Question

4. A steel sphere is to be passed through a circular
brass ring. At 20°C, the outer diameter of the
sphere is 25 cm 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 : Asteel = 1.2 x 10 c and
Obrass = 2.0 x 10-5•c-
Ans.525°C.​

Answer 1

Answer:

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 C