Question

A balloon of mass \(M\) is descending with a constant acceleration \(\alpha\). When a mass \(m\) is released from the balloon, it starts rising with the same acceleration \(\alpha\). Assuming that the volume of the balloon does not change, the value of \(m\) is:

• A. \(\left(\dfrac{\alpha}{\alpha+g}\right)M\)
• B. \(\left(\dfrac{2\alpha}{\alpha+g}\right)M\)
• C. \(\left(\dfrac{\alpha+g}{\alpha}\right)M\)
• D. \(\left(\dfrac{\alpha+g}{2\alpha}\right)M\)

Hint:

In buoyancy problems:
Buoyant force depends only on volume of displaced fluid
If volume is constant, buoyant force remains constant
Always write separate equations before and after mass change

Answer 1

The Correct Option is A

Step 1: Identify the forces acting on the balloon. Let the buoyant force acting on the balloon be \(B\). Since the volume does not change, \(B\) remains constant.
Step 2: Case I — Before releasing mass \(m\). The balloon (mass \(M\)) is descending with acceleration \(\alpha\). Taking downward direction as positive, applying Newton’s second law: \[ Mg - B = M\alpha \] \[ \Rightarrow B = M(g - \alpha) \quad \cdots (1) \]
Step 3: Case II — After releasing mass \(m\). Now the mass of the balloon becomes \((M - m)\) and it rises with acceleration \(\alpha\). Taking upward direction as positive: \[ B - (M-m)g = (M-m)\alpha \] \[ \Rightarrow B = (M-m)(g+\alpha) \quad \cdots (2) \]
Step 4: Equate buoyant forces from (1) and (2). \[ M(g-\alpha) = (M-m)(g+\alpha) \]
Step 5: Solve for \(m\). \[ Mg - M\alpha = Mg + M\alpha - mg - m\alpha \] \[ mg + m\alpha = 2M\alpha \] \[ m(g+\alpha) = M\alpha \] \[ m = \frac{M\alpha}{g+\alpha} \] Final Answer: \[ \boxed{m = \left(\dfrac{\alpha}{\alpha+g}\right)M} \]