The freezing point on a thermometer is marked as 20- and the boiling point at as 150- A temperature of 60- C on this thermometer will be read as A- 110-B- 40-C- 65-D- 98-

The correct option is D Temperature on any scale can be converted into other scale by X LFP/UFP LFP = Constant for all scales ∴ X 20^∘/150^∘ 20^∘ = C ...

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Standard X

Physics

Heating Curve of Ice

The freezing ...

Question

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

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Solution

The correct option is C
The melting point on the thermometer = $T_{1} = 20^{\circ} C$
boiling point, $T_{2} = 150^{\circ} C$
Let temperature on thermometer= $t^{\circ} C$
$\frac{t - 20}{130} = \frac{60 - 0}{100}$
$\frac{t - 20}{130} = \frac{3}{5}$
$5 t - 100 = 390$
$5 t = 490$
$t = 98^{\circ} C$

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

The correct option is C The melting point on the thermometer = T1=20∘C boiling point,T2=150∘C Let temperature on thermometer= t∘C t−20130=60−0100t−20130=355t−100=3905t=490t=98∘C

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

The correct option is C
The melting point on the thermometer = $T_{1} = 20^{\circ} C$
boiling point, $T_{2} = 150^{\circ} C$
Let temperature on thermometer= $t^{\circ} C$
$\frac{t - 20}{130} = \frac{60 - 0}{100}$
$\frac{t - 20}{130} = \frac{3}{5}$
$5 t - 100 = 390$
$5 t = 490$
$t = 98^{\circ} C$