Question

The active element of a certain laser is made of a glass rod 33.0 cm long and 1.30 cm in diameter. Assume the average coefficient of linear expansion of the glass is equal to 9.00 10-6 (°C)−1. The temperature of the rod increases by 70.0°C. (a) What is the increase in its length?(b) What is the increase in its diameter?
cm
(c) What is the increase in its volume?
cm³

Answer 1

L = 33cm , d =1.3 cm, r =0.65cm

$\alpha$ = 9x 10-6 (°C)−1

$\Delta$T = 70 ⁰C

(a)$\alpha$ =$\Delta$L/L($\Delta$T)

$\Delta$L = $\alpha$L$\Delta$T = (9x10^-6)(0.33)(70)

$\Delta$L =0.0208 cm

(b) $\alpha$ =$\Delta$d/d($\Delta$T)

$\Delta$d = $\alpha$d$\Delta$T = (9x10^-6)(0.013)(70)

$\Delta$d =0.000819 cm

(c) V =pi*r²L = (3.14*0.65*0.65*33) =43.78 cm^3

Coefficient of volume expansion $\gamma$ =3$\alpha$ = $\Delta$V/V($\Delta$T)

$\Delta$V= V(3$\alpha$)($\Delta$T) = (43.78x3x9x10^-6x70)

$\Delta$V=0.08274 cm³