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The length of a brass rod at 20^@C is o...

The length of a brass rod at `20^@C` is observed to be 0.8 m when measured by an iron scale, correct at `0^@C` . What will be the correct length of the rod at`0^@C` ? Take, `alpha_("iron") = 1.2 xx 10^(-5) ""^@C^(-1)`
`alpha_("brass") = 1.9 xx 10^(-5) ""^@C^(-1)`

Text Solution

Verified by Experts

Length of brass rod at `20^@C, l_1 = 0.8 m = 80 cm`
Length of brass rod at `0^@C, l_2 = ? `
It is given that the iron scale is correct at `0^@C` .
So at` 0^@C` , each division is of 1 cm.
At `20^@C` , each cm division is of `(1 + 1.2 xx 10^(-5) xx 20)`
= 1.00024 cm
True length of iron scale at `20^@C = 100 xx 1.00024 cm = 100.024 cm`
True length of brass rod at `20^@C = 80 xx 1.00024 cm`
` = 80.0192 cm`
So , `l_2 (1 + 19 xx 10^(-5) xx 20) = 80.0192 cm`
` rArr l_2(1+0.00038) = 80.0192 cm`
` rArr 1.00038 l_2 = 80.0192 `
` rArr l_2 = (80.0192)/(1.00038) =79.98 cm`
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