MHT CET · Physics · Laws of Motion
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]

- A \(\frac{\mathrm{mg}}{2 \mathrm{M}+\mathrm{m}}\)
- B \(\frac{\mathrm{Mg}}{\mathrm{M}+2 \mathrm{~m}}\)
- C \(\frac{\mathrm{Mg}}{2 \mathrm{M}+\mathrm{m}}\)
- D \(\frac{m g}{M+2 m}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{mg}}{2 \mathrm{M}+\mathrm{m}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& (\mathrm{M}+\mathrm{m}) \mathrm{g}-\mathrm{T}=(\mathrm{M}+\mathrm{m}) \mathrm{a} \\
& \mathrm{~T}-\mathrm{Mg}=\mathrm{Ma}
\end{aligned}\)
Adding the above equations,
\((\mathrm{M}+\mathrm{m}) \mathrm{g}-\mathrm{Mg}=(2 \mathrm{M}+\mathrm{m}) \mathrm{a}\)
\(\therefore \quad a=\frac{m g}{(2 M+m)}\)
& (\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,
\((\mathrm{M}+\mathrm{m}) \mathrm{g}-\mathrm{Mg}=(2 \mathrm{M}+\mathrm{m}) \mathrm{a}\)
\(\therefore \quad a=\frac{m g}{(2 M+m)}\)
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