Question

please answer this question​

Answer 1

Answer:

The best way to solve this type of questions is to follow the rule :

$\boxed{ \blue{ \bold{\dfrac{reading - lowest \: pt}{highest \: pt - lowest \: pt} = constant}}}$

The above equation is always true for any kind of temperature unit (irrespective of SI units or any other imaginary unit)

Now we shall be putting this equation for each of Celsius , Farenheit and Kelvin scale.

$\red{ = > \dfrac{C - 0}{100 -0 } = \dfrac{F - 32}{212 - 32} = \dfrac{K - 273}{373 - 273} }$

$\red{ = > \dfrac{C }{100 } = \dfrac{F - 32}{180} = \dfrac{K - 273}{100} }$

Cancelling the denominator by a common factor of 20 , we get :

$\red{ = > \dfrac{C }{5 } = \dfrac{F - 32}{9} = \dfrac{K - 273}{5} }$

So final answer is :

$\boxed{ \green{ \bold{\dfrac{C }{5 } = \dfrac{F - 32}{9} = \dfrac{K - 273}{5} }}}$

Correct option : b)

Answer 2

Solution :

⏭ To Find:

✏ The relation between Celcius, Fehreinheit and kelvin scale.

⏭ Formula:

✏ For any temperature reading scale,

$\bigstar \: \boxed{ \tt{ \pink{ \dfrac{SR - UFP}{UFP - LFP} = constant}}} \: \bigstar$

⏭ Terms indication:

• SR = Scale reading
• UFP = Upper Fix Point
• LFP = Lower Fix Point

⏭ Calculation:

✏ If C, F and K are the temperatures of a body on Celsius, Fehrenheit and Kelvin scales respectively, then,

$\displaystyle \dashrightarrow \sf \: \dfrac{C - 0}{100 - 0} = \dfrac{F - 32}{212 - 32} = \dfrac{K- 273}{373 - 273} \\ \\ \dashrightarrow \sf \: \dfrac{C}{100} = \dfrac{F - 32}{180} = \dfrac{K - 273}{100} \\ \\ \dashrightarrow \boxed{ \tt{ \purple{ \dfrac{C}{5} = \dfrac{F - 32}{9} = \frac{K - 273}{5} }}} \: \orange{ \bigstar}$

✒ Option-B)

⏭ Additional information:

• F = 9/5 (C) + 32
• K = C + 273.15
• K = 5/9 (F) + 255.37