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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 ___. |
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Answer» 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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