JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 27)
In the reported figure, heat energy absorbed by a system in going through a cyclic process is ___________ $$\pi$$J.
_20th_July_Morning_Shift_en_27_1.png)
Answer
100
Explanation
Consider the given diagram,
_20th_July_Morning_Shift_en_27_2.png)
Here, r1 = 10 $$\times$$ 103 m and r2 = 10 $$\times$$ 10-3 m
We know that for a complete cyclic process, change in internal energy, ($$\Delta$$U) = 0 .... (i)
According to 1st law of thermodynamics,
$$\Delta$$Q = $$\Delta$$U + W ..... (ii)
From Eqs. (i) and (ii), we get
$$\Rightarrow$$ $$\Delta$$Q = 0 + W
$$\Rightarrow$$ $$\Delta$$Q = W .... (iii)
$$\because$$ W = Area = $$\pi$$r1r2
= $$\pi$$ $$\times$$ (10 $$\times$$ 103) $$\times$$ (10 $$\times$$ 10$$-$$3)
W = 100 $$\pi$$ J .... (iv)
From Eqs. (iii) and (iv), we get
$$\Delta$$Q = 100 $$\pi$$ J
Here, r1 = 10 $$\times$$ 103 m and r2 = 10 $$\times$$ 10-3 m
We know that for a complete cyclic process, change in internal energy, ($$\Delta$$U) = 0 .... (i)
According to 1st law of thermodynamics,
$$\Delta$$Q = $$\Delta$$U + W ..... (ii)
From Eqs. (i) and (ii), we get
$$\Rightarrow$$ $$\Delta$$Q = 0 + W
$$\Rightarrow$$ $$\Delta$$Q = W .... (iii)
$$\because$$ W = Area = $$\pi$$r1r2
= $$\pi$$ $$\times$$ (10 $$\times$$ 103) $$\times$$ (10 $$\times$$ 10$$-$$3)
W = 100 $$\pi$$ J .... (iv)
From Eqs. (iii) and (iv), we get
$$\Delta$$Q = 100 $$\pi$$ J
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