Search any question & find its solution
Question:
Answered & Verified by Expert
A long cylindrical pipe of radius $20 \mathrm{~cm}$ is closed atits upper end and has an airtight piston of negligible mass as shown. When a $50 \mathrm{Kg}$ mass is attached to the other end of the pistion, it moves down. If the air in the enclosure is cooled from temperature $T$ to $T-\Delta T$, the piston moves back to its original position. Then $\Delta T / T$ is close to (Assuming air to be an ideal gas, $g=10 \mathrm{~m} / \mathrm{s}^{2}$, atmospheric pressure is $10^{5}$ Pascal),
Options:
Solution:
1304 Upvotes
Verified Answer
The correct answer is:
$0.04$
Initially pressure $=P_{0}$ When $50 \mathrm{~kg}$ mass is suspended
$\begin{array}{l}
\text { pressure }=P_{0}-\frac{m g}{A} \\
\text { (temp }=\text { constant }) \\
P_{0} V_{1}=\left(V_{0}-\frac{m g}{A}\right) V_{f} \\
\left(P_{0}-\frac{m g}{A}\right) V_{f}=n R T \\
\left(P_{0}-\frac{m g}{A}\right) V_{i}=n R(T-\Delta T) \\
=\frac{V_{f}}{V_{i}}=\frac{T}{T-\Delta T} \\
\frac{T-\Delta T}{T}=\frac{V_{i}}{V_{f}} \\
=\frac{\Delta T}{T}=\left(P_{0}-\frac{m g}{A}\right) \frac{1}{P_{0}} \\
\end{array}$
$\begin{aligned} \frac{\Delta \mathrm{T}}{\mathrm{T}} &=\frac{\mathrm{mg}}{\mathrm{AP}_{0}} \\ &=\frac{50 \times 10}{3.14 \times(0.2)^{2} \times 10^{5}} \\ &=\frac{500}{12.56 \times 10^{3}} \\ &=\frac{5}{125.6}=\frac{1}{25}=0.04 \end{aligned}$
$\begin{array}{l}
\text { pressure }=P_{0}-\frac{m g}{A} \\
\text { (temp }=\text { constant }) \\
P_{0} V_{1}=\left(V_{0}-\frac{m g}{A}\right) V_{f} \\
\left(P_{0}-\frac{m g}{A}\right) V_{f}=n R T \\
\left(P_{0}-\frac{m g}{A}\right) V_{i}=n R(T-\Delta T) \\
=\frac{V_{f}}{V_{i}}=\frac{T}{T-\Delta T} \\
\frac{T-\Delta T}{T}=\frac{V_{i}}{V_{f}} \\
=\frac{\Delta T}{T}=\left(P_{0}-\frac{m g}{A}\right) \frac{1}{P_{0}} \\
\end{array}$
$\begin{aligned} \frac{\Delta \mathrm{T}}{\mathrm{T}} &=\frac{\mathrm{mg}}{\mathrm{AP}_{0}} \\ &=\frac{50 \times 10}{3.14 \times(0.2)^{2} \times 10^{5}} \\ &=\frac{500}{12.56 \times 10^{3}} \\ &=\frac{5}{125.6}=\frac{1}{25}=0.04 \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.