\( [M^0 L^0 T^{1} A^0] \)
The dimensional formula represents the relationship between different physical quantities. To determine the dimensional formula for \( RC \), where \( R \) stands for resistance and \( C \) for capacitance, we first find the dimensions of \( R \) and \( C \) separately.
Resistance \((R)\): The dimensional formula for resistance can be derived from Ohm's Law \( V = IR \), where \( V \) (Voltage) has a dimensional formula \([M^1L^2T^{-3}A^{-1}]\) and \( I \) (Current) has a dimensional formula \([A^1]\). Solving for \( R \), it follows that:
\[R = \frac{V}{I} \Rightarrow [M^1L^2T^{-3}A^{-1}][A^{-1}] = [M^1L^2T^{-3}A^{-2}]\]
Capacitance \((C)\): Capacitance is defined by the relation \( Q = CV \), where \( Q \) (Charge) has a dimensional formula \([A^1T^1]\), and rearranging gives us:
\(C = \frac{Q}{V} \Rightarrow [A^1T^1][M^{-1}L^{-2}T^{3}A^{1}] = [M^{-1}L^{-2}T^{4}A^{2}]\)
Given \( RC \) is the product of resistance and capacitance, we multiply their dimensional formulas:
\[RC = [M^1L^2T^{-3}A^{-2}][M^{-1}L^{-2}T^{4}A^{2}] = [M^{1-1}L^{2-2}T^{-3+4}A^{-2+2}] = [M^0L^0T^1A^0]\]
Thus, correct consideration aligns with answer: \([M^0L^0T^{1}A^0]\).
Match List-I with List-II.
Choose the correct answer from the options given below :}
There are three co-centric conducting spherical shells $A$, $B$ and $C$ of radii $a$, $b$ and $c$ respectively $(c>b>a)$ and they are charged with charges $q_1$, $q_2$ and $q_3$ respectively. The potentials of the spheres $A$, $B$ and $C$ respectively are:
Two resistors $2\,\Omega$ and $3\,\Omega$ are connected in the gaps of a bridge as shown in the figure. The null point is obtained with the contact of jockey at some point on wire $XY$. When an unknown resistor is connected in parallel with $3\,\Omega$ resistor, the null point is shifted by $22.5\,\text{cm}$ towards $Y$. The resistance of unknown resistor is ___ $\Omega$. 
