A thin wire of length L and uniform linear mass density p is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XY is

Question

A thin wire of length L and uniform linear mass density p is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XY is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/53ba9944ea3d898f1a724a6e4363d920-.png" /> (A) (3 pL 3/8 π2) (B) (5 pL 3/16 π2) (C) ( pL 3/8 π2) (D) ( pL 3/16 π2)

Tardigrade Admin · Accepted Answer

The moment of inertia of a thin hoop about its diameter is

\frac{1}{2} M R^{2}

here,

M = L P

also, we have

2 π R = L

\Rightarrow R = \frac{L}{2 π}

So, we have

I = \frac{1}{2} M R^{2} = \frac{1}{2} L P (\frac{L}{2 π})^{2} = \frac{L ^{3} P}{8 π ^{2}}

Now using parallel axis theorem we have

L_{xx 1} = I_{c m} + M R^{2} = \frac{L ^{3} P}{8 π ^{2}} + L P (\frac{L}{2 π})^{2}

= \frac{L ^{3} P}{8 π ^{2}} + \frac{L ^{3} P}{4 π ^{2}} = \frac{3 L ^{3} P}{8 π ^{2}}

Q. A thin wire of length L and uniform linear mass density p is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XY is

$\frac{3 p L ^{3}}{8 π ^{2}}$

$\frac{5 p L ^{3}}{16 π ^{2}}$

$\frac{p L ^{3}}{8 π ^{2}}$

$\frac{p L ^{3}}{16 π ^{2}}$

Q. A thin wire of length L and uniform linear mass density p is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XY is

8π23pL3​

16π25pL3​

8π2pL3​

16π2pL3​

The moment of inertia of a thin hoop about its diameter is 21​MR2 here, M=LP also, we have 2πR=L ⇒R=2πL​ So, we have I=21​MR2=21​LP(2πL​)2=8π2L3P​ Now using parallel axis theorem we have Lxx1​=Icm​+MR2=8π2L3P​+LP(2πL​)2 =8π2L3P​+4π2L3P​=8π23L3P​

$\frac{3 p L ^{3}}{8 π ^{2}}$

$\frac{5 p L ^{3}}{16 π ^{2}}$

$\frac{p L ^{3}}{8 π ^{2}}$

$\frac{p L ^{3}}{16 π ^{2}}$