Answer:
T = 44.35 °C
Explanation:
d = 32cm
R = 16 cm
Lsteel = 10m
T1 = 20° C
Space = 0.3cm
The space between the sphere and the floor is represented by δL(total) after the temperature increases.
As the temperature increases, both will expand.
So,
0.3 x 10^(-2) = δL(steel) + δR(brass)
= {L(o) x α(steel) x δT} + {R(o) x {α(brass) x δT}
= {10 x 1.2 x 10^(-5) x (T-20)} + {0.16 x 2 x 10^(-5) x (T-20)}
= 12.32 x 10^(-5) x(T-20)
Therefore (T-20) = (0.3 x 10^(-2)) / {12.32 x 10^(-5)}
T = 20 + 24. 35 = 44.35 °C