Verified Solution • 2025–26 Session moderate

QJEEAdv10Ph0c6a: JEE Advanced 2010 Physics - Class 12 JEEAdv10Ph0c6a Solution: Incandescent bulbs are designed by keeping in mind that the…

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

Multiple Choice Options

  • A 1R100=1R40+1R60{1 \over {{R_{100}}}} = {1 \over {{R_{40}}}} + {1 \over {{R_{60}}}}
  • B R100=R40+R60{R_{100}} = {R_{40}} + {R_{60}}
  • C R100>R60>R40{R_{100}} > {R_{60}} > {R_{40}}
  • D 1R100>1R60>1R40{1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}

Step-by-Step Verified Solution

The power of the bulb is

P=V2RP = {{{V^2}} \over R}

Therefore,

100=V2R100⇒1R100=100V2100 = {{{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=V2R60⇒1R60=60V260 = {{{V^2}} \over {{R_{60}}}} \Rightarrow {1 \over {{R_{60}}}} = {{60} \over {{V^2}}} and 40=V2R40⇒1R40=40V240 = {{{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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