Three charges $Q$,$+q$,and $+q$ are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic potential energy of the configuration is zero if $Q$ is equal to

Which of the following reactions will yield $2-$propanol? Select the correct answer from $a, b, c$ and $d$.
$(I)$ $CH_2=CH-CH_3 + H_2O \xrightarrow{H^+} $
$(II)$ $CH_3CHO \xrightarrow[(ii) H_2O]{(i) CH_3MgI}$
$(III)$ $CH_2O \xrightarrow[(ii) H_2O]{(i) C_2H_5MgI}$
$(IV)$ $CH_2=CH-CH_3 \xrightarrow{\text{neutral } KMnO_4}$

For the circuit shown below,the Boolean polynomial is:

Sprain is caused due to stretching of

The value of $x$,where $x>0$ and $\tan \left(\sec ^{-1}\left(\frac{1}{x}\right)\right)=\sin \left(\tan ^{-1} 2\right)$ is

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