The temperature coefficient of resistance in equation R ( T )= R 0[1+ ( T T 0)]

Question

The temperature coefficient of resistance in equation R(T)=R0[1+(TT0)] equals the temperature coefficient of resistivity in equation (T)=0[1+(TT0)] only if the coefficient of thermal expansion is small. A cylindrical column of mercury is in a vertical glass tube. At 20 C, the length of the mercury column is 12.0 cm. The diameter of the mercury column is 1.6 mm and doesn't change with temperature because glass has a small coefficient of thermal expansion. The coefficient of volume expansion of the mercury is 1810^5K^1, its resistivity at 20 C is 9510^8m, and its temperature coefficient of resistivity is 0.00088 (C^)1.

(A) What is the change in its length?

The answer is not 43.2*10^-5

(B)Explain why the coefficient of volume expansion, rather than the coefficient of linear expansion, determines the change in length.

Explanation / Answer

A) inial temperature T = 200 C final temperature is not given say it is t

Length = 12 cm

Dia = 1.6 mm, radius r = 0.08 cm

Initial Volume Vi = 3.14 x 12 x 0.082 = 0.2412 cc

Coefficient of Volume expansion a = 18.1 x 10-5

Increase in volume Vd = 18.1 x 10 -5 x0.2412 x t

area = 3.14 x 0.64 remains constant

Increase in length 18.1 x 24.1 /(3.14x0.64) = 217.243 x t x 10-5 cm

Assume thetemperature increases by 1 degree then the chnage in length = 217 x 10-5 cm

B) In this case volume expansion is important is important because it is a liquid and the container can expand in which case linear expansion cannot be determind exactly

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