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Two masses A and B of masses 10 kg and 5kg respectively are connected with a string passing over a friction less pully as shown in fig. The coefficient of friction of A with table is 0.2. The minimum mass of C that may be placed on A to prevent it from moving equal to

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10 kg

15 kg

0 kg

1 kg

answer is A.

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Detailed Solution

$_{k}^{μ} (m_{1} + m_{2}) g = m_{3} g$
$\begin{array}{l} _{k}^{μ} (m_{1} + m_{2}) = m_{3} \\ 0.2 (10 + m_{2}) = 5 \\ 2 + 0.2 m_{2} = 5 \\ 0.2 m_{2} = 3 \\ m_{2} = \frac{3}{0.2} = 15 k g \end{array}$

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