Question

A thin rod of mass $96g$ and of length $1m$ is made to rotate about an axis perpendicular to the length of rod passing through a point at a distance $25cm$ from one end of rod. The moment of inertia of that rod is:

• A. $1.4\times {10}^{-2}kg{m}^2$
• B. $0.7\times {10}^{-2}kg{m}^2$
• C. $1.4\times {10}^{-3}kg{m}^2$
• D. $0.7\times {10}^{-3}kg{m}^2$

Answer 1

Answer: (A)

$1.4\times {10}^{-2}kg{m}^2$

$M=96gm$

$M=0.096kg.$

Ml of rod about the axis passing from the point which is 25 cm away from one end of the rod.

$\Rightarrow \frac{M{l}^2}{12}+M{r}^2$

$\Rightarrow \frac{0.096(1{)}^2}{12}+0.096(0.25{)}^2$

$\Rightarrow 0.096(\frac{1}{12}+\frac{1}{16})\Rightarrow 0.096\times \frac{7}{48}$

$\Rightarrow 14\times {10}^{-3}kg{m}^2$ or $1.4\times {10}^{-2}kg{m}^2$