The active element of a certain laser is an ordinary glass rod 19.1 cm long and

Question

The active element of a certain laser is an ordinary glass rod 19.1 cm long and 1.39 cm in diameter. If the temperature of the rod increases by 78.5°C, calculate its increase in length. (Use 9.0×10-6 °C-1 for the coefficient of linear expansion for glass. Use this value to find the coefficient of volume expansion in the later parts of the problem.) 1.35×10-2 cm You are correct. Your receipt no. is 169-8012 Help: Receipt Previous Tries Calculate its increase in diameter. 9.82×10-4 cm You are correct. Your receipt no. is 169-4346 Help: Receipt Previous Tries Calculate its increase in volume.

Explanation / Answer

PART A

Increase = l*d*deltaT

Increase = (19.1/100)*9.0×10-6*78.5 = 1.35*10-4 m

PART B

Ao = pi*d2/4

Af = Ao*(1+Beta*DeltaT)

pi*d2/4 =  pi*d2/4(1+ 18*10-6*78.5)

Solving that we have

d' =d + d*1.0007

Increase = 9.82*10-6 m

PART C

Increase in volumme = Vo*gamma*deltaT

= pi*(1.392)*19.1/(4*106)*27*10-6*78.5 = 6.143*10-8 m3

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.