Fluid Mechanics – Multiple Choice Practice Questions for AP Physics C and IIT-JEE

Questions served for AP Physics C exam. and IIT-JEE are generally not as simple as the questions served for other degree entrance exams. You require more practice with tougher questions to obtain higher scores in these exams. Here are a few multiple choice practice questions on fluid mechanics:

[There will be 5 options for AP Physics C exam., but 4 options only for IIT-JEE]

(1) A bowl has a small hole at the centre of its bottom. Water poured into the bowl drains through the hole and the height of water column at any instant t is H. The radius of the free surface of water is R (fig.) at the instant t. If the time rate of decrease of the height of water column is constant, how is H related to R?

(a) H α R^1/2

(b) H α R

(c) H α R²

(d) H α R³

(e) H α R⁴

The time rate of decrease of the height of water column is dH/dt and we have

dH/dt = constant, as given in the question.

Since the velocity of the water flowing out (velocity of efflux) through the hole is √(2gH), the rate of flow is av = a√(2gH) where a is the area of the hole.

Therefore, we have

a√(2gH) = πR²(dH/dt)

Since a, g, π and dH/dt are costants, H α R⁴ [Option (e)].

(2) A jar of uniform area of cross section A has a hole of area a at its bottom. If there is water column of height H₁ in the jar initially, what time is required for the height to become H₂?

(a) (A/a) √(4/g) (√H₁ – √H₂)

(b) (A/a) √(2/g) (√H₁ – √H₂)

(c) (A/a) (4/g) (H₁ –H₂)

(d) (A/a) √(1/g) (√H₁ – √H₂)

(e) (a/A) √(4/g) (√H₁ – √H₂)

Rate of flow of water through the hole is av where v is the efflux velocity given by

v =√(2gx) where x is the height of water column.

If the height of water column changes by dx (fig.) in a time dt, we have

– Adx (= avdt) = a√(2gx) dt

[The negative sign shows that x decreases].

Therefore, dt = – (A/a)[1/√(2g)] (dx/x^1/2)

Integrating between the limits x = H₁ and x = H₂ we obtain the required time t.

Thus t = (A/a)[1/√(2g)] _H1∫^H2 (–x^–1/2)dx

Or, t = (A/a)[1/√(2g)] _H2∫^H1 (x^–1/2)dx

This gives t = (A/a) √(2/g) (√H₁ – √H₂)

(3) A sphere of wax (density 900 kgm^–3) has a volume of 20 cm³. Iron nails are pierced into it so that it just gets submerged in water. If the volume of the iron nails is negligible compared to the volume of the sphere of wax, what is the mass of the iron nails in the sphere?

(a) 0.001 kg

(b) 0.002 kg

(c) 0.01 kg

(d) 0.02 kg

(e) 0.09 kg

Since the sphere of wax containing the iron nails gets just submerged in water, the mean density of the sphere must be equal to the density of water (1000 kgm^–3).

Mass of wax in the sphere is 20×10^–6×900 kg.

If m represents the mass of the nails, we have

(20×10^–6×900 + m) /(20×10^–6) = 1000

This gives m = 2×10^–3 kg = 0.002 kg.

You will find a few more multiple choice questions (with solution) in this section here.