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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.
| 501. |
Which of the two , velcity and acceleration, gives the direction of motion of the body . Explain it with the help of an illustration. |
| Answer» It is the volocity of body and bot the acceleration which gives the direction of motion of the body. When a body ism projected vertically upwards, its velocity and direction of motion are upwards but its acceleration is vertically downwards. | |
| 502. |
Is the acceleration of a car greater when when the accelerator is pushed to the floor or when brake is pushed hard ? |
| Answer» Acceleration of a car is greater when braken pedal is pushed hard, because car suddenly comes to rest, i.e. the rate if change of velocity of car is large. | |
| 503. |
Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.A. `tan^-1(1)`B. `tan^-1(2)`C. `tan^-1(3)`D. `tan^-1(4)` |
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Answer» Correct Answer - D (d) We find that `H = R` or `(u^2 sin^2 theta)/(2 g) = (u^2 2 sin theta cos theta)/(g)` or `tan theta = 4 or theta = tan^-1 (4)`. |
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| 504. |
A particle moves towards east with velocity `5m//s`. After `10 seconds` its direction changes towards north with same Velocity. The average acceleration of the particle is |
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Answer» Correct Answer - B |
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| 505. |
The equation of projectile is `y = 16 x - (5 x^2)/(4)`. Find the horizontal range. |
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Answer» Standard equation of projection motion. `y = x tan theta[1 - (x)/(R )]` Given equation : `y = 16 x - (5 x^2)/(4) = 16 x [1 -(x)/(64//5)]` By comparing above equations, `R = (64)/(5) = 12.8 m`. |
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| 506. |
A body starts from the origin with an acceleration of `6 ms^-1` along the x - axis and `8 ms^-1` along the y - axis. Find its distance from the origin after `4 s`. |
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Answer» Displacement along X - axis : `x = u_x t + (1)/(2)a_x t^2 = (1)/(2) xx 8 xx (4)^2 = 48m` Displacement along Y - axis : `y = u_y t + (1)/(2)a_y t^2 = (1)/(2) xx 8 xx (4)^2 = 64 m` Total distance from the origin =`sqrt(x^2 + y^2) = sqrt((48)^2 +(64)^2) = 80 m`. |
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| 507. |
Two cars cover the distance between two stations in 2 hours. The first car covers the first half of the distance at 20 kmph and the second half at 60 kmph and the second car covers the distance with uniform acceleration, starting with a velocity of 10 kmph.At what instant/instants will the two cars have the same velocity ? |
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Answer» Correct Answer - `1/2`hr, 2`1/2`hr |
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| 508. |
A point moves in the x-y plane according to the equations x=at, y=at-`bt^2`.Find (a)the equation of the trajectory, (b)acceleration as a function of t, (c )the instant `t_0` at which the velocity and acceleration are at `pi//4` . |
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Answer» Correct Answer - `y=x-b/ax^2,2b,a/b` |
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| 509. |
For the one dimensional motion, described by `x=t-sint`A. (a) ` x (t) gt o for all t gt 0`B. (b) ` v (t) gt0 for all t gt 0`C. (c ) ` a (t) gt 0 for all t gt 0`D. (d) ` v (t) ` lies between (0) and (2)` |
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Answer» Correct Answer - A::D ` x=t -sin t , velocity, ` v ( dx0/(dt) =1 t , acc. A = (dv)/(dt) = sin t` :. X (t) gt 0 ` fro all values ` t gto ` and ` v (t) can be zero for one value of (t) .(a) (t) can be zero for one value of (t) . If ` t=0, v 9t) = 1 - 1 =0`. if ` t= pi , v9t) = 1 - cos pi = 1 - (-1) =2`. |
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| 510. |
A car of mass `=m=1000 kg` is moving with constant speed `v=100 m//s` on a parabolic shaped bridge AFOE of span force applied by the bridge on the car when he car is at point E, is: A. `5000 sqrt(5/2) N`B. `5000/sqrt(2) N`C. `10000/sqrt(2) N`D. `5000 sqrt(2/5) N` |
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Answer» Correct Answer - A |
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| 511. |
In the arrangement of rigid links of equal length `l`, they can freely rotate about the joined ends as shown in the figure- 1.138. If the end U is pulled horizontally with constant speed 20 m/s, find the approx. speed of end `P` when the angle SUT is `90^(@)`. A. 5 m/sB. 10 m/sC. 7.1 m/sD. 14.12 m/s |
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Answer» Correct Answer - C |
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| 512. |
A ball is hit by a batsman at an angle of `37^(@)` as shown in figure. The man standing at `P` should run at what minimum velocity so that he catches the ball before it strikes the ground. Assume that height of man is negligible in comparison to maximum height of projectile. A. `3ms^(-1)`B. `5 ms^(-1)`C. `9ms^(-1)`D. `12ms^(-1)` |
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Answer» Correct Answer - B |
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| 513. |
A ball is hit by a batsman at an angle of `37^(@)` as shown in figure. The man standing at `P` should run at what minimum velocity so that he catches the ball before it strikes the ground. Assume that height of man is negligible in compariosn to maximum height of projectile. A. `3 ms^(-1)`B. `5 ms^(-1)`C. `9 ms^(-1)`D. `12 ms^(-1)` |
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Answer» Correct Answer - B `T = (2u_(y))/(a_(y)) rArr T = (2xx15 xx sin 37^(@))/(10) rArr T = 1.8 sec` `u_(min) = (dis tan ce)/(time) = (9)/(1.8) = 5m//s` |
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| 514. |
A ball is projected form top of a tower with a velocity of `5m//s` at an angle of `53^(@)` to horizontal. Its speed when it is at a height of `0.45m` from the point of projection is:A. `2 m//s`B. `3 m//s`C. `4 m//s`D. data insufficient. |
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Answer» Correct Answer - C From energy conservation `(1)/(2)mu^(2) - (1)/(2)mv^(2) = m gh` `v = sqrt(u^(2) -2gh) rArr v = sqrt((5)^(2) -2xx10xx4.5)` `rArr v = sqrt(25 -9) rArr v = 4m//s` |
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| 515. |
A boy dropped a ball from the top of a tower of height 125 m then the average velocity of the ball at the end of 5 second if it takes 5 s to reach the ground _______ ` m s^(-1)`A. 25B. 125C. 50D. 250 |
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Answer» Correct Answer - A Average velocity = `(" total displacement")/(" total time")` ` = 125/5 = 25 ms ^(-1)` |
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| 516. |
Unit of speed is ____A. ` m min^(-1)`B. `km h ^(-1)`C. `km s(-1)`D. All the above |
| Answer» Correct Answer - D | |
| 517. |
If body travels half of its path in the last second of its fall from rest, find the time and height of its fall. |
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Answer» Let (T) be the total o fall of vertical distance (S). As per question distanc travelled in (T) the second is ` S//2` and distance travelled in time (T-1) cecond is S//2. Here, u =0 , a= 9.8 ms^(-2) , D_T = S//2` As ` D_t = u + a/2 (2 T-10, we have `S//2 =0 + (9.8) /2 [2 T -1] = 4.9 ( 2T -1)` ...(i) Roe distance travelled in ` (T -1) secinds, we use the relation, ` S= ut + 1/2 at^2`, we gat `S/2 =0 + 1/2 xx 9.8 (T-1)^2 ` ..(iii) Equating (i) and (ii), we have ` 4.9 ( 2 T-1) = 9.8 (T-1)^2` or ` 2 T -1 = (T -10 ^2 = T^2 -2T + 1` or `T^2 -4T + 2 =0` :. ` T = (4 =- sqrt 16 -8)/2 = 2 +- sqrt 2 = (2 + 1.4140` or ` (- 1.414 0 = 3. 414 s or 0. 586 S` Time `0586 s` being less than ` 1 second` is not possible, since the total motion is for mote than ` 1 sectond`. Therefore ` T= 3.414 S` Potting this value in (i) , we get ` S= 2 xx 4.9 (2 xx 3 .414 -1) = 57 .11 m`. |
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| 518. |
Choose the correct statement :A. Every oscillatory motion is periodic in nature.B. Every periodic motion is oscillatory in nature.C. The motion of a pendulum bob is periodic in nature ( within its small amplitude)D. Both (a) and ( c) |
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Answer» Correct Answer - D (1) Every oscillatory motion is periodic in nature, i.e, every to and fro motion takes places in equal intervals of time. (2) Every periodic motion is not oscillatory in nature. For example, the rotation of the earth around the sun is periodicd in nature, but it is not an oscillatory in nature. |
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| 519. |
Choose the correct statementA. Unit of acceleration is `ms ^(-2)`B. Speed = distance/ time .C. Magnitude of displacement made by a body is less than or equal to the distance travelled by the body.D. All the above |
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Answer» Correct Answer - D (1) The unit o acceleration is m ` s^(-2)` (2) Speed = `("distance")/("time")` (3)Displacement is the shortest distance between the initial and final positions in a specified diretion . So, the magnitude of displacement travelled by a body is less than or equal to the distance travelled by the body. |
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| 520. |
A bike starts from rest and accelerates at `4 m//s^(2)` for 5.0 s. It then moves at contant velocity for 25.0s, and then decelerates at `2.0 m//s^(2)` unit it stops. Find the total distance that the motorcycle has moved. |
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Answer» Correct Answer - [650 m] |
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| 521. |
What is the rate of change of displacement called with respect to time? |
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Answer» Velocity is the rate of change of displacement called with respect to time. |
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| 522. |
How much should be the angle of a projection so that the body covers the maximum distance? |
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Answer» 45° angle of a projection. |
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| 523. |
Explain motion. |
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Answer» Concept of motion : If a body changes its place with time, then it is said to be in motion. An object is said to be in motion in absolute form when its speed is relative to a point in the universe which is stable. But there is no such point known in the world. Hence, nothing is in absolute rest or in absolute motion. Truly, rest and motion both are relative. If some object relative to the other changes its position as the time passes, then the object would be said moving relative to the other object. |
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| 524. |
A bus is moving with a velocity `10 ms^-1` on a straight road. A scooterist wishes to overtake the bus in `100 s`. If the bus is at a distance of `1 km` from the scooterist, with what velocity should the scooterist chase the bus ?A. `50ms^(-1)`B. `40ms^(-1)`C. `30ms^(-1)`D. `20ms^(-1)` |
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Answer» Correct Answer - D |
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| 525. |
Air is blowing and is providing a constant horizontal acceleration `a_(x)=g` to the particle as shown in the figure. Particle is projected from point `P` with a velocity `u` in upward direction. Let `Q` be the highest point of the particle. Speed of the particle at highest point `Q` is A. `sqrt(2)u`B. `u`C. `u//sqrt(2)`D. None of these |
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Answer» Correct Answer - B |
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| 526. |
The speedometer of a car moving eastward reads ` 50 km //h`. It passes another car which travels westward at `50 km//h`. (i) Do both the cars have same dpeed ? (ii) Do they have the same velocity ? (ii) What is the relative velocity of car ` Aw.r.t. car B`. |
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Answer» (i) Since, speed is a scalar quantity and speedometer reads instantaneous speed of the car, hence both the cars have same pseed. (ii) Since, velocity is a vector quantity, hence hothe the cars having opposite velocities as they are moving in opposite directions. (iii) Relativ e velocity of car `A w.r.t. car B` `=v_(A) + v_(B)` `= (50 km //h + 50 km //h)1 `= 100 km //h due east. |
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| 527. |
The variation ofvelocity ofa particle moving along a straight line is illustrated in the following figure-1.122. The distance covered by the particle in 4 seconds is : A. 60 mB. 25 mC. 55 mD. 30 m |
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Answer» Correct Answer - C |
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| 528. |
The following figure-1.105 shows the velocity-time graph ofa body. According to this, at the point B: A. The force is zeroB. The force is in the direction of the motionC. The force is in opposite direction of the motionD. It is only the gravitational force |
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Answer» Correct Answer - C |
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| 529. |
The position-time graoh of a unirom motion in one dimension of a body can have negative slope. When the speed of body decrease with tiem, the position-time graph of the moving body has begative slope .A. (a) Statement-1 is true , Statement-2 is true , Statvement -2 is correct explanation of Statement-1 .B. (b) Statement-1 is true , statement -2 is true , statement -2 is not coerrecrt explanation of Statement-1.C. (C ) Statement-1 is true , Statement-2 is false.D. (d) Statement-1 is false , Statement-2 is true. |
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Answer» Correct Answer - C The position -time graph of a moving body in one dimension can have negative slope if its velocity is negative. Here statement -2 is worng. |
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| 530. |
If ` vec A , vec B and vec C` are non-zero vectors and ` ec A. vec B=0 and vec B. vec C=0`, then find out the value ` vec A .vec B` |
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Answer» As ` vec A. vec B=0` , :. ` AB cos theta =0 or theta =0 or theta_(1) =90^(@)`, Similarly , ` vec B. vec C=0 , :. BC cos theta _(1) =0 or cos theta =0 or theta_(1) =0 or theta _(1) =90^(@), i.e ` vec A and vec C` are parallel to each other. :. vec A. vec C = AC cos 0^(@) = AC. |
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| 531. |
Three vectors ` vec A, vec B` and ` vec C` add up to zero . Find which is false.A. (a) `(vec Axx vec B) xx vec C` is not zero unless ` vec B, vec c` are prallelB. (b) ` (vec Axx vec B) xx vec C` is not zero unless ` vec B, vec c` are prallelC. (c ) If `vec A.vec B, vec C` define a plane ,` ( vec A xx vec B xx vec C)` is in that planeD. (d) `(vec A xx B) . vec C = | vec A||vec B|| vec C| rarr C^2 = A^2 + B^2` |
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Answer» Correct Answer - B::D Give , `vec A+ vec B+ vec C =0`, then ` vec A, vec B` and vec C` are in one plane and are represented by the three sides of a trianle taken in one order. (i) :. ` vec B xx ( vec A+ vec B+ vec C) = vec B xx 0=0 ` or ` vec Bxx vec A + vec B+ vec Bxx vec C=0` or ` vec B xx vec A+ 0+ vec Bxx vec C=0` or ` vec A xx vec B= vec B xx vec C` :. ( vec Axx vec B) xx vec C =( vec Bxx vec C)xx vec C, It cannot be zero. If ` vec B || vec C`, then ` vec Bxx vec C =0`, then (vec Bxx vec C) xx vec C=0` ltBrgt Thus , option (a) is correct. (ii) ( vec Axx vec B) . ( vec Bxx vec C) . ( vec C=0` whatever be the positions of ` vec A, vec B` and vec C` ` If ` vec B|| vec C`, the ` vec ` vec Bxx vec C =0`, the ( vec Bxx vec C) xx vec C=0` Thus, option (b) is false ( iii) ` (vec Axx vec B) = vec D= AB sin theta hat D`. The direction of ` hat D` is perpendicular to the plane cntaning ` vec A` and ` vec B`. ( vec A xx vec B) xx cvec C = vec D xx vec C`. Its direction is in the plane of ` vec A, vec B` and vec C`. Thus, option (c ) is correct. ( iv) If ` C^@ = A^2 + B^2`, the angle between ` vec A ` and vec B` is ` 90^@`. (vec A xx vec B).vec C= ( AB sin 90^@ hat D) . vec C= AB ( hat D. vec C) = ABC cos 8=90^@=0`. Thus, option (d) is fase . ( vec A xx vec B) xx cvec C = vec D xx vec C`. Its direction is in the plane of ` vec A, vec B` and vec C`. Thus, option (c ) is correct. ( iv) If ` C^@ = A^2 + B^2`, the angle between ` vec A ` and vec B` is ` 90^@`. |
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| 532. |
Write an example of zero vector. |
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Answer» The velocity vectors of a stationary object is a zero vectors |
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| 533. |
A projectile is fired with Kinetic energy 1 KJ. If the range is maximum, what is its Kinetic energy, at the highest point? |
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Answer» Here \(\frac{1}{2}\) mv =1kJ=1000 J, θ = 45° At the highest point, K.E. = \(\frac{1}{2}\) m (v cos0)2 = \(\frac{1}{2}\) \(\frac{mv^2}{2}\) = \(\frac{1000}{2}\) = 500 J. |
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| 534. |
If the instantaneous velocity of a particle is zero, will its instantaneous acceleration be necessarily zero ? |
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Answer» No. (highest point of vertical upward motion under gravity) |
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| 535. |
What is time of flight? |
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Answer» The time taken for the projectile to complete its trajectory or time taken by the projectile to hit the ground is called time of flight. |
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| 536. |
When does (i) height attained by a projectile maximum ? (ii) horizontal range is maximum ? |
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Answer» (i) Height is maximum at θ = 90 (ii) Range is maximum at θ = 45. |
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| 537. |
Explain projectile motion. |
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Answer» A projectile moves under the combined effect of two velocities.
There are two types of projectile motion:
There are two types of projectile motion:
To study the motion of a projectile, let us assume that,
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| 538. |
Write an example of zero vector. |
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Answer» The velocity vectors of a stationary object is a zero vectors. |
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| 539. |
A body projected horizontally moves with the same horizontal velocity although it moves under gravity Why? |
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Answer» Because horizontal component of gravity is zero along horizontal direction. |
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| 540. |
A body projected horizontallt moves with the same horizontal velocity throughout the motion although it is under the effect of forec of gravity. Why ? |
| Answer» In horizontal projection of a projectile, the force of gravity acts in a vertically downward direction which cannot affect the horizontal component velocity of the body. Due to it, the body move with a constant horizontal velocity in horizontally projected body. | |
| 541. |
What are the examples of projectile motion? |
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Answer» 1. An object dropped from window of a moving train. 2. A bullet fired from a rifle. 3. A ball thrown in any direction. 4. A javelin or shot put thrown by an athlete. 5. A jet of water issuing from a hole near the bottom of a water tank. |
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| 542. |
A boy throw a ball ertically upwards and catches it after 2 seconds. Which of the following is true regrading the motion of the ball? (a) The displacement is zero (b) Magnitude of acceleration is constantA. Only (a) is trueB. Only (b) is trueC. Both (a) and (b) are trueD. Both (a) and (b) are false |
| Answer» The initial and final positions of the body are same. Therefore is zero and acceleration of the body is equal to the acceleraton due to gravity and it is a constant | |
| 543. |
If a sporsts car at rest accelerates uniformaly to a speed of `144 km h^(-1)` in 5 s, it vovers a distance of ___________mA. 100B. 140C. 60D. 80 |
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Answer» `u=0` `v=144km h^(-1)=144xx(5)/(18)m s^(-1)=40 m s^(-1)` `t=5s` `a=(v-u)/(t)=(40)/(5)=8 m s^(-2)` `s=(1)/(2)xx8xx(5)^(2)=100m` |
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| 544. |
What is the angular velocity of the hour hand of a clock? |
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Answer» ω = \(\frac{2π}{12}\) = \(\frac{π}{6}\)rad h-1 |
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| 545. |
What is the angle between velocity and acceleration at the peak point of the projectile motion? |
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Answer» The angle between velocity and acceleration at the peak point of the projectile motion is 90°. |
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| 546. |
What is the angular velocity of the hour hand of a clock? |
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Answer» W = 2π/12 = π/6 rad h-1, |
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| 547. |
What is the average value of acceleration vector in uniform circular motion . |
| Answer» Null vector. | |
| 548. |
What is the source of centripetal acceleration for earth to go round the sun? |
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Answer» Gravitation force of the sun. |
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| 549. |
What is the source of centripetal acceleration for earth to go round the sun? |
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Answer» Gravitation force of the sun. |
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| 550. |
What is the source of centripetal acceleration for earth to go round the sun? |
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Answer» Gravitation force of sun. |
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