There is a long vertical tube of radius r containing air at atmospheric pressure. A steel ball is held at the mouth of the tube and dropped. The ball has radius r and it just fits inside the tube. The tube wall is perfectly smooth and no air can leak from the tube as the ball falls inside it. The ball falls through half the length of the tube before coming to rest. Assume that wall of the tube is perfectly conducting and temperature of the air inside the tube remains constant. Density of steel is d and atmospheric pressure is P₀ and take L >> r. Take air to be an ideal gas. At what depth from the top of the tube the ball will be in equilibrium?

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$L [1 - \frac{1}{4 \ln (2)}]$

$L [1 - \frac{1}{2 \ln (2)}]$

$L [2 - \frac{3}{2 \ln (2)}]$

$L [\frac{1}{2 \ln (2)}]$

answer is B.

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Detailed Solution

$V_{0} = AL; V = A (L - x) [A = {πr}^{2}]$

For isothermal compression of air
PV = P₀V₀
$PA [L - x] = P_{0} AL \Rightarrow P = P_{0} (\frac{L}{L - x})$ . Work done by the gas on the ball $= - P_{0} V_{0} \ln (\frac{L}{L - x})$
Work done by the gravity on the ball = mgx
If the ball comes to rest
$mgx - P_{0} V_{0} \ln (\frac{L}{L - x}) = 0 \begin{array}{l} Given x = \frac{L}{2} \\ mg \frac{L}{2} = P_{0} ALln (\frac{L}{L - \frac{L}{2}}) \\ ∴ \frac{4}{3} {πr}^{3} \cdot d \cdot g \cdot \frac{1}{2} = P_{0} {πr}^{2} \ln (2) \\ ∴ r = \frac{3}{2} \frac{P_{0} \ln (2)}{d \cdot g .} \end{array}$
In equilibrium,
$\begin{array}{l} {Pπr}^{2} = mg \\ P_{0} (\frac{L}{L - x}) {πr}^{2} = \frac{4}{3} {πr}^{3} \cdot d \cdot g \end{array} \begin{array}{l} L - x = \frac{3 P_{0} L}{4 rdg} = \frac{3}{4} L \cdot \frac{2}{3 \ln (2)} = \frac{L}{2 \ln (2)} \\ x = L [1 - \frac{1}{2 \ln (2)}] \end{array}$