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10-1.Thermometry, Thermal Expansion and Calorimetry
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A surveyor's $30$-$m$ steel tape is correct at some temperutre. On a hot day the tape has expanded to $30.01$ $m$. On that day, the tape indicates a distance of $15.52$ $m$ between two points. The true distance between these points is :-
A
$15.515 $ $m$
B
$15.520$ $ m$
C
$15.525$ $m$
D
$15$
Solution
$30 (1 + \alpha \theta ) = 30.01$
$15.52 (1 + \alpha \theta ) = x$
$x = 15.52 × \frac{30.01}{30}$
Standard 11
Physics
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