(4)

The galvanometer shows zero reading when no current passes through it.

Condition of balanced metre bridge, $\frac{R_1}{R_2}=\frac{x}{(l-x)}$

Given : $l=40 \text{cm}, R_1=8 \mathrm{cm}, R_2=12 \mathrm{cm}$

$\frac{8}{12}=\frac{x}{40-x}$

or   $320-8x=12x$

$\therefore$    $x=16$$\text{cm}$

From B,  $40-16=24$ $\text{cm}$

(2)

Unknown is $X,R=10$$\Omega$.

[IMAGE 200]------------------

Here, $\frac{l_1}{l_2}=\frac{3}{2} ; \frac{X}{R}=\frac{l_1}{l_2}$

$\Rightarrow X=\frac{3}{2}\times 10\Rightarrow X=15\Omega$

Thus, 1.5 m length has resistance $15\Omega$ hence, length of $1\Omega$ of the resistance wire $=\frac{1.5}{15}=0.1\text{m}=1.0\times {10}^{-1}\text{m}$
 

(4)

Yes, the bridge will work. For a balanced condition, the current drawn from the battery will be zero. Also, $P\propto l_1$ and $Q\propto l_2$. Therefore, the condition $\frac{P}{Q}=\frac{l_1}{l_2}$ will remain same after interchanging the cell and galvanometer.
 

(2)

In the first case,

[IMAGE 203]----------------------------

At balance point

[IMAGE 204]----------------------------

$\frac{5}{R}=\frac{l_1}{100-l_1}$                                                    ...(i)

In the second case,

At balance point

$\frac{5}{(R/2)}=\frac{1.6l_1}{100-1.6l_1}$                             ..(ii)

Divide eqn. (i) by eqn. (ii), we get

        $\frac{1}{2}=\frac{100-1.6l_1}{1.6(100-l_1)}$

or    $160-1.6l_1=200-3.2l_1$

        $1.6l_1= 40$ or  $l_1=\frac{40}{1.6}=25\mathrm{cm}$

Substituting this value in eqn. (i), we get:

        $\frac{5}{R}=\frac{25}{75}$   or  $R=\frac{375}{25}\Omega =15$$\Omega$