1.

A clock with a metallic pendulum at 15°C runs faster by 5 sec each day and at 30°C, runs slow by 10 sec. The coefficient of linear expansion of the metal is nxx10^(-6)//^(@)Cwhere the value of n in nearest integer is ___.

Answer»


Solution :From `T=2pisqrt((L(1+alphaT))/(G))`
We get , `(dt)/(T)=(1)/(2)alphadt`
Loss or gain per day `=dT=(1)/(2)adtxx86400`
SINCE `T=86400s` for each day ,
At `15^(@)C,5=(1)/(2)ALPHA(t-15)xx86400,"At "30^(@)C,10=(1)/(2)alpha(30-t)xx86400`
`therefore(30-t)/(t-15)=2rArr3t=60^(@)C,T=20^(@)C`
`thereforealpha=(10)/((t-15)xx86400)=(10)/(5xx86400)=0.0000023=2.3xx10^(-6)//^(@)C=2`


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