Question

Three identical rods each of mass 'm' and length ' are joined together at the ends at an angle 120 with each other. The moment of inertia about an axis passing through point a and perpendicular to the plane of the rods is :

Answer 1

Solution:

Given:

=> Velocity of sound in gas at 27°C = 330 m/s.

To Find:

=> Velocity at 327°C.

Let velocity at 327°C be x,

So,

\sf{\implies \dfrac{27}{330}=\dfrac{327}{x}}⟹33027=x327

\sf{\implies \dfrac{330 \times 327}{27} = x}⟹27330×327=x

\sf{\implies x = 3996.6\;m/s}⟹x=3996.6m/s

{\boxed{\boxed{\bf{So,\;velocity\;at\;327^{\circ}C=3996.6\;m/s}}}}So,velocityat327∘C=3996.6m/s

Answer 2

Answer:

The moment of inertia about an axis passing through point a and perpendicular to the plane of the rods is $=\frac{m l^2}{4}+\frac{m l^2}{4}+0=\frac{m l^2}{2}$.

Explanation:

Step : 1  The correct option is C $=\frac{\mathrm{ml}}{2}$

Given, m1=m2=m3=mkg

l1=l2=l3=ml

Moment of inertia of rod AB about axis along BC

$\begin{aligned}& I_{A B}=\frac{m l^2}{3} \sin ^2 60^{\circ} \\& =\frac{m l^2}{4}\end{aligned}$

(because, effective length of rod =lsin60∘)Similarly,

Step : 2 Moment of inertia of rod AC about axis along BC

$\begin{aligned}& \mathrm{I}_{\mathrm{AC}}=\frac{m \mathrm{l}^2}{3} \sin ^2 60^{\circ} \\& =\frac{m \mathrm{l}^2}{4}\end{aligned}$

Moment of inertia of rod BC about axis along BC

$\mathrm{I}_{\mathrm{BC}}=0$

Total moment of inertia $(1)=I_{A B}+I_{A C}+I_{B C}$

$=\frac{m l^2}{4}+\frac{m l^2}{4}+0=\frac{m l^2}{2}$

Step :3  The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis could be internal or exterior, fixed or not.

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