JEE Main 2025 Apr 7 Shift 2, Physics Q8: Properties of Fluids

JEE Main2025Apr 7, Shift 2Physics

A capillary tube of radius 0.1 mm is partly dipped in water (surface tension 70 dyn/cm and glass water contact angle $≃ 0^{\circ}$ ) with 30 $^{\circ}$ inclined with vertical. The length of water risen in the capillary is ______ cm.

(Take g = 9.8 m/s $^{2}$ )

A
$\frac{82}{5}$
B
$\frac{57}{2}$
C
$\frac{71}{5}$
D
$\frac{68}{5}$

Solution

The vertical height of capillary rise from Jurin's law is $h = \frac{2 T cos θ}{ρ g r}$ . Converting units: $T = 70$ dyn/cm $= 0.07$ N/m, $r = 0.1$ mm $= 1 0^{- 4}$ m, $ρ = 1 0^{3}$ kg/m $^{3}$ , and $cos 0^{\circ} = 1$ .

$h = \frac{2 \times 0.07 \times 1}{1 0 ^{3} \times 9.8 \times 1 0 ^{- 4}} = \frac{0.14}{0.98} \approx 0.1429$ m $= \frac{100}{7}$ cm.

When the tube is inclined at 30 $^{\circ}$ from the vertical, the water still rises to the same vertical height $h$ , but the length along the tube is longer: $ℓ = \frac{h}{cos 3 0 ^{\circ}} = \frac{h}{3 /2} = \frac{2 h}{3}$ .

$ℓ = \frac{2}{3} \times \frac{100}{7} = \frac{200}{7 3} \approx 16.49$ cm $\approx \frac{82}{5}$ cm.

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JEE Main 2025 Apr 7 Shift 2, Physics Q8: Properties of Fluids

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