In the formula $x=3y{z}^{2}$, x and z have dimensions of capacitance and magnetic induction field strength respectively. The dimensions of y in the system are:
A). ${M}^{-3}{L}^{-2}{T}^{4}{Q}^{4}$
B). ${M}^{-2}{L}^{-2}{T}^{2}{Q}^{2}$
C). ${M}^{-3}{L}^{-1}{T}^{4}{Q}^{4}$
D). ${M}^{2}{L}^{2}{T}^{-3}{Q}^{-1}$
Answer
632.7k+ views
Hint: To write the dimension of the capacitance and of magnetic induction field strength take the help of the units of those quantities. After finding the dimensions of capacitance and magnetic induction, substitute those values in the equation which is mentioned in the question. After solving that equation you will get the dimensions of y.
Complete step-by-step solution:
Given: x has dimensions of capacitance
z has dimensions of magnetic induction field
Unit of capacitance is ${A}^{2}{s}^{4}{kg}^{-1}{m}^{-2}$
Thus, it’s dimensions are ${M}^{-1}{L}^{-2}{T}^{4}{I}^{2}$
Dimensions of magnetic induction field strength is ${M}^{1}{T}^{-2}{I}^{-1}$
Now, let dimensions of y be ${M}^{\alpha}{L}^{\beta}{T}^{\gamma}{Q}^{\delta}$
$x=3y{z}^{2}$
But, numbers are dimensionless quantity so we are left with,
$x=y{z}^{2}$ ...(1)
Now, substituting the dimensions in above expression we get,
$y= \dfrac {{M}^{-1}{L}^{-2}{T}^{4}{I}^{2}}{{({M}^{1}{T}^{-2}{I}^{-1})}^{2}}$
$\therefore y= {M}^{-3}{L}^{-2}{T}^{8}{I}^{4}$ ...(2)
But, we know Q=IT
$\therefore {Q}^{4}= {I}^{4}{T}^{4}$
Substituting this value in the equation. (2) we get,
$y ={M}^{-3}{L}^{-2}{T}^{4}{Q}^{4}$
Thus, the dimensions of y in system is
${M}^{-3}{L}^{-2}{T}^{4}{Q}^{4}$.
Hence, the correct answer is option C i.e. ${M}^{-3}{L}^{-2}{T}^{4}{Q}^{4}$.
Note: To answer such types of questions you should remember the dimensions of at least the basic units. Or you can drive them if you know the units of quantities. While writing the dimensional quantity it should be expressed in its absolute units only i.e. S.I.units. Always try to break the physical quantities in fundamental quantities, as we know the dimensions of fundamental quantities hence it becomes easier to solve. There are few quantities of which you should remember the S.I. units and dimensional formula which includes energy, pressure, power, etc.
Complete step-by-step solution:
Given: x has dimensions of capacitance
z has dimensions of magnetic induction field
Unit of capacitance is ${A}^{2}{s}^{4}{kg}^{-1}{m}^{-2}$
Thus, it’s dimensions are ${M}^{-1}{L}^{-2}{T}^{4}{I}^{2}$
Dimensions of magnetic induction field strength is ${M}^{1}{T}^{-2}{I}^{-1}$
Now, let dimensions of y be ${M}^{\alpha}{L}^{\beta}{T}^{\gamma}{Q}^{\delta}$
$x=3y{z}^{2}$
But, numbers are dimensionless quantity so we are left with,
$x=y{z}^{2}$ ...(1)
Now, substituting the dimensions in above expression we get,
$y= \dfrac {{M}^{-1}{L}^{-2}{T}^{4}{I}^{2}}{{({M}^{1}{T}^{-2}{I}^{-1})}^{2}}$
$\therefore y= {M}^{-3}{L}^{-2}{T}^{8}{I}^{4}$ ...(2)
But, we know Q=IT
$\therefore {Q}^{4}= {I}^{4}{T}^{4}$
Substituting this value in the equation. (2) we get,
$y ={M}^{-3}{L}^{-2}{T}^{4}{Q}^{4}$
Thus, the dimensions of y in system is
${M}^{-3}{L}^{-2}{T}^{4}{Q}^{4}$.
Hence, the correct answer is option C i.e. ${M}^{-3}{L}^{-2}{T}^{4}{Q}^{4}$.
Note: To answer such types of questions you should remember the dimensions of at least the basic units. Or you can drive them if you know the units of quantities. While writing the dimensional quantity it should be expressed in its absolute units only i.e. S.I.units. Always try to break the physical quantities in fundamental quantities, as we know the dimensions of fundamental quantities hence it becomes easier to solve. There are few quantities of which you should remember the S.I. units and dimensional formula which includes energy, pressure, power, etc.
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