MHT CET · Physics · Mechanical Properties of Fluids
A ball rises to surface at a constant velocity in liquid whose density is 4 times greater than that of the material of the ball. The ratio of the force of friction acting on the rising ball and its weight is
- A \(3: 1\)
- B \(4: 1\)
- C \(1: 3\)
- D \(1: 4\)
Answer & Solution
Correct Answer
(A) \(3: 1\)
Step-by-step Solution
Detailed explanation
Frictional force \(=\) Viscous Force \(=V\left(\rho_1-\rho_2\right) g\)
Weight \(=\mathrm{mg}=\mathrm{V} \rho_1 \mathrm{~g}\)
Where \(\rho_1\) is the density of ball and \(\rho_2\) is density of liquid.
\(\begin{array}{ll}
& \rho_2=4 \rho_1 \\
\therefore \quad & V\left(\rho_1-4 \rho_1\right) g=V\left(-3 \rho_1\right) g . \\
& \text { Frictional force }=3 V\left(\rho_1 g\right) \\
\therefore \quad & \frac{\text { Frictional force }}{\text { Weight }}=\frac{3}{1}=3: 1 \text { ratio }
\end{array}\)
Weight \(=\mathrm{mg}=\mathrm{V} \rho_1 \mathrm{~g}\)
Where \(\rho_1\) is the density of ball and \(\rho_2\) is density of liquid.
\(\begin{array}{ll}
& \rho_2=4 \rho_1 \\
\therefore \quad & V\left(\rho_1-4 \rho_1\right) g=V\left(-3 \rho_1\right) g . \\
& \text { Frictional force }=3 V\left(\rho_1 g\right) \\
\therefore \quad & \frac{\text { Frictional force }}{\text { Weight }}=\frac{3}{1}=3: 1 \text { ratio }
\end{array}\)
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