Related InterviewSolutions

• The force `F_(1)` required to just moving a body up an inclined plane is double the force `F_(2)` required to just prevent the body from sliding down the plane. The coefficient of friction is `mu`. The inclination `theta` of the plane is :A. `tan ^(-1) mu`B. `tan^(-1) mu/2`C. `tan^(-1)2mu`D. `tan^(-1) 3mu`
• A curved section of a road is banked for a speed `v`. If there is no friction between the road and the tyres then.A. a car moving with speed v does not slip on the roadB. a car is more likely to slip on the road at speeds higher than v, than at speeds lower than vC. a car is more likely to slip on the road at speeds lower than v, than at speeds higher thab v.D. a car can remain stationary on the road without slipping.
• Repeat above Question 3, if the charge is negative and the angle made by the boundary with the velocity is `theta = pi/6.`A. `cos^(-1)((1)/(sqrt(3)))`B. `sin^(-1)((1)/(sqrt(3)))`C. `(pi)/(2)-sin^(-1)((1)/(sqrt(3)))`D. `(pi)/(2)-cos^(-1)((1)/(sqrt(3)))`
• The above question 26, the normal force acting, betweent :A. wedge and incline plane is `Mg cos theta`B. m and wedge is zeroC. m and wedge is zeroD. m and wedge is `mg sin theta`
• A small block of mass `m` is released from rest from point `A` inside a smooth hemisphere bowl of radius `R`, which is fixed on group such that `OA` is horizontal. The ratio `(x)` of magnitude of centripetal force and normal reaction on the block at any point `B` varies with `theta` as : A. B. C. D.
• In the following figure all surfaces are smooth. The system is released from rest , then : A. acceleration of wedge is greater then `g tan theta`B. acceleration of `m` is `g sqrt(1+2cos^(2) theta)`C. acceleration of m is gD. acceleration of wedge is `g sin theta`
• All surfaces shown in figure are smooth. System is released with the spring instretched. In equilibrium, compression in the spring will be A. `(2mg)/(k)`B. `((M+m)g)/(sqrt(2)k)`C. `(mg)/(sqrt(2)k)`D. `(mg)/(k)`
• An iron sphere weighing `10N` rests in a `V` shaped smooth trough whose sides form an angle of `60^(@)` as shown in the Then the reaction forces are .A. `R_(A)=10N and R_(B)=0` in case (i)B. `R_(A)=10 N and R_(3)=10N` in case (ii)C. `R_(A)=(20)/(sqrt(3))N and R_(B) ~= (10)/(sqrt(3))N` in case (iii)D. `R_(A)=10 N and R_(B)=10N` in all the three cases
• In the figure, the pulley P moves to the right with a constant speed `u`. The downward speed of A is `v_(A)`, and the speed of B to the right is `v_(B)` : A. `v_(B)=u+u_(A)`B. `v_(B)+u=v_(A)`C. `v_(A)=v_(B)`D. The two blocks have acceleration of the same magnitude
• a particle is moving in a circle of radius R in such a way that at any instant the normal and the tangential component of its acceleration are equal. If its speed at `t=0` is `v_(0)` then time it takes to complete the first revolution is `R/(alphav_(0))(1-e^(-betapi))`. Find the value of `(alpha+beta)`.A. `R//u_(0)`B. `u_(0)//R`C. `(R)/(u_(0))(1-e^(-2pi))`D. `(R)/(u_(0)) e^(-2pi)`