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$\frac{d?}{dt}=10t-3$            

at t = 2, V = 17

$i=\frac{V}{R}=\frac{17}{8}A$              

  $X=\overline{\left(A+\overline{B}\right)+\left(\overline{A}+B\right)}$

$\left(\overline{A+\overline{B}}\right)\cdot \left(\overline{\overline{A}+B}\right)$

$\left(\overline{A}\cdot \overline{B}\right)\cdot \left(\overline{A}\cdot \overline{B}\right)$

$\left(\overline{A}\cdot B\right)\cdot \left(A\cdot \overline{B}\right)=\overline{A}\cdot B\cdot A\cdot \overline{B}=0$  

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Order of reactivity towards electrophile

Strength of +M/+R : –OH > –CH₃ > –Cl

In case of halogens, their –I effect dominates over

+M hence –Cl is deactivating and  is lesser reactive than  for incoming electrophile.

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  $\frac{1}{f}=\left(\mu -1\right)\left(\frac{2}{R}\right)$                    f = 20 cm

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}+\frac{1}{10}=\frac{1}{20}$

CuSO₄ acts as catalyst in Kjeldahl's method of estimation of nitrogen.

Kindly go through the solution

Since $\frac{f}{2}<f_0$  

->current = 0

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