InterviewSolution
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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. |
A brick measures 12 cm `xx` 6 cm `xx` 3 cm. Initially, it is kept on the ground such that the rectangular surface with the smallest area is in contact with the ground and then it is placed such that the rectangular surface with the largest area touches the ground. Calculate the height of the centre of gravity from the ground in the two cases. Out of the two positions, in which case does the brick have greater stability and why? |
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Answer» (i) In first case Height of the brick = 12 cm Height fo CG from the ground = 6 cm (i) In second case Height of the brick = 3 cm Height of CG from the ground = 1.5 cm In the second case the brick has more stability because height of CG is small and also base area is more. |
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| 2. |
A ball is thrown in upwards direction then the force(s) acting on it is (are)A. gravitational forceB. mechanical forceC. fictional forceD. Both (a) and (c ). |
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Answer» Correct Answer - D When ball is thrown in upward direction, its weight always acts in downward direction and as the body is moving upwards, frictional force (due to air) acts in downward direction. |
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| 3. |
Two spheres of mass 10 g and 100 g each falls on the two pans of a table balance from a height of 40 cm and 10 cm, respectively. If both are brought to rest in 0.1 second, determine the force exerted by each sphere on the pans. |
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Answer» Find the velocity acquired by each sphere using `v^(2) = u^(2) + 2 as` Find the rate of change of momentum of a sphere `= F = (m(v-u))/(t)` The force exerted by a sphere = weight (w) + F. |
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| 4. |
We know, `L xx dL = E xx dE`, for a given type of lever, the effort arm is always less than loadarm, thenA. The V.R is less than oneB. The M.A is less than oneC. The given simple machine is a third order leverD. All the above. |
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Answer» Correct Answer - D `LxxdL = E xx dE implies (L)/(E) = (dE)/(dL)` Given `dE lt dL implies V.R. lt 1` and also `MA lt 1` This type of lever is a third class lever. |
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| 5. |
A bottle standing on its base is more stable than when it stands on its neck. This is so because when it stands on the base its ______.A. base area is moreB. centre of gravity is nearer to the baseC. centre of gravity is at the topD. Both (a) and (b). |
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Answer» Correct Answer - D A bottle standing on the base is more stable because its base area is more and the position of the centre of gravity is as low. |
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| 6. |
A force 50 N acts on a block A of mass 3 kg which is in contact with a block B of mass 2 kg, as shown in the figure. Find the forces acting on A and B. Will the same forces act on A and B if instead of applying the force to A, it is applied to B? |
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Answer» Both the blocks move with the same acceleration given by `a = ("Force applied on A")/("Mass of A + Mass of B")` The force on B is `F_(B) = m_(B) xx a` Then, B exerts an equal but opposite force on A. = 50 - `F_(B)` |
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| 7. |
A bullet of mass 50 g is fired from a gun. If the bullet acquires a velocity of 100 m `s^(1)` in 0.1 second, what is the recoil force on the gun? |
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Answer» Mass of bullet, m = 50 g = `50 xx 10^(-3)` kg Initial velocity of the bullet, u = 0 m `s^(-1)` Final velocity of the bullet, v = 100 m `s^(-1)` Time, t = 0.1 s Force exerted on the bullet, `F = ma = m[(v-u)/(t)]` `=50 xx 10^(-3) ((100-0)/(0.1)) = 50N` This is the action on the bullet Thus, reaction on the gun = 50 N `therefore` The recoil force on the gun = 50 N |
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| 8. |
2 joules of work is done in displacing a stone through 0.5 m. Find the force applied on the stone. |
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Answer» Given work, W = 2 J Displacement, s = 0.5 m W = F `xx` s 2 J = F `xx` 0.5 m `therefore F = (2)/(0.5) = 4 N` |
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| 9. |
What is the work done by a horse in displacing a cart through 5 m in the direction of the force if the force applied by the horse is 10 N? |
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Answer» Force = F = 10 N Displacement = 5 m Since the force and displacement are acting in the same direction, work done is given by `therefore` Work = `F xx s` ` = 10 xx 5` implies Work done by the horse = 50 J |
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| 10. |
Bottle lid opener is an example of ______ lever |
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Answer» Correct Answer - True |
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| 11. |
A physics student took a solid sphere of density 2 g `cm^(-3)` and radius 3.5 cm and applied a force on the sphere. If an acceleration fo 1.5 `cm s^(-2)` is produced in it, find the amount of force applied on it. |
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Answer» Given, acceleration, a = 1.5 cm `s(-2)` Radius of sphere, r = 3.5 cm Density of sphere = 2 g `cm^(-3)` Force, F = (mass) (acceleration) But mass, M = (volume) (density), Volume `V = (4)/(3) pi r^(3) = (4)/(3) xx (22)/(7) xx (3.5)^(3) cm^(3)` `thereforeM = (4)/(3)xx (22)/(7)xx ((7)/(2))^(3) xx 2=(1078)/(3)g` `therefore " Force, "F = (1078)/(3) xx 1.5 " dyne "=(1078)/(2) = 539` dyne. |
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| 12. |
Whenever a force is applied on a body, work is done. |
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Answer» Correct Answer - False |
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| 13. |
A car of mass 1000 kg is moving with a certain speed when a constant braking force 1000 N acts on it for 5s and the speed of the car reduces to half its original speed. Find the further time required to stop the car if the same constant force acts. |
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Answer» Find acceleration of the car by using F = ma. Find the initial speed, by using v = u + at To find the further time taken to stop the vehicle, use v = u + at |
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| 14. |
A car changes its speed from 20 km `h^(-1)` to 50 km `h^(-1)`. This is possible only if ______ is applied on the car. |
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Answer» Correct Answer - External force |
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| 15. |
If a force of 50 N is applied on a body and it is still at rest, then find the magnitude of static frictional force acting on it. |
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Answer» The magnitude of the force acting on the body (f) = 50 N. Due to application of this force, the body tends to move but is not set in motion, and hence, the applied force is equal in magnitude and opposite in direction to static frictional force. `therefore` The magnitude of static frictional force acting = 50 N. |
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| 16. |
A car changes its speed from 20 km `h^(-1)` to 50 km `h^(-1)` of mass 3600 kg in 5 s. Determine the net external force applied on the car. |
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Answer» `F = m((v-u)/(t)) = 3600 ((50-20)/(5)) xx (15)/(18) = 600 xx 30` 18000 N. |
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| 17. |
The work done to move an object of mass 2 kg to a height of 50 m from the ground is ______ J. `(g = 10 m s^(-2))`A. 100B. 1000C. 500D. 5000 |
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Answer» Correct Answer - B Work done = mgh `= 2xx10xx50 = 1000 J` |
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| 18. |
An object of mass 500 g moving with a speed of 10 m `s^(-1)` collides with another object of mass 250 g moving with a speed of 2 m `s^(-1)` in the opposite direction. After collision both the objects move in the same direction with common velocity. Write the proper order of steps to find out their common velocity after collision. (a) Calculate the total momentum of two bodies before collision as `m_(1) u_(1) + m_(2) u_(2)`. (b) Write down the given values of `m_(1)` and `m_(2)` in SI system (c ) Write the expression for their total momentum after collision as `(m_(1) + m_(2))` v where v is their common velocity. (d) Assign proper signs to the initial velocities `u_(1)` and `u_(2)`. (e) Using law of conservation of momentum Equate the above two expressions and determine the value of v as `(m_(1) u_(1) + m_(2) u_(2))/(m_(1) + m_(2))`A. b d a c eB. b d c e aC. a d b c eD. c d e b a |
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Answer» Correct Answer - A Given `m_(1) = 500 g = (1)/(2) kg, m_(2) = 250 g = (1)/(4)kg (b)` `u_(1) = 10 m s^(-1), u_(2) = -2 m s^(-1) (d)` Momentum before collision, `m_(1) u_(1) + m_(2) u_(2)` `=(1)/(2) xx 10 + (1)/(4) xx (-2), 5 - (1)/(2) = (9)/(2) Kg m s^(-1)` Momentum after collision, `(m_(1) + m_(2)) v = (3)/(4) v (c )` From law of conservation of momentum, `(9)/(2) = (3)/(4)v implies v = 6m s^(-1) (e)` |
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| 19. |
A crowbar is of length 3 m. Where should the fulcrum be placed along the length of the crowbar so that a boulder of 9 `kg_(wt)` be lifted with it by applying an effort of 10 N? `(g = 10 m s^(-2))` |
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Answer» Find the length of the crowbar. Find the value of the effort from given data Find the value of the load from the given data. (Convert it into S.I. units). Now, load `xx` load arm = effort X effort arm (1) Substitute the values of load and effort. Find the ratio of the load arm to effort arm from equation (1) Here, load arm + effort arm = 3 m. Now, find the position of the fulcrum from the load arm. Is load arm = `(1)/(10)` th part of 3 m? |
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| 20. |
Why does a professional cleaner or a sweeper prefers a broom stick with large handle than a small broom stick used for household purposes? |
| Answer» M.A. = `("Effort arm")/("Load arm")` | |
| 21. |
Why does an athlete run a certain distance before taking a long jump? |
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Answer» Does the velocity of an athlete increase by running a certain distance before taking a long jump? Then, will the momentum of the athlete increase? Will athlete develop inertia of motion? Then, the kinetic energy developed by athlete is used to execute the long jump. |
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| 22. |
An inclined plane of length 6 cm is used to lift a load of 42 N, by applying an effort of 7 N. Find the height of the plane. |
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Answer» Given, length of the plane, l = 6 m Load, L = 42 N effort, E = 7 N Let height of the plane = h We know, `(L)/(E) = (l)/(h)` `implies h = l xx (E)/(L) = 6 xx (7)/(42) = 1m ` |
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