Class NEET 2026 ALL Q #1906
COMPETENCY BASED
APPLY
4 Marks 2026 NTA-RE-NEET-2026 MCQ SINGLE
The temperature of a metallic sphere of radius R is increased by a small amount $\Delta T.$ If the linear coefficient of thermal expansion of the metal is $\alpha$, the approximate increase in the volume of the sphere is:
(A) $6\pi R^{3}\alpha\Delta T$
(B) $2\pi R^{3}\alpha\Delta T$
(C) $3\pi R^{3}\alpha\Delta T$
(D) $4\pi R^{3}\alpha\Delta T$
Correct Answer: D

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Detailed Solution

Step 1: Identify the initial volume

The volume $V$ of a sphere with radius $R$ is given by the formula: $$V = \frac{4}{3}\pi R^{3}$$

Step 2: Relate volume expansion to linear expansion

The coefficient of volume expansion $\gamma$ is related to the linear coefficient of thermal expansion $\alpha$ for isotropic solids by the relation: $$\gamma = 3\alpha$$

Step 3: Apply the thermal expansion formula

The change in volume $\Delta V$ for a temperature change $\Delta T$ is given by: $$\Delta V = V \gamma \Delta T$$

Step 4: Substitute and simplify

Substitute $V = \frac{4}{3}\pi R^{3}$ and $\gamma = 3\alpha$ into the equation: $$\Delta V = \left(\frac{4}{3}\pi R^{3}\right) (3\alpha) (\Delta T)$$ $$\Delta V = 4\pi R^{3}\alpha\Delta T$$

Final Answer: 4\pi R^{3}\alpha\Delta T

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must utilize the relationship between linear and volume expansion coefficients to solve for a specific physical geometry.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a sequence of mathematical steps involving differentiation or standard expansion formulas to derive the result.
Syllabus Audit: In the context of NEET, this is classified as COMPETENCY. It tests the conceptual understanding of thermal properties of matter, which is a core topic in the NEET Physics syllabus.