Thermal Expansion - NEET Physics Chapterwise MCQs & PYQs

NEET Thermal Expansion MCQs & PYQs

Question 1:

easy

When a metal rod is heated it expands because :

When a metal rod is heated, its atoms gain kinetic energy and vibrate more vigorously. As they vibrate, they tend to move slightly further apart because the increased energy weakens the attractive forces that hold them at a fixed distance. This increased atomic spacing results in the rod expanding in size.

Thus, the expansion of the metal rod occurs because the distance among its atoms increases with temperature. This phenomenon is the essence of thermal expansion.

Question 2:

easy

The volume of a metal sphere increases by 0.24% when its temperature is raised by 40ºC. The coefficient of linear expansion of the metal is :

The relationship between the coefficient of volume expansion (\( \beta \)) and the coefficient of linear expansion (\( \alpha \)) for a solid is:

\[
\beta = 3\alpha
\]

Given:
- Volume increase = 0.24%
- Temperature increase \( \Delta T = 40^\circ \text{C} \)

The coefficient of volume expansion \( \beta \) is given by:

\[
\beta = \frac{\text{Percentage increase in volume}}{\Delta T} = \frac{0.24}{40} = 0.006\% \, \text{per } ^\circ\text{C} = 6 \times 10^{-5} \, \text{per } ^\circ\text{C}
\]

Now, using \( \beta = 3\alpha \):

\[
\alpha = \frac{\beta}{3} = \frac{6 \times 10^{-5}}{3} = 2 \times 10^{-5} \, \text{per } ^\circ\text{C}
\]

Thus, the coefficient of linear expansion of the metal is \( 2 \times 10^{-5} \, \text{per } ^\circ\text{C} \).

Question 3:

easy

The percentage change in length of 1 m iron rod if its temperature changes by 100ºC is (\(\alpha\) for iron is \(2 \times 10^{-5}/\text{ºC}\))

The percentage change in length is given by \(\frac{\Delta L}{L} \times 100 = \alpha \Delta T \times 100 = (2 \times 10^{-5}) \times 100 \times 100 = 0.2%\).

Question 4:

easy

Thin copper wire of length \( L \) increases in length by 2% when heated from \( T_1 \) to \( T_2 \). If a copper cube having side \( 10L \) is heated from \( T_1 \) to \( T_2 \) then the percentage change in volume of the cube is

The percentage change in length is \( \frac{\Delta L}{L} \times 100 = 2% \). Since volume expansion coefficient is three times the linear expansion coefficient (\( \gamma = 3\alpha \)), the percentage change in volume is \( 3 \times 2% = 6% \).

Question 5:

easy

The percentage change in length of \( 1\text{ m} \) iron rod if its temperature changes by \( 100^circ\text{C} \) is (\( \alpha \) for iron is \( 2 \times 10^{-5}/^circ\text{C} \))

Using the formula for thermal expansion, \( \frac{\Delta L}{L} \times 100 = \alpha \Delta T \times 100 \). Plugging in the values: \( 2 \times 10^{-5} \times 100 \times 100 = 0.2% \).

Question 6:

easy

Assertion (A): Water is considered unsuitable for use in thermometers.


Reason (R): Thermal Expansion of water is non-uniform.


 

Assertion (A) is true. Water exhibits anomalous expansion between \(0^{\circ}\text{C}\) and \(4^{\circ}\text{C}\), making it unreliable for linear temperature scales. Reason (R) is true.


The non-uniform thermal expansion of water (especially its contraction then expansion) is why it's unsuitable for thermometers. (R) is the correct explanation for (A).

Question 7:

easy

Assertion (A): The temperature of a metallic rod is raised by a temperature \(\Delta t\) so that its length becomes double. The value of \(\alpha\) (coefficient of linear expansion) is given by \(\frac{\log_e (2)}{\Delta t}\).


Reason (R): Coefficient of linear expansion is defined as \(\frac{1}{l} \frac{dl}{dt}\).


 

Reason (R) is the correct definition for the instantaneous coefficient of linear expansion (assuming \(t\) is temperature), so it is true. Integrating \(dl/l = \alpha dT\) with constant \(\alpha\) gives \(ln(l/l_0) = \alpha \Delta T\), so \(l = l_0 e^{\alpha \Delta T}\). If \(l = 2l_0\), then \(2 = e^{\alpha \Delta T}\), leading to \(\alpha = \frac{ln 2}{\Delta T}\).


So Assertion (A) is true. (R) provides the foundational definition from which (A) is derived, thus it's the correct explanation.

Question 8:

easy

Assertion (A): Liquids usually expand more than solids.


Reason (R): The intermolecular forces in liquids are weaker than in solids.


 

Assertion (A) is true as liquids typically have higher coefficients of thermal expansion than solids. Reason (R) is true because weaker intermolecular forces in liquids allow molecules to move more freely and separate further upon heating.


(R) correctly explains (A), as the weaker forces enable greater thermal expansion.

Question 9:

easy

Assertion (A): Temperature of a rod is increased and again cooled to same initial temperature then its final length is equal to original length.


Reason (R): For a small temperature change, length of a rod varies as \( l = l_0 (1+\alpha \Delta T) \) provided \( \alpha \Delta T is small  \). Here symbol have their usual meaning.


 

Assertion is true as thermal expansion is reversible for elastic materials. Reason is the formula for linear expansion, \( l = l_0 (1+\alpha \Delta T) \), which confirms the assertion if \( \Delta T \) is reversed. Thus, both are true and (R) explains (A).

Question 10:

easy

Assertion (A): A temperature change which increases the length of a steel rod by ( 1% ) will increase its volume by ( 3% ).


Reason (R): The coefficient of volume expansion is nearly three times the coefficient of linear expansion.


 

Assertion (A) is true. If \( \frac{\Delta L}{L_0} = \alpha \Delta T = 0.01 \), then \( \frac{\Delta V}{V_0} = \gamma \Delta T \). Reason (R) is true, stating \( \gamma \approx 3\alpha \). Substituting, \( \frac{\Delta V}{V_0} \approx 3 (\alpha \Delta T) = 3(0.01) = 0.03 \), or ( 3% ). Thus, both are true and (R) correctly explains (A).