Class 12th

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation-

 total resistance = 2+8= 10 ohm

I= $\frac{6-4}{2+8}$ = 0.2A

The direction of flow of current is always from high potential to low potential

If VB $>$ VA

VB-4V-0.2 $*$ 8 = VA

V_B-V_A= 3.6V

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$\begin{array}{l}So,\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}+\frac{(-5)}{(x-2)}+\frac{4}{(x-3)}\\ Then,{\displaystyle \int \frac{3x-1dx}{(x-1)(x-2)(x-3)}}={\displaystyle \int \frac{dx}{x-1}}-5{\displaystyle \int \frac{dx}{x-2}}+4{\displaystyle \int \frac{dx}{x-3}}.\\ =log|x-1|-5log|x-2|+4log|x-3|+c\end{array}$

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation – in series combination of resistors current I =e/R+r

10I= $\frac{e}{R+\frac{r}{n}}$

$\frac{1+n}{1+\frac{1}{n}}$ $\Rightarrow$ 10 = ( $\frac{1+n}{1+n}$ )n $\Rightarrow$ n=10

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- according to the relation V=e-Ir and I=e/r+R

The graphical relation between voltage and resistance.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- If the value of R is increased, the current through the wire will decrease which in turn decreases the potential difference across AB, and  hence potential gradient (k) across AB decreases.
Since, at neutral point, for given emf of cell, l increases as potential gradient (k) across AB has decreased because E' = kl
Thus, with the increase of l, which will result in increase in balance length. So, jockey J will shift towards B