Sankalp NEET Full Test-3 Question-39 Solution

Question: 39. A sphere of radius R is cut from a larger solid sphere or radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

(1) \frac{7}{8}

(2) \frac{7}{40}

(3) \frac{7}{57}

(4) \frac{7}{64}

Answer: Option (3)

Explanation:

The moment of inertia of a solid sphere of mass M and

adius R about any diameter is given by:

I = \frac{2}{5}MR^2

Let the density of the material be uniform and equal to \rho.

Mass of the larger sphere of radius 2R:

M_L = \frac{4}{3}\pi (2R)^3 \rho = \frac{32}{3}\pi R^3 \rho

Mass of the smaller removed sphere of radius R:

M_S = \frac{4}{3}\pi R^3 \rho

The smaller sphere is removed from a distance R from the Y-axis,

so we use the parallel axis theorem.

Moment of inertia of the smaller sphere about its own center:

I_{SC} = \frac{2}{5}M_S R^2

Moment of inertia of the smaller sphere about the Y-axis:

I_S = I_{SC} + M_S R^2 = \frac{2}{5}M_S R^2 + M_S R^2 = \frac{7}{5}M_S R^2

Moment of inertia of the full large sphere about the Y-axis:

I_L = \frac{2}{5}M_L (2R)^2 = \frac{8}{5}M_L R^2

The moment of inertia of the remaining part is:

I_{rem} = I_L - I_S

Substitute values of M_L and M_S:

I_S = \frac{7}{5}\left(\frac{4}{3}\pi R^3 \rho\right)R^2 = \frac{28}{15}\pi \rho R^5 I_L = \frac{8}{5}\left(\frac{32}{3}\pi R^3 \rho\right)R^2 = \frac{256}{15}\pi \rho R^5 I_{rem} = \frac{256}{15}\pi \rho R^5 - \frac{28}{15}\pi \rho R^5 = \frac{228}{15}\pi \rho R^5

Now the required ratio is:

\frac{I_S}{I_{rem}} = \frac{\frac{28}{15}\pi \rho R^5}{\frac{228}{15}\pi \rho R^5} = \frac{28}{228} = \frac{7}{57}

Hence, the correct answer is Option (3).