Question

The active element of a certain laser is made of a glass rod 29.0 cm long and 1.40 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 62.0°C.

(a) What is the increase in its length? . cm

(b) What is the increase in its diameter?. cm

(c) What is the increase in its volume? cm3

Answer 1

Solution: length of the glass rod l_o = 29.0 cm

Diameter of the glass rod D_o = 1.40 cm

Coefficient of linear expansion α = 9.00*10^-6oC^-1

Change in the temperature ΔT = 62.0 ^oC

Part (a) The increase in the length Δl is given by,

Δl = l_o*α*ΔT

Δl = (29.0cm)*( 9.00*10^-6oC^-1)*(62.0^oC)

Δl = 0.016182 cm                               or

Δl = 1.6182*10^-2 cm

hence the answer is 1.6182*10^-2 cm

---------------------------------------------------------------------------------------------------------------------

Part (b) The change in the diameter can be considered as the linear expansion.

The increase in the diameter ΔD is given by,

ΔD = D_o*α*ΔT

ΔD = (1.40cm)*( 9.00*10^-6oC^-1)*(62.0^oC)

ΔD = 7.812*10^-4 cm.

Hence the answer is 7.812*10^-4 cm.

--------------------------------------------------------------------------------------------------------------------

Part (c) original radius of the rod, r = D_o/2 = 0.70 cm.

Original volume of the rod (cylinder)

V = length*face area

V = l*(π*r²)

V = (29cm)*(3.141*(0.70cm)²

V = 44.64203 cm³

After expansion, new length of the rod is l’ = l_o + Δl = 29cm + 1.6182*10^-2 cm

l’ = 29.016182 cm

After expansion, new diameter of the rod is D’ = D_o + ΔD = 1.4cm + 7.812*10^-4 cm

D’ = 1.4007812 cm

Thus new radius is r’ = 0.7003906 cm.

The volume of cylinder after expansion in length and diameter is given by,

V’ = l’*(π*r’²)

V’ = (29.016182cm)*(3.141*(0.7003906cm)²)

V’ = 44.71680cm³

Thus the increase in the volume of the rod is,

ΔV = V’ – V

ΔV = 44.71680cm³ - 44.64203 cm³

ΔV = 7.4774*10^-2 cm³

hence the answer is 7.4774*10^-2 cm³