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Question: Which of the following does not have the same dimension as Henry? A) \(\dfrac{{Joule}}{{{{(ampere)...

Which of the following does not have the same dimension as Henry?
A) Joule(ampere)2\dfrac{{Joule}}{{{{(ampere)}^2}}}
B) tesla−m2(ampere)2\dfrac{{tesla - {m^2}}}{{{{(ampere)}^2}}}
C) ohm−secondohm - {\text{second}}
D) 1farad−second\dfrac{1}{{farad - {\text{second}}}}

Explanation

Solution

Here, we have to define what henry signifies as a unit and find its dimensions using the fundamental quantities. Now, take every unit given in the options and find their dimensions as well using fundamental quantities. After deriving their dimensions, match them with the dimension of Henry to get the answer.

Complete answer:
Henry is the S.I. unit of inductance. When a voltage of one volt is produced by a variation of the inducing current of one ampere per second in a coil, then the self-induction of one henry is generated in that coil. Mathematically, it is represented by,
V(t)=LdIdtV(t) = L\dfrac{{dI}}{{dt}}
Where,
V(t)V(t) is the voltage across the circuit.
LL is the inductance.
II is the current.
tt is the time.
So, if we replace the quantities in the formula with the units we get,
volt=henry×amperetime......(1)volt = \dfrac{{henry \times ampere}}{{time}}......(1)
Dimension of volt =[ML2T−3I−1] = \left[ {M{L^2}{T^{ - 3}}{I^{ - 1}}} \right]
Dimension of ampere=[I] = \left[ I \right]
Dimension of time=[T] = \left[ T \right]
Putting the dimensions in equation (1) we get,
[ML2T−3I−1]=henry×[I][T]\left[ {M{L^2}{T^{ - 3}}{I^{ - 1}}} \right] = \dfrac{{henry \times \left[ I \right]}}{{\left[ T \right]}}
⇒henry=[ML2T−3I−1][T][I]\Rightarrow henry = \dfrac{{\left[ {M{L^2}{T^{ - 3}}{I^{ - 1}}} \right]\left[ T \right]}}{{\left[ I \right]}}
⇒henry=[ML2T−2I−2]\Rightarrow henry = \left[ {M{L^2}{T^{ - 2}}{I^{ - 2}}} \right]
Now, we will determine the dimensions of the other options.
A) Dimension of joule=[ML2T−2] = \left[ {M{L^2}{T^{ - 2}}} \right]
Joule(ampere)2=[ML2T−2][I2]\dfrac{{Joule}}{{{{(ampere)}^2}}} = \dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {{I^2}} \right]}}
⇒Joule(ampere)2=[ML2T−2I−2]\Rightarrow \dfrac{{Joule}}{{{{(ampere)}^2}}} = \left[ {M{L^2}{T^{ - 2}}{I^{ - 2}}} \right]

B) Dimension of tesla=[MT−2I−1] = \left[ {M{T^{ - 2}}{I^{ - 1}}} \right]
tesla−m2(ampere)2==[MT−2I−1][L2][I2]\dfrac{{tesla - {m^2}}}{{{{(ampere)}^2}}} = \dfrac{{ = \left[ {M{T^{ - 2}}{I^{ - 1}}} \right]\left[ {{L^2}} \right]}}{{\left[ {{I^2}} \right]}}
⇒tesla−m2(ampere)2=[ML2T−2I−3]\Rightarrow \dfrac{{tesla - {m^2}}}{{{{(ampere)}^2}}} = \left[ {M{L^2}{T^{ - 2}}{I^{ - 3}}} \right]

C) Dimension of ohm=[ML2T−3I−2] = \left[ {M{L^2}{T^{ - 3}}{I^{ - 2}}} \right]
ohm−second=[ML2T−3I−2][T]ohm - {\text{second}} = \left[ {M{L^2}{T^{ - 3}}{I^{ - 2}}} \right]\left[ T \right]
⇒ohm−second=[ML2T−2I−2]\Rightarrow ohm - {\text{second}} = \left[ {M{L^2}{T^{ - 2}}{I^{ - 2}}} \right]

D) Dimension of farad=[M−1L−2T4I2] = \left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{I^2}} \right]
1farad−second=1[M−1L−2T4I2][T]\dfrac{1}{{farad - {\text{second}}}} = \dfrac{1}{{\left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{I^2}} \right]\left[ T \right]}}
⇒1farad−second=1[M−1L−2T5I2]\Rightarrow \dfrac{1}{{farad - {\text{second}}}} = \dfrac{1}{{\left[ {{M^{ - 1}}{L^{ - 2}}{T^5}{I^2}} \right]}}
⇒1farad−second=[ML2T−5I−2]\Rightarrow \dfrac{1}{{farad - {\text{second}}}} = \left[ {M{L^2}{T^{ - 5}}{I^{ - 2}}} \right]
On comparing all the options we can conclude that option (B) and (D) does not have the dimension of Henry.

Note:
There are two types of physical quantities. They are fundamental quantities and derived quantities. The quantities which are independent and cannot be derived are called fundamental quantities. Mass, time, length are some examples of fundamental quantities. The quantities which are derived with the help of fundamental quantities are called derived quantities. Energy, magnetic field, force are derived quantities. Dimensions are always represented by fundamental quantities.