Two blocks are connected by a string passing over a pulley as shown. Wedge is fixed. M is sliding down with speed 12 m/s at an instant and suddenly the string breaks. The coefficient of friction between the blocks and wedge is $\frac{1}{\sqrt{2}}$ . Find the velocity (in m/s) of block M with respect to m after t = 2 seconds from the moment the string breaks. [Take $g = 10 m s^{- 2}$ and $\sqrt{\frac{3}{2}} = 1.2$ . Assume that the length of incline is sufficiently long.]

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

2.5

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$a_{M} = g \sin 30^{\circ} - \frac{g \cos 30^{\circ}}{\sqrt{2}} = 5 - 6 = - 1 m / s^{2}$

$a_{m} = - (g \sin 30^{\circ} + \frac{g \cos 30^{\circ}}{\sqrt{2}}) = - 11 m / s^{2}$

As limiting friction is greater than mgsin $θ$ , block ‘m’ will not move after $t = \frac{12}{11} \sec .$

$v_{M} = 12 - 1 \times 2 = 10 m / s$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE