Equal torque are applied about a central axis on two rings of same and same thickness but made up of different materials. If ratio of their densities is $4:1$then the ratio of their angular acceleration will be:-
A. $4:1$
B. $1:16$
C. $8:1$
D. $1:12$

Hint: As torques are equal, \[{I_1}{\alpha _1} = {I_2}{\alpha _2}\]​ and \[I{\text{ }} = {\text{ }}m{r^2}\]

Complete step by step solution:
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
\[{\alpha _2}/{\alpha _1} = {\rho _1}/{\rho _2} = 4:1\;\]

Hence, the correct option is A.

Note: Students must be careful while checking out whether the torque is applied about axis or diameter. If torque is applied about diameter, then the formula would be \[I = \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/
 \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ }}m{r^2}\].