1) When a resistor of $11 \Omega$ is connected in series with a electric cell. The current following in it is 0.5A .In stead when a resistor of $5 \Omega$ is connected to the same electric cell in series , the current increases by 0.4A. The internal resistance of the cell is A) $1.5 \Omega$ B) $2 \Omega$ C) $2.5 \Omega$ D) $3.5 \Omega$ Answer: Option CExplanation:Here, $i= \frac{E}{R+r}$ $0.5 =\frac{E}{11+r} \Rightarrow E=5.5+0.5 r$ $0.9 =\frac{E}{5+r}$ or $E=4.5+0.9r$ On solving we get r=2.5$\Omega$
