Two blocks A and B of masses 1.5 kg and 0.5 kg respectively are connected by a massless inextensible string passing over a frictionless pulley as shown in the figure. Block A is lifted until block B touches the ground and then block A is released. The initial height of block A is 80cm when block B just touches the ground. The max height reached by block B form the ground after block A falls on the ground is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

120cm

160cm

140cm

80cm

answer is B.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$Given, m_{A} = 1.5 kg, m_{B} = 0.5 kg$

The common acceleration of two masses,

$a = (\frac{m_{A} - m_{B}}{m_{A} + m_{B}}) g = \frac{1.5 - 0.5}{1.5 + 0.5} \times 10 = 5 m s^{- 2}$

So the block B is accelerated for 80cm till the block A hits the ground,

Velocity of B when A hits ground is given by $v = \sqrt{2 x 5 x 0.8} = 2 \sqrt{2} m / s$

So at that instant the block B is projected up with that velocity,

The further height reached by B $h = \frac{v^{2}}{2 g} = \frac{{(2 \sqrt{2})}^{2}}{2 x 10} = 0.4 = 40 cm$

So total height reached by block B= 80cm+40cm=120cm

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

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

Take Free Test