Question:
A balloon with mass 'm' is descending vertically with an acceleration 'a' (where a \(<\) g). The mass to be removed from the balloon, so that it starts moving vertically up with an acceleration 'a' is
A balloon with mass 'm' is descending vertically with an acceleration 'a' (where a \(<\) g). The mass to be removed from the balloon, so that it starts moving vertically up with an acceleration 'a' is
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1. Identify all forces acting on the body in each situation (buoyant force, weight).
2. Apply Newton's second law (\(F_{net} = \text{mass} \times \text{acceleration}\)) for each case.
3. The buoyant force depends on the volume of displaced air and remains constant if the balloon's volume doesn't change significantly.
4. Solve the system of equations to find the unknown mass to be removed.
Updated On: Jun 5, 2025
- \( \frac{ma}{g+a} \)
- \( \frac{ma}{g-a} \)
- \( \frac{2ma}{g+a} \)
- \( \frac{2ma}{g-a} \)
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The Correct Option is C
Solution and Explanation
Let \( F_B \) be the buoyant force (upthrust) acting on the balloon.
This force is constant.
Case 1: Balloon descending with acceleration 'a'.
The forces acting are: weight \(mg\) downwards, buoyant force \(F_B\) upwards.
Net downward force = \( mg - F_B \).
By Newton's second law: \( mg - F_B = ma \).
So, \( F_B = mg - ma = m(g-a) \cdots (1) \).
Case 2: Mass \( \Delta m \) is removed.
Let the new mass be \( m' = m - \Delta m \).
The balloon starts moving vertically up with acceleration 'a'.
The forces acting are: new weight \(m'g\) downwards, buoyant force \(F_B\) upwards.
Net upward force = \( F_B - m'g \).
By Newton's second law: \( F_B - m'g = m'a \).
So, \( F_B = m'g + m'a = m'(g+a) \cdots (2) \).
Equating the expressions for \( F_B \) from (1) and (2): \[ m(g-a) = m'(g+a) \] Substitute \( m' = m - \Delta m \): \[ m(g-a) = (m - \Delta m)(g+a) \] \[ mg - ma = mg + ma - \Delta m(g+a) \] \[ -ma = ma - \Delta m(g+a) \] \[ \Delta m(g+a) = ma + ma = 2ma \] \[ \Delta m = \frac{2ma}{g+a} \] The mass to be removed is \( \frac{2ma}{g+a} \).
This matches option (3).
This force is constant.
Case 1: Balloon descending with acceleration 'a'.
The forces acting are: weight \(mg\) downwards, buoyant force \(F_B\) upwards.
Net downward force = \( mg - F_B \).
By Newton's second law: \( mg - F_B = ma \).
So, \( F_B = mg - ma = m(g-a) \cdots (1) \).
Case 2: Mass \( \Delta m \) is removed.
Let the new mass be \( m' = m - \Delta m \).
The balloon starts moving vertically up with acceleration 'a'.
The forces acting are: new weight \(m'g\) downwards, buoyant force \(F_B\) upwards.
Net upward force = \( F_B - m'g \).
By Newton's second law: \( F_B - m'g = m'a \).
So, \( F_B = m'g + m'a = m'(g+a) \cdots (2) \).
Equating the expressions for \( F_B \) from (1) and (2): \[ m(g-a) = m'(g+a) \] Substitute \( m' = m - \Delta m \): \[ m(g-a) = (m - \Delta m)(g+a) \] \[ mg - ma = mg + ma - \Delta m(g+a) \] \[ -ma = ma - \Delta m(g+a) \] \[ \Delta m(g+a) = ma + ma = 2ma \] \[ \Delta m = \frac{2ma}{g+a} \] The mass to be removed is \( \frac{2ma}{g+a} \).
This matches option (3).
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