366611 A constant volume gas thermometer shows pressure reading 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

366612 If absolute zero is \( - 273.15^\circ C\) on Celsius temperature scale, then the absolute zero on the Fahrenheit scale is

366613 On a new scale of temperature (which is linear) and called the \(W\) scale, the freezing and boiling points of water are \(39^\circ W\) and \(239^\circ W\) respectively. What will be the temperature on the new scale, corresponding to a temperature of \(39^\circ W\) on the Celsius scale?

366614 A copper rod and a steel rod are heated. At \(0^\circ C\), the copper rod has a length \(L_{C}\) and the steel one has a length \(L_{S}\). When the rods are being heated or cooled, a difference of \(5.00\;cm\) is maintained between their lengths. Determine the value of \(L_{C}\).
\({\alpha _{Cu}} = 5 \times {10^{ - 5}}/\,^\circ C,{\alpha _S} = 2 \times {10^{ - 5}}/\,^\circ C.\)

366611 A constant volume gas thermometer shows pressure reading 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

366612 If absolute zero is \( - 273.15^\circ C\) on Celsius temperature scale, then the absolute zero on the Fahrenheit scale is

366613 On a new scale of temperature (which is linear) and called the \(W\) scale, the freezing and boiling points of water are \(39^\circ W\) and \(239^\circ W\) respectively. What will be the temperature on the new scale, corresponding to a temperature of \(39^\circ W\) on the Celsius scale?

366614 A copper rod and a steel rod are heated. At \(0^\circ C\), the copper rod has a length \(L_{C}\) and the steel one has a length \(L_{S}\). When the rods are being heated or cooled, a difference of \(5.00\;cm\) is maintained between their lengths. Determine the value of \(L_{C}\).
\({\alpha _{Cu}} = 5 \times {10^{ - 5}}/\,^\circ C,{\alpha _S} = 2 \times {10^{ - 5}}/\,^\circ C.\)

366611 A constant volume gas thermometer shows pressure reading 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

366612 If absolute zero is \( - 273.15^\circ C\) on Celsius temperature scale, then the absolute zero on the Fahrenheit scale is

366613 On a new scale of temperature (which is linear) and called the \(W\) scale, the freezing and boiling points of water are \(39^\circ W\) and \(239^\circ W\) respectively. What will be the temperature on the new scale, corresponding to a temperature of \(39^\circ W\) on the Celsius scale?

366614 A copper rod and a steel rod are heated. At \(0^\circ C\), the copper rod has a length \(L_{C}\) and the steel one has a length \(L_{S}\). When the rods are being heated or cooled, a difference of \(5.00\;cm\) is maintained between their lengths. Determine the value of \(L_{C}\).
\({\alpha _{Cu}} = 5 \times {10^{ - 5}}/\,^\circ C,{\alpha _S} = 2 \times {10^{ - 5}}/\,^\circ C.\)

366611 A constant volume gas thermometer shows pressure reading 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

366612 If absolute zero is \( - 273.15^\circ C\) on Celsius temperature scale, then the absolute zero on the Fahrenheit scale is

366613 On a new scale of temperature (which is linear) and called the \(W\) scale, the freezing and boiling points of water are \(39^\circ W\) and \(239^\circ W\) respectively. What will be the temperature on the new scale, corresponding to a temperature of \(39^\circ W\) on the Celsius scale?

366614 A copper rod and a steel rod are heated. At \(0^\circ C\), the copper rod has a length \(L_{C}\) and the steel one has a length \(L_{S}\). When the rods are being heated or cooled, a difference of \(5.00\;cm\) is maintained between their lengths. Determine the value of \(L_{C}\).
\({\alpha _{Cu}} = 5 \times {10^{ - 5}}/\,^\circ C,{\alpha _S} = 2 \times {10^{ - 5}}/\,^\circ C.\)