MHT CET · Physics · Mechanical Properties of Fluids
A metal ball of radius \(9 \times 10^{-4} \mathrm{~m}\) and density \(10^4 \mathrm{~kg} / \mathrm{m}^3\) falls freely under gravity through a distance 'h' and erfters a tank of water. Considering that the metal ball has constant velocity, the value of h is [coefficient of viscosity of water \(=8.1 \times 10^{-4} \mathrm{pa}-\mathrm{s}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) density of water \(\left.=10^3 \mathrm{~kg} / \mathrm{m}^3\right]\)
- A 20 m
- B 18 m
- C 15 m
- D 12 m
Answer & Solution
Correct Answer
(A) 20 m
Step-by-step Solution
Detailed explanation
Velocity of the ball when the ball hits the surface of water
\(v=\sqrt{2 g h}\)
Terminal velocity of the ball,
\(v=\frac{2}{9} r^2 g \frac{(P-\sigma)}{\eta}\)
Equating (i) and (ii),
\(\sqrt{2 g h}=\frac{2}{9} \frac{r^2 g}{\eta}(P-\sigma)\)
Solving for h ,
\(h=\frac{2}{81} r^4\left(\frac{P-\sigma}{\eta}\right)^2 g\)
Substituting the given values in the above equation, we have
\(\mathrm{h}=\frac{2}{81} \times\left(9 \times 10^{-4}\right)^4 \times\left(\frac{10^4-10^3}{8.1 \times 10^{-4}}\right)^2 \mathrm{~g}=20 \mathrm{~m}\)
\(v=\sqrt{2 g h}\)
Terminal velocity of the ball,
\(v=\frac{2}{9} r^2 g \frac{(P-\sigma)}{\eta}\)
Equating (i) and (ii),
\(\sqrt{2 g h}=\frac{2}{9} \frac{r^2 g}{\eta}(P-\sigma)\)
Solving for h ,
\(h=\frac{2}{81} r^4\left(\frac{P-\sigma}{\eta}\right)^2 g\)
Substituting the given values in the above equation, we have
\(\mathrm{h}=\frac{2}{81} \times\left(9 \times 10^{-4}\right)^4 \times\left(\frac{10^4-10^3}{8.1 \times 10^{-4}}\right)^2 \mathrm{~g}=20 \mathrm{~m}\)
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