1.

If the gap between steel sails on railway track of 66m long is 3.63 cm at 10^(@)C. Then at what value of temperature will be just touch of steel is 11xx10^(-6)""^(@)C

Answer»

SOLUTION :`L_(0)=66 m`
`L_(o)=6600 cm`
`alpha=11xx10^(-6)""^(@)C`
`alpha=(DeltaL)/(L_(o) DeltaT)`
`L_(t)-L_(o)=3.63 cm`
`DeltaL=3.63 cm`
`t_(1)=10^(@)C t_(2) =?`
so `DeltaT=(DeltaL)/(L_(o)xxalpha)`
`DeltaT=(3.63)/(6600xx11xx10^(-6))`
`t_(2)-t_(1)=50`
`t_(2)=10=50`
so final temperature `t_(2)=50+10=60^(@)C`


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