
The volume of a metal sphere increases by $0.24\%$ when its temperature is raised by ${40^0}C$. What is the coefficient of linear expansion of the metal?
A) $2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$
B) $6 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$
C) $18 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$
D) $1.2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$
Answer
126.3k+ views
Hint: When a matter is subjected to heat or a temperature change, its atoms gain energy and move very fast, leading to thermal expansion. When we heat things, not only their length changes, but there is also a change in their area and volume. This gives rise to the need to define three different coefficients of thermal expansion, namely, linear, areal, and volumetric coefficients of thermal expansion.
Complete step by step solution:
For a small value of $\Delta V$:
$\Delta V \propto \Delta T$ and $\Delta V \propto V$
$\Rightarrow \Delta V \propto V\Delta T$
$\therefore \Delta V = \gamma V\Delta T$……………….. Equation (1)
where,
$\gamma =$ coefficient of volumetric expansion
$V =$ volume of body
$\Delta T =$ change in temperature
$\Delta V =$ a small change in volume [as $\Delta V \ll V$]
Now, $\Delta V = {V_T} - {V_0}$ as
${V_0} =$ the initial volume of the body and
${V_T} =$ the volume of the body after increasing temperature, i.e., at the temperature $T$
$\Rightarrow {V_T} - {V_0} = \gamma {V_0}\Delta T$
$\Rightarrow {V_T} = {V_0} + \gamma {V_0}\Delta T$
Therefore, the change in the volume of a body is given by:
$\Rightarrow {V_T} = {V_0}(1 + \gamma \Delta T)$
Here, we have $\dfrac{{\Delta V}}{V} = 0.24\% = \dfrac{{0.24}}{{100}}$ and $\Delta T = 40{}^ \circ C$
Using equation (1) and rearranging it, we get
$\Rightarrow \gamma = \dfrac{{\Delta V}}{{V\Delta T}}$
Substituting the values to find the coefficient of volumetric expansion:
$\Rightarrow \gamma = \dfrac{{0.24}}{{100}} \times \dfrac{1}{{40}}$
$\Rightarrow \gamma = 0.06 \times {10^{ - 3}}$
$\Rightarrow \gamma = 6 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$
Just like we derived the coefficient of volumetric expansion, similarly according to equation (1), linear expansion of a body is expressed as:
$\Delta L = \alpha L\Delta T$……………………….Equation (2)
where $\alpha =$ coefficient of linear expansion
Let us assume a cube of the side $L$ whose volume is ${L^3}$.
Now, $\Delta V = {(L + \Delta L)^3} - {L^3}$
Using the mathematical formula, ${(a + b)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}$
We get $\Delta V = [{L^3} - 3L{(\Delta L)^2} + 3{L^2}.\Delta L - {(\Delta L)^3}] - {L^3}$
Since, $\Delta L \ll L$,${(\Delta L)^3} \approx 0$ and ${(\Delta L)^2} \approx 0$
$\Rightarrow \Delta V = {L^3} + 3{L^2}.\Delta L - {L^3}$
$\Rightarrow \Delta V = 3{L^2}.\Delta L$
Substituting the value of $\Delta L$ from equation (2);
$\Rightarrow \Delta V = 3{L^2}(\alpha L\Delta T)$
$\Rightarrow \Delta V = 3\alpha {L^3}\Delta T$……….Equation (3)
Comparing equations (1) and (3), we get
$\Rightarrow \gamma = 3\alpha$
$\Rightarrow \alpha = \dfrac{\gamma }{3}$
$\Rightarrow \alpha = \dfrac{{6 \times {{10}^{ - 5}}{}^ \circ {C^{ - 1}}}}{3}$
$\Rightarrow \alpha = 2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$
The correct answer is [A], $2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$.
Note: Every material has its unique coefficient of volumetric expansion, but all three different coefficients of thermal expansion are interdependent, further indicating that these coefficients are also a characteristic property of every material (mostly metal).
Complete step by step solution:
For a small value of $\Delta V$:
$\Delta V \propto \Delta T$ and $\Delta V \propto V$
$\Rightarrow \Delta V \propto V\Delta T$
$\therefore \Delta V = \gamma V\Delta T$……………….. Equation (1)
where,
$\gamma =$ coefficient of volumetric expansion
$V =$ volume of body
$\Delta T =$ change in temperature
$\Delta V =$ a small change in volume [as $\Delta V \ll V$]
Now, $\Delta V = {V_T} - {V_0}$ as
${V_0} =$ the initial volume of the body and
${V_T} =$ the volume of the body after increasing temperature, i.e., at the temperature $T$
$\Rightarrow {V_T} - {V_0} = \gamma {V_0}\Delta T$
$\Rightarrow {V_T} = {V_0} + \gamma {V_0}\Delta T$
Therefore, the change in the volume of a body is given by:
$\Rightarrow {V_T} = {V_0}(1 + \gamma \Delta T)$
Here, we have $\dfrac{{\Delta V}}{V} = 0.24\% = \dfrac{{0.24}}{{100}}$ and $\Delta T = 40{}^ \circ C$
Using equation (1) and rearranging it, we get
$\Rightarrow \gamma = \dfrac{{\Delta V}}{{V\Delta T}}$
Substituting the values to find the coefficient of volumetric expansion:
$\Rightarrow \gamma = \dfrac{{0.24}}{{100}} \times \dfrac{1}{{40}}$
$\Rightarrow \gamma = 0.06 \times {10^{ - 3}}$
$\Rightarrow \gamma = 6 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$
Just like we derived the coefficient of volumetric expansion, similarly according to equation (1), linear expansion of a body is expressed as:
$\Delta L = \alpha L\Delta T$……………………….Equation (2)
where $\alpha =$ coefficient of linear expansion
Let us assume a cube of the side $L$ whose volume is ${L^3}$.
Now, $\Delta V = {(L + \Delta L)^3} - {L^3}$
Using the mathematical formula, ${(a + b)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}$
We get $\Delta V = [{L^3} - 3L{(\Delta L)^2} + 3{L^2}.\Delta L - {(\Delta L)^3}] - {L^3}$
Since, $\Delta L \ll L$,${(\Delta L)^3} \approx 0$ and ${(\Delta L)^2} \approx 0$
$\Rightarrow \Delta V = {L^3} + 3{L^2}.\Delta L - {L^3}$
$\Rightarrow \Delta V = 3{L^2}.\Delta L$
Substituting the value of $\Delta L$ from equation (2);
$\Rightarrow \Delta V = 3{L^2}(\alpha L\Delta T)$
$\Rightarrow \Delta V = 3\alpha {L^3}\Delta T$……….Equation (3)
Comparing equations (1) and (3), we get
$\Rightarrow \gamma = 3\alpha$
$\Rightarrow \alpha = \dfrac{\gamma }{3}$
$\Rightarrow \alpha = \dfrac{{6 \times {{10}^{ - 5}}{}^ \circ {C^{ - 1}}}}{3}$
$\Rightarrow \alpha = 2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$
The correct answer is [A], $2 \times {10^{ - 5}}{}^ \circ {C^{ - 1}}$.
Note: Every material has its unique coefficient of volumetric expansion, but all three different coefficients of thermal expansion are interdependent, further indicating that these coefficients are also a characteristic property of every material (mostly metal).
Recently Updated Pages
JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Classification of Elements and Periodicity in Properties Chapter For JEE Main Chemistry

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (January 24th Shift 2) Chemistry Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Login 2045: Step-by-Step Instructions and Details

Class 11 JEE Main Physics Mock Test 2025

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
