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Question: The radius of a ring is R and its coefficient of linear expansion is \(\alpha \). If the temperature...

The radius of a ring is R and its coefficient of linear expansion is α\alpha . If the temperature ring increases byθ\theta then its circumference will increase by
A. πR2αθ\pi {R^2}\alpha \theta
B. 2πRθ2\pi R\theta
C. πRαθ2\pi {R^{}}\alpha \dfrac{\theta }{2}
D. πRαθ4\pi {R^{}}\alpha \dfrac{\theta }{4}

Explanation

Solution

Hint – In such questions, we need to remember the basic concept of linear expansion of a rod on change in temperature and then convert the length into the circumference of the circle.
Formula used - ΔLL=αΔT\dfrac{{\Delta L}}{L} = \alpha \Delta T

Complete step-by-step solution -
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The ring is generally metallic so when we heat the ring its diameter increases and its not only about a ring. If you heat a metallic disc its diameter increases and it becomes thick as well. This is the thermal expansion of the ring.
Given,
Radius of ring=R
Coefficient of linear expansion=α\alpha
Temperature change=θ\theta
We know that, L=2πRL = 2\pi R
ΔLL=αΔT\dfrac{{\Delta L}}{L} = \alpha \Delta T
ΔLL=αΔT\dfrac{{\Delta L}}{L} = \alpha \Delta T, this is the formula for linear expansion where ΔLand ΔT\Delta L\,{\text{and }}\Delta T is the change in length and change in temperature. This formula is helpful in most of the problems.
ΔL=LαΔT =2πRαΔT=2πRθ  \Delta L = L\alpha \Delta T \\\ = 2\pi R\alpha \Delta T = 2\pi R\theta \\\
=2πRθ= 2\pi R\theta
Hence the correct answer is 2πRθ2\pi R\theta .
Hence, the correct option is B.

Note - In such a type of question we must take care in the application of basic formula of linear expansion and good conceptual knowledge is also required for determining the change in the length of the ring. Doing this will solve your problem and will give you the right answer.