Question

A steel sphere is to be passed through a circular brass ring. At 20 degree celcius, the outer diameter of the sphere is 25cm and the inner diameter of the ring is 24.9 cm. If both are heated together, find the temperature at which the sphere will just pass through the ring. Given:- alpha steel = 1.2 x 10 ^-5 degree centigrade^-1
Alpha brass = 2 x 10^-5

Answer 1

Answer:

525

Step-by-step explanation:

Given A steel sphere is to be passed through a circular brass ring. At 20 degree celcius, the outer diameter of the sphere is 25cm and the inner diameter of the ring is 24.9 cm. If both are heated together, find the temperature at which the sphere will just pass through the ring. Given:- alpha steel = 1.2 x 10 ^-5 degree centigrade^-1

Alpha brass = 2 x 10^-5

alpha steel = 1.2 x 10^-5, D steel = 25 cm

alpha brass = 2 x 10 ^-5, D brass = 24.9 cm

t = 20 degree celsius

D = Do(1 + α Δt)

Do brass ( 1 + αbrassΔt) = Do steel(1 + α steelΔt)

 24.9(1 + 2 x 10^-5(t - 20) = 25(1 + 1.2 x 10^-5(t - 20)

24.9 / 25 = 1 + 1.2 x 10^-5(t - 20) / 1 + 2 x 10^-5(t - 20)

 0.996 = 1 - 24 x 10^-5 + 1.2 x 10^-5t / 1 - (40 x 10^-5) + 2 x 10^-5t

 0.996 = 1.2 x 10^-5t + 0.99976 / 2 x 10^-5t + 0.9996

 1.2 x 10^-5t - 1.992 x 10^-5t = 0.9956016

 t = 0.0041584 / 0.792 x 10 ^-5

t = 525.05 degree celsius.

Answer 2

Step-by-step explanation:

here alpha of ring is Greater than alpha of the sphre