(C) Let $R$ be the upward buoyant force acting on the balloon.Case $1$: The balloon is moving down with acceleration $a_0$.The equation of motion is:

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(C) Let $R$ be the upward buoyant force acting on the balloon.Case $1$: The balloon is moving down with acceleration $a_0$.The equation of motion is: $Mg - R = Ma_0$ ....$(i)$Case $2$: After removing mass $m$,the balloon moves up with acceleration $2a_0$.The equation of motion is: $R - (M - m)g = (M - m)(2a_0)$ ....(ii)Adding equations $(i)$ and (ii):$(Mg - R) + (R - (M - m)g) = Ma_0 + (M - m)(2a_0)$$Mg - Mg + mg = Ma_0 + 2Ma_0 - 2ma_0$$mg = 3Ma_0 - 2ma_0$$mg + 2ma_0 = 3Ma_0$$m(g + 2a_0) = 3Ma_0$$m = \frac{3Ma_0}{g + 2a_0}$

Class 11 Physics Newton's Laws of Motion and Friction Motion of Body (or Connected Bodies in horizontal or vertical) (by String or Contact)

Medium

$A$ balloon carrying some sand of mass $M$ is moving down with a constant acceleration $a_0$. The mass $m$ of sand to be removed so that the balloon moves up with double the acceleration $2a_0$ is:

AP EAMCET 2024 All AP EAMCET

A
$m=\frac{2 M a_0}{a_0+g}$
B
$m=\frac{2 M a_0}{a_0-g}$
C
$m=\frac{3 M a_0}{g+2 a_0}$
D
$m=\frac{3 M a_0}{g-2 a_0}$

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Class 11

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Chapter

Newton's Laws of Motion and Friction

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Motion of Body (or Connected Bodies in horizontal or vertical) (by String or Contact)

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$A$ balloon carrying some sand of mass $M$ is moving down with a constant acceleration $a_0$. The mass $m$ of sand to be removed so that the balloon moves up with double the acceleration $2a_0$ is:

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In the diagram shown,the normal reaction force between the $2\,kg$ and $1\,kg$ blocks is (Consider the surface to be smooth) $.........N$ (Given $g = 10\,ms^{-2}$)

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