A round uniform body of radius R, mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle \theta with the horizontal. Then its acceleration is

g\,\sin \theta \over 1 + I/MR^2

g\,\sin \theta \over 1 - MR^2/I

g\,\sin \theta \over 1 + MR^2/I

g\,\sin \theta \over 1 - I/MR^2

Rotational Motion PYQ — JEE Main 2007 · Shift 0 · Q95 with Solution

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Correct answer: B

A body of mass $M$ rolls down an incline of angle $\theta$ without slipping.

Along the plane, the forces are:

So the translational equation is $Mg\sin\theta - f = Ma \quad ...(1)$

The friction provides torque for rotation: $fR = I\alpha \quad ...(2)$

Since the body rolls without slipping, $a = \alpha R \quad \Rightarrow \quad \alpha = \frac{a}{R}$

Substitute into (2): $fR = I\frac{a}{R}$ $f = \frac{Ia}{R^2} \quad ...(3)$

Using (3) in (1): $Mg\sin\theta - \frac{Ia}{R^2} = Ma$

Bring the $a$ terms together: $Mg\sin\theta = a\left(M + \frac{I}{R^2}\right)$

Therefore, $a = \frac{Mg\sin\theta}{M + I/R^2}$

Divide numerator and denominator by $M$ : $a = \frac{g\sin\theta}{1 + I/(MR^2)}$

This corresponds to: $\boxed{\frac{g\sin\theta}{1 + I/(MR^2)}}$

So the correct option is B.