Physics - NLM Question with Solution | TestHub

Let us assume that ${\mathrm{m}}_2$ move downwards with acceleration 'a' \& ${\mathrm{m}}_1$ moves up the plane with some acceleration. Drawing F.B.D. of ${\mathrm{m}}_1\&{\mathrm{m}}_2$, we have

$$\begin{array}{rlr}& \mathrm{T}-{\mathrm{m}}_1\mathrm{g}\sin 37^{\circ }=\mathrm{ma}& \mathrm{T}-{\mathrm{m}}_1\mathrm{g}\frac{3}{5}={\mathrm{m}}_1\mathrm{a}\end{array}$$

$${\mathrm{m}}_2\mathrm{g}-\mathrm{T}={\mathrm{m}}_2\mathrm{a}$$

From (1) \& (2), $\left({\mathrm{m}}_2-\frac{3{\mathrm{m}}_1}{5}\right)\mathrm{g}=\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{a}$

$$\mathrm{a}=\frac{\left({\mathrm{m}}_2-\frac{3{\mathrm{m}}_1}{5}\right)\mathrm{g}}{{\mathrm{m}}_1+{\mathrm{m}}_2}$$

(A) If ${\mathrm{m}}_1=\frac{{\mathrm{m}}_1}{2}$, then a becomes negative, then direction of a will be opposite to what we have assumed, thus ${\mathrm{m}}_1$ will move up the plane.

(B) If ${\mathrm{m}}_1=\frac{3}{4}{\mathrm{m}}_1$, then a becomes positive, and then as shown, ${\mathrm{m}}_1$ will move up the plane.

(C) If ${\mathrm{m}}_2={\mathrm{m}}_1$, then a becomes positive and ${\mathrm{m}}_2$ will move downwards as shown.

(D) If ${\mathrm{m}}_2=\frac{3}{5}{\mathrm{m}}_1$, then a becomes zero, and system will remain at rest. Hence all options are correct.