A clock with a metallic pendulum gains 5 seconds each day at 22^@C and loses 10 seconds each day at 45^@C . Then,

A clock with a metallic pendulum gains 5 seconds each day at 22^@C and

1.	A clock with a metallic pendulum gains 5 seconds each day at 22^@C and loses 10 seconds each day at 45^@C . Then,
Answer» clock keeps correct time at t`79/3 ""^@C` clock keeps correct time at t=`89/3 ""^@C` coefficient of linear expansion of metallic pendulum is `2.1 xx 10^(-5) ""^@C^(-1)` coefficient of linear expansion of metallic pendulum is `1.5 xx 10^(-5) C^(-1)` Solution :Time taken for one oscillation of pendulum is ` T= 2pi sqrt(L/G)` or`T^2 = 4pi^2 xx L/g `…(i) Partially differentiating (i), we get ` 2T Delta T = 4pi^2 (Delta L)/(g)`….(ii) Dividing (i) by (i) , we get `(Delta T)/(T) = (Delta L)/(2L) = (alpha L Delta t)/(2L ) = 1/2 alpha Delta t` T = 1 DAY = 24 hours = 86400 s `therefore Delta T = 1/2 alpha Delta t xx 86400 ` sec If at this temperature ,clock keeps correct time , then at `22^@C` , gain in time is ` 5 = 1/2 alpha(t - 22) xx 86400 `...(iii) At `45^@C` LOSS in time is ` 10= 1/2 alpha (45- t) xx 86400`...(iv) Dividing (iv) by (iii) , we have `10/5 = (45 - t)/(t - 22)` ` rArr 2(t - 22) = 45 - t` ` 2t - 44 = 45 - t ` ` 3t = 89` ` t = 89/3 ""^@C` substituting this in (iii) , we have ` 5 = 1/2 alpha( 89/3 - 22) xx 86400` `rArr alpha = (5 xx 2 xx 3)/(23 xx 86400) = (30)/(23 xx 0.864) xx 10^(-5)` `= 1.5 xx 10^(-5) ""^@C^(-1)`