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Physics Question on thermal properties of matter

A steel tape is calibrated at 20∘C20^{\circ} C. when the temperature of the day is −10∘C-10^{\circ} C , the percentage error in the measurement with the tape is α=12×10−6 /∘C)\alpha=12 \times 10^{-6}\,/^{\circ} C)

A

0.036

B

0.0036

C

0.0018

D

0.00036

Answer

0.00036

Explanation

Solution

We know, l2=l1(1+αdt) where dt= change in temperature =30∘Cl_{2}=l_{1}(1+\alpha dt ) \text { where } dt =\text { change in temperature }=30^{\circ} C The percentage error in the measurement will be,  (dl/l) X10o =α dt 100=12×30×100×10−6 =36×10−3=0.036%\begin{array}{l} \text { (dl/l) X10o }=\alpha \text { dt } 100=12 \times 30 \times 100 \times 10^{-6} \\\ =36 \times 10^{-3}=0.036 \% \end{array}