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(A) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance of the filament.Since $V$ is c

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$100 \, W$ bulb has a thicker filament

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(A) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance of the filament.Since $V$ is constant for both bulbs,$P \propto \frac{1}{R}$.The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $L$ is the length and $A$ is the cross-sectional area (thickness).Given that the lengths $L$ are the same and the material (resistivity $\rho$) is the same,$R \propto \frac{1}{A}$.Substituting this into the power relation,we get $P \propto \frac{1}{1/A} \implies P \propto A$.Therefore,the bulb with higher power $(100 \, W)$ must have a larger cross-sectional area (thicker filament) compared to the bulb with lower power $(60 \, W)$.

The electric bulbs have tungsten filaments of same length. If one of them gives $60 \, W$ and the other $100 \, W$,then:

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