A narrow but tall cabin is falling freely near the earth's surface. Inside the cabin, two small stones $A$ and $B$ are released from rest (relative to the cabin). Initially $A$ is much above the centre of mass and $B$ much below the centre of mass of the cabin. A close observation of the motion of $A$ and $B$ will reveal that -

• A

  both $A$ and $B$ continue to be exactly at rest relative to the cabin
• Correct answer
  B

  A moves slowly upward and $B$ moves slowly downward relative to the cabin
• C

  both $A$ and $B$ fall to the bottom of the cabin with constant acceleration due to gravity
• D

  $A$ and $B$ move slightly towards each other vertically

Solution

$a_{B}>a_{C}>a_{A}$ $a_{B}=g$ $\vec{a}_{\Lambda / C M}=\vec{a}_{\Lambda}-\vec{a}_{C M}(\uparrow)$ $\vec{a}_{B / C M}=\vec{a}_{B}-\vec{a}_{C M}(\downarrow)$

Hence, the correct answer is option: 'B'.