A spherical shell of outer radius R and inner radius R / 2 contains a solid sphere of radius R / 2 (see figure). The density of the material of the solid sphere is and that of the shell is / 2 . What is the average mass

Correct answer: B. 9 / 16 aligned & Average density = Total mass Total volume & =( _1 V_1+ _2 V_2)/(V) & = × (4)/(3) π( R 2 )^3+( )/(2)[(4)/(3) π R^3-(4)/(3) π( R 2 )^3] (4)/(3) π R^3 & = × (4)/(3) π R^3 8 +( )/(2)[(4)/(3) π R^3-(4)/(3) π R^3 8 ] (4)/(3) π R^3 & =( )/(8)+( )/(2)[1-(1)/(8)] & =( )/(8)+( )/(2)[(7)/(8)] & =( )/(8)+(7 )/(16) & =(2 +7 )/(16)=(9 )/(16). aligned

NDA 2024

Exam: NDA
Year: 2024
Subject: General Ability
Chapter: General Science

NDA 2024 · General Ability · General Science

A spherical shell of outer radius R and inner radius R / 2 contains...

A spherical shell of outer radius

R

and inner radius

\mathrm{R} / 2

contains a solid sphere of radius

\mathrm{R} / 2

(see figure). The density of the material of the solid sphere is

\rho

and that of the shell is

\rho / 2

. What is the average mass density of the larger sphere thus formed? A spherical shell of outer radius R and inner radius R / 2 contains a solid sphere of radius R / 2 (see figure). The

A spherical shell of outer radius R and inner radius R / 2 contains a solid sphere of radius R / 2 (see figure). The

Options

A $3 \rho / 4$
B $9 \rho / 16$
C $7 \rho / 8$
D $5 \rho / 8$

Correct answer

B. $9 \rho / 16$

Step-by-step solution

\begin{aligned} & \text{ Average density }=\frac{\text{ Total mass }}{\text{ Total volume }} \\ & =\frac{\rho_1 V_1+\rho_2 V_2}{V} \\ & =\frac{\rho \times \frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)^3+\frac{\rho}{2}\left[\frac{4}{3} \pi \mathrm{R}^3-\frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)^3\right]}{\frac{4}{3} \pi \mathrm{R}^3} \\ & =\frac{\rho \times \frac{4}{3} \pi \frac{\mathrm{R}^3}{8}+\frac{\rho}{2}\left[\frac{4}{3} \pi \mathrm{R}^3-\frac{4}{3} \pi \frac{\mathrm{R}^3}{8}\right]}{\frac{4}{3} \pi \mathrm{R}^3} \\ & =\frac{\rho}{8}+\frac{\rho}{2}\left[1-\frac{1}{8}\right] \\ & =\frac{\rho}{8}+\frac{\rho}{2}\left[\frac{7}{8}\right] \\ & =\frac{\rho}{8}+\frac{7 \rho}{16} \\ & =\frac{2 \rho+7 \rho}{16}=\frac{9 \rho}{16}. \end{aligned}

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