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Question: A quantity X is given by \({{{\varepsilon }}_{{o}}}{{L}}\dfrac{{{{\Delta V}}}}{{{{\Delta T}}}}\)wher...

A quantity X is given by εoLΔVΔT{{{\varepsilon }}_{{o}}}{{L}}\dfrac{{{{\Delta V}}}}{{{{\Delta T}}}}where ε∘{\varepsilon _ \circ }is the permittivity of the free space, L is length. ΔV\Delta V is the potential difference and ΔT\Delta T is the time interval. The dimensional formula for X is the same as that of
A) Resistance
B) Charge
C) Voltage
D) Current

Explanation

Solution

Reduce the given formula of X to the maximum limit where we may get definite quantity and then find dimensional formula of that quantity to compare with the formula given by
X=εoLΔVΔT{{X = }}{{{\varepsilon }}_{{o}}}{{L}}\dfrac{{{{\Delta V}}}}{{{{\Delta T}}}}
Where,
ε∘{\varepsilon _ \circ }=permittivity of the free space
L= length,
ΔV\Delta V = potential difference
ΔT\Delta T= time interval
After converting we can easily compare the dimensional formula of X with the given options.

Complete step by step solution:
According to question
Now, firstly we will try to reduce the formula of X into the formula of a dimensionally valid quantity.
X=εoLΔVΔT{{X = }}{{{\varepsilon }}_{{o}}}{{L}}\dfrac{{{{\Delta V}}}}{{{{\Delta T}}}}
In the above equation the term ε∘L{\varepsilon _ \circ }L represent the capacitance because we know that
ε∘L=C  {\varepsilon _ \circ }L{{ }} = C\;
Now, on putting value of ε∘L{\varepsilon _ \circ }L in formula of X, we get
X=CΔVΔT{{X = C}}\dfrac{{{{\Delta V}}}}{{{{\Delta T}}}}
The term CΔVC\Delta V can be replaced by charge ΔQ\Delta Qbecause we know that
CΔV=ΔQ{{C\Delta V = \Delta Q}}
Now, putting value ofCΔVC\Delta V in formula of X we get
X=ΔQΔT{{X = }}\dfrac{{{{\Delta Q}}}}{{{{\Delta T}}}}
We know that ΔQΔT\dfrac{{{{\Delta Q}}}}{{{{\Delta T}}}} is equal to current.
∴  X=ΔQΔT=⇒X=I\therefore {{\;X = }}\dfrac{{{{\Delta Q}}}}{{{{\Delta T}}}}{{ = }} \Rightarrow X{{ = I}}
Where I{{I}} stands for current

∴\therefore Quantity XX is dimensionally equal to the current. Option (D) is correct.

Note:
We can also solve this question by using the dimensions of each quantity and then reducing it by putting in the formula of X. In this way we get the dimension of X. then also we need to find the dimension of each quantity given in the option. After all this we finally compare the dimension of X with any other physical quantity.