JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 23)
A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc
varies as $$\left( {{{{\sigma _0}} \over r}} \right)$$
, then the radius of gyration of the disc about its axis passing through the centre is:
_12th_April_Morning_Slot_en_23_1.png)
$$\sqrt {{{{a^2} + {b^2} + ab} \over 2}} $$
$$\sqrt {{{a + b} \over 3}} $$
$$\sqrt {{{{a^2} + {b^2} + ab} \over 3}} $$
$$\sqrt {{{a + b} \over 2}} $$
Explanation
dI = (dm)r2
= ($$\sigma $$ dA)r2
= $$\left( {{{{\sigma _0}} \over r}2\pi dr} \right){r^2} = {\sigma _0}2\pi {r^2}dr$$
$$I = \int {DI} = \int\limits_a^b {{\sigma _0}2\pi {r^2}dr} $$
$$ = {\sigma _0}2\pi \left( {{{{b^3} - {a^3}} \over 3}} \right)$$
m = $$\int {dm = \int {\sigma dA = {\sigma _0}2\pi } } \int\limits_a^b {dr} $$
= ($$\sigma $$ dA)r2
= $$\left( {{{{\sigma _0}} \over r}2\pi dr} \right){r^2} = {\sigma _0}2\pi {r^2}dr$$
$$I = \int {DI} = \int\limits_a^b {{\sigma _0}2\pi {r^2}dr} $$
$$ = {\sigma _0}2\pi \left( {{{{b^3} - {a^3}} \over 3}} \right)$$
m = $$\int {dm = \int {\sigma dA = {\sigma _0}2\pi } } \int\limits_a^b {dr} $$
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