A solid steel ball of diameter 3.6 mm acquired terminal velocity 2.45 × 10 -2 m / s while falling under gravity through an oil of density 925 kg m -3 . Take density of steel as 7825 kg m -3 and g as 9.8 m / s 2 . The viscosity of the oil in SI unit is

Physics · Fluid Mechanics

JEE Main 2025 — 3 April, Evening Shift — Question 63

A solid steel ball of diameter 3.6 mm acquired terminal velocity $2.45 \times 10^{-2} \mathrm{~m} / \mathrm{s}$ while falling under gravity through an oil of density $925 \mathrm{~kg} \mathrm{~m}^{-3}$ . Take density of steel as $7825 \mathrm{~kg} \mathrm{~m}^{-3}$ and $g$ as $9.8 \mathrm{~m} / \mathrm{s}^{2}$ . The viscosity of the oil in SI unit is

Option A:
2.18
Option B:
1.68
Option C:
2.38
Option D:
1.99
Correct

Answer: D

Step-by-step solution

$v=\frac{2}{9} \frac{r^{2} g(\sigma-\rho)}{\eta}$

\eta=\frac{2}{9} \times \frac{1.8 \times 1.8 \times 10^{-6} \times 10}{2.45 \times 10^{-2}}\{7825-925\}

\approx 2

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Evening Shift
Subject: Physics
Chapter: Fluid Mechanics
Topic: Bernoulli's Equation and its Applications