In a meter bridge experiment $\mathrm{S}$ is a standard resistance. $\mathrm{R}$ is a resistance wire. It is found that balancing length is $l=25 \;\mathrm{cm} .$ If $\mathrm{R}$ is replaced by a wire of half length and half diameter that of $\mathrm{R}$ of same material, then the balancing distance $\left.l^{\prime} \text { (in } \mathrm{cm}\right)$ will now be

If the balance point is obtained at the $35^{th} cm$ in a meter bridge the resistances in the left and right gaps are in the ratio of

In a meter bridge (Figure), the null point is found at a distance of $33.7 \;cm$ from A. If now a reststance of $12\; \Omega$ is connected in parallel with $S$, the null potnt occurs at $51.9 \;cm$. Determine the values of $R$ and $S$.

Fig. shows rough sketch of meter bridge. $(G)$ deflects zero at length $\ell \, cm$. Now $R_1$ and $R_2$ are interchanged then balancing length increases by $25\, cm$. Find $R_1/R_2$

The resistances of the four arms $P, Q, R$ and $S$ in a Wheatstone's bridge are $10\,ohm$, $30\,ohm$, $30\,ohm$ and $90\,ohm$, respectively. The e.m.f. and internal resistance of the cell are $7\,volt$ and $5\,ohm$ respectively. If the galvanometer resistance is $50\,ohm$, the current drawn from the cell will be …………… $A$