Answer
64.8k+ views
Hint First of all get the mass $dm$ of the disc by using the expression:
$dm = \left( {\dfrac{m}{{{V_H}}}} \right) \times {V_D}$
where, $m$ is the mass of hemisphere
${V_H}$ is the volume of hemisphere
${V_D}$ is the volume of disc
Now, use the formula
${I_{yy'}} = {I_{cm}} + m{r^2}$
Then, integrate it with the limit from $0$ to $R$
After solving we will get the value of ${I_{yy'}}$ then, use the value of $\rho = \dfrac{{3m}}{{2\pi {R^3}}}$
Then, if $I$ displace them parallel to the axis and again join them to complete a sphere
$2I = \dfrac{2}{5}m{R^2}$
Complete Step by Step Solution
Suppose that hemisphere is constructed by small discs
Mass $dm$of disc= $dm = \left( {\dfrac{m}{{{V_H}}}} \right) \times {V_D}$
$
\therefore dm = \left( {\dfrac{m}{{\dfrac{2}{3}\pi {R^3}}}} \right)\left( {\pi {y^2}dx} \right) \\
dm = \rho \left( {\pi {y^2}dx} \right) \\
$
Because, $\rho = \dfrac{m}{{\dfrac{2}{3}\pi {R^3}}}$
According to parallel and perpendicular axis theorem we get
${I_{yy'}} = {I_{cm}} + m{r^2}$
$\therefore d{I_{yy'}} = (dm)\dfrac{{{y^2}}}{4} + (dm){x^2}$
Now, integrating both sides
$
{I_{yy'}} = \int\limits_0^R {(dm)\dfrac{{{y^2}}}{4} + } \int\limits_0^R {dm} {x^2} \\
\Rightarrow \int\limits_0^R {(\rho \pi {y^2}dx)\dfrac{{{y^2}}}{4} + } \int\limits_0^R {(\rho \pi {y^2}dx){x^2}} \\
$
Now, put the limits
$
{I_{yy'}} = \dfrac{{\rho \pi }}{4}\left[ {\dfrac{{{R^5}}}{5} + {R^5} - 2{R^2}\dfrac{{{R^3}}}{3}} \right] + \left[ {\rho \pi {R^5}\left( {\dfrac{{{R^3}}}{3}} \right) - \rho \pi \left( {\dfrac{{{R^5}}}{5}} \right)} \right] \\
= \rho \pi {R^5}\left[ {\dfrac{1}{4} + \dfrac{1}{{20}} - \dfrac{1}{4} \times \dfrac{2}{3}} \right] + \rho \pi {R^5}\left[ {\dfrac{1}{3} - \dfrac{1}{5}} \right] \\
= \rho \pi {R^5}\left( {\dfrac{4}{{15}}} \right) \\
$
Now, putting the value of $\rho $ in the above equation
$
{I_{yy'}} = \dfrac{{3m}}{{2\pi {R^3}}}\pi {R^5}\dfrac{4}{{15}} = \dfrac{2}{5}m{R^2} \\
\therefore I = \dfrac{2}{5}m{R^2} \\
$
But if $I$displaces the parallel to axis and join them to complete the sphere then,
$2I = \dfrac{2}{5}m{R^2}$
After cancelling 2 on both sides we get
$I = \dfrac{{m{R^2}}}{5}$
Hence, option (A) is the correct answer
Note Moment of Inertia of continuous bodies is $\dfrac{2}{5}m{R^2}$
Moment of Inertia of disc about centre of mass is $\dfrac{{m{R^2}}}{2}$
Moment of inertia of sphere about centre of mass is $\dfrac{2}{5}m{R^2}$
$dm = \left( {\dfrac{m}{{{V_H}}}} \right) \times {V_D}$
where, $m$ is the mass of hemisphere
${V_H}$ is the volume of hemisphere
${V_D}$ is the volume of disc
Now, use the formula
${I_{yy'}} = {I_{cm}} + m{r^2}$
Then, integrate it with the limit from $0$ to $R$
After solving we will get the value of ${I_{yy'}}$ then, use the value of $\rho = \dfrac{{3m}}{{2\pi {R^3}}}$
Then, if $I$ displace them parallel to the axis and again join them to complete a sphere
$2I = \dfrac{2}{5}m{R^2}$
Complete Step by Step Solution
Suppose that hemisphere is constructed by small discs
Mass $dm$of disc= $dm = \left( {\dfrac{m}{{{V_H}}}} \right) \times {V_D}$
$
\therefore dm = \left( {\dfrac{m}{{\dfrac{2}{3}\pi {R^3}}}} \right)\left( {\pi {y^2}dx} \right) \\
dm = \rho \left( {\pi {y^2}dx} \right) \\
$
Because, $\rho = \dfrac{m}{{\dfrac{2}{3}\pi {R^3}}}$
According to parallel and perpendicular axis theorem we get
${I_{yy'}} = {I_{cm}} + m{r^2}$
$\therefore d{I_{yy'}} = (dm)\dfrac{{{y^2}}}{4} + (dm){x^2}$
Now, integrating both sides
$
{I_{yy'}} = \int\limits_0^R {(dm)\dfrac{{{y^2}}}{4} + } \int\limits_0^R {dm} {x^2} \\
\Rightarrow \int\limits_0^R {(\rho \pi {y^2}dx)\dfrac{{{y^2}}}{4} + } \int\limits_0^R {(\rho \pi {y^2}dx){x^2}} \\
$
Now, put the limits
$
{I_{yy'}} = \dfrac{{\rho \pi }}{4}\left[ {\dfrac{{{R^5}}}{5} + {R^5} - 2{R^2}\dfrac{{{R^3}}}{3}} \right] + \left[ {\rho \pi {R^5}\left( {\dfrac{{{R^3}}}{3}} \right) - \rho \pi \left( {\dfrac{{{R^5}}}{5}} \right)} \right] \\
= \rho \pi {R^5}\left[ {\dfrac{1}{4} + \dfrac{1}{{20}} - \dfrac{1}{4} \times \dfrac{2}{3}} \right] + \rho \pi {R^5}\left[ {\dfrac{1}{3} - \dfrac{1}{5}} \right] \\
= \rho \pi {R^5}\left( {\dfrac{4}{{15}}} \right) \\
$
Now, putting the value of $\rho $ in the above equation
$
{I_{yy'}} = \dfrac{{3m}}{{2\pi {R^3}}}\pi {R^5}\dfrac{4}{{15}} = \dfrac{2}{5}m{R^2} \\
\therefore I = \dfrac{2}{5}m{R^2} \\
$
But if $I$displaces the parallel to axis and join them to complete the sphere then,
$2I = \dfrac{2}{5}m{R^2}$
After cancelling 2 on both sides we get
$I = \dfrac{{m{R^2}}}{5}$
Hence, option (A) is the correct answer
Note Moment of Inertia of continuous bodies is $\dfrac{2}{5}m{R^2}$
Moment of Inertia of disc about centre of mass is $\dfrac{{m{R^2}}}{2}$
Moment of inertia of sphere about centre of mass is $\dfrac{2}{5}m{R^2}$
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)