `beta`. The muzzle velocity of the shell fired at `t=y_(0)` from the cannon is `u`,such that the shell moves perpendicular to the inclined just after the firing. the difference in range of the shell relative to the trolley car and ground is: A. `(u^(2)sin 2theta)/(g cos beta)`B. `(u^(2)cos theta)/(2g sin beta)`C. `(u^(2) sin theta sin beta)/(2g)`D. `(2U^(2) sin theta cos (theta-beta))/(g cos^(2) g)` |
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Answer» Correct Answer - D Distance moved by shell `=(V_(0)cos theta)t+1/2g sin betat^(2)` `t=(2U sin theta)/(g cos beta)` on solving `x=(2U^(2) sin theta)/(g cos^(2) beta)cos(theta-beta)` |
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| 33. |
A cannon is fixed with a smooth massive trolley car at an angle `theta` as shown in the figure.The trolley car slides from rest down the inclined plane of angle of inclination `beta`. The muzzle velocity of the shell fired at `t=y_(0)` from the cannon is `u`,such that the shell moves perpendicular to the inclined just after the firing. The value of `t_(0)` is: A. `(u cos theta)/g`B. `(u cos theta)/(g cos theta)`C. `(u cos theta)/(g sin beta)`D. `(u sin theta)/(g cos theta)` |
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Answer» Correct Answer - C Velocity along tralley is zero i.e., `V_(x)=U_(x)+a_(x)t` `0=U cos theta-(g sin beta)t, t=(U cos theta)/(g sin beta)` |
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| 34. |
A cannon is fixed with a smooth massive trolley car at an angle `theta` as shown in the figure.The trolley car slides from rest down the inclined plane of angle of inclination `beta`. The muzzle velocity of the shell fired at `t=y_(0)` from the cannon is `u`,such that the shell moves perpendicular to the inclined just after the firing. after what time should the shell be fired such that it will go vertically up? A. `(u cos theta)/(g sin beta)`B. `(u sin (theta+beta))/(g cos theta sin beta)`C. `(u cos (theta+beta))/(g cos beta)`D. `(u cos (theta+beta))/(g sin beta cos beta)` |
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Answer» Correct Answer - D `V_(x)=U_(x),((a_(x))cos beta)t= U cos (theta+beta)` `g sin beta cos betat=U cos (theta+beta),t=(U cos (theta+beta))/(g sin beta cos beta)` |
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| 35. |
The v-t plot of a moving object ios shown in the figure. The average velocity of the object during the first 10 s is A. zeroB. `2.5ms^(-1)`C. `5 ms^(-1)`D. `2 ms^(-1)` |
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Answer» Correct Answer - A Since, total displacement is zero, hence average velocity is also zero. |
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| 36. |
An athlete complete one round of a circular track of diameter `200 m` in `40 s`. What will be the distance covered and the displacement at the end of `2` minutes `20 s` ? |
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Answer» Diameter of circular track, d = 200 m Circumference of circular track `=2pi r=pi xx (d)` `=(22)/(7)xx200=(4400)/(7)m` Time for completing one round = 40 s. |
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| 37. |
A point size body is moving along a circle at an angular velocity `2.8rads^(-1)`.If centripetal acceleration of body is `7ms^(-2)` then its speed isA. `1.25ms^(-1)`B. `2.5ms^(-1)`C. `3.5ms^(-1)`D. `7ms^(-1)` |
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Answer» Correct Answer - B Centripetal acceleration `a=vomega rArr v=a/omega` |
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| 38. |
ABCDEF is a regular hexagon with point O as centre. The value of `vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)` is |
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Answer» From the diagram `vec(AB)=vec(-DE),vec(BC)=vec(-EF)` `vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)`. `=vec(AB)+(vec(AB)+vec(BC))+vec(AD)+(vec(AD)+vec(DE))+(vec(AD)+vec(DE)+vec(EF))` `=vec(3AD)=3(vec(2AO))=6vec((AO))` |
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| 39. |
A river `400 m` wide is flowing at a rate of `2.0 m//s.` A boat is sailing at a velocity of `10.0 m//s` with respect to the water In a direction perpendicular to the river. (a) Find the time taken by the boat to reach the opposite bank. (b) How far from the point directly opposite to the starting point does the boat reach the opposite bank?A. `40 sec,80 m`B. `30 sec,40 m`C. `20 sec,20 m`D. `35 sec,80 m` |
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Answer» Correct Answer - A As it is given that boat is sailing in a direction normal to current. Crossing velocity of boat is `=10 m//s`. So time taken by the boat to reach the other bank is `400/10=40s`.Drift due to flow of river is = Drift velocity `xx` time to cross the river Here boat is sailing in normal direction so direction so drift velocity is the river current velocity.Thus, drift is `x=2.0xx40=80m` |
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| 40. |
A particle `A` is projected from the ground with an initial velocity of `10m//s` at an angle of `60^(@)` with horizontal. From what height should an another particle `B` be projected horizontally with velocity `5m//s` so that both the particles collide in ground at point `C` if both are projected simultaneously `g=10 m//s^(2)`. A. `10 m`B. `15 m`C. `20 m`D. `30 m` |
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Answer» Correct Answer - B `t=(2U sin theta)/g=sqrt3 , S=1/2"gt"^(2) h=15m` |
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| 41. |
If `vecA xx vecB = vecC+ vecD`, them select the correct alternative:A. `vecB` is parallel to `vecC + vecD.`B. `vecA` is perpendicular to `vecC.`C. Component of `vecC` along `vecA` = component of `vecD` along `vecA`D. Component of `vecC` along `vecA =` - component of `vecD` along `vecA` |
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Answer» Correct Answer - D According to definition of cross-product `vec(C)+vec(D)` is perpendicular to both `vec(A)andvec(B)` i.e., `vec(A).(vec(C)+vec(D))=0orvec(A). vec(C)+vec(A).vec(D)=0` or A (component of `vec(C)` along `vec(A)`) + A(component of `vce(D)` along `vec(A)`)=0 or component of `vec(C)` along `vec(A)` =- component of `vec(D)` along `vec(A)` |
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| 42. |
सिद्ध कीजिए कि `|vecA xx vecB|^(2) + |vecA * vecB|^(2) = (AB)^(2).`A. zeroB. `A^(2)B^(2)`C. ABD. `sqrt(AB)` |
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Answer» Correct Answer - B Let `theta` be angle between vectors `vec(A)andvce(B)` `:.|vec(A)xxvec(B)|=ABsinthetaand|vec(A).vec(B)|=ABc ostheta` `:.|vec(A)xxvec(B)|^(2)+|vec(A).vec(B)|^(2)=(ABsintheta)^(2)+(ABcostheta)^(2)` `=A^(2)B^(2)sin^(2)cos^(2)theta=A^(2)B^(2)` |
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| 43. |
If `vecA` and `vecB` are two vectors, then which of the following is wrong?A. `vecA + vecB = vecB + vecA`B. `vecA * vecB = vecB * vecA`C. `vecA xx vecB = vecB xx vecA`D. `vecA - vecB = - (vecB - vecA)` |
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Answer» Correct Answer - C Vector product of two vectors is anti-commutative. i.e., `vec(A)xxvec(B)=-vec(B)xxvec(A)` |
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| 44. |
When we analyse the projectile motion from any accelerated frame `O` as `vecr_(o),vecu_(o)` and `veca_(o)` respectively, express the following terms, `vecr_(pO)+veca_(p)-vecr_(O),vecu_(pO)=vecu_(p)-vecu_(O)` and `veca_(pO)=veca_(p)-veca_(O)` where `P` stands for projectile. Then using the following kinematical equations of the projectile (For constant acceleration) relative to the accelerating frame ,we have `vecS_(pO)=vecu_(pO)t+1/2veca_(pO)t^(2),vecv_(pO)` `=vecu_(pO)+veca_(pO)t` and `v_(pO)^(2)=u_(pO)^(2)+2veca.vecs_pO` Using the above expressions, anwer the following question:A projectile has initial velocity `v_(0)` realtive to the large plate which is moving with a constant upward acceleration a. Which of the following remains equal for the observers `A` and `B`? A. Maximum heightB. RangeC. Time of flightD. Angle of projection |
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Answer» Correct Answer - D Angle of projection remains equal for both `A&B`. |
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| 45. |
When we analyse the projectile motion from any accelerated frame `O` as `vecr_(o),vecu_(o)` and `veca_(o)` respectively, express the following terms, `vecr_(pO)+veca_(p)-vecr_(O),vecu_(pO)=vecu_(p)-vecu_(O)` and `veca_(pO)=veca_(p)-veca_(O)` where `P` stands for projectile. Then using the following kinematical equations of the projectile (For constant acceleration) relative to the accelerating frame ,we have `vecS_(pO)=vecu_(pO)t+1/2veca_(pO)t^(2),vecv_(pO)` `=vecu_(pO)+veca_(pO)t` and `v_(pO)^(2)=u_(pO)^(2)+2veca.vecs_pO` Using the above expressions, anwer the following question:A projectile has initial velocity `v_(0)` realtive to the large plate which is moving with a constant upward acceleration a. Refering to Q.1, velocity of the projectile relative to `B` ofter some time A. `ltv_(0) "at an angle" thetalttheta_(0)`B. `gtv_(0) "at an angle" thetagttheta_(0)`C. `gtv_(0) "at an angle" theta=theta_(0)`D. `v_(0) "at an angle" theta=theta_(0)` |
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Answer» Correct Answer - D Here`B` is ground frame, w.r.t `B` projection velocity will be greater than `V_(0)`. Hence when `thetagttheta_(0)` projectile velocity will be greater than `V_(0)`. |
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| 46. |
Which of the following represents uniformly accelerated motion ?A. `x=sqrt((t+a)/(b))`B. `x=(t+a)/(b)`C. `t=sqrt((x+a)/(b))`D. `x=sqrt(t+a)` |
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Answer» Correct Answer - C `t=sqrt((x+a)/(b)) or (x+a)=bt^(2) or x=-a+bt^(2)` Comparing this equation with genral equation of uniformly accelerated motion, `s=s_(i)+ut+(1)/(2)at^(2)` we see that `s_(i)=-a, u=0` and acceleration = 2b. |
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| 47. |
The displacement (x) of a particle depends on time t as `x=alpha t^(2)-beta t^(3)`. Choose the incorrect statements from the following.A. The particle never returns to its starting pointB. The particle comes to rest after time `(2alpha)/(3beta)`C. The initial velocity of the particle is zeroD. The initial acceleration of the particle is zero |
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Answer» Correct Answer - D x= 0 at t = 0 and `t = alpha//beta`. So, the particle returns to starting point at `t=alpha//beta`. `upsilon=(dx)/(dt)=2alpha t-3beta t^(2)` At t = 0, `upsilon = 0` i.e., initial velocity of particle is zero. `upsilon=0` at t = 0 and `t=(2alpha)/(3beta)` Thus, the particle comes to rest after time `t=2alpha//3beta rArr a=(d upsilon)/(dt)=2alpha-6beta t` At `t=0, a=2, a ne 0` |
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| 48. |
Calculate the angular speed of flywheel making 420 revolutions per minute. |
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Answer» Given v = 420 revolutions per minute. ` 420/60` revoltions per second v = 7 revolutions- per second. Therefore angular speed ` omega = 2 piv` ` 2 xx 22/7 xx 7` `omega` = 44 rad/s. |
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| 49. |
A circular disc is rotating about its own axis at uniform rate completes `30` rotations in one minute.The angular velocity of disc in rad `s^(-1)` isA. `2pi`B. `pi`C. `pi/2`D. `pi/4i` |
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Answer» Correct Answer - B `omega=2 pi n ,` where n=no. of revolutions per sec |
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| 50. |
A circular plate is rotating about its own axis at an angular velocity `100` revolutions per minute. The linear velocity of a particle `P` of plate at a distance `4.2 cm` from axis of rotation isA. `0.22 m//s`B. `0.44 m//s`C. `2.2 m//s`D. `4.4 m//s` |
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Answer» Correct Answer - B Linear velocity `v=romega` |
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