## A block starts moving up an inclined plane of inclination 30° with an initial velocity vo . It comes back to its initial position ……

Q: A block starts moving up an inclined plane of inclination 30° with an initial velocity vo . It comes back to its initial position with velocity vo/2 . The value of coefficient of kinetic friction between the block and inclined plane is close to I/1000 . The nearest integer to I is …….

Click to See Solution :
Ans: 346

Sol:

When block is moving up the plane ,

Retardation a1 = gsin30° + μ g cos30°

$\displaystyle a_1 = \frac{g}{2} + \frac{\mu g \sqrt{3}}{2}$

$\displaystyle v^2 = u^2 + 2 a s$

$\displaystyle 0 = v_o^2 – 2 (\frac{g}{2} + \frac{\mu g \sqrt{3}}{2}) s$

$\displaystyle v_o^2 = g(1 + \sqrt{3}\mu)$ …(i)

When block is moving down the plane ,

Retardation , $\displaystyle a_2 = gsin30° – μ g cos30°$

$\displaystyle a_2 = \frac{g}{2} – \frac{\mu g \sqrt{3}}{2}$

writing equation for down the plane ,

$\displaystyle (\frac{v_o}{2})^2 = 2 (\frac{g}{2} – \frac{\mu g \sqrt{3}}{2}) s$

$\displaystyle \frac{v_o^2}{4} = 2 (\frac{g}{2} – \frac{\mu g \sqrt{3}}{2}) s$

$\displaystyle v_o^2 = 4 g(1 – \sqrt{3}\mu)$ ….(ii)

From (i) & (ii)

$\displaystyle g(1 + \sqrt{3}\mu) = 4 g(1 – \sqrt{3}\mu)$

$\displaystyle (1 + \sqrt{3}\mu) = 4(1 – \sqrt{3}\mu)$

$\displaystyle 5 \sqrt{3}\mu = 3$

$\displaystyle \mu = \frac{3}{5 \sqrt{3}}= \frac{\sqrt{3}}{5}$

$\displaystyle \mu = \frac{1.732}{5} \times \frac{200}{200}$

$\displaystyle \mu = \frac{346.2}{1000}$

Therefore I = 346

## A small block starts slipping down from a point B on an inclined plane AB , which is making an angle θ with the horizontal …..

Q: A small block starts slipping down from a point B on an inclined plane AB , which is making an angle θ with the horizontal section BC is smooth and remaining section CA is rough with a coefficient of friction μ . It is found that the block comes to rest as it reaches the bottom (Point A) of the inclined plane . If BC = 2AC , the coefficient of friction is given by μ = k tanθ . The value of k is ……

Click to See Answer :
Ans: k =
Sol: Let AC = x then BC = 2x

When block moves from B to C ,

$\displaystyle V_C^2 = V_B^2 + 2 (g sin\theta)2x$

$\displaystyle V_C^2 = 0 + 2 (g sin\theta)2x$ ….(i)

When block moves from C to A ,

$\displaystyle V_A^2 = V_C^2 + 2 (g sin\theta – \mu g cos\theta )x$

$\displaystyle 0 = V_C^2 + 2 (g sin\theta – \mu g cos\theta ) x$

$\displaystyle V_C^2 = – 2 (g sin\theta – \mu g cos\theta ) x$ ….(ii)

From (i) & (ii)

$\displaystyle 2 (g sin\theta)2x = – 2 (g sin\theta – \mu g cos\theta ) x$

μ = 3 tanθ

Therefore , k = 3

## Shown in the figure is rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends ….

Q. Shown in the figure is rigid and uniform one meter long rod AB held in horizontal position by two strings tied to its ends and attached to the ceiling . The rod is of mass ‘ m ‘ and has another weight of mass 2m hung at a distance of 75 cm from A . The tension in the string at A is

(a) 0.5 m g

(b) 2 m g

(c) 0.75 m g

(d) 1 m g

Click to See Answer :
Ans: (d)

Sol:Let tension in the string at A is T

Moment of force about point B = 0

-(2 m g × 0.25 ) -(m g × 0.50 ) + (T × 1) = 0

T = (2 m g × 0.25 ) + (m g × 0.50 )

T = 1 m g

## A spring mass system (mass m , spring constant k , natural length l) rests in equilibrium on a horizontal disc …..

Q: A spring mass system (mass m , spring constant k , natural length l) rests in equilibrium on a horizontal disc . The free end of the spring is fixed at the centre of disc . If the disc together with spring mass system , rotates about its axis with an angular velocity ω , (k >> m ω2 ) the relative change in the length of the spring is best given by option

(a) $\displaystyle \frac{2 m \omega^2}{k}$

(b) $\displaystyle \frac{ m \omega^2}{3 k}$

(c) $\displaystyle \frac{ m \omega^2}{k}$

(d) $\displaystyle \sqrt{\frac{2}{3}} \frac{ m \omega^2}{k}$

Click to See Answer :
Ans: (c)
Sol: Let Δl be the extension in the spring

Therefore , k Δl = m ω2 (l + Δl )

k Δl – m ω2 Δl = m ω2 l

Δl ( k – m ω2) = m ω2 l

$\displaystyle \frac{\Delta l}{l} = \frac{m \omega^2 }{k-m \omega^2}$

Using , k >> m ω2

$\displaystyle \frac{\Delta l}{l} = \frac{m \omega^2 }{k}$

## A particle of mass m is fixed to one end of a spring having force constant k and unstretched length l ….

Q: A particle of mass m is fixed to one end of a spring having force constant k and unstretched length l . The other end is fixed . The system is given an angular speed ω about the fixed end of the spring such that it rotates in a circle in gravity free space . Then stretch in the spring is

(a) $\displaystyle \frac{m l \omega^2}{k- m \omega}$

(b) $\displaystyle \frac{m l \omega^2}{k- m \omega^2}$

(c) $\displaystyle \frac{m l \omega^2}{k + m \omega^2}$

(d) $\displaystyle \frac{m l \omega^2}{k + m \omega}$

Click to See Answer :
Ans: (b)

Sol: Let extension in the spring is Δl

Therefore , k Δl = m ω2 (l + Δl )

k Δl = m ω2 l + m ω2 Δl

k Δl – m ω2 Δl = m ω2 l

$\displaystyle \Delta l = \frac{m l \omega^2}{k- m \omega^2}$