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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]

  1. A \(\frac{\mathrm{mg}}{2 \mathrm{M}+\mathrm{m}}\)
  2. B \(\frac{\mathrm{Mg}}{\mathrm{M}+2 \mathrm{~m}}\)
  3. C \(\frac{\mathrm{Mg}}{2 \mathrm{M}+\mathrm{m}}\)
  4. D \(\frac{m g}{M+2 m}\)
Verified Solution

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,
\((\mathrm{M}+\mathrm{m}) \mathrm{g}-\mathrm{Mg}=(2 \mathrm{M}+\mathrm{m}) \mathrm{a}\)
\(\therefore \quad a=\frac{m g}{(2 M+m)}\)
From MHT CET
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