MHT CET · Physics · Mechanical Properties of Fluids
A lead sphere of mass ' \(m\) ' falls in viscous liquid with terminal velocity \(\mathrm{V}_0\). Another lead sphere of mass ' 8 m ' but of same material will fall through the same liquid with terminal velocity
- A \(\mathrm{V}_0\)
- B \(8 \mathrm{~V}_0\)
- C \(4 \mathrm{~V}_0\)
- D \(64 \mathrm{~V}_0\)
Answer & Solution
Correct Answer
(C) \(4 \mathrm{~V}_0\)
Step-by-step Solution
Detailed explanation
The formula for terminal velocity for a spherical body
\(V=\frac{2 g(\rho-\sigma) r^2}{2 \eta}\)
If mass is made 8 times, material being same volume also becomes 8 times \(\because\) density is same. So, radius of sphere becomes twice.
\(\therefore \quad\) The new terminal velocity
\(\begin{aligned}
& \frac{V_0}{V_2}=\left(\frac{r_1}{r_2}\right)^2 \\
& \frac{V_0}{V_2}=\left(\frac{I}{2}\right)^2 \\
& V_2=4 V_\theta
\end{aligned}\)
\(V=\frac{2 g(\rho-\sigma) r^2}{2 \eta}\)
If mass is made 8 times, material being same volume also becomes 8 times \(\because\) density is same. So, radius of sphere becomes twice.
\(\therefore \quad\) The new terminal velocity
\(\begin{aligned}
& \frac{V_0}{V_2}=\left(\frac{r_1}{r_2}\right)^2 \\
& \frac{V_0}{V_2}=\left(\frac{I}{2}\right)^2 \\
& V_2=4 V_\theta
\end{aligned}\)
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