Start free

Physics · Mechanical Properties of Matter

JEE Main 2024 — 9 April, Shift 2 — Question 45

A spherical ball of radius 1×10−4 m1 \times 10^{-4} \mathrm{~m} and density 10510^{5} kg/m3\mathrm{kg} / \mathrm{m}^{3} falls freely under gravity through a distance hh before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of hh is approximately : (The coefficient of viscosity of water is 9.8×10−69.8 \times 10^{-6} Ns/m2)\left.\mathrm{N} \mathrm{s} / \mathrm{m}^{2}\right)

  1. Option A:

    2296 m

  2. Option B:

    2249 m

  3. Option C:

    2518 m

    Correct
  4. Option D:

    2396 m

Answer: C

Step-by-step solution

Since the speed does not change after entering water, the entry speed equals the terminal speed in water.

Terminal velocity (Stokes’ law):

vt=29r2(ρ−ρw)gη≈29(10−4)2(105)×9.89.8×10−6≈222 m s−1v_t=\frac{2}{9}\frac{r^2(\rho-\rho_w)g}{\eta} \approx \frac{2}{9}\frac{(10^{-4})^2(10^{5})\times9.8}{9.8\times10^{-6}} \approx 222\ \text{m s}^{-1}

Speed after free fall:

v2=2gh⇒h=vt22g≈(222)22×9.8≈2.52×103 mv^2=2gh \Rightarrow h=\frac{v_t^2}{2g} \approx \frac{(222)^2}{2\times9.8} \approx 2.52\times10^{3}\ \text{m}   ⟹  h≈2518 m\implies h\approx 2518\ \text{m}

Answer key and solution verified before publishing.

Practise Mechanical Properties of Matter

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2024
Paper
9 April, Shift 2
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Viscosity

More Mechanical Properties of Matter questions from this paper

A spherical ball of radius 1 × 10 -4 m and density 10 5 kg / m 3… | JEE Main 2024 PYQ with Solution · DhiX AI