Question

give me answer within 10 minutes plz​

Answer 1

$\huge\red\bigstar\huge\mathcal{\underline{ \underline{SOLUTION}}}\bigstar$

the moment of inertia of a spherical shell about it's geometrical axis is ....

$I_{sphere}= \frac{2}{3} M_{sphere}R {}^{2} \: \: \\ where \: R = radius \: of \: the \: sphere...$

now the moment of inertia of the disc of radius r about it's tangential axis is...

I_disc=M_discr

$I_{disc}= \frac{5}{4} M_{disc}r {}^{2}$

now ...as the sphere has been melted into disc so

and their M.I. are same about the respective axes so...

$M_{sphere}=M_{disc}=M\\I_{sphere}=I_{disc}=I$

therefore....

$\implies\frac{2}{3} M_{sphere}R {}^{2} = \frac{5}{4} M_{disc} r {}^{2} \\ \implies \frac{2}{3} MR {}^{2} = \frac{5}{4} Mr {}^{2} \\ \implies r = \sqrt{ \frac{8}{15}R {}^{2} } = \sqrt{ \frac{8}{15} } R \: \\ \implies \boxed{ r = \frac{2 \sqrt{2} }{ \sqrt{15} } R \: }$