A balloon has $5 g$ of air. A small hole is pierced into it. The air escapes at a uniform rate with a velocity of $4 c m / s$ . If the balloon shrinks completely in $2.5$ seconds, then the mean force acting on the balloon is:

2 d y n e

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50 d y n e

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8 d y n e

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8 N

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Solution

The correct option is C $8 d y n e$
$M e a n f o r c e = \frac{c h a n g e i n m o m e n t u m}{t i m e}$
Units are given in CGS unit, Hence force will be in $d y n e$
hence $M e a n f o r c e = \frac{5 \times 4}{2.5} = 8 d y n e$

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A balloon has 5g of air. A small hole is pierced into it. The air escapes at a uniform rate with a velocity of 4cm/s. If the balloon shrinks completely in 2.5 seconds, then the mean force acting on the balloon is:

The correct option is C 8dyneMeanforce=change in momentumtimeUnits are given in CGS unit, Hence force will be in dynehence Mean force=5×42.5=8 dyne

A balloon has $5 g$ of air. A small hole is pierced into it. The air escapes at a uniform rate with a velocity of $4 c m / s$ . If the balloon shrinks completely in $2.5$ seconds, then the mean force acting on the balloon is:

The correct option is C $8 d y n e$
$M e a n f o r c e = \frac{c h a n g e i n m o m e n t u m}{t i m e}$
Units are given in CGS unit, Hence force will be in $d y n e$
hence $M e a n f o r c e = \frac{5 \times 4}{2.5} = 8 d y n e$