Three blocks A, B, and C are suspended as shown in fig. Mass of each of blocks A and B is m. If the system is in equilibrium, and the mass of C is M, then:

M > 2 m

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M = 2 m

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M < 2 m

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None of these

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Solution

The correct option is A $M < 2 m$
Given,
Mass of block A & B $= m$
Mass of block C $= M$
Let
Mass of A $= m_{A}$ & Mass of B $= m_{B}$
Tension in string is $= T$
At equilibrium, $T = m_{A} g = m_{B} g = m g$
Weight of block C is = $M g$

Forces on block C
$2 T cos θ = M g$
$cos θ = \frac{M g}{2 T} = \frac{M g}{2 m g} = \frac{M}{2 m}$
If $0 < θ < 90^{0}$ then $1 > cos θ > 0$
$1 > \frac{M}{2 m} > 0$
$2 m > M$

Hence, $2 m > M$

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