The Balloon Knows

Classical Mechanics Level 2

The whole setup about to be described has been done with respect to surrounding air.

The weight of an empty balloon on a spring balance is $W_{1}$ . The weight becomes $W_{2}$ when the balloon is filled with air. Let the weight of the air itself be $w$ .

Neglect the balloon's thickness when it is filled with air and neglect the difference in the densities of air inside and outside the balloon.

Then which of the following options could be a plausible situation:

A. $W_{1} = W_{2}$

B. $W_{2} < W_{1} + w$

C. $W_{2} = W_{1} + w$

D. $W_{2} > W_{1}$

A, B A, C, D C, D

1 solution

Aryan Sanghi
Mar 2, 2021

Let the density of air be $d$ .

The apparent weight of the balloon after being filled with air $W_2 = W_1 + w - B$ where $B = \text{Buoyant Force}$

$B = Vdg$ $B = \bigg(\frac{w}{dg}\bigg)dg = w$

Therefore, The apparent weight of the balloon after being filled with air $W_2 = W_1 + w - B = W_1$

$\boxed{W_2 = W_1}$

Thank you! I don't know why, I am always confused in these types of problems :(

Vinayak Srivastava - 3 months, 1 week ago

Keep practicing and you'll solve any problem. :)

Aryan Sanghi - 3 months, 1 week ago

Ok, thanks for your answer as well as comment! I will give more time to fluids next time I study them. :)

Vinayak Srivastava - 3 months, 1 week ago

Did you mention me @Vinayak Srivastava

Siddharth Chakravarty - 3 months, 1 week ago

Yeah, I thought the answer was wrong, but I was wrong. I had mentioned both of you for an explanation.

Vinayak Srivastava - 3 months, 1 week ago

The Balloon Knows

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