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I tried to solve this question through the first law Since it is cyclic ∆U over a complete cycle is 0 Hence ∆Q=∆W=Area under graph. Now since in first circle process is going clockwise work will be positive, in smaller circle work will be negative Hence w=area of bigger circle-area of smaller circle. On doing this i am getting 3πJ but the answer is 2πJ . Can you pls tell me where i am wrong

I tried to solve this question through the first law 
Since it is cyclic ∆U over a complete cycle is 0
Hence ∆Q=∆W=Area under graph. Now since in first circle process is going clockwise work will be positive, in smaller circle work will be negative
Hence w=area of bigger circle-area of smaller circle. On doing this i am getting 3πJ but the answer is 2πJ . Can you pls tell me where i am wrong

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Grade:11

1 Answers

Mahima Kanawat
1010 Points
5 years ago
Dear student 
Your calculations for area is wrong
For big circle... Area = pi * 3/2 *3/2 = 9pi/4
For small circle... Area= pi*1/2*1/2 = pi/4
Difference is 8pi/4 = 2pi
WITH REGARDS 
MAHIMA 
ASKIITIANS FORUM EXPERT 

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