Question

From a given sample of uniform wire, two circular loops p and q are made, P of radius r and Q of radius nr. If the M.I of Q about its axis is 4 times than that of P, then value of n is:

• A. $4^{\frac{1}{3}}$
• B. $4^{\frac{2}{3}}$
• C. $4^{\frac{1}{4}}$
• D. $4^{\frac{1}{2}}$

Answer 1

The correct option is A

$4^{\frac{1}{3}}$

Step 1: Given

Radius of P is r

Radius of Q is nr

The M.I of Q about its axis is 4 times than that of P, $I_Q=4I_P$

Step 2: Determine the Moment of Inertia of ring P

Let the mass of ring P be M

Then the moment of inertia of P will be $I_P=M{r}^2$

Step 3: Determine the Moment of Inertia of ring Q

As ring Q is n times longer than ring P, it'll weigh n time more than ring P.

Hence, the mass of ring Q will be mM

Then the moment of inertia of Q will be $I_Q=\left(nM\right){\left(nr\right)}^2=M{n}^3{r}^2$

Step 4: Compare both the values

$$\begin{gathered} Given,I_Q=4I_P \\ M{n}^3{r}^2=4M{r}^2 \\ n=4^{\frac{1}{3}} \end{gathered}$$

Hence, option A is the correct option.