Hint: For a balanced Wheatstone bridge, $\frac{P}{Q} = \frac{R}{S}$ $or$ $\frac{P}{R} = \frac{Q}{S}$ .

Step 1: Find the initial balanced condition.

$Let$ $resistance$ $for$ $segments$ $of$ $meterbridge$ $wire$ $be,$
$R_{1}$ $=$ ${kl}_{1}$ $and$ $R_{2}$ $=$ ${kl}_{2}$
$Initially,$
$\frac{P}{Q}$ $=$ $\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}}$

Step 2: Interchange the cell and the galvanometer.

When cell and galvanometer are interchanged,

$Resistance$ $of$ $branch$ $ABD,$ $R_{B}$ $=$ $P$ $+$ $R_{1}$
                                                         $=$ $\frac{{Ql}_{1}}{l_{2}} + {kl}_{1}$      $. . . (1)$
$Resistance$ $of$ $branch$ $ACD,$ $R_{C}$ $=$ $Q$ $+$ $R_{2}$
                                                         $=$ $Q$ $+$ ${kl}_{2}$         $. . . (2)$
$From$    $(1)$ $and$ $(2) . . . . .$
       $R_{B}$    $=$ $\frac{l_{1}}{l_{2}} R_{C}$
$Step 3 :$ $Find$ $the$ $potential$ $drop$ $across$ $the$ $galvanometer .$
$Current$ $i$ $by$ $battery$ $distributes$ $in$ $branches$ $ABD$ $and$ $ACD :$
      $i_{1}$ $=$ $\frac{R_{C} i}{R_{B}}$ $+$ $R_{C} = \frac{{il}_{2}}{l_{1} + l_{2}}$     $and$     $i_{2}$ $=$ $\frac{R_{B} i}{R_{B}}$ $+$ $R_{C} = \frac{{il}_{1}}{l_{1} + l_{2}}$
$\frac{V_{AB}}{V_{AC}} = \frac{i_{1} P}{i_{2} Q} = \frac{l_{2}}{l_{1}} \times \frac{l_{1}}{l_{2}} = 1$
$\Rightarrow$ $V_{AB} = V_{AC}$
$So,$ $V_{B} = V_{C}$    $\Rightarrow$ $zero$ $deflection$ $in$ $galvanometer$

1.	Mitochondria	2.	Golgi complex
3.	Endoplasmic reticulum	4.	Lysosomes

NEET 2019 Exam Paper (Odisha)