When the temperature of a body increases from T to T - T- its moment of inertia increases from I to I - I- If - is the coeffficient of linear expansion of the material of the body- then - I-I is neglect higher orders of -

When the temperature of a rod increases from $t$ to $t + Δ t$ , the moment of inertia of the rod increases from $I$ to $I + Δ I$ . If the coefficient of linear expansion of the rod is $α$ , the ratio $\frac{Δ I}{I}$ is

Q. When the temperature of a rod increases from

T

T + Δ T

, the moment of inertia of the rod about an axis increases from

I

I + Δ I

. If the coefficient of linear expansion of rod is

α

, then find the ratio

\frac{Δ I}{I}

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When the temperature of a body increases from T to T+ΔT, its moment of inertia increases from I to I+ΔI. If α is the coeffficient of linear expansion of the material of the body, then ΔII is (neglect higher orders of α):

The correct option is D 2αΔTInitial moment of inertia , I1=I=MR21 and final MOI, I2=I+ΔI=MR22Also R2=R1[1+α(T+ΔT−T)]=R1[1+αΔT]....(1)Now I2−I1I1=R22−R21R21so, ΔII=[1+αΔT]2−1 using (1)or ΔII=1+2αΔT−1=2αΔT (neglecting higher order)

When the temperature of a body increases from $T$ to $T$ $+ Δ$ $T$ , its moment of inertia increases from $I$ to $I$ $+ Δ$ $I$ . If $α$ is the coeffficient of linear expansion of the material of the body, then $\frac{Δ I}{I}$ is (neglect higher orders of $α$ ):