If $C, R, L$ and $I$ denote capacity,resistance,inductance and electric current respectively,the quantities having the same dimensions of time are :
$(1)$ $C R$
$(2)$ $\frac{L}{R}$
$(3)$ $\sqrt{L C}$
$(4)$ $L I^2$

$AB$ crystallizes in a body-centred cubic lattice with edge length '$a$' equal to $387 \ pm$. The distance between two oppositely charged ions in the lattice is ............. $pm$.

The value of the integral $\int_0^4 \frac{dx}{1+x^2}$ obtained by using the Trapezoidal rule with $h=1$ is

The voltage of an $AC$ source varies with time according to the equation $V = 100 \sin(100\pi t) \cos(100\pi t)$,where $t$ is in seconds and $V$ is in volts. Then:

$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is equal to :

$A$ volume of $x \ mL$ of $5 \ M \ NaHCO_3$ solution was mixed with $10 \ mL$ of $2 \ M \ H_2CO_3$ solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of $235.3 \ mV$, then the value of $x = . . . . . . \ mL$ (nearest integer).
$Sn_{(s)} \mid Sn(OH)_6^{2-}(0.5 \ M) \mid HSnO_2^{-}(0.05 \ M) \mid OH^{-} \mid Bi_2O_{3(s)} \mid Bi_{(s)}$
Consider upto one place of decimal for intermediate calculations.

Given :	$E^{\circ}_{HSnO_2^{-} \mid Sn(OH)_6^{2-}} = -0.9 \ V$
	$E^{\circ}_{Bi_2O_3 \mid Bi} = -0.44 \ V$
	$pKa_{(H_2CO_3)} = 6.11$
	$\frac{2.303 \ RT}{F} = 0.059 \ V$
	$Antilog(1.29) = 19.5$