If the linear expansivity of a metal rod is 4 \(\times\) 10\(^{-5}\) per \(^{\circ}\)C, what will be the new length of the rod if it is heated from 15\(^{\circ}\)C t...
Question
If the linear expansivity of a metal rod is 4 \(\times\) 10\(^{-5}\) per \(^{\circ}\)C, what will be the new length of the rod if it is heated from 15\(^{\circ}\)C to 95\(^{\circ}\)C from its original length of 20cmOptions
A)
0.064 cm
B)
0.64 cm
C)
20.64 cm
D)
20.064 cm
Related Lesson: Polarization | Electric Charges and Fields
The correct answer is D.
Explanation:
Coefficient of linear eapsnsivity,\( \alpha=4\times10^-⁵ \)
Change in temperature, \( \Delta\theta=95°C-15°C
=80° C
\)
Original length, l¹=20cm
Final length, l²=xcm
\( \alpha=(l²-l¹)÷l¹\Delta\theta
4\times10^–⁵=(x-20)÷(20\times80)
x=20.064cm \)
Change in temperature, \( \Delta\theta=95°C-15°C
=80° C
\)
Original length, l¹=20cm
Final length, l²=xcm
\( \alpha=(l²-l¹)÷l¹\Delta\theta
4\times10^–⁵=(x-20)÷(20\times80)
x=20.064cm \)
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Coefficient of linear eapsnsivity,\( \alpha=4\times10^-⁵ \)
Change in temperature, \( \Delta\theta=95°C-15°C
=80° C
\)
Original length, l¹=20cm
Final length, l²=xcm
\( \alpha=(l²-l¹)÷l¹\Delta\theta
4\times10^–⁵=(x-20)÷(20\times80)
x=20.064cm \)
Thank you for providing an explanation for the question!
Alpha = L2 - L1/ ∆temp. (L1)
4*10^-5 = L2 - 20/ 95-15 (20)
4*10^-5 = L2- 20/ 80 (20)
4*10^-5 = L2- 20/ 1600
4*10^-5 * 1600 = L2 - 20
0.064 + 20 = L2
Therefore, L2 = 20.064cm
I think A is the correct answer