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A clock with a metallic pendulum gains 5 s each day at a temperature of `15^@C` and loses 10 s each day at a temperature of `30^@C`. Find the coefficient of thermal expansion of the pendulum metal. |
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Answer» Loss or gain per day `=dT=(1)/(2)alphadtxx86400` 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` `(30-t)/(t-15)=2implies3t=60^@Cimpliest=20^@C` `alpha=(10)/((t-15)xx86400)=(10)/(5xx86400)=0.000023` `2.3xx(10^-5^@)/(C)=2` |
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