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Question

Answers

A. $4:1$

B. $1:16$

C. $8:1$

D. $1:12$

Answer
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The given values are

\[Torque1\; = Torque{\text{ }}2\]

We know that

\[Torque{\text{ }} = {\text{ }}I{\text{ }}\alpha \;\]

Where, \[\alpha = angular\] acceleration

\[I = \] moment of inertia

Now equating both torques

\[{I_1}{\alpha _1} = {I_2}{\alpha _{2}}\]

Moreover moment of inertia of ring is given by

\[I = m{r^2}\]

Where \[m = mass\] and \[r = radius\]of the inner circle of the ring.

Now as given, both the ring have same thickness,

Assuming the radius to be equal to for both the rings, we get

\[{m_1}{r^2}\alpha 1 = {m_2}{r^2}\alpha 2\;\]

We know that mass,

m=V ρ

Where \[\rho = density\] and \[V = volume\]

Volume V is also same for both the rings as radius and thickness are same

\[{\rho _1}V{\text{ }}r{\text{ }}{\alpha _1} = {\rho _2}V{\text{ }}r{\text{ }}{\alpha _{2}}\]

Equation gets reduced to

\[{\alpha _2}/{\alpha _1} = {\rho _1}/{\rho _2} = 4:1\;\]

\kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}m{r^2}\].