Question

The moment of inertia of a ring of mass $10g$ and radius $1cm$ about a tangent to the ring and normal to its plane is :

• A. $2\times {10}^{-6}kg{m}^2$
• B. $5\times {10}^{-7}kg{m}^2$
• C. $5\times {10}^{-8}kg{m}^2$
• D. $4\times {10}^{-6}kg{m}^2$

Answer 1

Answer: (A)

$2\times {10}^{-6}kg{m}^2$

The moment of inertia about a tangent to the ring and normal to its plane is given as,

$I=2M{R}^2$

$=2\times 10\times {10}^{-3}{\left(1\times {10}^{-2}\right)}^2$

$=2\times {10}^{-6}kg{m}^2$

Thus, the moment of inertia about a tangent to the ring and normal to its plane is $2\times {10}^{-6}kg{m}^2$.