KVPY SA Exam 05-Nov-2017 Question Paper

Question : 30 of 80

Marks: +1, -0

A long cylindrical pipe of radius 20 cm is closed at its upper end and has an airtight piston of negligible mass as shown. When a 50 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 – ∆T, the piston moves back to its original position. Then ∆T/T is close to (Assuming air to be an ideal gas, g = 10 m/s

^{2}

, atmospheric pressure is 10

^{5 }

Pascal),

0.01
0.02
0.04
0.09

Validate

Solution:

Initially pressure =

P_0

When 50 kg mass is suspended
pressure =

P_0-{mg}/A

(temp = constant)

P_0V_i=(V_0-{mg}/A)V_\f

(P_0-{mg}/A)V_\f = nRT

(P_0-{mg}/A)V_\i = nR(T-∆T)

V_\f/V_\i=T/{T-∆T}

{T-∆T}/T=V_\i/V_\f

1-{∆T}/T=(P_0-{mg}/A)1/P_0

{∆T}/T={mg}/{AP_0}

={50 ×10}/{ 3.14× (0.2) ^2×10^5}

{500}/{12.56× 10^3}

5/125.6=1/25=0.04

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