JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 1)
For the given cyclic process CAB as shown for a gas, the work done is :
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1 J
10 J
5 J
30 J
Explanation
Since P$$-$$V indicator diagram is given, so work done by gas is area under the cyclic diagram.
$$ \therefore $$ $$\Delta $$W = Work done by gas = $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ 5 J
= 10 J
$$ \therefore $$ $$\Delta $$W = Work done by gas = $${1 \over 2}$$ $$ \times $$ 4 $$ \times $$ 5 J
= 10 J
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