The position of a particle moving on the $x-$axis is given by $x(t) = A \sin t + B \cos^2 t + Ct^2 + D$,where $t$ is time. The dimension of $\frac{ABC}{D}$ is $-$

If the time period $t$ of the oscillation of a drop of liquid of density $d$,radius $r$,vibrating under surface tension $s$ is given by the formula $t = \sqrt{r^{2b} s^c d^{a/2}}$. It is observed that the time period is directly proportional to $\sqrt{\frac{d}{s}}$. The value of $b$ should therefore be

Let force $F = A \sin(Ct) + B \cos(Dx)$,where $x$ and $t$ are displacement and time respectively. The dimensions of $\frac{C}{D}$ are the same as the dimensions of

If $10 \ g \ cm \ s^{-1} = x \ N \ s$,then the number $x$ is

Consider a spongy block of mass $m$ floating on a flowing river. The maximum mass of the block is related to the speed of the river flow $v$, acceleration due to gravity $g$, and the density of the block $\rho$ such that $m_{\max} = k v^x g^y \rho^z$ ($k$ is a constant). The values of $x, y$, and $z$ should then respectively be:
(Mass of the spongy block is assumed to vary due to absorption of water by it)

The refractive index of the medium is $\mu = A + \frac{B}{\lambda^{2}}$,where $A$ and $B$ are constants and $\lambda$ is the wavelength of light. The dimensions of $B$ are the same as that of