The length of a brass rod at 20^@Cis 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 at0^@C ?Take, alpha_("iron") = 1.2 xx 10^(-5) ""^@C^(-1) alpha_("brass") = 1.9 xx 10^(-5) ""^@C^(-1)

The length of a brass rod at 20^@Cis observed to be 0.8 m when measure

1.	The length of a brass rod at 20^@Cis 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 at0^@C ?Take, alpha_("iron") = 1.2 xx 10^(-5) ""^@C^(-1) alpha_("brass") = 1.9 xx 10^(-5) ""^@C^(-1)
Answer» Solution : 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`