Practice free Thermal Properties of Matter (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
At what temprature the centigrade (celsius) and Fahrenheit readings at the same.
$ { C \over 5} = { F -32 \over 9 } $
Mercury thermometers can be used to measure tempratures up to
The boiling point of mercuryis $400 ^\circ C$ . Therefore the mercury thermeter can be used to measure the range upto $ 360 ^\circ C.$
When the room temprature becomes equal to the dew point the relative humidity of the room is
Relative humidity is defined as the ratio of the current absolute humidity to the highest possible absolute humidity (which depends on the current air temperature). When the room temperature becomes equal to the dew point, the air is fully saturated with water vapor, and thus the relative humidity is 100%.
If the length of a cylinder on heating increases by 2% the area of its base will increase by.
$ A \alpha L^2 \Rightarrow { \triangle A \over A } = 2 { \triangle L \over L } $
Density of substance at $0 ^\circ C $ is 10 gm/cc and at $ 100 ^ \circ C $ its density is 9.7 gm/CC. The coefficient of linear expansion of the substance will be
Coeffcient of volume expansion $ r = { \triangle \rho \over \rho . \triangle T } = { \rho_1 - \rho_2 \over \rho ( \triangle \theta ) } $ Hence, cofficent of linear expansion
An iron bar of length 10mis heated from 00C to 1000C. If the coefficient of linear thermal expansion of iron is ${ 10 \times 10 ^ {-8} \over C } $ the increase in the length of bar is
Increase in length $ \triangle L = Lo \alpha \triangle \theta $
Melting point of ice.....
Melting point of ice decreases with increase in pressure.
A vessel contains 110 g of water the heat capacity of the vessel is equal to 10 g of water The initial temprature of water in vessel is $10 ^\circ C$ If 220 g of hot water at $70 ^\circ C$ is poured in the vessel the Final temperature neglecting radiation loss will be
Let final temperature of water $ bc \theta $ heat taken = Heat given $ 110 \times 1 ( \theta - 10) + 10 ( \theta - 10 ) =220 \times 1 ( 70 - \theta) $ $ \Rightarrow \theta = 48.8 ^\circ C \approx 50 ^\circ C $
The temprature at which the vapour pressure of a liquid becomes equals of the external pressure is its.
At boiling point vapour pressure becomes equal to the external pressure
10 g of ice at $0 ^\circ C$ is mixed with 100 g of water at $50 ^\circ C $ what is the resultant temprature of mixture.
$ \theta_{mix} = { mw \theta w - { miLi \over cw } \over mi + mw } $
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