Embibe Experts Solutions for Exercise 4: Assignment

If $y$ represents pressure and $x$ represents velocity gradient, then the dimensions of $\frac{d^{2}y}{d{x}^2}$ are

The dimensions of $\frac{\alpha }{\beta }$ in the equation $F=\frac{\alpha -t^2}{\beta v^2}$, where $F$ is the force, $v$ is velocity and $t$ is time, is

Even if a physical quantity depends upon three quantities, out of which two are dimensionally same, then the formula cannot be derived by the method of dimensions. This statement

In a new system of units energy $\left(E\right)$, density $\left(d\right)$ and power $\left(P\right)$ are taken as fundamental units, then the dimensional formula of universal gravitational constant $G$ will be

In equation $y=x^2{\cos }^{2}2\pi \frac{\beta \gamma }{\alpha }$, the units of $x,\alpha ,\beta$ are $\mathrm{m}, {\mathrm{s}}^{-1}$ and ${\left({\mathrm{ms}}^{-1}\right)}^{-1}$ respectively. The units of $y$ and $\gamma$ are

Let $P$ represent radiation pressure, $c$ represent speed of light and $I$ represent radiation energy striking a unit area per second, then $P^x{I}^y{c}^z$ will be dimensionless for

The number of particles crossing per unit area perpendicular to $Z$ axis per unit time is given by $N=-D\frac{\left(N_2-N_1\right)}{\left(Z_2-Z_1\right)}$, where $N_2$ and $N_1$ are the number of particles per unit volume at $Z_2$ and $Z_1$ respectively. What is the dimensional formula for $D$ ?

If energy $E$, velocity $V$ and time $T$ are taken as fundamental units, the dimensional formula for surface tension is