The freezing point on a thermometer is marked as $20^{o}$ and the boiling point as $150^{o}$ . A temperature of $(60^{o} C)$ on this thermometer will be read as:

40^{o}

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65^{o}

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98^{o}

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None of these

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Solution

The correct option is C $98^{o}$
$\begin{matrix} B y u sin g t h e f o r m u l a \frac{T_{t} - T_{f P}}{T_{B P} - T_{f p}} = \frac{T - 0}{100 - 0} \frac{T_{t} - 20}{150 - 20} = \frac{60}{100} T_{t} - 20 = \frac{60 \times 130}{100} T_{t} = 78 + 20 = 98^{\circ} H e n c e, t h e o p t i o n C i s t h e c o r r e c t a n s w e r . \end{matrix}$

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The freezing point on a thermometer is marked as 20o and the boiling point as 150o. A temperature of (60oC)on this thermometer will be read as:

The correct option is C 98oByusingtheformulaTt−TfPTBP−Tfp=T−0100−0Tt−20150−20=60100Tt−20=60×130100Tt=78+20=98∘Hence,theoptionCisthecorrectanswer.

The freezing point on a thermometer is marked as $20^{o}$ and the boiling point as $150^{o}$ . A temperature of $(60^{o} C)$ on this thermometer will be read as: