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 -A- 110- CB- 40- CC- 60- CD- 98- C

The correct option is D 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. nbsp; So, x 20 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

110^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60^{\circ} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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, $\frac{x - 20}{150 - 20} = \frac{60 - 0}{100 - 0}$
$\Rightarrow x = 98^{\circ} C$

Suggest Corrections

Similar questions

Q. If a thermometer on

C -

scale reads freezing point of water as

20 C

and boiling point as

150 C

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

60^{\circ} 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

Q. 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.

Join BYJU'S Learning Program

Explore more

Scale Conversions

Standard XII Physics

Join BYJU'S Learning Program

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, $\frac{x - 20}{150 - 20} = \frac{60 - 0}{100 - 0}$
$\Rightarrow x = 98^{\circ} C$