A spherical body of mass $m$ and radius $r$ is allowed to fall in a medium of viscosity $\eta $. The time in which the velocity of the body increases from zero to $0.63$ times the terminal velocity $(v)$ is called time constant $(\tau )$. Dimensionally $\tau $ can be represented by
$\frac{{m{r^2}}}{{6\pi \eta }}$
$\sqrt {\left( {\frac{{6\pi mr\eta }}{{{g^2}}}} \right)} $
$\frac{m}{{6\pi \eta rv}}$
None of the above
If the dimensions of length are expressed as ${G^x}{c^y}{h^z}$; where $G,\,c$ and $h$ are the universal gravitational constant, speed of light and Planck's constant respectively, then
If velocity of light $c$, Planck’s constant $h$ and gravitational constant $G$ are taken as fundamental quantities, then express time in terms of dimensions of these quantities.
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 $\left[ T ^{-1}\right]$.
Consider following statements
$(A)$ Any physical quantity have more than one unit
$(B)$ Any physical quantity have only one dimensional formula
$(C)$ More than one physical quantities may have same dimension
$(D)$ We can add and subtract only those expression having same dimension
Number of correct statement is
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length $(l)$, mass of the bob $(m)$ and acceleration due to gravity $(g)$. Derive the expression for its time period using method of dimensions.