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### IIT-JEE 2010 Paper 1 Offline

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

A
$${1 \over {{R_{100}}}} = {1 \over {{R_{40}}}} + {1 \over {{R_{60}}}}$$
B
$${R_{100}} = {R_{40}} + {R_{60}}$$
C
$${R_{100}} > {R_{60}} > {R_{40}}$$
D
$${1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}$$

## 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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