Question

if a thermometer reads freezing point of water 10° and boiling point 140°, then what will it read when actual temperature of water is 60°c​

Answer 1

Given:

Upper Fixed Point (UFP) = 140°

Lower Fixed Point (LFP) = 10°

Actual temperature of water (C) = 60°C

To Find:

Reading of faulty thermometer

Answer:

By using the following relation we get:

$\bf \leadsto \dfrac{C - 0}{100 - 0} = \dfrac{Reading - LFP}{UFP - LFP} \\ \\ \rm \leadsto \dfrac{6 \cancel{0}}{10 \cancel{0 }} = \dfrac{Reading - 10}{140 - 10} \\ \\ \rm \leadsto \dfrac{Reading - 10}{130} = \dfrac{3}{5} \\ \\ \rm \leadsto Reading - 10 = 130 \times \dfrac{3}{5} \\ \\ \rm \leadsto Reading - 10 =26 \times 3 \\ \\ \rm \leadsto Reading - 10 =78 \\ \\ \rm \leadsto Reading =78 + 10 \\ \\ \rm \leadsto Reading =88 \degree$

$\therefore$ Reading of faulty thermometer = 88°

Answer 2

Explanation:

Given : -

• if a thermometer reads freezing point of water 10° .

• boiling point 140°,

To Find : -

• what will it read when actual temperature of water is 60°c

Solution : -

let x = reading of new thermometer ,

then,

➻ ( x - 10)/(140 - 10) = C/100

➻ (x -10)/13 = C/10

➻ ( x - 10)/13 = 60/10

➻ ( x - 10)/13 = 6

➻ x - 10 = 78

➻ x = 78 + 10

➻ x = 88