JEE Main 2025 Jan 28 Shift 2, Physics Q20: Newton's Laws Of Motion

JEE Main2025Jan 28, Shift 2Physics

A balloon and its content having mass $M$ is moving up with an acceleration $a$ . The mass that must be released from the content so that the balloon starts moving up with an acceleration $3 a$ will be : (Take $g$ as acceleration due to gravity)

A
$\frac{3 M a}{2 a - g}$
B
$\frac{3 M a}{2 a + g}$
C
$\frac{2 M a}{3 a + g}$
D
$\frac{2 M a}{3 a - g}$

Solution

Let the buoyant (upthrust) force on the balloon be $F$ , assumed unchanged by mass release.

Before release: $F - M g = M a \Rightarrow F = M (a + g)$ .

After releasing mass $x$ , the balloon's mass is $M - x$ and acceleration is $3 a$ :

$F - (M - x) g = (M - x) 3 a$ .

Substituting $F$ : $M (a + g) - (M - x) g = 3 a (M - x)$ , i.e. $M a + xg = 3 M a - 3 a x$ .

Solving: $x (g + 3 a) = 2 M a \Rightarrow x = \frac{2 M a}{3 a + g}$ .

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JEE Main 2025 Jan 28 Shift 2, Physics Q20: Newton's Laws Of Motion

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