ExamBro
ExamBro
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

  1. A \(g / 2\) upwards, \(g\) downwards
  2. B \(g\) upwards, \(g / 2\) downwards
  3. C \(g\) upwards, \(2 g\) downwards
  4. D \(2 g\) upwards, \(g\) downwards
Verified Solution

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 \(2 m\).
\(
a_m=\frac{m g}{m}=g \text { (downwards) and }
\)
\(
a_{2 m}=\frac{m g}{2 m}=\frac{g}{2} \text { (upwards) }
\)
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app