MCQ
The heat absorbed by a system in going through the given cyclic process is:


- ✓$61.6 \mathrm{~J}$
- B$431.2 \mathrm{~J}$
- C$616 \mathrm{~J}$
- D$19.6 \mathrm{~J}$

$\Delta \mathrm{Q}=\mathrm{W}=\text { area of } \mathrm{P}-\mathrm{V} \text { curve. }$
$=\pi \times\left(140 \times 10^3 \mathrm{~Pa}\right) \times\left(140 \times 10^{-6} \mathrm{~m}^3\right)$
$\Delta \mathrm{Q}=61.6 \mathrm{~J}$
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|
Column $-I$ Angle of projection |
Column $-II$ |
| $A.$ $\theta \, = \,{45^o}$ | $1.$ $\frac{{{K_h}}}{{{K_i}}} = \frac{1}{4}$ |
| $B.$ $\theta \, = \,{60^o}$ | $2.$ $\frac{{g{T^2}}}{R} = 8$ |
| $C.$ $\theta \, = \,{30^o}$ | $3.$ $\frac{R}{H} = 4\sqrt 3 $ |
| $D.$ $\theta \, = \,{\tan ^{ - 1}}\,4$ | $4.$ $\frac{R}{H} = 4$ |
$K_i :$ initial kinetic energy
$K_h :$ kinetic energy at the highest point
