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Question

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 :

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$4\pi R^3 \alpha \Delta T$

Thermal Expansion Volume Increase Calculation

This problem asks for the approximate increase in the volume of a metallic sphere when its temperature increases.

Key Concepts

  • The initial volume ($V_0$) of a sphere with radius $R$ is given by the formula $V_0 = \frac{4}{3}\pi R^3$.
  • The linear coefficient of thermal expansion is $\alpha$.
  • For isotropic materials like metals, the volume expansion coefficient ($\beta$) is approximately three times the linear expansion coefficient: $\beta \approx 3\alpha$.
  • The change in volume ($\Delta V$) due to a temperature change ($\Delta T$) is calculated as $\Delta V = V_0 \beta \Delta T$.

Volume Increase Calculation

  1. Initial Volume: Calculate the initial volume of the sphere:

    $ V_0 = \frac{4}{3}\pi R^3 $

  2. Volume Expansion Coefficient: Determine the volume expansion coefficient:

    $ \beta \approx 3\alpha $

  3. Change in Volume: Apply the formula for volume expansion:

    $ \Delta V = V_0 \beta \Delta T $

    Substitute the expressions for $V_0$ and $\beta$:

    $ \Delta V = \left(\frac{4}{3}\pi R^3\right) (3\alpha) (\Delta T) $

    Simplify the expression:

    $ \Delta V = 4\pi R^3 \alpha \Delta T $

Conclusion

The approximate increase in the volume of the sphere is $4\pi R^3 \alpha \Delta T$. This corresponds to Option B.

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