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Dear student, Weight #160;of #160;balloon #160;= #160;wMass #160;of #160;balloon #160;, #160;m=wgBalloon #160;moving #160;with #160;accele ...

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Standard XII

Physics

Newton's Laws for System of Particles

A balloon of ...

Question

A balloon of gross weight w newton is falling vertically downward with a constant acceleration a(< g). The magnitude of the air resistance is:
(a) w (b) $w (1 + \frac{a}{g})$ (c) $w (1 - \frac{a}{g})$ (d) $w \frac{a}{g}$

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Solution

Dear student,
$W e i g h t o f b a l l o o n = w M a s s o f b a l l o o n, m = \frac{w}{g} B a l l o o n m o v i n g w i t h a c c e l e r a t i o n a m / s^{2} N e t f o r c e = W e i g h t - r e s i s \tan c e m a = w - r e s i s \tan c e r e s i s \tan c e = w - \frac{w a}{g} = w (1 - \frac{a}{g})$
Regards

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A balloon of gross weight w newton is falling vertically downward with a constant acceleration a(< g). The magnitude of the air resistance is: (a) w (b) w 1+ag (c) w 1-ag (d) wag

Dear student, Weight of balloon = wMass of balloon , m=wgBalloon moving with acceleration a m/s2Net force =Weight-resistancema=w-resistanceresistance = w-wag=w1-ag Regards

A balloon of gross weight w newton is falling vertically downward with a constant acceleration a(< g). The magnitude of the air resistance is:
(a) w (b) $w (1 + \frac{a}{g})$ (c) $w (1 - \frac{a}{g})$ (d) $w \frac{a}{g}$