## A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the plane is μ and…

Q: A block of mass m is on an inclined plane of angle θ. The coefficient of friction between the block and the plane is μ and tan θ > μ . The block is held stationary by applying a force P parallel to the plane.The direction of force pointing up the plane is taken to be positive. As P is varied from P1 = mg (sinθ – μ cos θ) to P2 = mg (sin θ + μ cos θ), the frictional force f versus P graph will look like (a) (b) (c) (d) Ans: (a)

## A piece of wire is bent in the shape of a parabola y=kx^2 (y-axis vertical) with a bead of mass m on it. The bead can slide…

Q: A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point slide of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

(a)a/gk

(b)a/2gk

(c)2a/gk

(d)a/4gk

Ans: (b)

Sol: Draw FBD

by applying pseudo force

$\large Nsin\theta = mg$

$\large Ncos\theta = ma$

on dividing

$\large tan\theta = \frac{g}{a}$

## Two particular of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface…

Q: Two particular of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance a from the centre P (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the seperation between them becomes 2x is

(a) $\large \frac{F}{2m}\frac{a}{\sqrt{a^2 – x^2}}$

(b) $\large \frac{F}{2m}\frac{x}{\sqrt{a^2 – x^2}}$

(c) $\large \frac{F}{2m}\frac{x}{a}$

(d) $\large \frac{F}{2m}\frac{\sqrt{a^2 – x^2}}{x}$

Ans: (b)

Sol: When string is pulled by a force F , let tension created is T

$\large 2 T cos\theta = F$

$\large T = \frac{F}{2}sec\theta$

Acceleration of the particle is

$\large a = \frac{T sin\theta}{m}$

$\large a = \frac{F}{2m}tan\theta$

$\large = \frac{F}{2m}\frac{x}{\sqrt{a^2-x^2}}$

## System shown in figure is in equilibrium and at rest. The spring and string are massless, now the string is cut…

Q: System shown in figure is in equilibrium and at rest. The spring and string are massless, now the string is cut. The acceleration of mass 2m and m just after the string is cut will be (a)g/2 upwards, g downwards

(b)g upwards, g/2 downwards

(c)g upwards, 2g downwards

(d)2g upwards, g downwards

Ans: (a)

Sol: At equilibrium F = 3 mg

When string is cut , acceleration of mass m is

am = g (downward)

Acceleration of mass 2m is

$\large a_{2m} = \frac{F – 2mg}{2m}$

$\large = \frac{3mg – 2mg}{2m}$

a2m = g/2 (upwards)

## The pulley and string shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium…

Q: The pulley and string shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be (a)0°

(b)30°

(c)45°

(d)60°

Ans: (c)
Sol: Let T = tension in the string

T = mg

For equilibrium of √2 mg

$\large 2T cos\theta = \sqrt{2}mg$

$\large cos\theta = \frac{1}{\sqrt{2}}$