A faulty thermometer has its lower fixed poin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The relationship between a faulty scale $(X)$ and the Celsius scale $(C)$ is given by the formula: $\frac{X - L_X}{U_X - L_X} = \frac{C - L_C}{U_C

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The relationship between a faulty scale $(X)$ and the Celsius scale $(C)$ is given by the formula: $\frac{X - L_X}{U_X - L_X} = \frac{C - L_C}{U_C - L_C}$.Here,the lower fixed point of the faulty scale $L_X = -10^\circ$ and the upper fixed point $U_X = 110^\circ$.For the Celsius scale,the lower fixed point $L_C = 0^\circ$ and the upper fixed point $U_C = 100^\circ$.Given the temperature on the faulty scale $X = 62^\circ$,we substitute the values into the formula:$\frac{62 - (-10)}{110 - (-10)} = \frac{C - 0}{100 - 0}$$\frac{62 + 10}{110 + 10} = \frac{C}{100}$$\frac{72}{120} = \frac{C}{100}$$C = \frac{72}{120} 	imes 100 = 0.6 	imes 100 = 60^\circ C$.

$A$ faulty thermometer has its lower fixed point marked as $-10^\circ C$ and upper fixed point marked as $110^\circ C$. If the temperature of a body shown on this scale is $62^\circ$,the temperature shown on the Celsius scale is ........ $^\circ C$.

Similar Questions

Is the bulb of a thermometer made of a diathermic or adiabatic wall?

Write the relation between Celsius and Fahrenheit scales with a graph.

What is the base of the absolute temperature scale?

On a new temperature scale, the melting point of ice is $20^{\circ} X$ and the boiling point of water is $110^{\circ} X$. $A$ temperature of $40^{\circ} C$ would be indicated on this new temperature scale as (in $^{\circ} X$)

Calculate the temperature which has the same numerical value on the Celsius and Fahrenheit scales.

Vedclass Test Series

Exam Paper Generator

Online Exam Module