InterviewSolution
Saved Bookmarks
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. |
Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.A. `theta=45^(@)`B. `theta=tan^(-1)(1/4)`C. `theta=tan^(-1)(4)`D. `theta=tan^(-1)(2)` |
|
Answer» Correct Answer - c Given `H=R, (u^(2) sin^(2) theta)/(2g)=(2u^(2) sin theta cos theta)/g` `implies tan theta=4 implies theta=tan^(-1)(4)` |
|
| 2. |
The horizontal range of a projectile is `4 sqrt(3)` times its maximum height. Its angle of projection will beA. `45^(@)`B. `60^(@)`C. `90^(@)`D. `30^(@)` |
| Answer» Correct Answer - D | |
| 3. |
Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.A. `45^(@)`B. `theta = tan^(-1) (0.25)`C. `theta = tan^(-1) 4` or `(theta = 76^(@))`D. `60^(@)` |
| Answer» Correct Answer - C | |
| 4. |
The centripetal acceleration is given byA. `v2//r`B. v2rC. vrD. `v//r` |
| Answer» Correct Answer - A | |
| 5. |
A cycle wheel of radius 0.4 m completes one revolution in one second then the acceleration of a point on the cycle wheel will beA. `0.8 m//s`B. `0.4 m//s`C. `1.6 pi^(2) m//s^(2)`D. `0.4 pi^(2) m//s^(2)` |
| Answer» Correct Answer - C | |
| 6. |
The angular speed of a fly wheel making 120 revolutions / minute isA. `2pi rad//s`B. `4pi^(2)pi//rad`C. `pi rad//s`D. `4pirad//s` |
| Answer» Correct Answer - D | |
| 7. |
A wheel completes 2000 revolutions to cover the 9.5 km. distance. then the diameter of the wheel isA. `1.5m`B. `1.5 cm`C. `7.5 cm`D. `7.5 m` |
| Answer» Correct Answer - A | |
| 8. |
A car is moving in a circular horizonta track of radius 10m with a constant speed of 10 m/s. A pendulum bob is suspended from the roof of the cat by a light rigid rod of length 1.00m. The angle made by the rod with track isA. ZeroB. `30^(@)`C. `45^(@)`D. `60^(@)` |
| Answer» Correct Answer - C | |
| 9. |
Assertion : Improper banking of roads causes wear and tear of tyres. Reason : The necessary centripetal force is provided by the force of friction between the tyres and the road.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
| Answer» Correct Answer - A | |
| 10. |
A stone tied to the a string of `80 cm` long is whirled in a horizontal circle with a constant speed. If the stone makes `25` revolutions in `14 s` then, magnitude of acceleration of the same will be:A. `650 cm//s^(2)`B. `680 cm//s^(2)`C. `750 cm//s^(2)`D. `990 cm//s^(2)` |
|
Answer» Correct Answer - d If angular velocity is `omega` and radius of circular path is r then, acceleration `alpha=omega^(2)r` `T=("Number of revolution")/("Time taken")=25/14=1.78s` and `omega=(2pi)/1.78=3.52 rad//s` Hence, `alpha=(3.52)^(2)xx80=991.23~~990cm//s^(2)` |
|
| 11. |
A stone tied to a string of length `L` is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at lowest position and has a speed `u` . Find the magnitude of the change in its velocity as it reaches a position, where the string is horizontal.A. `sqrt(u^(2)-2gL)`B. `sqrt(2gL)`C. `sqrt(u^(2)- gl)`D. `sqrt(2(u^(2)-gL))` |
| Answer» Correct Answer - D | |
| 12. |
A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal circle. The maximum tension the string can withstand is 16 Newton . The maximum velocity of revolution that can be given to the stone without breaking it, will beA. `20 ms^(-1)`B. `16 ms^(-1)`C. `14 ms^(-1)`D. `12 ms^(-1)` |
| Answer» Correct Answer - D | |
| 13. |
A shell fired from the ground is just able to cross in a horizontal direction the top of a wall 90m away and 45m high. The direction of projection of the shell will be:A. `25^(@)`B. `30^(@)`C. `60^(@)`D. `45^(@)` |
|
Answer» Correct Answer - d `h=(u^(2)sin^(2) theta)/(2g) and R=(u^(2)sin2 theta)/g` `45/180=1/4 tan theta or theta=45^(@)` |
|
| 14. |
A body of mass 0.4 kg is whirled in a vertical circle making 2 rev/sec . If the radius of the circle is 2 m , then tension in the string when the body is at the top of the circle, isA. `41.56 N`B. `89.86 N`C. `109.86 N`D. `122.2 N` |
|
Answer» Correct Answer - D T=mw²r-mg for 2 rev/sec we need to find w(omega) i.e. angular velocity |
|
| 15. |
In a vertical circle of radius r , at what point in its path a particle has tension equal to zero if it is just able to complete the vertical circleA. Highest pointB. Lowest pointC. Any pointD. At a point horizontally from the centre of circle of radius r |
| Answer» Correct Answer - A | |
| 16. |
A hollow sphere has radius 6.4 m. Minimum velocity required by a motor cyclist at bottom to complete the circle will be.A. `17.7 m//s`B. `10.2 m//s`C. `12.4 m//s`D. `16.0 m//s` |
| Answer» Correct Answer - A | |
| 17. |
The tension in the string revolving in a vertical circle with a mass m at the end which is at the lowest positionA. `(m v^(2))/(r)`B. `(m v^(2))/(r)-mg`C. `(m v^(2))/(r) + mg`D. `mg` |
| Answer» Correct Answer - C | |
| 18. |
The maximum velocity at the lowest point, so that the string just slack at the highest point in a vertical circle of radius l.A. `sqrt(gl)`B. `sqrt(3 gl)`C. `sqrt(5 gl)`D. `sqrt(7 gl)` |
| Answer» Correct Answer - C | |
| 19. |
A pendulum bob on a 2 m string is displaced `60^(@)` from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its pathA. `sqrt(2)m//s`B. `sqrt(9.8) m//s`C. `4.43 m//s`D. `1//sqrt(2) m//s` |
| Answer» Correct Answer - C | |
| 20. |
The string of pendulum of length l is displaced through `90^(@)` from the vertical and released. Then the minimum strength of the string in order to withstand the tension, as the pendulum passes through the mean position isA. `mg`B. `3mg`C. `5 mg`D. `6 mg` |
| Answer» Correct Answer - B | |
| 21. |
A particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance `h` below the highest points, such that `h` is equal toA. `h = R`B. `h = (R)/(3)`C. `h = (R)/(2)`D. `h = (2R)/(3)` |
| Answer» Correct Answer - B | |
| 22. |
A Particle is kept at rest at the top of a sphere of diameter `42m`.when disturbed slightly, it slides down. At what height `h` from the bottom, the particle will leave the sphereA. `14 m`B. `28 m`C. `35 m`D. `7m` |
| Answer» Correct Answer - C | |
| 23. |
Three ships A, B and C are in motion. The motion of A as seen by B is with speed v towards north-east . The motion. Of B as seen by C is with speed v towards the north-west. Then as seen by A, C will be moving towardsA. northB. southC. eastD. west |
|
Answer» Correct Answer - b `vecv_(A)-vecv_(B)=v((hati+hatj)/(sqrt(2))), vecv_(B)-vecv_(C)=v((-hati+hatj)/(sqrt(2)))` Adding: `implies vecv_(A)-vecv_(C)=(2v)/(sqrt(2))hatj` `implies vecv_(C)-vecv_(A)=sqrt(2) vhatj =sqrt(2) (-hatj)` So C will be moving towards south as seen by A. or `a_(p) gt 10m//s^(2)` |
|
| 24. |
STATEMENT-1: When a particle moves in a circle with a uniform speed, it velocity and acceleration both changes. STATEMENT-2: The centripetal acceleration in circular motion is depdendent on angular velocity of the body.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
| Answer» Correct Answer - B | |
| 25. |
A particle moves in a plane with constant acceleration in a direction different from the initial velocity. The path of the particle will beA. A straight lineB. An arc of a circleC. A parabolaD. An ellipse |
| Answer» Correct Answer - C | |
| 26. |
A particle moves in a circle of radius 5 cm with constant speed and time period `0.2pis`. The acceleration of the particle isA. `25m//s^(2)`B. `36m//s^(2)`C. `5m//s^(2)`D. `15m//s^(2)` |
|
Answer» Correct Answer - c Given `r=5 cm =5xx10^(-2)m and T= 0.2 pis` we know that acceleration `a=r omega^(2)=(4pi^(2))/(T^(2))r` `=(4xxpi^(2)xx5xx10^(-2))/((0.2pi)^(2))=5 ms^(-2)` |
|
| 27. |
Three identical particles are joined together by a thread as shown in figure All the partical are moving in a horizontal plane If the vertical of the outermost particle is `v_(0)` then the ratio of tension in the three sections of the string `(T_(1):T_(2):T_(3) = ?)` isA. `3:5:7`B. `3:4:5`C. `7:11:6`D. `3:5:6` |
| Answer» Correct Answer - D | |
| 28. |
A sphere is suspended by a thread of length l. What minimum horizontal velocity has to be imparted the ball for it to reach the height of the suspension?A. `gl`B. `2gl`C. `sqrt(gl)`D. `sqrt(2 gl)` |
| Answer» Correct Answer - D | |
| 29. |
A ball of mass 0.1 Kg. is whirled in a horizontal circle of radius 1 m . by means of a string at an initial speed of 10 R.P.M. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed isA. 5r.p.m.B. 10 r.p.m.C. 20 r.p.mD. 14 r.p.m. |
| Answer» Correct Answer - A | |
| 30. |
A point mass m is suspended from a light thread of length l, fixed at O, is whirled in a horizontal circle at constant speed as shown. From your point of view, stationary with respect to the mass, the forces on the mass areA. B. C. D. |
| Answer» Correct Answer - C | |
| 31. |
If a cyclist moving with a speed of `4.9 m//s` on a level road can take a sharp circular turn of radius 4 m , then coefficient of friction between the cycle tyres and road isA. 0.41B. 0.51C. 0.61D. 0.71 |
| Answer» Correct Answer - C | |
| 32. |
A body of mass `2 kg` has an initial velocity of `3 m s^-1` along `OE` and it is subjected to a force of `4 N` in `O F` direction perpendicular to `O E`. Find the distance of the body from `O` after `4 s`. .A. `12m`B. `28m`C. `20m`D. `48m` |
|
Answer» Correct Answer - c `u_(x)=3ms^(-1), as=0, u_(y)=2m//se^(2) and t=4sec`. If `s_(x) and s_(y)` be the displacement alont x-axis and y-axis respectively then `s_(x)=u_(x)t+1/2a_(x)t^(2)` `=3xx4+1/2xx0xx(4)^(2)=12m` and `s_(y)=u_(y)t+1/2a_(y)t^(2)` `=0xx4+1/2xx2xx(4)^(2)=16m` `s=sqrt(s_(x)^(2)+s_(y)^(2))=sqrt((12)^(2)+(16)^(2))` `=sqrt(144+256)=sqrt(400)=20m` |
|
| 33. |
A body is projected at an angle of `30^(@)` with the horizontal and with a speed of `30ms^(-1)`. What is the angle with the horizontal after 1.5 second? `(g=10ms^(-2))`A. `0^(@)`B. `30^(@)`C. `60^(@)`D. `90^(@)` |
|
Answer» Correct Answer - a The time of light is given by `T=(2u sin theta)/g=(2xx30xx1)/(10xx2)=3 sec` Thus, after 1.5 sec the body is at the highest point. As the direction ofmotion is horizontal after 5 sec the angle with the horizontal is `0^(@)`. |
|
| 34. |
If the body is moving in a circle of radius r with a constant speed v , its angular velocity isA. `v^(2)//r`B. vrC. v/rD. r/v |
| Answer» Correct Answer - C | |
| 35. |
A body of mass 5 kg is moving in a circle of radius m 1 with an angular velocity of 2 radian/sec . The centripetal force isA. 10 NB. 20NC. 30 ND. 40 N |
| Answer» Correct Answer - B | |
| 36. |
A ball of mass 0.25 kg attached to the end of a string of length 1.96 m is moving in a horizontal circle. The string will break if the tension is more than 25 N . What is the maximum speed with which the ball can be movedA. `14 m//s`B. `3m//s`C. `3.92 m//s`D. `5m//s` |
| Answer» Correct Answer - A | |
| 37. |
Two stones are projected with the same speed but making different angles with the horizontal. Their reanges are equal. If the angles of projection of one is `pi//3` and its maximum height is `h_(1)` then the maximum height of the other will be:A. `3h_(1)`B. `2h_(1)`C. `h_(1)//2`D. `h_(1)//3` |
|
Answer» Correct Answer - d `(h_(2))/(h_(1))=(u^(2)sin^(2) theta_(2))/(2g)xx(2g)/((u^(2)sin^(2) theta_(2))` `=(sin^(2) theta_(2))/(sin^(2) theta_(1))=(sin^(2) pi//6)/(sin^(2)pi//6)=1/3` `h_(2)=(h_(1))/3` |
|
| 38. |
Four bodies A,B,C and D are projected with equal velocities having angles of projection `15^(@),30^(@),45^(@) and 60^(@)` with the horizontal respectively. The body having the shortest range isA. AB. BC. CD. D |
|
Answer» Correct Answer - a Range, `R=(u^(2)sin 2 theta)/g` For a given initial velocity `R prop sin 2 theta` `R_(2)=(u^(2)sin (90^(@)+2 theta))/g=(u^(2)cos 2 theta)/g=R_(1)` |
|
| 39. |
At what point of a projectile motion acceleration and velocity are perpendicular to each otherA. At the point of projectionB. At the point of dropC. At the topmost pointD. Any where in between the point of projection and topmost point |
| Answer» Correct Answer - C | |
| 40. |
Two bodies are thrown up at angles of `45^(@)` and `60^(@)`, respectively, with the horizontal. If both bodies attain same vertical height, then the ratio of velocities with which these are thrown isA. `sqrt((2)/(3))`B. `(2)/(sqrt(3))`C. `sqrt((3)/(2))`D. `sqrt(3)/(2)` |
| Answer» Correct Answer - C | |
| 41. |
A string of length `L` is fixed at one end and carries a mass `M` at the other end. The string makes `2//pi` revolution per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is. A. `ML`B. `2ML`C. `4 ML`D. `16 ML` |
| Answer» Correct Answer - D | |
| 42. |
A bomb is dropped from an aeroplane moving horizontally at constant speed. When air resistance is taken into consideration, the bombA. Falls to earth exactly below the aeroplaneB. Fall to earth behind the aeroplaneC. Falls to earth ahead of the aeroplaneD. Flies with the aeroplane |
| Answer» Correct Answer - B | |
| 43. |
A bullet is dropped from the same height when another bullet is fired horizontally. They will hit the groundA. One after the otherB. SimultaneouslyC. Depends on the observerD. None of the above |
| Answer» Correct Answer - B | |
| 44. |
The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by `y = (8t-5t^(2))` meter and `x = 6t` meter , where t is in second. The velocity with which the projectile is projected isA. `8m//sec`B. `6m//sec`C. `10 m//sec`D. Not obtainable from the data |
| Answer» Correct Answer - C | |
| 45. |
A sphere of mass m is tied to end of a string of length l and rotated through the other end along a horizontal circular path with speed v . The work done in full horizontal circle is |
| Answer» Correct Answer - A | |
| 46. |
A stone tied to a string is rotated in a circle. If the string is cut, the stone flies away from the circle becauseA. A centripetal force acts on the stoneB. A centripetal force acts on the stoneC. Of its inertiaD. Reaction of the centripetal force |
| Answer» Correct Answer - C | |
| 47. |
If `a_(r)` and `a_(t)` represent radial and tangential accelerations, the motion of a particle will be uniformly circular ifA. `a_(r) = 0` and `a_(t) = 0`B. `a_(r) = 0` but `a_(t) !=0`C. `a_(r) != 0` but `a_(t) =0`D. `a_(r) != 0` but `a_(t) !=0` |
| Answer» Correct Answer - C | |
| 48. |
A string breaks if its tension exceeds 10 newtons . A stone of mass 250 gm tied to this string of length 10 cm is rotated in a horizontal circle. The maximum angular velocity of rotation can beA. `20 rad//s`B. `40 rad//s`C. `100 rad//s`D. `200 rad//s` |
| Answer» Correct Answer - A | |
| 49. |
A paticle of mass `m` is executing uniform circular motion on a path of radius `r`. If `p` is the magnitude of its linear momentum, then the radial force acting on the particle isA. pmrB. `(rm)/(p)`C. `(mp^(2))/(r )`D. `(p^(2))/(rm)` |
| Answer» Correct Answer - D | |
| 50. |
The magnitude of the centripetal force acting on a body of mass m executing uniform motion in a circle of radius r with speed v isA. mvrB. `mv^(2)//r`C. `v//r^(2)m`D. `v//rm` |
| Answer» Correct Answer - B | |