Answer
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Hint: To find the dimension of the resistance, first break it into its fundamental quantities. Express resistance in terms of voltage and current. Find the dimension of voltage by defining the dimensions of electric field and distance. Find the dimension of current. Then we will get our answer.
Formula used: For the dimension of resistance we should use the formula
$\text{Resistance=}\dfrac{\text{Voltage}}{\text{current}}$
Step by step answer:
we know that the relation of resistance with voltage and current is,
$\text{Resistance=}\dfrac{\text{Voltage}}{\text{current}}$or $\text{R=}\dfrac{\text{V}}{\text{I}}.............(1)$
We know that,
\[\text{Voltage=electric field }\!\!\times\!\!\text{ }\!\!~\!\!\text{ distance}\]
$\text{V=E }\!\!\times\!\!\text{ d}...........\text{(2)}$
The relation of an electric field with force is
Electric field$=\dfrac{\text{force}}{\text{charge}}$
$\text{E=}\dfrac{\text{F}}{\text{Q}}...............(3)$
Put the equation (3) into equation (2)
$\text{V=}\dfrac{\text{F }\!\!\times\!\!\text{ d}}{\text{Q}}.................\text{(4)}$
Now put the equation (4) into equation (1)
$\text{R=}\dfrac{\text{F }\!\!\times\!\!\text{ d}}{\text{Q }\!\!\times\!\!\text{ I}}..................(5)$
We know that
$\text{Q=I }\!\!\times\!\!\text{ t}.................\text{(6)}$
Put equation (6) into equation (5)
$\text{R=}\dfrac{\text{F }\!\!\times\!\!\text{ d}}{{{\text{I}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ t}}..............(7)$
As we know that
$Force=Mass\times acceleration$
And the dimension of force is, $\left[ \text{ML}{{\text{T}}^{\text{-2}}} \right]$
Put the dimension of force in the equation (7) and also convert other quantities of the equation (7) in dimension form.
$\text{R=}\dfrac{\left[ \text{ML}{{\text{T}}^{\text{-2}}} \right]\left[ \text{L} \right]}{\left[ {{\text{I}}^{\text{2}}} \right]\left[ \text{T} \right]}$
The dimension of resistance is $\left[ \text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-3}}}{{\text{I}}^{\text{-2}}} \right]$
The answer is (D)
Additional Information:
To find the dimensions of resistance we can use another formula of resistance too that is shown below.
$R=\rho \dfrac{l}{A}$
Where $\rho =$ Resistivity of the wire
$l=$ Length of the wire
$\text{A=}$ the cross-section area of the wire
$R=\dfrac{V}{\sigma EA}$
Where $V=$ Voltage
$\sigma =$ conductivity of resistance
$E=$ Electric field intensity
$A=$ The cross-section area of a wire
Note: Mostly in this type of question where a quantity (like resistance) has more than one formula, the student gets confused about which formula will be used for getting the answer in a very easy way. So, when we have that kind of confusion just remember all the formulas you know. And then the analysis in which formula you know the dimensions of all the quantities present in the formula.
Like, suppose you use $R=\rho \dfrac{l}{A}$ a formula for finding the dimension of resistance but if you don't know the dimension of$\rho$ then you cannot get the answer so avoid the use of this formula.
Formula used: For the dimension of resistance we should use the formula
$\text{Resistance=}\dfrac{\text{Voltage}}{\text{current}}$
Step by step answer:
we know that the relation of resistance with voltage and current is,
$\text{Resistance=}\dfrac{\text{Voltage}}{\text{current}}$or $\text{R=}\dfrac{\text{V}}{\text{I}}.............(1)$
We know that,
\[\text{Voltage=electric field }\!\!\times\!\!\text{ }\!\!~\!\!\text{ distance}\]
$\text{V=E }\!\!\times\!\!\text{ d}...........\text{(2)}$
The relation of an electric field with force is
Electric field$=\dfrac{\text{force}}{\text{charge}}$
$\text{E=}\dfrac{\text{F}}{\text{Q}}...............(3)$
Put the equation (3) into equation (2)
$\text{V=}\dfrac{\text{F }\!\!\times\!\!\text{ d}}{\text{Q}}.................\text{(4)}$
Now put the equation (4) into equation (1)
$\text{R=}\dfrac{\text{F }\!\!\times\!\!\text{ d}}{\text{Q }\!\!\times\!\!\text{ I}}..................(5)$
We know that
$\text{Q=I }\!\!\times\!\!\text{ t}.................\text{(6)}$
Put equation (6) into equation (5)
$\text{R=}\dfrac{\text{F }\!\!\times\!\!\text{ d}}{{{\text{I}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ t}}..............(7)$
As we know that
$Force=Mass\times acceleration$
And the dimension of force is, $\left[ \text{ML}{{\text{T}}^{\text{-2}}} \right]$
Put the dimension of force in the equation (7) and also convert other quantities of the equation (7) in dimension form.
$\text{R=}\dfrac{\left[ \text{ML}{{\text{T}}^{\text{-2}}} \right]\left[ \text{L} \right]}{\left[ {{\text{I}}^{\text{2}}} \right]\left[ \text{T} \right]}$
The dimension of resistance is $\left[ \text{M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-3}}}{{\text{I}}^{\text{-2}}} \right]$
The answer is (D)
Additional Information:
To find the dimensions of resistance we can use another formula of resistance too that is shown below.
$R=\rho \dfrac{l}{A}$
Where $\rho =$ Resistivity of the wire
$l=$ Length of the wire
$\text{A=}$ the cross-section area of the wire
$R=\dfrac{V}{\sigma EA}$
Where $V=$ Voltage
$\sigma =$ conductivity of resistance
$E=$ Electric field intensity
$A=$ The cross-section area of a wire
Note: Mostly in this type of question where a quantity (like resistance) has more than one formula, the student gets confused about which formula will be used for getting the answer in a very easy way. So, when we have that kind of confusion just remember all the formulas you know. And then the analysis in which formula you know the dimensions of all the quantities present in the formula.
Like, suppose you use $R=\rho \dfrac{l}{A}$ a formula for finding the dimension of resistance but if you don't know the dimension of$\rho$ then you cannot get the answer so avoid the use of this formula.
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