Two blocks A and B of masses 1.5 kg and i.5 kg, respectively are connected by a massless inextensible string passing over a frictionles pulley as shown in the figure. Block A is lifted until block B touches the ground and then block B touches the ground and then block A is released. the initial height of block A is 80 cm when block B just touches the ground. the maximum height reached by block B from the ground after the block A falls on the ground is

Options

A. 80 cm
B. 120 cm
C. 140 cm
D. 160 cm

Answer

B

Explanation

According to the question,
Given, `m_(A) = 1.5 kg,m_(B) = 0 . 5 kg`
The common acceleration of two masses,
`a = ((m_(A) - m_(s))/(m_(A) +m_(S)))g`
`= (1.5 - 0 .5)/(1.5 =0.5) xx 10`
or `a = 5 ms^(-2)`
According to the question, block A is lifted until block B touches the ground a shown in the figure below,
When block A is relesed then it falling freely under gravity for h height , therefore from third equation of the motion,
`v^(2) = u^(2) +2gh or v^(2) = 0 +2 xx 10 h = 20 h`
But for hight 0.8 m block A falls
with common acceleration, therefore from third equation of the motion ,
`v_(1)^(2) = u_(1)^(2)- 2axx 0 . 8`
here, `v_(1) = 0,u_(1)^(2) = v^(2) = 20 h`
`rArr 20 h - 2xx5 xx0.8 " "[:.a = 5 ms^(-2)]`
`rArr h = 04m = 40 cm`
Hence, the maximum height reached by block B
= 80 + 40 = 120 cm