MHT CET · Physics · Thermal Properties of Matter
On an imaginary linear scale of temperature (called ' \(\mathrm{W}\) ' scale) the freezing and boiling points of water are \(39^{\circ} \mathrm{W}\) and \(239^{\circ} \mathrm{W}\) respectively. The temperature on the new scale corresponding to \(39^{\circ} \mathrm{C}\) temperature on Celsius scale will be
- A \(139^{\circ} \mathrm{W}\)
- B \(78^{\circ} \mathrm{W}\)
- C \(117^{\circ} \mathrm{W}\)
- D \(200^{\circ} \mathrm{W}\)
Answer & Solution
Correct Answer
(C) \(117^{\circ} \mathrm{W}\)
Step-by-step Solution
Detailed explanation
In Celsius scale the freezing and boiling point are \(0^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\). In the given imaginary scale the freezing and boiling points are \(39^{\circ} \mathrm{W}\) and \(239^{\circ} \mathrm{W}\). Hence, we can write
\(
\frac{\mathrm{C}-0}{100}=\frac{\mathrm{W}-39}{200}
\)
For \(\mathrm{C}=39^{\circ}, \frac{39}{100}=\frac{\mathrm{W}-39}{200}\)
Solving, \(\mathrm{W}=117\)
\(
\frac{\mathrm{C}-0}{100}=\frac{\mathrm{W}-39}{200}
\)
For \(\mathrm{C}=39^{\circ}, \frac{39}{100}=\frac{\mathrm{W}-39}{200}\)
Solving, \(\mathrm{W}=117\)
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