If voltage across a bulb rated $$220\,V - 100\,W$$ drops by $$2.5\% $$ of its rated value, the percentage of the rated value by which the power would decrease is
A.
$$20\% $$
B.
$$2.5\% $$
C.
$$5\% $$
D.
$$10\% $$
Answer :
$$5\% $$
Solution :
Power, $$\left( P \right) = \frac{{{V^2}}}{R}$$
For small variation,
$$\frac{{\Delta P}}{P} \times 100 = 2 \times \frac{{\Delta V}}{V} \times 100 = 2 \times 2.5 = 5\% $$
Therefore, power would decrease by $$5\% .$$
Releted MCQ Question on Electrostatics and Magnetism >> Electric Current
Releted Question 1
The temperature coefficient of resistance of a wire is 0.00125 per $$^ \circ C$$ At $$300\,K,$$ its resistance is $$1\,ohm.$$ This resistance of the wire will be $$2\,ohm$$ at.
The electrostatic field due to a point charge depends on the distance $$r$$ as $$\frac{1}{{{r^2}}}.$$ Indicate which of the following quantities shows same dependence on $$r.$$
A.
Intensity of light from a point source.
B.
Electrostatic potential due to a point charge.
C.
Electrostatic potential at a distance r from the centre of a charged metallic sphere. Given $$r$$ < radius of the sphere.