In the arrangement shown, each of the blocks A and B has mass ‘m’ coefficient of friction between block B and table top is 0.5. The pulley is light then magnitude of reaction at the spindle of the pulley is _____

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$\frac{3 m g}{2 \sqrt{2}}$

$\frac{m g}{2 \sqrt{2}}$

$\frac{2 \sqrt{2} m g}{3}$

$\frac{3 \sqrt{2} m g}{5}$

answer is C.

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Detailed Solution

$\begin{array}{l} A c c e l e r a t i o n, a = \frac{m g - 0.5 \times m g}{2 m} = g / 4 \\ ∴ t e n s i o n s t r i n g, ∴ T = m g - m \times \frac{g}{4} = \frac{3 m g}{4} \\ ∴ R e a c t i o n a t s p i n d l e = \sqrt{T^{2} + T^{2}} = \frac{3 m g}{2 \sqrt{2}} \end{array}$

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