Physics - NLM Question with Solution | TestHub

(1)

Let distance travelled by block be s

Now ${\mathrm{v}}^2-{\mathrm{u}}^2=2{\mathrm{a}}_1\mathrm{s}$

${\mathrm{O}}^2-{\mathrm{v}}_1{}^2=2\times (-\mu \mathrm{g})\mathrm{s}$

$\Rightarrow {\mathrm{v}}_1=\sqrt{2\mu \mathrm{gs}}$ ........(i)

Here ${\mathrm{a}}_2=\mathrm{g}\sin \theta +\mu \mathrm{g}\cos \theta$

Now, ${\mathrm{v}}^2-{\mathrm{u}}^2=2{\mathrm{a}}_2\mathrm{s}$

${\mathrm{O}}^2-{\mathrm{v}}_2{}^2=2[-(g\sin \theta +\mu g\cos \theta )]\mathrm{s}$

${\mathrm{v}}_2=\sqrt{2\mathrm{s}[\mathrm{g}\sin \theta +\mathrm{ug}\cos \theta ]}$ ............(ii)

Here ${\mathrm{a}}_3=-[\mathrm{g}\sin \theta +\mu \mathrm{g}\cos \theta ]$

Now, ${\mathrm{v}}^2-{\mathrm{u}}^2=2{\mathrm{a}}_3\mathrm{s}$

$\mathrm{O}-{\mathrm{v}}_3=2[-(\mathrm{g}\sin \theta -\mu \mathrm{g}\cos \theta )]\mathrm{s}$

${\mathrm{v}}_3=\sqrt{2\mathrm{s}[\mathrm{g}\sin \theta -\mathrm{ug}\cos \theta ]}$ ...........(iii)

From (i), (ii) & (iii) we can see that ${\mathrm{v}}_3<{\mathrm{v}}_1<{\mathrm{v}}_2$. So option (c) is correct.