Four blocks A, B, C and D of masses 6 kg, 3 kg, 6 kg and 1kg respectively are connected by light strings passing over frictionless pulleys as shown in the figure. The strings P and Q are horizontal. The coefficient of friciton between the horizontal surface and the block B is 0.2 and the blocks A and B move together. If the system is released from rest then the tension in string Q is (Acceleration due to gravity, `g = 10 ms^(-2)`)

Options

A. 48 N
B. 24 N
C. 12 N
D. 6 N

Answer

C

Explanation

Given, `m_(A) = 6 kg, m_(B) = 3 kg, m_(C ) = 6kg` and `m_(D) = 1 kg` and coefficient of friction between block B and horizontal surface, `mu = 0.2`
Now, the acceleration of complete system,
`a = (m_(C )g - m_(D) g - mu (m_(A) + m_(B))g)/(m_(A) + m_(B) + m_(C ) + m_(D))`
`= (6 xx 10 - 1 xx 10 - 0.2 (6+3) xx 10)/(6 + 3 + 6 + 1)`
`= (60 - 10 - 18)/(16)`
`= (32)/(16) = 2 ms^(-2)`
Now, the tension in the string Q is given by

`m_(D)a = T - m_(D)g`
`T = m_(D) a + m_(D) g`
`T = 1 xx 2 + 1 xx 10 = 12 N`
Hence, the correct answer is (C ).