The Question is from neet-physics thermal-properties-of-matter , and it was created by the author NAGA SAI

Question

A solid sphere of radius 'R' made of a material with a volume expansion coefficient '$\gamma$' is heated by $\Delta T$. The fractional change in its moment of inertia about a diameter is:

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Thermal Expansion and Moment of Inertia
	**Related Topics**: Moment of Inertia of a sphere, Volume expansion
	**Prerequisites**:  Relationship between moment of inertia and radius, Volume expansion formula
## SOLUTION PATHWAY
**Given Information**:
	Radius of the sphere: 'R'
	Volume expansion coefficient: '$\gamma$'
	Change in temperature: '$\Delta T$'
**Method & Steps**:
	1. Find the change in radius due to thermal expansion.
		Reasoning: When the sphere is heated, its radius increases.
		Formula/Rule: $\Delta V = V \gamma \Delta T$, where $\Delta V$ is the change in volume, $V$ is the original volume, $\gamma$ is the volume expansion coefficient, and $\Delta T$ is the change in temperature.
	2. Relate the change in volume to the change in radius.
		Working: For a sphere, $V = \frac{4}{3}\pi R^3$.  Therefore, $\Delta V = 4\pi R^2 \Delta R$. Substituting this into the volume expansion formula, we get $4\pi R^2 \Delta R = (\frac{4}{3}\pi R^3) \gamma \Delta T$. Simplifying gives $\frac{\Delta R}{R} = \frac{1}{3} \gamma \Delta T$.
		Key Point: The fractional change in radius is one-third the fractional change in volume.
	3. Find the fractional change in moment of inertia.
		Working: The moment of inertia of a solid sphere about a diameter is $I = \frac{2}{5}MR^2$.  The fractional change in moment of inertia is $\frac{\Delta I}{I} = \frac{\frac{2}{5}M(2R\Delta R)}{\frac{2}{5}MR^2} = \frac{2\Delta R}{R}$.  Substituting $\frac{\Delta R}{R} = \frac{1}{3} \gamma \Delta T$, we get $\frac{\Delta I}{I} = 2(\frac{1}{3} \gamma \Delta T) = \frac{2}{3} \gamma \Delta T$.
		Result:  $\frac{2}{3} \gamma \Delta T$
		Verification: The fractional change in the moment of inertia is proportional to the fractional change in $R^2$.
 ## OPTION ANALYSIS
[A] :x: This option corresponds to using the fractional change in radius directly, not the fractional change in $R^2$.
[B] ✓: This is the correct answer, derived by considering the change in $R^2$.
[C] :x: This is the fractional change in radius, not the fractional change in moment of inertia.
[D] :x: This option doesn't have any physical significance in this context.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Using linear expansion coefficient instead of volume expansion coefficient: Remember that volume changes, and hence the volume expansion coefficient is relevant.
		2. Forgetting to consider the change in $R^2$:  The moment of inertia depends on $R^2$, so the fractional change needs to reflect this.
	**Quick Tips**:
		1. Remember the relationship between fractional change in volume and radius for a sphere: $\frac{\Delta R}{R} = \frac{1}{3} \frac{\Delta V}{V}$.
		2.  The fractional change in a quantity proportional to $x^n$ is approximately $n$ times the fractional change in $x$, for small changes. In this case, $I \propto R^2$, so $\frac{\Delta I}{I} \approx 2 \frac{\Delta R}{R}$.

NAGA SAI · Answer

(2/5)γΔT

NAGA SAI · Answer

The correct Option is "(2/3)γΔT"

NAGA SAI · Answer

(1/3)γΔT

NAGA SAI · Answer

(1/5)γΔT

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