Answer
Verified
355k+ views
Hint: Moment of Inertia (M.I.) of the solid sphere along its diameter is $I = \dfrac{{2M{R^2}}}{5}$.As this sphere is recast into 8 smaller spheres hence the mass of smaller spheres is \[\dfrac{M}{8}\]. As the material of both the materials is the same thus density remains the same.
Formula used
Moment of Inertia (M.I.) of the solid sphere along its diameter is $I = \dfrac{{2M{R^2}}}{5}$
$\rho = \dfrac{M}{V}$ where$\rho $ is the density, $M$ is the mass, $V$ is the volume.
$V = \dfrac{{4\pi {R^3}}}{3}$ Where $R$is the radius and $V$ is the volume.
Complete step by step solution:
Let Mass and radius of the bigger sphere be $M$ and$R$.
So the moment of inertia is $I = \dfrac{{2M{R^2}}}{5}$
As this sphere is recast into 8 smaller spheres hence the mass of smaller spheres is \[\dfrac{M}{8}\] and let radius be $r$ .
As the material of both the materials is the same thus density remains the same from this we can calculate the radius r of the new smaller sphere formed.
From,
($\rho $ is the density, $M$ is the mass, $V$ is the volume)
$\rho = \dfrac{M}{V}$
Volume of a sphere of radius R is \[ {V_R} = \dfrac{{4\pi {R^3}}}{3}\]
For a sphere of mass $M$ and radius$R$,
$\rho = \dfrac{M}{{\dfrac{{4\pi {R^3}}}{3}}}$
Volume of a sphere of radius r is \[ {V_r}= \dfrac{{4\pi {r^3}}}{3}\]
For a sphere of mass \[\dfrac{M}{8}\] and radius $r$
$\rho = \dfrac{{\dfrac{M}{8}}}{{\dfrac{{4\pi {r^3}}}{3}}}$
From the above two equation and as both spheres have density we can assert that,
$\rho = \dfrac{M}{{\dfrac{{4\pi {R^3}}}{3}}} = \dfrac{{\dfrac{M}{8}}}{{\dfrac{{4\pi {r^3}}}{3}}}$
$ \Rightarrow \dfrac{M}{{\dfrac{{4\pi {R^3}}}{3}}} = \dfrac{{\dfrac{M}{8}}}{{\dfrac{{4\pi {r^3}}}{3}}}$
$ \Rightarrow {r^3} = \dfrac{{{R^3}}}{8}$
$ \Rightarrow r = \dfrac{R}{2}$
As the moment of inertia of solid sphere along its diameter is $I = \dfrac{{2M{R^2}}}{5}$
So the moment of inertia of the smaller sphere whose mass (M) is \[\dfrac{M}{8}\] and radius(R) is$\dfrac{R}{2}$
${I_{smaller-sphere}}= \dfrac{{2 \times \dfrac{M}{8} \times {{(\dfrac{R}{2})}^2}}}{5}$
$ \Rightarrow {I_{smaller-sphere}}= \dfrac{{2M{R^2}}}{{5 \times 32}}$
As $I = \dfrac{{2M{R^2}}}{5}$
$ \Rightarrow {I_{smaller-sphere}} = \dfrac{I}{{32}}$
Hence the answer to this question is (D) $\dfrac{I}{{32}}$
Note:
Always remember that $I = \dfrac{{2M{R^2}}}{5}$is the moment of inertia of solid sphere along its diameter and not $I = \dfrac{{2M{R^2}}}{3}$ which is the moment of inertia of hollow sphere along its diameter also be careful about the mentioned axis about which the moment of inertia is being written these small checks while attempting a question can save you from silly mistakes in the exam.
Formula used
Moment of Inertia (M.I.) of the solid sphere along its diameter is $I = \dfrac{{2M{R^2}}}{5}$
$\rho = \dfrac{M}{V}$ where$\rho $ is the density, $M$ is the mass, $V$ is the volume.
$V = \dfrac{{4\pi {R^3}}}{3}$ Where $R$is the radius and $V$ is the volume.
Complete step by step solution:
Let Mass and radius of the bigger sphere be $M$ and$R$.
So the moment of inertia is $I = \dfrac{{2M{R^2}}}{5}$
As this sphere is recast into 8 smaller spheres hence the mass of smaller spheres is \[\dfrac{M}{8}\] and let radius be $r$ .
As the material of both the materials is the same thus density remains the same from this we can calculate the radius r of the new smaller sphere formed.
From,
($\rho $ is the density, $M$ is the mass, $V$ is the volume)
$\rho = \dfrac{M}{V}$
Volume of a sphere of radius R is \[ {V_R} = \dfrac{{4\pi {R^3}}}{3}\]
For a sphere of mass $M$ and radius$R$,
$\rho = \dfrac{M}{{\dfrac{{4\pi {R^3}}}{3}}}$
Volume of a sphere of radius r is \[ {V_r}= \dfrac{{4\pi {r^3}}}{3}\]
For a sphere of mass \[\dfrac{M}{8}\] and radius $r$
$\rho = \dfrac{{\dfrac{M}{8}}}{{\dfrac{{4\pi {r^3}}}{3}}}$
From the above two equation and as both spheres have density we can assert that,
$\rho = \dfrac{M}{{\dfrac{{4\pi {R^3}}}{3}}} = \dfrac{{\dfrac{M}{8}}}{{\dfrac{{4\pi {r^3}}}{3}}}$
$ \Rightarrow \dfrac{M}{{\dfrac{{4\pi {R^3}}}{3}}} = \dfrac{{\dfrac{M}{8}}}{{\dfrac{{4\pi {r^3}}}{3}}}$
$ \Rightarrow {r^3} = \dfrac{{{R^3}}}{8}$
$ \Rightarrow r = \dfrac{R}{2}$
As the moment of inertia of solid sphere along its diameter is $I = \dfrac{{2M{R^2}}}{5}$
So the moment of inertia of the smaller sphere whose mass (M) is \[\dfrac{M}{8}\] and radius(R) is$\dfrac{R}{2}$
${I_{smaller-sphere}}= \dfrac{{2 \times \dfrac{M}{8} \times {{(\dfrac{R}{2})}^2}}}{5}$
$ \Rightarrow {I_{smaller-sphere}}= \dfrac{{2M{R^2}}}{{5 \times 32}}$
As $I = \dfrac{{2M{R^2}}}{5}$
$ \Rightarrow {I_{smaller-sphere}} = \dfrac{I}{{32}}$
Hence the answer to this question is (D) $\dfrac{I}{{32}}$
Note:
Always remember that $I = \dfrac{{2M{R^2}}}{5}$is the moment of inertia of solid sphere along its diameter and not $I = \dfrac{{2M{R^2}}}{3}$ which is the moment of inertia of hollow sphere along its diameter also be careful about the mentioned axis about which the moment of inertia is being written these small checks while attempting a question can save you from silly mistakes in the exam.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
According to classical free electron theory A There class 11 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main