In India, electricity is supplied for domestic use at $$220\,V.$$ It is supplied at $$110\,V$$ in USA. If the resistance of a $$60\,W$$ bulb for use in India is $$R,$$ the resistance of a $$60\,W$$ bulb for use in USA will be
A.
$$R$$
B.
$$2\,R$$
C.
$$\frac{R}{4}$$
D.
$$\frac{R}{2}$$
Answer :
$$\frac{R}{4}$$
Solution :
Since, power rating of bulb in both the countries India and USA is same, so
$$\eqalign{
& \frac{{V_1^2}}{{{R_1}}} = \frac{{V_2^2}}{{{R_2}}}\,\,\left[ {P = \frac{{{V^2}}}{R}} \right] \cr
& \Rightarrow \frac{{220 \times 220}}{{{R_1}}} = \frac{{110 \times 110}}{{{R_2}}} \cr
& \Rightarrow \frac{{{R_2}}}{{{R_1}}} = \frac{{110 \times 110}}{{220 \times 220}} \cr
& \Rightarrow {R_2} = \frac{R}{4}\,\,\left( {\because {R_1} = R} \right) \cr} $$
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.