Question

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 thickness of the balloon when it is filled with air. Also neglect the difference in the densities of the air inside and outside the balloon.
(A) \[{{W}_{2}}={{W}_{1}}+W\]
(B) \[{{W}_{2}}=\sqrt{{{W}_{1}}W}\]
(C) \[{{W}_{2}}={{W}_{1}}\]
(D) \[{{W}_{2}}={{W}_{1}}-W\]

Answer 1

Hint: We are given with a balloon. First, it is weighed on the balance and it is noted. Then it is filled with air and again weighed and noted. We know air occupies space and has weight. So, the second weight must be greater than the first.

Complete step by step answer:
the weight of the empty balloon= \[{{W}_{1}}\]
the weight of the air which is being pumped into the balloon= W
the weight with air inside= \[{{W}_{2}}\]
therefore, \[{{W}_{1}}\]+W=\[{{W}_{2}}\]
given that we have to Neglect the thickness of the balloon when it is filled with air and neglect the difference in the densities of the air inside and outside the balloon.
Therefore, the relation between the weight can be stated as \[{{W}_{2}}={{W}_{1}}-W\]
Thus, the correct option is (D)
Additional Information: The spring balance is simply a spring fixed at one end with a hook to attach an object at the other. It works on the principle of Hooke's law. When a load is attached to the spring, it stretches and the spring's extension is proportional to the weight of the object attached to it.

Note:weight is the force with which the earth pulls the body towards itself and it is measured in Newtons. Air occupies space and has weight and so when air is inside the body its weight increases.