Respuesta :
Answer:
The temperature must be increased by 471.19° to increases the time period from 1.500 s to 1.506 s.
Explanation:
Given that,
Time period = 1.500 sec
Linear thermal expansion [tex]\alpha= 1.90\times10^{-5}\ /C^{\circ}[/tex]
Increases period = 1.506 sec
We need to calculate the initial length
Using formula of time period
[tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]
Put the value into the formula
[tex]1.500=2\pi\sqrt{\dfrac{l}{9.8}}[/tex]
On squaring both side
[tex](1.500)^2=4\pi^2\times\dfrac{l}{9.8}[/tex]
[tex]l=\dfrac{1.500^2\times9.8}{4\pi^2}[/tex]
[tex]l=0.5585\ m[/tex]
We need to calculate the new length
For period to be 1.506
[tex]1.506=2\pi\sqrt{\dfrac{l'}{9.8}}[/tex]
[tex]l'=\dfrac{1.506^2\times9.8}{4\pi^2}[/tex]
[tex]l'=0.5635\ m[/tex]
We need to calculate the temperature
Using formula of thermal expansion
[tex]l'=l(1+\alpha T)[/tex]
[tex]T=\dfrac{l'-l}{l\alpha}[/tex]
Put the value into the formula
[tex]T=\dfrac{0.5635-0.5585}{0.5585\times 1.90\times10^{-5}}[/tex]
[tex]T=471.19^{\circ}[/tex]
Hence, The temperature must be increased by 471.19° to increases the time period from 1.500 s to 1.506 s.
