A thermometer has wrong calibration (of course at equal distance and the capillary is of uniform diameter). It reads the melting point of ice as $- 10 ° C$ . It reads $60 ° C$ in place of $50 ° C$ . The boiling point of water on this scale is

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100 ° C

130 ° C

110 ° C

120 ° C

answer is B.

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Detailed Solution

faulty reading =

60 ° C

LFP = $- 10 ° C$ , C = 50

$\frac{R - LFP}{UFP - LFP} = \frac{C - LFP}{UFP - LFP}$

$\Rightarrow \frac{60 - (- 10)}{UFP - (- 10)} = \frac{C}{100} \Rightarrow \frac{60 + 10}{UFP + 10} = \frac{50}{100}$

$\Rightarrow \frac{70}{UFP + 10} = \frac{1}{2} \Rightarrow UFP + 10 = 140$

$∴ UFP = 130 ° C$

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