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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Fluid Mechanics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Fluid Mechanics

A spring balance measures the weight of an empty balloon to be W_1...

A spring balance measures the weight of an empty balloon to be W1W_1. After filling the balloon with air, its weight becomes W2W_2. Let ww denote the weight of the air itself. Disregard the thickness of the inflated balloon. Furthermore, ignore any difference between the densities of the air inside and outside the balloon.

Options

  1. AW2<W1+wW_2 < W_1 + w
  2. BW2=W1+wW_2 = W_1 + w
  3. CW2=W1W_2 = W_1
  4. DW2>W1W_2 > W_1

Correct answer

C. W2=W1W_2 = W_1

Step-by-step solution

Let the true weight of the empty balloon material be W0W_0 and its volume be vv.

The buoyant force acting on the empty balloon due to the surrounding air is Fb1=ρairvgF_{b1} = \rho_{air} v g.

The spring balance measures the apparent weight. Thus, the apparent weight of the empty balloon is:
W1=W0−ρairvgW_1 = W_0 - \rho_{air} v g

When the balloon is inflated, let the volume of the air inside be VV. The true weight of the air inside is given as ww, so w=ρairVgw = \rho_{air} V g (since the densities of air inside and outside are the same).

The total true weight of the inflated balloon is W0+wW_0 + w.

The new buoyant force acting on the inflated balloon is due to the total volume displaced, which is v+Vv + V.
Fb2=ρair(v+V)g=ρairvg+ρairVg=Fb1+wF_{b2} = \rho_{air} (v + V) g = \rho_{air} v g + \rho_{air} V g = F_{b1} + w

The apparent weight of the inflated balloon is:
W2=(W0+w)−Fb2W_2 = (W_0 + w) - F_{b2}

Substituting Fb2F_{b2} into the equation:
W2=W0+w−(Fb1+w)W_2 = W_0 + w - (F_{b1} + w)
W2=W0−Fb1W_2 = W_0 - F_{b1}

Since W0−Fb1=W1W_0 - F_{b1} = W_1, we get:
W2=W1W_2 = W_1

Since the weight of the air w>0w > 0, it strictly follows that:
W2<W1+wW_2 < W_1 + w

Thus, both W2=W1W_2 = W_1 and W2<W1+wW_2 < W_1 + w are correct statements.
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