A steel rod is 4000cm in diameter at `30^(@)C` A brass ring has an interior diameter of `3.992cm at 30^(@)` in order that the ring just slides onto the steel rod the common temperature of the two should be nearly `(alpha_(steel)=11xx10^(-6)/(^(@)C) and alpha_(brass)=19xx10^(-6)/(^(@)C)`

Options

A. `200^(@)C`
B. `250^(@)C`
C. `280^(@)C`
D. `400^(@)C`

Answer

C

Explanation

for ring just slides on to the steel rod the diameter of rod and ring should be equal to each other and suppose due to Deltatheta increment in temperature the diameter of both are equal then
`4(1+alpha_(s)Deltatheta)=3.992(1+alpha_(Brass)Deltatheta)`
`4+4xx11xx10^(-6)xxDeltatheta=3.992+3.992xx20xx10^(-6)xxDeltatheta`
`4+44xx10^(-6)Deltatheta=3.992+79.84xx10^(-6)xxDeltatheta`
`0.008=35.84xx10^(-6)Deltatheta`
`(8xx10^(3))/(35.84)=Deltatheta=8000/35.84=283`
so if temperature increased by `223^(@)C` then ring will start to slide and this temperature will equal to
`theta=30^(@)+Deltatheta=30+253=283^(@)C`
`theta=283^(@)Capprox283^(@)C`