JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Fluid Mechanics

WhatsApp Email

Two narrow bores of diameter 5.0 mm and 8.0 mm are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube.
[Take surface tension of water

T = 7.3 \times 1 0^{- 2} Nm^{- 1}

, angle of contact

= 0, g = 10 ms^{- 2}

and density of water

= 1.0 \times 1 0^{3} kg m^{- 3}

]

(A)2.19 mm✓ Correct
(B)4.97 mm
(C)5.34 mm
(D)3.62 mm

Answer: (A)

Solution

The angle of contact is zero, so water rises in each bore by

h = \frac{2 T cos θ}{ρ g r} = \frac{2 T}{ρ g r}

The radii are

r_{1} = 2.5 \times 1 0^{- 3}

m and

r_{2} = 4.0 \times 1 0^{- 3}

m, and

2 T = 0.146

N/m, so

h_{1} = \frac{0.146}{1 0 ^{3} \times 10 \times 2.5 \times 1 0 ^{- 3}} = 5.84 \times 1 0^{- 3} m, h_{2} = \frac{0.146}{1 0 ^{3} \times 10 \times 4.0 \times 1 0 ^{- 3}} = 3.65 \times 1 0^{- 3} m

The limbs therefore stand

h_{1} - h_{2} = 2.19 \times 1 0^{- 3} m = 2.19 mm

apart, the narrower bore holding the higher column.

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Fluid Mechanics →

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Fluid Mechanics

Solution

More JEE Main questions on Fluid Mechanics