A) 540 cal
B) 40 cal
C) 500 cal
D) zero
Correct Answer: C
Solution :
Work done \[[\exp (7.62)=2038.6,k=1.4\times {{10}^{-24}}J/k]\] \[2037.6\times {{10}^{-3}}A\] \[203.76\times {{10}^{-3}}A\] \[20.376\times {{10}^{-3}}A\] From first law of thermodynamics \[2.0376\times {{10}^{-3}}A\] \[\mu F\]You need to login to perform this action.
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