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MCQEasyJEE 2026Viscosity & Stoke's Law

JEE Physics 2026 Question with Solution

A small metallic sphere of diameter 2 mm2 \, \text{mm} and density 10.5 g/cm310.5 \, \text{g/cm}^3 is dropped in glycerine having viscosity 10 Poise10 \, \text{Poise} and density 1.5 g/cm31.5 \, \text{g/cm}^3 respectively. The terminal velocity attained by the sphere is ___ cm/s\text{cm/s}. (π=227\pi = \frac{22}{7} and g=10 m/s2g = 10 \, \text{m/s}^2)__

  • A

    1.51.5

  • B

    2.02.0

  • C

    3.03.0

  • D

    1.01.0

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: diameter of sphere = 2 mm2 \, \text{mm}, density of sphere = 10.5 g/cm310.5 \, \text{g/cm}^3, viscosity of glycerine = 10 Poise10 \, \text{Poise}, density of glycerine = 1.5 g/cm31.5 \, \text{g/cm}^3.

Find: terminal velocity of the sphere.

Use Stokes' law for terminal velocity:

vt=2r2g(ρ−σ)9ηv_t = \frac{2r^2 g (\rho - \sigma)}{9\eta}

Convert all quantities to CGS units as indicated in the solution.

r=1 mm=0.1 cmr = 1 \, \text{mm} = 0.1 \, \text{cm} g=1000 cm/s2g = 1000 \, \text{cm/s}^2 η=10 Poise\eta = 10 \, \text{Poise}

Substitute the values:

vt=2(0.1)2(1000)(10.5−1.5)9(10)v_t = \frac{2(0.1)^2(1000)(10.5 - 1.5)}{9(10)}

Now simplify:

vt=2(0.01)(1000)(9)90=18090v_t = \frac{2(0.01)(1000)(9)}{90} = \frac{180}{90} vt=2 cm/sv_t = 2 \, \text{cm/s}

Therefore, the terminal velocity is 2 cm/s2 \, \text{cm/s} and the correct option is B.

Unit-Consistency Focus

The key step is to keep all quantities in the CGS system because density is given in g/cm3\text{g/cm}^3 and viscosity is given in Poise.

If SI and CGS units are mixed, the numerical answer becomes incorrect.

With

r=0.1 cm,ρ−σ=10.5−1.5=9 g/cm3,g=1000 cm/s2,η=10 Poiser = 0.1 \, \text{cm}, \quad \rho - \sigma = 10.5 - 1.5 = 9 \, \text{g/cm}^3, \quad g = 1000 \, \text{cm/s}^2, \quad \eta = 10 \, \text{Poise}

we get

vt=2×0.01×1000×99×10=18090=2 cm/sv_t = \frac{2 \times 0.01 \times 1000 \times 9}{9 \times 10} = \frac{180}{90} = 2 \, \text{cm/s}

Hence, the answer is 2.02.0, which corresponds to option B.

Common mistakes

  • Using g=10 m/s2g = 10 \, \text{m/s}^2 directly with CGS quantities is incorrect because the formula here is evaluated in the CGS system. Convert it to 1000 cm/s21000 \, \text{cm/s}^2 before substitution.

  • Taking the radius as 1 cm1 \, \text{cm} instead of 1 mm=0.1 cm1 \, \text{mm} = 0.1 \, \text{cm} is wrong. The given diameter is 2 mm2 \, \text{mm}, so the radius is half of that.

  • Using the full density of the sphere instead of the density difference is a conceptual error. In Stokes' law here, use (ρ−σ)(\rho - \sigma), that is 10.5−1.510.5 - 1.5, not 10.510.5 alone.

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