${\mathrm{h}=\mathrm{s}=\frac{1}{2} \mathrm{g} \sin \theta \mathrm{t}^{2}}$
${\text { or } \mathrm{t}=\mathrm{t_s}=\frac{1}{\sin \theta} \sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}}$
${\text { and } \mathrm{u}_{\mathrm{s}}=\sqrt{2 \mathrm{gh}}=\sqrt{2(\mathrm{g} \sin \theta) \mathrm{s}}}$
while in case of free fall,
$\mathrm{t}_{\mathrm{F}}=\sqrt{\frac{2 \mathrm{h}}{\mathrm{g}}}$ and $\mathrm{u}_{\mathrm{F}}=\sqrt{2 \mathrm{gh}}$
$\frac{\mathrm{t}_{\mathrm{F}}}{\mathrm{t}_{\mathrm{s}}}=\sin \theta<1 \mathrm{i.e.}, \mathrm{t}_{\mathrm{F}}<\mathrm{t}_{\mathrm{s}}$
i.e., falling body reaches the ground first.
Also $\quad \frac{\mathrm{u}_{\mathrm{F}}}{\mathrm{u}_{\mathrm{s}}}=1,$ i.e., $\mathrm{u}_{\mathrm{F}}=\mathrm{u}_{\mathrm{s}}$
i.e., both reach the ground with same speed (not velocity as for falling body direction is vertical while for sliding body along the plane downwards).
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| Column $I$ | Column $II$ |
| $(A)$ Process $A \rightarrow B$ | $(p)$ Internal energy decreases. |
| $(B)$ Process $B \rightarrow C$ | $(q)$ Internal energy increases. |
| $(C)$ Process $C \rightarrow D$ | $(r)$ Heat is lost. |
| $(D)$ Process $D \rightarrow A$ | $(s)$ Heat is gained. |
| $(t)$ Work is done on the gas. |
