A thermometer has wrong calibration- It reads the melting point of ice as -10- - If it reads 60- in place of 50- C - what is the temperature of the boiling point of water on this scale-

The correct option is C 130^∘ For any two thermometric scales, T_1 LFP_1UFP_1 LFP_1=T_2 LFP_2UFP_2 LFP_2 ∴T_celsius 0100=T ( 10)UFP ( 10) ⇒ ...

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Byju's Answer

Standard VII

Physics

Inter-Conversion of Temperature in Different Scales

A thermometer...

Question

A thermometer has wrong calibration. It reads the melting point of ice as $- 10^{\circ}$ . If it reads $60^{\circ}$ in place of $50^{\circ} C$ , what is the temperature of the boiling point of water on this scale.

130^{\circ}

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90^{\circ}

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110^{\circ}

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120^{\circ}

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Solution

The correct option is C $130^{\circ}$
For any two thermometric scales,
$\frac{T_{1} - L F P_{1}}{U F P_{1} - L F P_{1}} = \frac{T_{2} - L F P_{2}}{U F P_{2} - L F P_{2}}$

$∴ \frac{T_{c e l s i u s} - 0}{100} = \frac{T - (- 10)}{U F P - (- 10)}$

$\Rightarrow U F P + 10 = 140$
$∴ U F P = 130^{o}$

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A thermometer has wrong calibration. It reads the melting point of ice as −10∘ . If it reads 60∘ in place of 50∘C , what is the temperature of the boiling point of water on this scale.

The correct option is C 130∘For any two thermometric scales,T1−LFP1UFP1−LFP1=T2−LFP2UFP2−LFP2∴Tcelsius−0100=T−(−10)UFP−(−10)⇒UFP+10=140∴UFP=130o

A thermometer has wrong calibration. It reads the melting point of ice as $- 10^{\circ}$ . If it reads $60^{\circ}$ in place of $50^{\circ} C$ , what is the temperature of the boiling point of water on this scale.