An alumimum plate fixed in a horizontal position has a hole of diameter 2.000 cm. A steel sphere of diameter 2.005 cm rests on this hole. All the lengths refer to a temperature of `10^C`.The temperature of the entire system is slowly increased. At what temperature will the ball fall down? coefficient of linear expansion of aluminium is `23 xx 10^(-6) o^C(-1)` and that of steel is `11 xx 10^(-6) 0^C^(-1)`.

Explanation

`Given d_st=2.005cm`
`d_Al=2.000cm`
`alpha_st=11 xx 10^-6/^0C`
`alpha_Al= 23 xx 10^-6/^0C`
we know,
`d'_st = 2.005(1+alpha_st DeltaT)`
`(where Delta T is change in temperature)`
`rArr d'_st=2.005+2.005 xx11`
`xx 10^-6 xx Delta T`
`d'_Al =2(1+alpha_Al xx Delta T)`
`=(2+2 xx 23 xx 10^-6 xx Delta T)`
The steel ball will fall when both the diameters become equal.
so,
`rArr(2.005+2.005 xx 11 xx 10^6 Delta T)`
`=(2+2 xx 23 xx 10^6 Delta T)`
`rArr ((46-22.055) xx 10^-6 Delta T) = 0.005`
`rArr (Delta T) = (0.005 xx 10^6)/23.945 = 208.81`
Now `Delta T = T_2-T_1= T_2 -10^0C`
`[:. T_1=10^0C given]`
`rArr T_2 = Delta T + T_1 =208.81 +10`
`=218.81= ~219^0C`