A cylindrical steel plug is inserted into a circular hole of diameter 2.60 cm in a brass plate. When the plug and the plates are at a temperature of `20^(@)C,` the diameter of the plug is 0.010 cm cmaller than that of the hole. The temperature at which the plug will just fit in it is
`("Given",alpha_(steel)=(11xx10^(-6))/("^(@)C)and alpha_(bress)=(19xx10^(-6))/C)`

Options

A. `-48^(@)C`
B. `-20^(@)C`
C. `-10^(@)C`
D. `-485^(@)C`

Answer

D

Explanation

Diameter of brass plate = 2.6 cm and diameter of steel plate
`" "=(2.6-0.01)cm=2.59cm`
Now, `" "(d+Deltad)_(b)=(d+Deltad)_(s)`
or `" "d_(b)(1+alphaDeltatheta)_(b)=d_(s)(1+alphaDeltatheta)_(s)`
`therefore" "2.6 (1+19xx10^(-6)Delta theta )=2.59 (1+11 xx10^(-6)Deltatheta)`
`therefore" "Deltatheta=-478^(@)C or theta_(f)=-458^(@)C`