(4)

[IMAGE 163]----------------------------

Circuit ABCDEF form Wheatstone bridge,

So, Resistance across arm AD, $R_{AD}=6 \Omega$ can be eliminated.

[IMAGE 164]----------------------------

From circuit ABCDEF:

Resistance $5 \Omega$ and $3 \Omega$ are in series and $2.5 \Omega$ and $1.5 \Omega$ are in series. Then both the series combinations are in parallel with each other. Thus equivalent resistance is

              $\frac{1}{R_{\text{eq}}}=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}$

$\therefore$   $R_{\text{eq}}=\frac{8}{3} \Omega$

Now, the given circuit can be redrawn as

[IMAGE 165]----------------------------

Current through battery,

$I=\frac{5}{10}=0.5$$\text{A}$

(2)

Given, voltage, $\left(V\right)=(200\pm 4)\text{V}$

Current, $\left(I\right)=(20\pm 0.2)A$

Resistance, $\left(R\right)=\frac{V}{I}=\frac{200}{20}=10\Omega$

Now, error in $R$ is given by,

$\frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}=\frac{4}{200}+\frac{0.2}{20}=0.03$

$\therefore$   $\Delta R=R\times 0.03=10\times 0.03=0.3\Omega$

So, the value of Resistance is $(10\pm 0.3)\Omega$

(2)

Let wire $A$ have cross sectional area $A$, radius $r$, length $l$ and diameter $d$.

$R\to$Resistance of wire

For wire B : $L\text{'}=L;$ $d\text{'}=3d;$ $r\text{'}=3r$

Area, $A\text{'}=9A$

As, $R=\frac{\rho L}{A} \text{and} R\text{'}=\frac{\rho L\text{'}}{A\text{'}}=\frac{\rho L}{9A}=\frac{R}{9}$

Given : $R=81\Omega ,$ $R\text{'}=\frac{81 \Omega }{9}=9\Omega$

(4)

Net emf of the circuit is

$E=10-5=5$$V$

Net resistance, $R=1+2+7=10\Omega$

$\therefore$ $\text{Current, }I=\frac{E}{R}=\frac{5}{10}=0.5\text{A}$

Current will be directed from A to B through E.

(2)

The resistance of a wire of length $l$ and area $A$ and resistivity $\rho$ is given as

          $R=\frac{\rho l}{A}$

Given, $l\text{'}=nl$

As the volume of the wire remains constant

$\therefore$    $A\text{'}l\text{'}=Al \text{or} A\text{'}=\frac{Al}{l\text{'}}=\frac{Al}{nl} \text{or} A\text{'}=\frac{A}{n}$

$\therefore$   $R\text{'}=\frac{\rho l\text{'}}{A\text{'}} \text{or} R\text{'}=\frac{\rho nl}{\frac{A}{n}}=\frac{n^2\rho l}{A}=n^{2}R$