Question

if anyone solve this questions
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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:

$\begin{gathered}\bf \leadsto \dfrac{C - 0}{100 - 0} = \dfrac{Reading - LFP}{UFP - LFP} \\ \\ \rm \leadsto \dfrac{6 \cancel{0}}{10 \cancel{0 }} = \dfrac{Reading - 10}{140 - 10} \\ \\ \rm \leadsto \dfrac{Reading - 10}{130} = \dfrac{3}{5} \\ \\ \rm \leadsto Reading - 10 = 130 \times \dfrac{3}{5} \\ \\ \rm \leadsto Reading - 10 =26 \times 3 \\ \\ \rm \leadsto Reading - 10 =78 \\ \\ \rm \leadsto Reading =78 + 10 \\ \\ \rm \leadsto Reading =88 \degree\end{gathered} </p><p>⇝ </p><p>100−0</p><p>C−0</p><p> </p><p> = </p><p>UFP−LFP</p><p>Reading−LFP</p><p> </p><p> </p><p>⇝ </p><p>10 </p><p>0</p><p> </p><p> </p><p>6 </p><p>0</p><p> </p><p> </p><p> </p><p> = </p><p>140−10</p><p>Reading−10</p><p> </p><p> </p><p>⇝ </p><p>130</p><p>Reading−10</p><p> </p><p> = </p><p>5</p><p>3</p><p> ⇝Reading−10=130× </p><p>5</p><p>3</p><p> </p><p> </p><p>⇝Reading−10=26×3</p><p>⇝Reading−10=78</p><p>⇝Reading=78+10</p><p>⇝Reading=88°</p><p> </p><p> </p><p></p><p>\therefore∴ Reading of faulty thermometer = 88°$

Answer 2

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:

⇝

100−0

C−0

=

UFP−LFP

Reading−LFP

⇝

10

0

6

0

=

140−10

Reading−10

⇝

130

Reading−10

=

5

3

⇝Reading−10=130×

5

3

⇝Reading−10=26×3

⇝Reading−10=78

⇝Reading=78+10

⇝Reading=88°

⇝100−0C−0=UFP−LFPReading−LFP⇝10060=140−10Reading−10⇝130Reading−10=53⇝Reading−10=130×53⇝Reading−10=26×3⇝Reading−10=78⇝Reading=78+10⇝Reading=88°∴∴Readingoffaultythermometer=88°