Solution

Given that,

Power of first bulb,$${{P}_{1}}=100\,W$$

Power of second bulb, $${{P}_{2}}=25\,W$$

Power is given by,

$$ P=\dfrac{{{V}^{2}}}{R} $$

$$ R=\dfrac{{{V}^{2}}}{P} $$

$$\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{\left( \dfrac{{{V}^{2}}}{{{P}_{1}}} \right)}{\left( \dfrac{{{V}^{2}}}{{{P}_{2}}} \right)}=\dfrac{{{P}_{2}}}{{{P}_{1}}}........(1)$$

And the resistance in terms of resistivity is given by:

$$ \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{\dfrac{\rho l}{\pi {{({{d}_{1}}/2)}^{2}}}}{\dfrac{\rho l}{\pi {{({{d}_{2}}/2)}^{2}}}}=\dfrac{d_{2}^{2}}{d_{1}^{2}} $$

$$ so, $$

$$ \dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{d_{2}^{2}}{d_{1}^{2}} $$

$$ \dfrac{{{d}_{2}}}{{{d}_{1}}}=\sqrt{\dfrac{{{P}_{2}}}{{{P}_{1}}}} $$

$$ \dfrac{{{d}_{2}}}{{{d}_{1}}}=\sqrt{\dfrac{25}{100}} $$

$$ \dfrac{{{d}_{2}}}{{{d}_{1}}}=\dfrac{1}{2} $$

The ratio of the diameter of the 100 W bulb to the that of the 25 W bulbs is 2:1