Class 11Physics10-1.Thermometry, Thermal Expansion and CalorimetryTemperature and Temperature scales
EasyMercury boils at $367^{\circ}C$. However,mercury thermometers are made such that they can measure temperatures up to $500^{\circ}C$. This is achieved by:
- AMaintaining a vacuum above the mercury column in the stem of the thermometer
- BFilling nitrogen gas at high pressure above the mercury column
- CFilling nitrogen gas at low pressure above the mercury level
- DFilling oxygen gas at high pressure above the mercury column
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Similar Questions
$Assertion :$ In a pressure cooker,water is brought to a boil. The cooker is then removed from the stove. Now,upon removing the lid of the pressure cooker,the water starts boiling again.
$Reason :$ The impurities in water bring down its boiling point.
$Reason :$ The impurities in water bring down its boiling point.
MediumAIIMS 2004
View SolutionThe temperature of a substance increases by $27^{\circ}C$. On the Kelvin scale,this increase is equal to ........ $K$.
Easy
View SolutionOn a new temperature scale, the melting point of ice is $20^{\circ} X$ and the boiling point of water is $110^{\circ} X$. $A$ temperature of $40^{\circ} C$ would be indicated on this new temperature scale as (in $^{\circ} X$)
The surface of the lake is at $2^{\circ}C$. Find the temperature of the bottom of the lake in $^{\circ}C$.
Easy
View SolutionThe electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
$R = R_{0} [1 + \alpha (T - T_{0})]$
The resistance is $101.6 \; \Omega$ at the triple-point of water $(273.16 \; K)$,and $165.5 \; \Omega$ at the normal melting point of lead $(600.5 \; K)$. What is the temperature (in $K$) when the resistance is $123.4 \; \Omega$?
$R = R_{0} [1 + \alpha (T - T_{0})]$
The resistance is $101.6 \; \Omega$ at the triple-point of water $(273.16 \; K)$,and $165.5 \; \Omega$ at the normal melting point of lead $(600.5 \; K)$. What is the temperature (in $K$) when the resistance is $123.4 \; \Omega$?
Medium
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