Question

-24.) A copper rod 30 cm long has cross-sectional area 0.8 cm2. What tensile force
would be required to stretch the rod by the same amount as produced by heating
it through 10 °C? [Y= 1.17 x 109 Pa, &c. = 1.7 x 10 -5 °C -1]​

Answer 1

A copper rod 30 cm long has cross-sectional area 0.8 cm^2. The tensile force

that would be required to stretch the rod by the same amount as produced by heating  it through 10 °C is 15.912 N.

1) We know, change in length of a rod or wire due to tensile force (P) is

dl=(PL/AY)

2) Again, change in length due to change of temperature is

dl = L * c * dT (c is the coefficent of thermal expansion, dT is the change in temperature)

3) So equating the above two relations we get,

(P * L/ A * Y) = L * c * dT

4) Putting the values given in the question,

(P * 0.3) / (0.8 * 10^(-4) * 1.17 * 10^9)  =  0.3 * 1.7 * 10^(-5) * 10

So, on calculating, the value of P = 15.912 N