A copper ring has a diameter of exactly `25 mm` at its temperature is `0^@ C`. An aluminium sphere has a diameter is exactly `25.05 mm` and its temperature it `100^@ C`. The sphere is placed on top of the rind and two allowed to come to thermal equilibrium. The ratio of the mass of the sphere and ring is (given `alpha_(cu) = 17 xx 10^(-6).^@C^(-1)alpha_(Al) = 2.3 xx 10^(-5) .^@C^(-1)` specific heat of `Cu` is `0.0923 cal//g^@C` and for `Al` is `0.215 cal//g^@C`.
A
`1//5`
B
`23//108`
C
`23//54`
D
`216//23`
Text Solution
Verified by Experts
The correct Answer is:
C
A equilibrium `(MsDelta t)_(Cu) = (MsDelta t)_(Al)` When sphere passing through ring then Radius of sphere = Radius ring `(l_(2))_(Cu) = (l_(2))_(Al)` `25(1+alpha_(Cu) Deltat)=25.05(1+ alpha_(Al) Delta t)`.
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