## A cylinder of mass M and radius R is resting on two corner edges A and B as shown in figure….

Q: A cylinder of mass M and radius R is resting on two corner edges A and B as shown in figure. The normal reaction at the edges A and B are : (Neglect friction)

(A) $N_A = \sqrt{2} N_B$

(B) $N_B = \sqrt{3} N_A$

(C) $N_A = \frac{M g}{2}$

(D) $N_B = \frac{2\sqrt{3} M g}{5}$

Click to See Solution :
Ans: (B) , (C)
Sol: For equilibrium NA cos 60° + NB cos 30° = Mg
and NA sin 60° = NB sin 30°

On Solving , $N_B = \sqrt{3} N_A$

$N_A = \frac{M g}{2}$

## Two blocks each of mass m lie on a smooth table. They are attached to two other masses as shown in the figure….

Q: Two blocks each of mass m lie on a smooth table. They are attached to two other masses as shown in the figure. The pulleys and strings are light. An object O is kept at rest on the table. The sides AB and CD of the two blocks are made reflecting. The acceleration of two images formed in those two reflecting surfaces w.r.t. each other

(A) 5g/6

(B) 5g/3

(C) g/3

(D) 17g/6

Click to See Solution :
Ans: (D)
Sol: Acceleration of block AB $= \frac{3 m g}{3m + m} = \frac{3}{4}g$

Acceleration of block CD $= \frac{2 m g}{2 m + m} = \frac{2}{3}g$

Acceleration of image in mirror AB

= 2 acceleration of mirror

$= 2 (-\frac{3 g}{4}) = -\frac{3}{2}g$

Acceleration of image in mirror CD $= 2 (\frac{2 g}{3}) = \frac{4 g}{3}$

Acceleration of the two image w.r.t. each other

$= \frac{4 g}{3} – (- \frac{3 g}{2}) = \frac{17 g}{6}$

## In the figure shown all the surface are smooth.All the blocks A, B and C are movable x-axis is horizontal and y-axis vertical as shown ….

Q: In the figure shown all the surface are smooth.All the blocks A, B and C are movable, x-axis is horizontal and
y-axis vertical as shown. Just after the system is released from the position as shown.

(A)Acceleration of ‘A’ relative to ground is in negative y-direction

(B) Acceleration of ‘A’ relative to B is in positive x-direction

(C) The horizontal acceleration of ‘B’ relative to ground is in negative x-direction.

(D) The acceleration of ‘B’ relative to ground along the inclined surface of ‘C’ is greater than g sinθ

Click to See Solution :
Ans: (A), (B) , (C) , (D)
Sol: There is no horizontal force on block A, therefore it does not move in x-direction, whereas there is net
downward force (mg ñ N) is acting on it, making its acceleration along negative y-direction. Block B moves downward as well as in negative x- direction. Downward acceleration of A and B will be equal due to constrain, thus w.r.t. B,A moves in positive x-direction.Due to the component of normal exterted by C on B, it moves in negative x-direction.

The force acting vertically downward on block B are mg and NA (normal reaction due to block A). Hence the component of net force on block B along the inclined surface of Bis greater than mg sinθ . Therefore the acceleration of ‘B’ relative to ground directed along the inclined surface of ‘C’ is greater than g sinθ

## A bead of mass m is attached to one end of a spring of natural length R and spring constant k ….

Q: A bead of mass m is attached to one end of a spring of natural length R and spring constant $k = \frac{(\sqrt{3}+1)mg}{R}$ . The other end of the spring is fixed at point A on a smooth vertical ring of radius R as shown in figure. The normal reaction at B just after it is released to

(a) $\displaystyle \frac{m g}{2}$

(b) $\displaystyle \sqrt{3} m g$

(c) $\displaystyle 3 \sqrt{3} m g$

(d) $\displaystyle \frac{3 \sqrt{3} m g}{2}$

Click to See Solution :
Ans: (d)

Sol: The extension is spring is x = 2R cos 30° – R

x = (√3 – 1 )R

Applying Newton’s second law to the bead normal to circular ring at point B

N = k (√3 – 1 )R cos 30° + m g cos 30°

N = $k = \frac{(\sqrt{3}+1)mg}{R}$ (√3 – 1 )R cos 30° + m g cos 30°

$\displaystyle N = \frac{3 \sqrt{3} m g}{2}$