JEE MAIN - Physics (2025 - 28th January Evening Shift - No. 7)
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 ‘3a’ will be
(Take ‘g’ as acceleration due to gravity)
$ \frac{3Ma}{2a + g} $
$ \frac{2Ma}{3a + g} $
$ \frac{3Ma}{2a - g} $
$ \frac{2Ma}{3a - g} $
Explanation
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$$\begin{aligned} &\begin{aligned} & F-m g=m a \\ & F=m a+m g \\ & F-(m-x) g=(m-x) 3 a \end{aligned}\\ &\text { Put F }\\ &\begin{aligned} & \mathrm{Ma}+\mathrm{mg}-\mathrm{mg}+\mathrm{xg}=3 \mathrm{ma}-3 \mathrm{xa} \\ & \mathrm{x}=\frac{2 \mathrm{ma}}{\mathrm{~g}+3 \mathrm{a}} \end{aligned} \end{aligned}$$
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