Question

# 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 $I = \raise.5ex\hbox{\scriptstyle 1}\kern-.1em/ \kern-.15em\lower.25ex\hbox{\scriptstyle 2} {\text{ }}m{r^2}$.