Four blocks \(A, B, C\) and \(D\) of masses \(6 \mathrm{~kg}, 3 \mathrm{~kg}\), \(6 \mathrm{~kg}\) and…
Four blocks \(A, B, C\) and \(D\) of masses \(6 \mathrm{~kg}, 3 \mathrm{~kg}\), \(6 \mathrm{~kg}\) and \(\mathrm{I} \mathrm{kg}\) 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 friction 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 the string \(Q\) is (Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\))
\(48 \mathrm{~N}\)
\(24 \mathrm{~N}\)
\(12 \mathrm{~N}\)
\(6 \mathrm{~N}\)
Solution
Given, \(m_A=6 \mathrm{~kg}, m_B=3 \mathrm{~kg}, m_C=6 \mathrm{~kg}\), and \(m_D=1 \mathrm{~kg}\) and coefficient of friction between block \(B\) and horizontal surface, \(\mu=0.2\)
Now, the acceleration of complete system,
\(\begin{aligned}
a & =\frac{m_C g-m_D g-\mu\left(m_A+m_B\right) g}{m_A+m_B+m_C+m_D} \\
& =\frac{6 \times 10-1 \times 10-0.2(6+3) \times 10}{6+3+6+1} \\
& =\frac{60-10-18}{16} \\
& =\frac{32}{16}=2 \mathrm{~ms}^{-2}
\end{aligned}\)
Now, the tension in the string \(Q\) is given by
\(\begin{aligned}
m_D a & =T-m_D g \\
T & =m_D a+m_D g \\
T & =1 \times 2+1 \times 10=12 \mathrm{~N}
\end{aligned}\)
Hence, the correct answer is (c).
