JEE Advanced · Physics · 5. Laws of Motion
System shown in figure is in equilibrium and at rest. The spring and string are massless, now the string is cut. The acceleration of mass \(2 m\) and \(m\) just after the string is cut will be

- A \(g / 2\) upwards, \(g\) downwards
- B \(g\) upwards, \(g / 2\) downwards
- C \(g\) upwards, \(2 g\) downwards
- D \(2 g\) upwards, \(g\) downwards
Answer & Solution
Correct Answer
(A) \(g / 2\) upwards, \(g\) downwards
Step-by-step Solution
Detailed explanation
Initially under equilibrium of mass ' \(m\) '
\(
T=m g
\)
Now, the string is cut. Therefore, \(T=m g\) force is decreased on mass \(m\) upwards and downwards on mass \(2 m\).
\(
a_m=\frac{m g}{m}=g \text { (downwards) and }
\)
\(
a_{2 m}=\frac{m g}{2 m}=\frac{g}{2} \text { (upwards) }
\)
\(
T=m g
\)
Now, the string is cut. Therefore, \(T=m g\) force is decreased on mass \(m\) upwards and downwards on mass \(2 m\).
\(
a_m=\frac{m g}{m}=g \text { (downwards) and }
\)
\(
a_{2 m}=\frac{m g}{2 m}=\frac{g}{2} \text { (upwards) }
\)
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