366632
Assertion : The temperature at which Centigrade and Fahrenheit thermometers read the same is \(-40^{\circ}\). Reason : There is no relation between Fahrenheit and Centigrade temperature.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The relation between \(F\) and \(C\) scale is, \(\dfrac{C}{5}=\dfrac{F-32}{9}\). If \(F=C \Rightarrow C=-40^{\circ} C\), i.e., at \(-40^{\circ}\) the Centrigrade and Fahrenheit thermometers reads the same. So option (3) is correct.
PHXI11:THERMAL PROPERTIES OF MATTER
366633
For measurments of very high temperature say around \(3000^\circ C\), one can use:
1 Gas thermometer
2 Platinum resistance thermometer
3 Vapour pressure thermometer
4 Pyrometer( Radiation thermometer)
Explanation:
Conceptual Question
PHXI11:THERMAL PROPERTIES OF MATTER
366634
On a temperature scale ' \(X\) ', the boiling point of water is \(65^\circ X\) and the freezing point is \( - 15^\circ X\). Assume that the \(X\) scale is linear. The equivalent temperature corresponding to \( - 95^\circ X\) on the Farenheit scale would be
1 \( - 48^\circ F\)
2 \( - 63^\circ F\)
3 \( - 112^\circ F\)
4 \( - 148^\circ F\)
Explanation:
\(\dfrac{X-X_{\text {freeze }}}{X_{\text {boil }}-X_{\text {freeze }}}=\dfrac{t-32^{\circ}}{212^{\circ}-32^{\circ}}\) as \(\dfrac{\text { Reading on scale }- \text { Lower fixed point }}{\text { Upper fixed point }- \text { Lower fixed point }}\) \(=\text { constant for any scale. }\) \(\dfrac{-95^{\circ}-\left(-15^{\circ}\right)}{65^{\circ}-\left(-15^{\circ}\right)}=\dfrac{t-32^{\circ}}{180^{\circ}} ; \dfrac{-80^{\circ}}{80^{\circ}}=\dfrac{t-32^{\circ}}{180^{\circ}}\) \(t=-180^{\circ}+32^{\circ} ; t=-148^{\circ} F\)
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PHXI11:THERMAL PROPERTIES OF MATTER
366632
Assertion : The temperature at which Centigrade and Fahrenheit thermometers read the same is \(-40^{\circ}\). Reason : There is no relation between Fahrenheit and Centigrade temperature.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The relation between \(F\) and \(C\) scale is, \(\dfrac{C}{5}=\dfrac{F-32}{9}\). If \(F=C \Rightarrow C=-40^{\circ} C\), i.e., at \(-40^{\circ}\) the Centrigrade and Fahrenheit thermometers reads the same. So option (3) is correct.
PHXI11:THERMAL PROPERTIES OF MATTER
366633
For measurments of very high temperature say around \(3000^\circ C\), one can use:
1 Gas thermometer
2 Platinum resistance thermometer
3 Vapour pressure thermometer
4 Pyrometer( Radiation thermometer)
Explanation:
Conceptual Question
PHXI11:THERMAL PROPERTIES OF MATTER
366634
On a temperature scale ' \(X\) ', the boiling point of water is \(65^\circ X\) and the freezing point is \( - 15^\circ X\). Assume that the \(X\) scale is linear. The equivalent temperature corresponding to \( - 95^\circ X\) on the Farenheit scale would be
1 \( - 48^\circ F\)
2 \( - 63^\circ F\)
3 \( - 112^\circ F\)
4 \( - 148^\circ F\)
Explanation:
\(\dfrac{X-X_{\text {freeze }}}{X_{\text {boil }}-X_{\text {freeze }}}=\dfrac{t-32^{\circ}}{212^{\circ}-32^{\circ}}\) as \(\dfrac{\text { Reading on scale }- \text { Lower fixed point }}{\text { Upper fixed point }- \text { Lower fixed point }}\) \(=\text { constant for any scale. }\) \(\dfrac{-95^{\circ}-\left(-15^{\circ}\right)}{65^{\circ}-\left(-15^{\circ}\right)}=\dfrac{t-32^{\circ}}{180^{\circ}} ; \dfrac{-80^{\circ}}{80^{\circ}}=\dfrac{t-32^{\circ}}{180^{\circ}}\) \(t=-180^{\circ}+32^{\circ} ; t=-148^{\circ} F\)
366632
Assertion : The temperature at which Centigrade and Fahrenheit thermometers read the same is \(-40^{\circ}\). Reason : There is no relation between Fahrenheit and Centigrade temperature.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The relation between \(F\) and \(C\) scale is, \(\dfrac{C}{5}=\dfrac{F-32}{9}\). If \(F=C \Rightarrow C=-40^{\circ} C\), i.e., at \(-40^{\circ}\) the Centrigrade and Fahrenheit thermometers reads the same. So option (3) is correct.
PHXI11:THERMAL PROPERTIES OF MATTER
366633
For measurments of very high temperature say around \(3000^\circ C\), one can use:
1 Gas thermometer
2 Platinum resistance thermometer
3 Vapour pressure thermometer
4 Pyrometer( Radiation thermometer)
Explanation:
Conceptual Question
PHXI11:THERMAL PROPERTIES OF MATTER
366634
On a temperature scale ' \(X\) ', the boiling point of water is \(65^\circ X\) and the freezing point is \( - 15^\circ X\). Assume that the \(X\) scale is linear. The equivalent temperature corresponding to \( - 95^\circ X\) on the Farenheit scale would be
1 \( - 48^\circ F\)
2 \( - 63^\circ F\)
3 \( - 112^\circ F\)
4 \( - 148^\circ F\)
Explanation:
\(\dfrac{X-X_{\text {freeze }}}{X_{\text {boil }}-X_{\text {freeze }}}=\dfrac{t-32^{\circ}}{212^{\circ}-32^{\circ}}\) as \(\dfrac{\text { Reading on scale }- \text { Lower fixed point }}{\text { Upper fixed point }- \text { Lower fixed point }}\) \(=\text { constant for any scale. }\) \(\dfrac{-95^{\circ}-\left(-15^{\circ}\right)}{65^{\circ}-\left(-15^{\circ}\right)}=\dfrac{t-32^{\circ}}{180^{\circ}} ; \dfrac{-80^{\circ}}{80^{\circ}}=\dfrac{t-32^{\circ}}{180^{\circ}}\) \(t=-180^{\circ}+32^{\circ} ; t=-148^{\circ} F\)