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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.
| 101. |
A bird is sitting in a large closed cage which is placed on a spring balance. It records a weight of 25 N . The bird (mass m = 0.5 kg ) flies upward in the cage with an acceleration of `2m//s^(2)` . The spring balance will now record a weight ofA. 24 NB. 25 NC. 26 ND. 27 N |
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Answer» Correct Answer - B |
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| 102. |
A rifle of `20kg` mass can fire 4 "bullets"//"s". The mass of each bullet is `35 xx 10^(-3) kg` and its final velocity is `400ms^(-1)`., Then what force must be applied on the rifle so that it does not move backwards while firing the bullets ? . |
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Answer» Law of conservation of momentum `MV +4mv =0` `rArrV = - (4mv)/(M) = -(4 xx35xx10^(-3)xx400)/(20) =-2.8ms^(-1)` Force applied on the rifle `F =(MV)/(t) = - (20xx2.8)/(1) = - 56 N` . |
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| 103. |
A machine gun is mounted on a 2000kg car on a harizontal frictionless surface. At some instant the gun fires bullets of mass 10gm with a velocity of `500m/sec` with respect to the car. The number of bullets fired per second is ten. The average thrust on the system isA. 550 NB. 50 NC. 250 ND. 250 dyne |
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Answer» Correct Answer - B |
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| 104. |
A gun of mass kg 10 fires 4 bullets per second. The mass of each bullet is 20 g and the velocity of the bullet when it leaves the gun is `300ms^(-1)`. The force required to hold the gun while firing isA. 6 NB. 8 NC. 24 ND. 240 N |
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Answer» Correct Answer - C |
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| 105. |
N bullet each of mass mkg are fired with a velocity `vms(-2)` at the rate of n bullets per second upon a wall. The reaction offered by the wall to the bullets is given byA. nmvB. `(Nmv)/(n)`C. `n(Nm)/(v)`D. `n(Nv)/(m)` |
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Answer» Correct Answer - A Total mass of bullets=Nm, time `t(N)/(n)` . Momentum of the bullets striking the wall=Nmv Rate of change of momentum (Force) `=(Nmv)/(t)=nmv` . |
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| 106. |
In the figure, the pulley P moves to the right with a constant speed `v_(0)`. If the velocity of B is v in downward direction, find the velocity of A. |
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Answer» Let B moves downward by y, the pulley moves to right in time t, a distance `v_(0)` t, the block will move to right by x. `x=v_(0)t+y` `(dx)/(dt)=v_(0)+(dy)/(dt)` |
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| 107. |
An elevator weighing 6000 kg is pulled upward by a cable with an acceleration of `5ms^(-2)` . Taking g to be `10 ms^(-2)` , then the tension in the cable isA. 6000 NB. 9000 NC. 60000 ND. 90000 N |
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Answer» Correct Answer - D |
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| 108. |
A man of mass 90kg is standing in an elevator whose cable broke suddenly. If the elevator falls, apparent weight of the man is:A. 90NB. 90g NC. 0ND. any negative value |
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Answer» Correct Answer - C Pseudo force balances weight |
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| 109. |
A pendulum bob of mass 50 gm is suspended from the ceiling of an elevator. The tension in the string if the elevator goes up with uniform velocity is approximatelyA. `0.30`NB. `0.40`NC. `0.42` ND. `0.50` N |
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Answer» Correct Answer - D |
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| 110. |
If a force of 250 N act on body, the momentum acquired is 125 `kg -m//s` . What is the period for which force acts on the bodyA. `0.5` secB. `0.2` secC. `0.4` secD. `0.25` sec |
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Answer» Correct Answer - A |
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| 111. |
The force-time`( F – t )` curve of a particle executing linear motion is as shown in the figure. The momentum acquired by the particle in time interval from zero to 8 second will be A. `-2` N-sB. `+4` N-sC. 6 N-sD. Zero |
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Answer» Correct Answer - D |
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| 112. |
If the tension in the cable of 1000 kg elevator is 1000 kg weight, the elevatorA. Is accelerating upwardsB. Is accelerating downwardsC. May be at rest or acceleratingD. May be at rest or in uniform motion |
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Answer» Correct Answer - D |
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| 113. |
A body of mass 3 kg is acted on by a force which varies as shown in the graph below. The momentum acquired is given by A. ZeroB. 5 N-sC. 30 N-sD. 50 N-s |
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Answer» Correct Answer - D |
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| 114. |
The angle of inclination of an inclined plane is `60^(@)`. Coefficient of friction between `10kg` body on it and its surface is `0.2, g= 10ms^(-2)`. The acceleration of the body down the plane in `ms^(-2)` is . In the above problem the frictional force on the body is .A. `56.6 N`B. `66.6N`C. `76.6N`D. `86.6N` |
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Answer» Correct Answer - C `F =ma, a = g (sin theta -mu_(k) cos theta)` |
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| 115. |
The angle of inclination of an inclined plane is `60^(@)`. Coefficient of friction between `10kg` body on it and its surface is `0.2, g= 10ms^(-2)`. The acceleration of the body down the plane in `ms^(-2)` is . In the above problem the minimum force on the body is .A. ZeroB. `5N`C. `7.5N`D. `10N` |
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Answer» Correct Answer - D `f_(k) = mu_(k) N,f = mu_(k) mg cos theta` |
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| 116. |
The angle of inclination of an inclined plane is `60^(@)`. Coefficient of friction between `10kg` body on it and its surface is `0.2, g= 10ms^(-2)`. The acceleration of the body down the plane in `ms^(-2)` is .A. `5.667`B. `6.66`C. `7.66`D. Zero |
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Answer» Correct Answer - C `F =ma, a = g (sin theta -mu_(k) cos theta)` |
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| 117. |
Two blocks of masses 3 kg and 6 kg are connected by a string as shown in the figure over a frictionless pulley. The acceleration of the system is A. `4m//s^(2)`B. `2m//s^(2)`C. zeroD. `6m//s^(2)` |
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Answer» Correct Answer - C `a=(3g-6gsin30^(@))/(3+6)=0` |
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| 118. |
In the system shown in figure, all surfaces are smooth, pulley and strings are massless. Mass of both A and B are equal. The system is released from rest. (a) Find the `veca_(A). veca_(B)` immediately after the system is released. `veca_(A) "and" veca_(B)` are accelerations of block A and B respectively. (b) Find `veca_(A)` immediately after the system is released. |
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Answer» Correct Answer - `veca_(A). Veca_(B) = 0` immediately after release (b) `veca_(A) = (g)/(2) (leftarrow)` |
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| 119. |
In the system shown in the figure all surfaces are smooth, pulley and string are massless. The string between the two pulleys and between pulley and block of mass 5 m is parallel to the incline surface of the block of mass 4 m. The system is released from rest. Find the acceleration of the block of mass 4 m . `["tan" 37^(@) = (3)/(4)]` |
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Answer» Correct Answer - `(44g)/(205)` |
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| 120. |
The force required to stretch a spring varies with the distance a shown in the figure. If the experiment is performed with the above spring of half length, the line OA willA. Rotate clockwiseB. Rotate anticlockwiseC. Remain as it isD. Become double in length |
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Answer» Correct Answer - B |
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| 121. |
The elevator shown in fig. is descending with an acceleration of `2ms^(-2)`. The mass of the block `A=0.5kg`. Find the force (in Newton) exerted by block A on block B. A. 2NB. 4NC. 6ND. 8N |
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Answer» Correct Answer - B `N=0.5xx10-0.6xx2=4N` |
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| 122. |
In an elevator moving vertically up with an acceleration g, the force exerted on the floor by a passanger of mass M isA. MgB. `(1)/(2)`MgC. ZeroD. 2 Mg |
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Answer» Correct Answer - D |
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| 123. |
The spring balance A reads 2 kg with a block m suspended from it. A balance B reads 5 kg when a beaker filled with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid as shown in figure. In this situation A. The balance A will read more than 2 kgB. The balance B will read more than 5 kgC. The balance A will read less than 2 kg and B will read more than 5 kgD. The balances A and B will read 2 kg and 5 kg respectively |
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Answer» Correct Answer - B::C |
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| 124. |
In previous question, if we pull the block by the force `F` making an angle theta and the block remains stationary, the value of friction force is. A. `mumg`B. `Fcostheta`C. `(mumg)/(sintheta+mucostheta)`D. `(mumg)/sqrt(1+mu^(2))` |
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Answer» Correct Answer - B `f_(s)gtFcostheta` (Since block is stationary) `f=Fcostheta` . |
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| 125. |
In doubling the mass and acceleration of the mass, the force acting on the mass with respect to the previous valueA. Decreases to halfB. Decreases to halfC. Increases two timesD. Increases four times |
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Answer» Correct Answer - D |
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| 126. |
In previous question, if `mu=0.3` the acceleration of the block will be:A. zeroB. `(g)/(10)uarr`C. `(g)/(4)darr`D. `(g)/(5)darr` |
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Answer» Correct Answer - D If `mu=0.3` ltbr. Maximum value of resisting force will be `F_("lim")=muN=0.3mg` But have driving force is `mg//2` . Therefoer the friction will be kinetic nature, the block will slid down. Hence acceleration of the block `a=((mg)/(2)-0.3mg)/(m)=0.2g=(g)/(5)darr` . |
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| 127. |
A small disc A is placed on an inclined plane forming an angle `alpha` with the horizontal and is imparted an initial velocity `v_(0)`. Find how the velocity of the disc depends on the angle `theta`, shown in figure-2.209, if the friction coefficient `mu = tan alpha` and at the initial moment `theta=pi//2` |
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Answer» Correct Answer - `[v=(v_(0))/(1+cosphi)]` |
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| 128. |
A small disc P is placed on an inclined plane forming an angle `theta` with the horizontal and imparted an initial velocity `v_(0)`. Find how the velocity of disc depends on the angle `phi` which its velocity vector makes with the x axis (see figure). The coefficient of friction is `mu= "tan" theta` and initially `phi_(0) = (pi)/(2)` |
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Answer» Correct Answer - `(v_(0))/(1 + "cos"phi)` |
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| 129. |
On a stationary sail-boat, air is blown at the sails from a fan attached to the boat. The boat willA. Remain stationaryB. spin aroundC. Move in a direction opposite to that in which air is blownD. Move in the direction in which the air is blown |
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Answer» Correct Answer - A The force exerted by the air of fan on the bot is internal and for motion external force is required. |
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| 130. |
In order to lift a heavy block A, an engineer has designed a wedge system as shown. Wedge C is fixed. A horizontal force F is applied to B to lift block A. Wedge B itself has negligible mass and mass of A is M. The coefficient of friction at all surfaces is `mu`. Find the value of applied force F at which the block A just begins to rise. |
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Answer» Correct Answer - `F = (Mg)/(1 - mu^(2))[mu + (mu"cos"theta + "sin"theta)/("cos"theta - mu"sin"theta)]` |
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| 131. |
Three blocks A , B and C weighing 1, 8 and 27 kg respectively are connected as shown in the figure with an inextensible string and are moving on a smooth surface. `T_(3)` is equal to 36 N . Then `T_(2)` is A. 18 NB. 9 NC. `3.375` ND. `1.20` N |
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Answer» Correct Answer - B |
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| 132. |
A 140g ball, in horizontal flight with a speed `v_(1)` of 39.0 m/s. is sturck by a bat. After leaving the bat, the ball travel in the opposite direction with speed `v_(2)=39.0 m//s`. If the impact time `Deltat` for the ball-bat collision is 1.20ms, what average net force acts on the ballA. 1308NB. 1090NC. 9100ND. 980N |
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Answer» Correct Answer - C Here mass of ball, m=140g=0.14kg initial velocity `v_(1)=39m//s`, final velocity `v_(2)=-39m//s`, impact time `Deltat=1.20ms` impulse, I=change in momentum. `Deltap=mv_(2)-mv_(1)=m(v_(2)-v_(1))` `=0.14(-39-39)` `=-10.92 kg m//s` `therefore F_("mv")=(I)/(Deltat)=(10.92kg m//s)/(1.20xx10^(-3)s)=91xx10^(2)N` |
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| 133. |
A car is driven round a curved path of radius `18m` without the danger of skidding The coefficeient of friction between the tyres of the car and the surface of the curved path is 0.2 What is the maximum speed in kmph of the car for safe driving ? `(g =10 ms^(-2))` . |
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Answer» Maximum speed. `v =sqrt(mu_(s)gr)` `v =sqrt(0.2 xx 10 xx 18) =sqrt36 = 6ms^(-1)` . |
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| 134. |
Indicate the direction of frictional force on a car which is moving along the curved path with non zero tangential acceleration, in anti-clock direction:A. B. C. D. |
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Answer» Correct Answer - C As car is moving in anticlockwise direction and have tangential acceleration as well as radial acceleration. :. Friction componet should be along tangential and radial acceleration. |
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| 135. |
Assertion:Mass is a measure of inertia of the body in linear motion. Reason: Greater the mass, greater is the force required to change its state of rest or of uniform motion.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 assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - A |
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| 136. |
Assertion : No force is required by the body to remain in any state. Reason : In uniform linear motion, acceleration has a finite value.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 assertionC. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - C |
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| 137. |
A man of mass `m` stands on a platform of equal mass `m` and pulls himself by two ropes passing over pulleys as shown in figure.If he pulls each rope with a force equal to half his weight ,his upwards acceleration would be A. `(g)/(2)`B. `(g)/(4)`C. `g`D. zero |
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Answer» Correct Answer - D `F =4((mg)/(2))rarr upwards` `W =2mgrarrdownwards` `F = W, :. a =0` . |
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| 138. |
A man with mass 85 kg stands on a platform with mass 25 kg. He pulls on the free end of a rope that runs over a pulley on the ceiling and has its other end fastened to the platform. The mass of the rope and the mass of the pulley can be neglected, and the pulley is frictionless. With what force does he have to pull so that he and the platform have an upward acceleration of `2.2m/s^(2)`. (Take `g= 10m/s^(2)`) |
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Answer» Correct Answer - [15 N, 270 N] |
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| 139. |
A given object takes 3 times as much time to slide downa `45°` rough incline as it takes to slide down a perfectly smooth `45°` incline. The coefficient of friction between the object and the incline is :A. `1//8`B. `8//9`C. `1//2sqrt(2)`D. `2sqrt(2)//3` |
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Answer» Correct Answer - B |
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| 140. |
A smooth block is released at rest on a `45^(@)` incline and then slides a distance `d`. The time taken to slide is n times as much to slide on rough incline than on a smooth incline. The coefficient of friction isA. `mu_(k)=1-(1)/(n^(2)`B. `mu_(k) =sqrt((1-(1)/(n^(2)))`C. `mu_(k) = (1)/(1 -n^(2))`D. `mu_(k)=sqrt((1)/(1-n^(2))` |
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Answer» Correct Answer - A `mu_(k) = tan theta(1-(1)/(n^(2)))` |
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| 141. |
In the given figure, `a=15m//s^(2)` represents the total acceleration of a particle moving in the clockwise direction in a circle of radius `R=2.5m` at a given instant of time. The speed of the particle is. A. `5.7m//s`B. `6.2m//s`C. `4.5m//s`D. `5.0m//s` |
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Answer» Correct Answer - A `acos30^(@)=a_(c)=(v^(2))/(R)` implies `v^(2)=aRxx(sqrt(3))/(2)` implies `v^(2)=32.47` implies `v~=5.7m//s` |
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| 142. |
A motorcycle is travelling on a curved track of radius 500 m. If the coefficient of friction between road and tyres is 0.5, the speed avoiding skidding will beA. 50 m/sB. 75m/sC. 25m/sD. 35m/s |
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Answer» Correct Answer - A `mumg=(mv^(2))/(r)` `v=sqrt(mugr)=sqrt(0.5xx10xx500)=50 m//s` |
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| 143. |
A body of mass 2kg is placed on a horizontal surface having kinetic friction `0.4` and static friction `0.5` . If the force applied on the body is `2.5N` , then the frictional force acting on the body will beA. 8NB. 10NC. 20ND. `2.5N` |
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Answer» Correct Answer - D `f_(k)=0.4xx2xx10N=8N` `f_(ms)=0.5xx2xx10N=10N` Since the appiled force is less than force of friction, therefore the force of friction is equal to the applied force i.e., `2.5N` . |
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| 144. |
A car is travelling along a curved road of radius r. If the coefficient of friction between the tyres and the road is `mu` the car will skid if its speed exceeds .A. `2sqrt(murg)`B. `sqrt(3murg)`C. `sqrt(2murg)`D. `sqrt(murg)` |
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Answer» Correct Answer - D `(mv^(2))/(r) = mumg,v_(max) =sqrt(mu_(s)rg` |
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| 145. |
A car starts from rest to cover a distance s. the coefficient of friction between the road and the tyres is `mu`. The minimum time in which the car can cover the distance is proportional toA. `mu`B. `sqrt(mu)`C. `1//mu`D. `1//sqrt(mu)` |
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Answer» Correct Answer - D |
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| 146. |
Two block (A) 2kg and (B) 5kg rest one over the other on a smooth horizontal plane. The cofficient of static and dynamic friction between (A) and (B) is the sae and equal to 0.60. The maximum horizontal force that can be applied to (B) in order that both (A) and (B) do not have any relative motion: A. 42NB. 42kgfC. 5.4kgfD. 1.2N |
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Answer» Correct Answer - A `a=(F)/(7)` for block A `2xx(F)/(7)=0.6xx2xx10` F=42N |
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| 147. |
A car is moving along a straight horizontal road with a speed `v_(0)` . If the coefficient of friction between the tyres and the road is `mu` , the shortest distance in which the car can be stopped isA. `(V_(0)^(2))/(2mug)`B. `(V_(0)^(2))/(mug)`C. `((V_(0))/(mug))^(2)`D. `(2V_(0)^(2))/(mug)` |
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Answer» Correct Answer - A `O^(2)-V_(0)^(2)=2(-mug)xxs` `s=(V_(0)^(2))/(2mug)` |
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| 148. |
A car is moving along a stight horizontal road with a speed `v_(0)` . If the coefficient of friction between the tyres and the road is mu, the shortest distance in which the car can be stopped is.A. `(v_(0)^(2))/(2mug)`B. `(v_(0))/(mug)`C. `((v_(0))/(mug))^(2)`D. `(v_(0))/(mu)` |
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Answer» Correct Answer - A Retarding force `F=ma=muR=mumg` :. `a=mug` Now from equation of motion `v^(2)=u^(2)-2as` implies `0=u^(2)-2as` implies `s=(u^(2))/(2a)=(u^(2))/(2mug)` implies `s=(v_(0)^(2))/(2mug)` |
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| 149. |
The lengths of smooth & rough inclined planes of inclination `45^(@)` is same Times of sliding of a body on two surface `t_(1)t_(2)` and `mu =0.75` then `t_(1):t_(2)=` .A. `2:1`B. `2:3`C. `1:2`D. `3:2` |
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Answer» Correct Answer - C `t_("smooth")=sqrt((2l)/(g sin theta)),t_(rough) = sqrt((2l)/(g (sin theta-mu_kcostheta)))` |
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| 150. |
A vehicle of mass `M` is moving on a rough horizontal road with a momentum P If the coefficient of friction between the tyres and the road is `mu` is then the stopping distance is .A. `(P)/(2muMg)`B. `(P^(2))/(2muMg)`C. `(P^(2))/(2muM^(2)g)`D. `(P^(2))/(2muM^(2)g)` |
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Answer» Correct Answer - C `P =mv, v^(2) -u^(2) =2as, a =mu_(k)g` . |
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