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A steel tape 1m long is correctly calibr...

A steel tape 1m long is correctly calibrated for a temperature of `27^(@)C`. The length of a steel rod measured by this tape is found to be `63.0cm` on a hot day when the temperature is `45.0^(@)C`. What is the acutual length of the steel rod on that day? what is the length of the same steel rod on a day when the temperature is `27.0^(@)C` ? coefficient of linear expansion of steel ` = 1.20xx10^(-5) .^(@)C^(-1)`.

Text Solution

Verified by Experts

Length of the steel tape at `27^(@)C` is `100cm` i.e., `L=100cm and T = 27^(@)C`
The length of steel tape at `45^(@)C` is `L ' = L+Delta L = L + alpha L Delta T`
`=100+(1.20xx10^(-5))xx100xx(45^(@)-27^(@))=100.0216cm`
Length of `1cm` mark at `27^(@)C` on this scale, at `45^(@)C` =`100.0216//100cm`
lenght of `63cm` measured by this tape at `45^(@)C` will be = `(100.0216)/(100)xx63 =63.0136 cm`
Lenght of the same steel rod on a day when the temperature is `27^(@)=63xx1 = 63cm`.
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Knowledge Check

  • A steel tape 1 m long is correctly calibrated for a temperature of 27.0^(@)C . The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45^(@)C . Coefficient of linear expansion of steel = 1.20 xx 10^(-5) //K . what is the actual length of the steel rod on that day?

    A
    `63.0136 cm`
    B
    `63.2134 cm`
    C
    `63.1526 cm`
    D
    `63.3136 cm`

  • If a cylinder of diameter 1.0cm at 30^(@)C is to be slid into a hole of diameter 0.9997 cm in a steel plate at the same temperature, the minimum required rise in the temperature of the plate is: (Coefficient of linear expansion of steel = 12 xx 10^(-6//@)C )

    A
    `25^(@)C`
    B
    `35^(@)C`
    C
    `45^(@)C`
    D
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  • A steel rod of length 1 m is heated from 25^@ "to" 75^@ C keeping its length constant. The longitudinal strain developed in the rod is ( Given, coefficient of linear expansion of steel = 12 xx 10^-6//^@ C ).

    A
    `6 xx 10^-4`
    B
    `-6 xx 10^-5`
    C
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    D
    zero

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