Two identical blocks each of mass ' $M$ ' attached to the ends of a massless inextensible string which…

Two identical blocks each of mass ' $M$ ' attached to the ends of a massless inextensible string which passes over a pulley with a fixed axis as shown below. A small mass ' $m$ ' is now placed on the block B. The acceleration with which the two blocks move together is [ $\mathrm{g}=$ gravitational acceleration]
  1. $\frac{\mathrm{mg}}{2 \mathrm{M}+\mathrm{m}}$
  2. $\frac{\mathrm{Mg}}{\mathrm{M}+2 \mathrm{~m}}$
  3. $\frac{\mathrm{Mg}}{2 \mathrm{M}+\mathrm{m}}$
  4. $\frac{m g}{M+2 m}$

Solution

$\begin{aligned} & (\mathrm{M}+\mathrm{m}) \mathrm{g}-\mathrm{T}=(\mathrm{M}+\mathrm{m}) \mathrm{a} \\ & \mathrm{~T}-\mathrm{Mg}=\mathrm{Ma} \end{aligned}$ Adding the above equations, $(\mathrm{M}+\mathrm{m}) \mathrm{g}-\mathrm{Mg}=(2 \mathrm{M}+\mathrm{m}) \mathrm{a}$ $\therefore \quad a=\frac{m g}{(2 M+m)}$ ~

Asked in: MHT CET 2024 (02 May Shift 1)

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