If a thermometer reads freezing point of water as 20-C and boiling point as 150-C- how much thermometer would read when the actual temperature is 60-C-

Temperature on any scale can be converted into other scale by x L.F.PU.F.P L.F.P = Constant for all scales. nbsp; So, x 20150 20 = 60 0100 0 ⇒ x = 98^ ...

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Standard XIII

Physics

Scale Conversions

If a thermome...

Question

If a thermometer reads freezing point of water as $20^{\circ} C$ and boiling point as $150^{\circ} C,$ how much thermometer would read when the actual temperature is $60^{\circ} C$ ?

98^{\circ} C

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110^{\circ} C

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40^{\circ} C

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60^{\circ} C

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Solution

The correct option is A $98^{\circ} C$
Temperature on any scale can be converted into other scale by $\frac{x - L . F . P}{U . F . P - L . F . P} =$ Constant for all scales.
So, $\frac{x - 20}{150 - 20} = \frac{60 - 0}{100 - 0}$
$\Rightarrow x = 98^{\circ} C$

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Scale Conversions

Standard XIII Physics

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If a thermometer reads freezing point of water as 20∘C and boiling point as 150∘C, how much thermometer would read when the actual temperature is 60∘C?

The correct option is A 98∘CTemperature on any scale can be converted into other scale by x−L.F.PU.F.P−L.F.P= Constant for all scales. So, x−20150−20=60−0100−0 ⇒x=98∘C

If a thermometer reads freezing point of water as $20^{\circ} C$ and boiling point as $150^{\circ} C,$ how much thermometer would read when the actual temperature is $60^{\circ} C$ ?

The correct option is A $98^{\circ} C$
Temperature on any scale can be converted into other scale by $\frac{x - L . F . P}{U . F . P - L . F . P} =$ Constant for all scales.
So, $\frac{x - 20}{150 - 20} = \frac{60 - 0}{100 - 0}$
$\Rightarrow x = 98^{\circ} C$