Answer
Verified
37.2k+ views
Hint: Begin by writing the expression for moment of inertia of a rod about its perpendicular axis passing through the centre. Use the coefficient of linear expansion to find the length of the rod after increase in temperature. Then find the moment of inertia of the rod at the final increased temperature using expanded length and subtract it from the moment of inertia at initial temperature.
Complete step by step solution
The moment of inertia of a rod about its perpendicular bisector is I.
$ \Rightarrow I = \dfrac{{M{L^2}}}{{12}}$
Where M is the mass of the rod and L is the length of the rod.
When the temperature of the rod is increased by$\Delta T$, there is some expansion in the length of the rod which is given by: $\Delta L = \alpha L\Delta T$.
$ \Rightarrow $ The length of the rod will now become, $L + \Delta L$ .
Now, the moment of inertia of the rod about the perpendicular bisector will be
\[
\Rightarrow I' = \dfrac{{M{{(L + \Delta L)}^2}}}{{12}} \\
\Rightarrow I\prime = \dfrac{M}{{12}}\left[ {{L^2} + {{(\Delta L)}^2} + 2L\Delta L} \right] \\
\]
Since $\Delta L$ is a very small quantity, its square will be even smaller and almost negligible.
$ \Rightarrow I\prime = \dfrac{M}{{12}}\left[ {{L^2} + 2L\Delta L} \right]$
Now substituting the value of \[\Delta L\]in the above equation,
\[ \Rightarrow I\prime = \dfrac{M}{{12}}\left[ {{L^2} + 2L\left( {\alpha L\Delta T} \right)} \right]\]
\[ \Rightarrow I\prime = \dfrac{{M{L^2}}}{{12}} + 2\alpha \Delta T\dfrac{{M{L^2}}}{{12}}\]
Therefore, the increase in moment of inertia of the rod is $I' - I$ ,
$
\Rightarrow I' - I = \dfrac{{M{L^2}}}{{12}} + 2\alpha \Delta T\dfrac{{M{L^2}}}{{12}} - \dfrac{{M{L^2}}}{{12}} \\
\Rightarrow I' - I = 2\alpha \Delta T\dfrac{{M{L^2}}}{{12}} \\
\Rightarrow I' - I = 2\alpha I\Delta T \\
$
So, option (B) is correct.
Note: One must not get confused between the expressions of moment of inertia of the rod through its perpendicular bisector (i.e. through its centre of mass) and through one end of the rod.
Moment of inertia of the rod through an axis passing through the centre is $\dfrac{{M{L^2}}}{{12}}$. This can be derived from the centre of mass expression.
Moment of inertia of the rod through an axis passing through one end of the rod is $\dfrac{{M{L^2}}}{3}$. This can be derived using the parallel axis theorem.
Complete step by step solution
The moment of inertia of a rod about its perpendicular bisector is I.
$ \Rightarrow I = \dfrac{{M{L^2}}}{{12}}$
Where M is the mass of the rod and L is the length of the rod.
When the temperature of the rod is increased by$\Delta T$, there is some expansion in the length of the rod which is given by: $\Delta L = \alpha L\Delta T$.
$ \Rightarrow $ The length of the rod will now become, $L + \Delta L$ .
Now, the moment of inertia of the rod about the perpendicular bisector will be
\[
\Rightarrow I' = \dfrac{{M{{(L + \Delta L)}^2}}}{{12}} \\
\Rightarrow I\prime = \dfrac{M}{{12}}\left[ {{L^2} + {{(\Delta L)}^2} + 2L\Delta L} \right] \\
\]
Since $\Delta L$ is a very small quantity, its square will be even smaller and almost negligible.
$ \Rightarrow I\prime = \dfrac{M}{{12}}\left[ {{L^2} + 2L\Delta L} \right]$
Now substituting the value of \[\Delta L\]in the above equation,
\[ \Rightarrow I\prime = \dfrac{M}{{12}}\left[ {{L^2} + 2L\left( {\alpha L\Delta T} \right)} \right]\]
\[ \Rightarrow I\prime = \dfrac{{M{L^2}}}{{12}} + 2\alpha \Delta T\dfrac{{M{L^2}}}{{12}}\]
Therefore, the increase in moment of inertia of the rod is $I' - I$ ,
$
\Rightarrow I' - I = \dfrac{{M{L^2}}}{{12}} + 2\alpha \Delta T\dfrac{{M{L^2}}}{{12}} - \dfrac{{M{L^2}}}{{12}} \\
\Rightarrow I' - I = 2\alpha \Delta T\dfrac{{M{L^2}}}{{12}} \\
\Rightarrow I' - I = 2\alpha I\Delta T \\
$
So, option (B) is correct.
Note: One must not get confused between the expressions of moment of inertia of the rod through its perpendicular bisector (i.e. through its centre of mass) and through one end of the rod.
Moment of inertia of the rod through an axis passing through the centre is $\dfrac{{M{L^2}}}{{12}}$. This can be derived from the centre of mass expression.
Moment of inertia of the rod through an axis passing through one end of the rod is $\dfrac{{M{L^2}}}{3}$. This can be derived using the parallel axis theorem.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
f a body travels with constant acceleration which of class 1 physics JEE_Main
A hollow sphere of mass M and radius R is rotating class 1 physics JEE_Main
If the beams of electrons and protons move parallel class 1 physics JEE_Main
Two radioactive nuclei P and Q in a given sample decay class 1 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Other Pages
Lowering in vapour pressure is highest for A 02 m urea class 11 chemistry JEE_Main
An electric bulb has a power of 500W Express it in class 11 physics JEE_Main
Which of the following Compounds does not exhibit tautomerism class 11 chemistry JEE_Main
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
The nitride ion in lithium nitride is composed of A class 11 chemistry JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main