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. $1:12$

Answer 1

Answer: (A)

$4:1$

Given torque${\tau }_1={\tau }_2$-----------(1)

We know $\tau =Fr=I\alpha$ where $\alpha$=angular acceleration;I= moment of inertia

Now equation 1 can be written as,

$I_1{\alpha }_1=I_2{\alpha }_2$

also I of ring $I_{ring}=M{R}^2$ where M= mass and R= radius of the inner circle of the ring.

Now as mentioned in the question both the ring have same thickness ,and assuming the radius to be equal for both the rings, we get

$M_1{R}_1{\alpha }_1=M_2{R}_2{\alpha }_2$

Also mass$M=\frac{\rho }{V}$ where $\rho$=density and $V$=volume

${\rho }_1{V}_1{R}_1{\alpha }_1={\rho }_2{V}_2{R}_2{\alpha }_2$

As parameters of both the rings are same the component of volume and radius gets cancel out.

$\frac{{\alpha }_1}{{\alpha }_2}=\frac{{\rho }_2}{{\rho }_1}=4:1$