Distance of image of $A$ 'is $v_{A}$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v_{A}}+\frac{1}{-60}=\frac{1}{60} \Rightarrow v_{A}=-30 \mathrm{cm}$
Distance of image of $B^{\prime}$ is $v_{B}$
$\frac{1}{v_{B}}+\frac{1}{-90}=\frac{1}{60}$
$\frac{1}{v_{B}}=\frac{1}{60}+\frac{1}{90} \Rightarrow \frac{1}{v_{B}}=\frac{5}{180}$
$v_{B}=36 \mathrm{cm}$
$m=\frac{\text { length of image }}{\text { length of object }}=\frac{6}{30}=\frac{1}{5}$
Distance of image increases with convex mirror.
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The internal resistance of each cell is.................$\Omega$
| S.No. | $R$ | $l$ | $100-l$ | $S = \left( {\frac{{100 - l}}{l}} \right)R$ |
| $1$ | $20\,\Omega $ | $43$ | $57$ | $26.51\,\Omega $ |
| $2$ | $30\,\Omega $ | $51$ | $49$ | $28.82\,\Omega $ |
| $3$ | $40\,\Omega $ | $59$ | $41$ | $27.80\,\Omega $ |
| $4$ | $60\,\Omega $ | $70$ | $30$ | $25.71\,\Omega $ |