A steel tape is correctly calibrated at `20^(@)C` and is used to measure the length of a table at `30^(@)C`. Find the percentage error in the measurement of length. `[alpha_(steel) = 11 xx 10^(-5)//.^(@)C]`

A steel tape is callibrated at 20^(@)C . When the temperature of the day is -10^(@)C , the percentage error in the measurement with the tape is (alpha=12" "10^(-6)//""^(@)C)

A steel taps is calibrated at 20^(0)C . When the temperature of the day is -10^(0)C , the percentage error in the measurement with the tap is (alpha = 12 xx 10^(-6 //0)C)

A steel scale gives correct reading at 10^(@)C . If the length of a rod is measured with the help of that scale at 30^(@)C then the measured length will be

A metal tape gives correct measurement at 15^(@)C . It is used to measure a distance of 100m at 45^(@)C . The error in the measurement, it alpha=12xx10^(-6)//1^(@)C is

A brass scale correctly calibrated at 15°c is employed to measure a length at a temperature of 35°C. It the scale measures a reading of 75 cm ( alpha_("brass") = 2 xx 10^(-5) //^(@) C)

A steel tape is callibrated at 20^@ C . On a cold day when the temperature is -15^@ C , what will be the percentage error in the tape ?

A steel tape is callibrated at 20^@ C . On a cold day when the temperature is -15^@ C , what will be the percentage error in the tape ?

A precise steel tape measure has been calibrated at 20^(@)C . At 40^(@)C a. will it read high or low, and b. what will be the percentage error?

A second's pendulum clock has a steel wire. The clock is calibrated at 20^@ C . How much time does the clock lose or gain in one week when the temperature is increased to 30^@ C ? alpha_(steel) = 1.2 xx 10^(-5).^@C^(-1) .