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

40^{o}

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65^{o}

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98^{o}

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110^{o}

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Solution

The correct option is D $98^{o}$
Given: Freezing point on thermometer $= 20$ , Boiling point on thermometer $= 150$ ,
Hence, number of equal parts on thermometer $= 150 - 20 = 130$ , temperature on celsius scale $= 20^{o} C$
Let temperature on thermometer $= t$ ,
Therefore ,
$\frac{t - 20}{130} = \frac{60 - 0}{100}$ ,
$\Rightarrow$ $t - 20 = 78$
$\Rightarrow$ $t = 78 + 20 = 98^{o}$

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

The correct option is D 98oGiven: Freezing point on thermometer =20 , Boiling point on thermometer =150 ,Hence, number of equal parts on thermometer =150−20=130, temperature on celsius scale =20oCLet temperature on thermometer =t ,Therefore , t−20130=60−0100 ,⇒ t−20=78 ⇒t=78+20=98o

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