Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A thin circular plate of mass M and radius R has its density varying as $\rho (r)=\rho\, _{0}r$ with $\rho _{0}$ as constant and r is the distance from its center. The moment of Inertia of the circular plate and passing through its edge is $I=a\; MR^{2}$ . The value of the coefficient a is :

Option 1)
$3/5$

Option 2)
$1/2$

Option 3)
$8/5$

Option 4)
$3/2$

Answers (1)

best_answer

take a small circular ring at a distance x from centre of thickness $dx.$

$dI=dmx^{2}+dm\; R^{2}$

$dI=dm (x^{2}+R^{2})$

& $dm=\rho _{o}.2\pi x.dx$

$=\rho _{o} x.2\pi x.dx$

$dm=\rho _{o} .2\pi x^{2}.dx$

$m=\int dm=\rho _{o}2\pi \int x^{2}dx=\rho _{o}\times2\pi \left [ \frac{R^{3}}{3} \right ]$

$I = \int dI = \varrho _{o}\times2\pi\int (x^{2}+R^{2})x^{2}dx$

$=\rho _{o}\times2\pi\left [ \int x^{4}dx +R^{2}\int x^{2}dx\right ]$

$I=\rho _{o}\times2\pi\left \{ \left [ \frac{x^{5}}{5} \right ]_{o}^{R}+R^{2} \left [ \frac{x^{3}}{3} \right ]_{o}^{R}\right \}$

$I=\rho _{o}\times2\pi\left [ \frac{R^{5}}{5}+\frac{R^{5}}{3} \right ]$

$I=\frac{16\pi}{15}\rho _{o}\pi R^{5}, M=2\pi \rho _{o}(\frac{R^{3}}{3})$

So $I = \frac{8}{5}mR^{2}\Rightarrow a=\frac{8}{5}$

Option 1)

$3/5$

Option 2)

$1/2$

Option 3)

$8/5$

Option 4)

$3/2$

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs

anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question