Question

Two disc one of density 7.2 g/cm and the other of density 8.9 g/cm are of same mass and thickness. Their
moments of inertia are in the ratio :-
​

Answer 1

Explanation:

Bro please mark as brainliest it's my humble request to you

Answer 2

${ \bf{ \underline{ \red{ \underline{Question:-}}}}}$

Two disc one of density 7.2 g/cm³ and the other of density 8.9 g/cm³ are of same mass and thickness. Their moments of inertia are in the ratio :-

${ \bf{ \underline{ \red{ \underline{Solution:-}}}}}$

{According to question}

${ \dagger{ \sf{ \: let \: { \red{ \rho_{1} }}} \: be \: the \: density \: of \: first \: disc}}$

${{ \sf{ \: and \: { \red{ \rho_{2} }}} \: be \: the \: density \: of \: second \: disc}}$

${ \dagger{ \sf \: { \red{ \rho_{1} }} = 7.2 \: \: g/cm³}}$

${ \dagger{ \sf \: { \red{ \rho_{2} }} = 8.9 \: \: g/cm³}}$

${ \dagger{ \sf{ \: let \: { \red{ m_{1}, \: t_1}}} \: and \: {{ \red{ m_{2} , \: t_1}}}}{ \sf{\: be \: the \: mass \: and \: thikness \: }}}$

${{ \sf{ \: of \: first \: disc \: and \:second \: disc \: repectively.}}}$

According to given mass & thickness are same of both disc

${ \bf{ \dashrightarrow{ \red{ m_{1} = \red{ m_{2}}}}}}{ \sf{ \: and}}{ \bf{ { \: \: \red{ t_{1} = \red{ t_{2}}}}} \: \: \: \: .....(1)}$

Hence,

We know formula of density is

${ \sf{ \dashrightarrow{ \huge{ \purple{ \rho} = \frac{m}{v} }}}}$

where,

—› m = mass & v = volume

Hence,

${ \sf{ \dashrightarrow{ \huge{ m = \purple{ \rho} \times v }}} \: \: ....(2)}$

from ( 1 )

${ \sf{ \huge{ \dashrightarrow{ m _{1} = m _{2} }}}}$

from ( 2 )

${ \bf{ { \dashrightarrow{ { \rho}_{1} \times v _{1} = { \rho}_{2} \times v _{2} }}}}$

we know volume of disc = πr²h but we take πr²t

where t is thickness of disc

—› volume of disc = πr²t

Now,

${ \bf{ { \dashrightarrow{ { \rho}_{1} \times \pi \times {r}^{ \gray2} _{1} \times {t} _{1} = {{ \rho}_{2} \times \pi \times {r}^{ \gray2} _{2} \times {t} _{2}}}}}}$

From ( 1 )

${ \bf{ { \dashrightarrow{ { \rho}_{1} \times {r}^{ \gray2} _{1} = {{ \rho}_{2} \times {r}^{ \gray2} _{2} }}}}}$

Now,

${ \bf{ \huge { \dashrightarrow{ \frac{{r}^{ \gray2} _{1}}{{r}^{ \gray2} _{2}} = { \frac{{ \rho}_{2} }{{ \rho}_{1} } }}}}}$

Now we use formula of moment of inertia of disc .

${ \bf{ \green{ \dashrightarrow{ \purple{ I_{(disc)} = \frac{1}{2}m {r}^{2} }}}}}$

From ( 1 )

${ \bf{ \green{ \dashrightarrow{ \purple{ I_{(disc)} \: { \gray{\alpha} } \: \: {r}^{2} }}}}}$

Hence,

${ \bf{ \huge{ \green{ \dashrightarrow{ \purple{ \frac{I_{(disc \: 1)} }{I_{(disc \: 2)} } \: = \: \: \frac{{r}^{ \gray2} _{1}}{{r}^{ \gray2} _{2}} }}}}}}$

We know

${ \bf{ \huge { \dashrightarrow{ \frac{{r}^{ \gray2} _{1}}{{r}^{ \gray2} _{2}} = { \frac{{ \rho}_{2} }{{ \rho}_{1} } }}}}}$

Hence,

—› I (disc 1 ) : I (disc 2) = 8.9 : 7.2

Hence ratio of moment of inertia of disc's are 8.9 : 7.2

i hope it helps you.