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A smooth ring of mass m can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley B and carries a block C of mass 2m as shown below. At the instant ring and the block just starts sliding

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The tension in the string is

T = {ma}_{A} \sec θ

The relation between magnitudes of accelerations of A and C is

a_{C} = a_{A} \cos θ

The The relation between magnitudes of acceleration of A and c is

a_{A} = a_{C} \cos θ

The acceleration of the block is

\frac{2 g \cos^{2} θ}{1 + 2 \cos^{2} θ}

answer is A, C, D.

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Detailed Solution

From fig we write

a_{A} \cos θ = a_{C}

And $T \cos θ = {ma}_{A}$

$T = {ma}_{A} \sec θ$

Now $2 mg - T = 2 {ma}_{C}$

$2 mg - \frac{{ma}_{C}}{\cos^{2} θ} = 2 {ma}_{C}$

$2 mg = 2 {ma}_{C} + \frac{{ma}_{C}}{\cos^{2} θ}$

$2 g = a_{C} [\frac{1}{\cos^{2} θ} + 2]$

$2 g = a_{C} [\frac{{1+2cos}^{2} θ}{\cos^{2} θ}]$

$∴ a_{C} = \frac{2 g \cos^{2} θ}{1 + 2 \cos^{2} θ}$

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