
Photons are quantum of radiation with energy $ M{L^2}{T^{ - 2}} $ y $ E = hv $ , where $ v $ is the frequency and $ h $ is Planck's constant. The dimensions of $ h $ are the same as that of:
(A) Linear impulse
(B) Angular impulse
(C) Linear momentum
(D) Angular momentum
Answer
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Hint: find the dimension of $ h $ that is planck's constant with the help of a given relation. Find the dimension individually of linear impulse, angular impulse, linear momentum, angular momentum. Then compare them with the dimension of the planck's constant, by that we will find whose dimension is the same as that of the planck's constant.
Complete Step By Step Answer:
dimension of energy is given as $ [M{L^2}{T^{ - 2}}] $
Dimension of frequency is given as $ [{T^{ - 1}}] $
Therefore dimension of $ h $ that is planck's constant is given as $ h = \dfrac{E}{v} = \dfrac{{[M{L^2}{T^{ - 2}}]}}{{[{T^{ - 1}}]}} $
$ \Rightarrow h = [M{L^2}{T^{ - 1}}] $
Now find the dimensions of the given options
Linear impulse is given as
$ linear\;impulse = force \times time $
$ \Rightarrow [ML{T^{ - 2}}][T] $
So, dimension formula of linear impulse $ \Rightarrow [ML{T^{ - 1}}] $
Angular impulse is given as a product of torque and time.
$ angular\;impulse = \tau \times t $
$ \Rightarrow [M{L^2}{T^{ - 2}}][T] $
So, dimension formula of angular impulse $ \Rightarrow [M{L^2}{T^{ - 1}}] $
Linear momentum is given as the product of mass and velocity,
$ linea\; momentum = mv $
$ \Rightarrow [M][L{T^{ - 1}}] $
So, dimension formula of linear momentum $ \Rightarrow [MLT{}^{ - 1}] $
Angular momentum is given as the product of mass, velocity, radius,
$ angular\;momentum = mvr $
$ \Rightarrow [M][L{T^{ - 1}}][L] $
So, dimension formula of angular momentum $ \Rightarrow [M{L^2}{T^{ - 1}}] $
Therefore we conclude that angular impulse and angular momentum have the same dimensional formula as planck's constant.
Hence, option B) angular impulse and option D) angular momentum are the correct options.
Note:
The power to which the basic quantities are elevated to represent that quantity is called the physical quantity's dimensions. Any physical quantity's dimensional formula is an expression that describes how and which of the basic quantities are included in that quantity. A dimensional equation is the equation derived by equating a physical quantity with its dimensional formula.
Complete Step By Step Answer:
dimension of energy is given as $ [M{L^2}{T^{ - 2}}] $
Dimension of frequency is given as $ [{T^{ - 1}}] $
Therefore dimension of $ h $ that is planck's constant is given as $ h = \dfrac{E}{v} = \dfrac{{[M{L^2}{T^{ - 2}}]}}{{[{T^{ - 1}}]}} $
$ \Rightarrow h = [M{L^2}{T^{ - 1}}] $
Now find the dimensions of the given options
Linear impulse is given as
$ linear\;impulse = force \times time $
$ \Rightarrow [ML{T^{ - 2}}][T] $
So, dimension formula of linear impulse $ \Rightarrow [ML{T^{ - 1}}] $
Angular impulse is given as a product of torque and time.
$ angular\;impulse = \tau \times t $
$ \Rightarrow [M{L^2}{T^{ - 2}}][T] $
So, dimension formula of angular impulse $ \Rightarrow [M{L^2}{T^{ - 1}}] $
Linear momentum is given as the product of mass and velocity,
$ linea\; momentum = mv $
$ \Rightarrow [M][L{T^{ - 1}}] $
So, dimension formula of linear momentum $ \Rightarrow [MLT{}^{ - 1}] $
Angular momentum is given as the product of mass, velocity, radius,
$ angular\;momentum = mvr $
$ \Rightarrow [M][L{T^{ - 1}}][L] $
So, dimension formula of angular momentum $ \Rightarrow [M{L^2}{T^{ - 1}}] $
Therefore we conclude that angular impulse and angular momentum have the same dimensional formula as planck's constant.
Hence, option B) angular impulse and option D) angular momentum are the correct options.
Note:
The power to which the basic quantities are elevated to represent that quantity is called the physical quantity's dimensions. Any physical quantity's dimensional formula is an expression that describes how and which of the basic quantities are included in that quantity. A dimensional equation is the equation derived by equating a physical quantity with its dimensional formula.
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