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In the arrangement the masses of bodies are equal to $m_{0}$ , $m_{1}$ , and $m_{2}$ , the masses of the pulley and the threads are negligible, and there is no friction in the pulley. Find the acceleration ' $a$ ' with which the body $m_{0}$ comes down, and the tension of the thread binding together the bodies $m_{1}$ and $m_{2}$ if the coefficient of friction between the bodies and the horizontal surface is equal to $μ$ .

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$a = [\frac{m_{0} - μ (m_{1} + m_{2})}{m_{0}}] g, T = [\frac{m_{0} m_{2} g (1 + μ)}{m_{1} + m_{2}}]$

$a = [\frac{m_{0} - μ (m_{1} + m_{2})}{m_{0} + m_{1} + m_{2}}] g, T = [\frac{m_{0} m_{2} g (1 + μ)}{m_{0}}]$

$a = [\frac{m_{0} - μ (m_{1} + m_{2})}{m_{0} + m_{1} + m_{2}}] g, T = [\frac{m_{0} m_{2} g (1 + μ)}{m_{0} + m_{1} + m_{2}}]$

$a = [\frac{m_{0} - μ (m_{1} + m_{2})}{m_{0}}] g, T = [\frac{m_{0} m_{2} g}{m_{0} + m_{1} + m_{2}}]$

answer is A.

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Detailed Solution

If $a$ is the acceleration of the blocks, then

$m_{0} g - T_{1} = m_{0} a$ …(i)

$T_{1} - (T_{2} + μ m_{1} g) = m_{1} a$ …(ii)

$T_{2} + μ m_{2} g) = m_{2} a$ …(iii)

After solving above equations, we get

$a = [\frac{m_{0} - μ (m_{1} + m_{2})}{m_{0} + m_{1} + m_{2}}] g, T = [\frac{m_{0} m_{2} g (1 + μ)}{m_{0} + m_{1} + m_{2}}]$

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