How does a balloon filled with helium rise in air? The weight of the balloon with helium is less than the force of buoyancy on the balloon and hence the balloon rises. Here is a question high lighting this principle:

**A balloon of mass 6 kg is to be filled with helium so as to lift an instrument weighing 24 kg. The minimum volume of helium to be filled in the balloon is nearly (Density of air = 1.29 kg m**^{–3}, density of **helium = 0.18 kg m**^{–3 }approximately)

**(a) ) 22 m ^{3} (b) 27 m^{3} (c) 30 m^{3} (d) ) 133 m^{3} (e) 166 m^{3} **

The balloon will start rising when the force of buoyancy on the balloon just exceeds the weight of the balloon with helium and the instrument. The force of buoyancy is equal to the weight of displaced air. Therefore, the minimum volume V of helium to be filled is given by

(6 + 0.18 V + 24)g = 1.29 Vg.

This gives V = 27 m^{3} approximately.

[The weight of air displaced by the instrument and the empty balloon is negligible and therefore not considered here].

Now consider the following MCQ:

**A block of cork floats on the surface of water (in a container) with 30 cm outside water. If the system is placed on the moon where there is no atmosphere and the value of acceleration due to gravity is approximately one sixths that on the earth, the portion outside water will be**

**(a) 5 cm **

**(b) slightly less than 5 cm **

**(c) slightly less than 30 cm **

**(d) slightly greater than 30 cm **

**(e) slightly greater than 5 cm **

If the acceleration due to gravity has a non-zero value, the exposed portion will remain unchanged since the *weight* of the floating body and the *weight *of displaced fluid are directly proportional to the value of the acceleration due to gravity. So, the exposed portion (or, the immersed portion) will remain the same whether you take the system to the moon or mars or inside a coal mine, when you consider the effect of ‘g’ only. But, you have to consider the effect of the atmosphere also. Since there is no atmosphere on the moon, there is no force of buoyancy due to the atmosphere and the net force of buoyancy (which is due to water in the container and due to the air above) is slightly reduced. Therefore, the block of cork gets immersed a little more and the exposed portion is slightly less than 30 cm.

[Note that if ‘g’ is zero, there is no *weight *and therefore* *no question of floatation].

You will find similar questions (with solution) at AP Physics Resources: Fluid Mechanics- Questions on Force of Buoyancy

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