1. Given data

• Mass on left incline: $M = 10\,\text{kg}$
• Mass on right incline: $m = ?$
• Coefficient of friction: $\mu = 0.25$
• Acceleration of $M$: $a = 2\,\text{m/s}^2$ downward along its plane
• $g = 10\,\text{m/s}^2$
• $\tan 37^\circ = \frac{3}{4}$

From the figure of the standard doubly inclined plane setup, the inclinations are $37^\circ$ and $53^\circ$. Using $\tan 37^\circ=\frac{3}{4}$, we get

and hence

2. Decide friction directions

Since $M$ moves down its incline, the other block $m$ moves up its incline.

So friction opposes motion:

• On $M$: friction acts up the plane.
• On $m$: friction acts down the plane.

3. Equation for block $M=10\,\text{kg}$

Take motion of $M$ down the plane as positive.

Forces along the plane:

• Down the plane: $Mg\sin 53^\circ$
• Up the plane: tension $T$
• Up the plane: friction $f_M = \mu Mg\cos 53^\circ$

Thus,

Substitute values:

4. Equation for block $m$

Block $m$ moves up its plane, so take upward along plane as positive.

Forces along the plane:

• Up the plane: tension $T$
• Down the plane: $mg\sin 37^\circ$
• Down the plane: friction $f_m = \mu mg\cos 37^\circ$

So,

Substitute values:

5. Check with options

Thus the correct option is:

6. Comparison with stored answer

Stored correct answer: A

Our derived answer is also A, so they agree.