Physics Circle home Student Login

Home›Chapters›Viscosity PYQs›Terminal Velocity›Question 10

JEE Main 2026 (23 Jan, Shift 2) · Terminal Velocity

Terminal Velocity question from JEE Main 2026 (23 Jan, Shift 2)

Q10JEE Main 2026 · 23 JAN · SHIFT 2Medium

Terminal Velocity

A small metallic sphere of diameter 2 mm and density 10.5 g/cm310.5\text{ g/cm}^{3} is dropped in glycerine having viscosity 10 Poise and density 1.5 g/cm31.5\text{ g/cm}^{3} respectively. The terminal velocity attained by the sphere is ‾ cm/s\underline{\hspace{3em}}\text{ cm/s}.

(π=227\left(\pi = \frac{22}{7}\right. and g=10 m/s2)g = 10\text{ m/s}^{2})

Show answer and solution

Answer: (d)

  1. vt=2r2(ρ−σ)g9ηv_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta} (CGS units).
  2. r=0.1r = 0.1 cm, ρ−σ=9 g/cm3\rho - \sigma = 9\ \text{g/cm}^3, g=1000 cm/s2g = 1000\ \text{cm/s}^2, η=10\eta = 10 P
  3. vt=2×0.01×9×10009×10=2v_t = \dfrac{2 \times 0.01 \times 9 \times 1000}{9 \times 10} = 2 cm/s
  4. So the answer is (d) 2.0.
Report a mistake

More questions on Terminal Velocity