
$0=10 \mathrm{\,E}-14(10 \mathrm{\,r})$ ..........$(2)$
So, $6=10 \mathrm{\,E}-12\left[10 \times \frac{\mathrm{E}}{14}\right]$
$E=\frac{42}{10}=4.2 \mathrm{\,V}$


$Y :$ The resistivity of a semiconductor decreases with increases of temperature.
$Z :$ In a conducting solid, the rate of collision between free electrons and ions increases with increase of temperature.
Select the correct statement from the following :

