On a new scale of temperature, which is linear and called the \(\mathrm{W}\) scale, the freezing and boiling points of water are \(39^\circ ~\mathrm{W}\)and \(239^\circ ~\mathrm{W}\) respectively. What will be the temperature on the new scale corresponding to a temperature of \(39^\circ ~\mathrm{C}\) on the Celsius scale?
1. \(78^\circ ~\mathrm{C}\)
2. \(117^\circ ~\mathrm{W}\)
3. \(200^\circ ~\mathrm{W}\)
4. \(139^\circ ~\mathrm{W}\)
If a graph is plotted between temperature of a body on degree Celsius (along y-axis) and degree Fahrenheit [along x-axis] at different temperatures, then the slope of the graph will be:
1.
2.
3.
4.
A temperature of \(100^{\circ}\mathrm {F}\) (Fahrenheit scale) is equal to T K (Kelvin scale). The value of T is:
1. 310.9
2. 37.8
3. 100
4. 122.4
The temperature of a body on the Kelvin scale is found to be x K. When it is measured by a Fahrenheit thermometer, it is found to be xF, then the value of x is:
1. 40
2. 313
3. 574.25
4. 301.25
The triple points of neon and carbon dioxide are \(24.57\) K and \(216.55\) K respectively. The value of these temperatures on Fahrenheit scales will be:
1. | \(-415.44^\circ ~\mathrm{F} ,~-69.88^\circ ~\mathrm{F}\) |
2. | \(-248.58^\circ ~\mathrm{F} ,~-56.60^\circ~ \mathrm{F}\) |
3. | \(315.44^\circ ~\mathrm{F} ,~-69.88^\circ ~\mathrm{F}\) |
4. | \(415.44^\circ ~\mathrm{F} ,~-79.88^\circ~ \mathrm{F}\) |