Match the following two coloumns
Column $-I$ | Column $-II$ |
$(A)$ Electrical resistance | $(p)$ $M{L^3}{T^{ - 3}}{A^{ - 2}}$ |
$(B)$ Electrical potential | $(q)$ $M{L^2}{T^{ - 3}}{A^{ - 2}}$ |
$(C)$ Specific resistance | $(r)$ $M{L^2}{T^{ - 3}}{A^{ - 1}}$ |
$(D)$ Specific conductance | $(s)$ None of these |
$A \to q, B \to s, C \to r, D \to p$
$A \to q, B \to r, C \to p, D \to s$
$A \to p, B \to q, C \to s, D \to r$
$A \to p, B \to r, C \to q, D \to s$
The frequency $(v)$ of an oscillating liquid drop may depend upon radius $(r)$ of the drop, density $(\rho)$ of liquid and the surface tension $(s)$ of the liquid as : $v=r^{ a } \rho^{ b } s ^{ c }$. The values of $a , b$ and $c$ respectively are
Consider two physical quantities A and B related to each other as $E=\frac{B-x^2}{A t}$ where $E, x$ and $t$ have dimensions of energy, length and time respectively. The dimension of $A B$ is
The potential energy of a particle varies with distance $x$ from a fixed origin as $U=\frac{A \sqrt{x}}{x^2+B}$, where $A$ and $B$ are dimensional constants then dimensional formula for $A B$ is
The equation of a circle is given by $x^2+y^2=a^2$, where $a$ is the radius. If the equation is modified to change the origin other than $(0,0)$, then find out the correct dimensions of $A$ and $B$ in a new equation: $(x-A t)^2+\left(y-\frac{t}{B}\right)^2=a^2$.The dimensions of $t$ is given as $\left[ T ^{-1}\right]$.