JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 19)
Mass per unit area of a circular disc of radius $$a$$ depends on the distance r from its centre as $$\sigma \left( r \right)$$ = A + Br
. The moment of inertia of the disc about the axis, perpendicular to the plane and
assing through its centre is:
$$2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)$$
$$\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)$$
$$2\pi {a^4}\left( {{{aA} \over 4} + {B \over 5}} \right)$$
$$2\pi {a^4}\left( {{A \over 4} + {B \over 5}} \right)$$
Explanation
_7th_January_Evening_Slot_en_19_1.png)
dI = dm(r2)
and dm = $$\sigma $$2$$\pi $$rdr
$$ \therefore $$ dI = $$\sigma $$2$$\pi $$rdr(r2) = $$\sigma $$2$$\pi $$r3dr
Given $$\sigma \left( r \right)$$ = A + Br
$$ \therefore $$ dI = 2$$\pi $$(A + Br)r3dr
$$\int {dI = \int\limits_0^a {2\pi {r^3}\left( {A + Br} \right)dr} } $$
$$ \Rightarrow $$ I = $$2\pi \left[ {{{A{r^4}} \over 4} + {{B{r^5}} \over 5}} \right]$$
= $$2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)$$
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