The freezing point on a thermometer is marked as 20- and the boiling point as 150- - A temperature of 60- C on this thermometer will be read as-

The correct option is D 98^∘ For any two thermometric scales, T_1 LFP_1UFP_1 LFP_1=T_2 LFP_2UFP_2 LFP_2 where LFP and UFP are Lower Fixed Poin ...

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Byju's Answer

Standard VII

Physics

Inter-Conversion of Temperature in Different Scales

The freezing ...

Question

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

40^{\circ}

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65^{\circ}

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98^{\circ}

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110^{\circ}

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Solution

The correct option is D $98^{\circ}$
For any two thermometric scales,
$\frac{T_{1} - L F P_{1}}{U F P_{1} - L F P_{1}} = \frac{T_{2} - L F P_{2}}{U F P_{2} - L F P_{2}}$
where $L F P a n d U F P$ are Lower Fixed Point and Upper Fixed Point respectively.

$∴ \frac{T_{c e l s i u s} - 0}{100^{o}} = \frac{T - 20^{o}}{130^{o}}$

$\Rightarrow T = 78^{o} + 20^{o} = 98^{o}$

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Similar questions

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

Q. If a thermometer reads freezing point of water as

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and boiling point as

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how much thermometer would read when the actual temperature is

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Q. If a thermometer on

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and boiling point as

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, how much the thermometer would read when the actual temperature is

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The freezing point on a thermometer is marked as 20∘ and the boiling point as 150∘ . A temperature of 60∘C on this thermometer will be read as.

The correct option is D 98∘For any two thermometric scales,T1−LFP1UFP1−LFP1=T2−LFP2UFP2−LFP2where LFP and UFP are Lower Fixed Point and Upper Fixed Point respectively.∴Tcelsius−0100o=T−20o130o⇒T=78o+20o=98o

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