In the arrangement shown, the springs are light and have stiffness $k_{1} = 100 N / m$ and $k_{2} = 200 N / m$ and the pulleys are ideal. An $8.0 k g$ block suspended from the lower pulley is initially held at rest maintaining the strings straight and springs relaxed. Now the force supporting the block is gradually reduced to zero. Acceleration due to gravity is $10 m / s^{2}$ . If the block descends by a distance $(5 n)$ cm in the process of reduction of the force, find ‘n’.

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

answer is 3.

(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

$2 k_{1} x_{1} = k_{2} x_{2} = \frac{m g}{2} \frac{(\frac{x_{1}}{2} + x_{2})}{2} = x, x = 15$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE