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Q.

Figure shows a block of mass $m_{3}$ placed on a smooth wedge of mass $m_{2} .$ (Assume all conditions as ideal). Find the minimum value of $m_{1}$ and the corresponding tension (T) in the string, so that block $m_{3}$ will move vertically downward with an acceleration of $10 m s^{- 2} .$ (Take $g = 10 m s^{- 2}$ )

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a

$m_{1} = \frac{m_{2} \cot θ}{1 - \cot θ}$

b

$m_{1} = \frac{m_{2} \tan θ}{1 - \tan θ}$

c

$T = \frac{m_{2} g}{\tan θ}$

d

$T = \frac{m_{2} g}{\cot θ}$

answer is A, C.

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Detailed Solution

$h = \frac{1}{2} g t^{2}; x = \frac{1}{2} a t^{2}; \frac{h}{x} = \frac{g}{a}$
$\tan θ = \frac{g}{a}; a = g \cot θ$
$m_{1} g - T = m_{1} g \cot θ$
$T = m_{1} g (1 - \cot θ);$
$T = m_{2} g \cot θ$
$T = \frac{m_{2} g}{\tan θ}; m_{1} g (1 - \cot θ) = m_{2} g \cot θ$
$m_{1} = \frac{m_{2} \cot θ}{1 - \cot θ}$
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Figure shows a block of mass m3 placed on a smooth wedge of mass m2. (Assume all conditions as ideal). Find the minimum value of m1 and the corresponding tension (T) in the string, so that block m3 will move vertically downward with an acceleration of 10ms−2. (Take g=10ms−2)