Q : 1

A cyclist is going on over bridge with constant speed. The value of friction force on the cycle is.

(a) Decreasing than increasing.

(b) Increasing than decreasing

(c) Decreasing

(d) Increasing

Answer Option : (a)

Explanation : Speed of the cyclist is constant, It means that friction force is used to balance the mgsin θ e  component of the weight of cyclist. so when cyclist is going up the over bridge , the  mgsin θ is decreasing so the friction force will also decrease. when cyclist will go down the over bridge , the mgsin θ will increase so friction will also increase so option (1) is correct.

Q : 2. A car starts accelerating with a constant acceleration a_o on a horizontal circular  road and take its speed up to max. allowed speed on the circular road to move without slipping ------>

(a) The magnitude of friction is constant

(b) The direction of friction is constant

(c) Friction is constant

(d) The value of a_o cant be greater than g

Answer Option : (d)

Explanation :


Here speed is increasing, So the net friction force required will act along the   direction. means friction force will help to accelerate (speed up) the car as well as provide it centripetal acceleration.

=> (fs)_max  =N = mg

        a_max = g

here

=> As is a component of , so can not be greater than .

Q : 3 Suppose you are in Merry go round. you place a coin  on the rod (which join the center and your seat) and you feel that coin is coming back to you It means that

(a) There is a physical force toward you which is acting on the coin

(b) The friction force is not sufficient for that coin.

(c) What you are feeling is not actually happening.

(d) The person on ground tells that coin is in circular motion.

Answer Option : (b)

Explanation : => You may understand this problem that there is a centrifugal force which act on the coin and thats why it comes back to you. correct but centrifugal force is a virtual force there is no physical existence of centrifugal force. It is a pseudo force.

=> It seems to us as we are in merry go round that the coin is coming back to us. But the main reason is the max.value of frictional force is not able to provide required centripetal force to the coin thats why it slips with respect to the rod of merry go round.

=> The person on the ground is in inertial frame according to him the coin is slipping with respect to the rod so it is not in circular motion.

Q : 4 If a bob suspended with a light string is given a velocity v at the lowest points such that the string slack just before the highest point

(a) The bob will complete the circular path.

(b) The sting will again become taut at the highest point.

(c) Just before the slacking, the tension T was greater than the mg cos θ ( the component of mg towards center at that point)

(d) At the point where string become slack, 

Answer Option : (d)
Explanation :

At a point A----->
As  we move towards B the mv² the mgcos θ is increase and  is increasing

=> If string slacks at B it means that the mg cos is able to provide required centripetal force at that point. so there is no need of Tensions at point B and string slacks.

=> Now As bob reach at C from B the is decreased and mg cos θ become mg at the highest point

=> You can understand that mg will be greater than the required centripetal force at the highest point. So bob will continue the circular motion and string will not become taut again at the highest point

=> A t D ----> Just before the point B, the most of the part of required centripetal force is provide by mg cos θ As you know that at very next moment tension become zero.

So A t D  -----> mg cos θ>T

Q : 5

           

(a) When the block leave the surface

(b) When the block leave the surface

(c) At the point just before leaving the surface N = 0.

(d) When the block leave the surface

Answer Option : (c,d)
Explanation : From the equation it is clear that, as block comes down mg sin θ is decreasing and is increasing.

As shown in the figure at point A ---->

= Force  by which block will press the spherical surface =w ------------------------(1)

From newton III^rd law ---->

----------------------(2)

So one will point come when mg cos θ is totally used to provide required centripetal force and at this point block will not push the spherical surface.

=> So at this point w = 0 ,  so N = 0 --------------------(3)

=> But block will not leave the surface at this point and note that block is still in touch with surface but N=0

=> At the very next point mg cos θ will not able to provide required centripetal force and block will leave the contact from the surface.

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