Assertion & Reason type questions Read the assertion and reason carefully to mark the correct option out of the options given below Assertion : A bubble comes from the bottom of a lake to the top. Reason : Its radius increases.
Physics MCQs for NEET — Practice Questions with Answers
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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.
Amount of heat required to raise the temprature of a body through 1k is called it is
$ \theta = m.c \triangle \theta $ ; if $ \triangle \theta = 1k $ then $ \theta = mc Thermal capacity$
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 $
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