JEE Advance - Physics (2010 - Paper 1 Offline - No. 11)
Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100, 60 and 40 W bulbs have filament resistances R100, R60 and R40 respectively, the relation between these resistances is
$${1 \over {{R_{100}}}} = {1 \over {{R_{40}}}} + {1 \over {{R_{60}}}}$$
$${R_{100}} = {R_{40}} + {R_{60}}$$
$${R_{100}} > {R_{60}} > {R_{40}}$$
$${1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}$$
Explanation
The power of the bulb is
$$P = {{{V^2}} \over R}$$
Therefore,
$$100 = {{{V^2}} \over {{R_{100}}}} \Rightarrow {1 \over {{R_{100}}}} = {{100} \over {{V^2}}}$$
where R100 is the resistance (at any temperature) corresponds to 100 W. Similarly,
$$60 = {{{V^2}} \over {{R_{60}}}} \Rightarrow {1 \over {{R_{60}}}} = {{60} \over {{V^2}}}$$ and $$40 = {{{V^2}} \over {{R_{40}}}} \Rightarrow {1 \over {{R_{40}}}} = {{40} \over {{V^2}}}$$
From these equations, we get
$${P_{100}} > {P_{60}} > {P_{40}} \Rightarrow {1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}$$
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