The graph between two temperature scales P an | vedclass

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

Question

(B) The general formula for converting between two temperature scales is given by:$\frac{	ext{Reading on scale} - 	ext{Lower fixed point}}{	ext{Upp

Vedclass · Accepted Answer

$\frac{t_Q}{100}=\frac{t_P-30}{150}$

Vedclass · Answer

(B) The general formula for converting between two temperature scales is given by:$\frac{	ext{Reading on scale} - 	ext{Lower fixed point}}{	ext{Upper fixed point} - 	ext{Lower fixed point}} = 	ext{Constant}$From the graph,for scale $P$,the lower fixed point is $30$ and the upper fixed point is $180$ (since $180 - 30 = 150$ divisions).For scale $Q$,the lower fixed point is $0$ and the upper fixed point is $100$ (since $100 - 0 = 100$ divisions).Applying the formula:$\frac{t_P - 30}{180 - 30} = \frac{t_Q - 0}{100 - 0}$Simplifying the expression:$\frac{t_P - 30}{150} = \frac{t_Q}{100}$

The graph between two temperature scales $P$ and $Q$ is shown in the figure. Between the upper fixed point and the lower fixed point,there are $150$ equal divisions on scale $P$ and $100$ divisions on scale $Q$. The relationship for conversion between the two scales is given by:

Similar Questions

The resistance of a thermometer is $100 \Omega$ at the triple point of water $(273 \ K)$ and is $300 \Omega$ at the melting point of gold $(873 \ K)$. The temperature at which the resistance of the thermometer is $200 \Omega$ is: (in $K$)

Explain the study of thermal properties of matter by giving definitions of temperature and heat.

$A$ constant volume gas thermometer shows pressure readings of $50 \, cm$ and $90 \, cm$ of mercury at $0^{\circ} C$ and $100^{\circ} C$ respectively. When the pressure reading is $60 \, cm$ of mercury,the temperature is ......... $^{\circ} C$.

No other thermometer is as suitable as a platinum resistance thermometer to measure temperature in the entire range of

The temperature of a body shown by a faulty Celsius thermometer is $49^{\circ} C$ and by a correct Fahrenheit thermometer is $122^{\circ} F$. The correction to be applied to the faulty thermometer is

Vedclass Test Series

Exam Paper Generator

Online Exam Module