JEE Main 2021 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 26 August 2021, Shift 2, Physics Q17

Question 17 of 90 in this shift, Physics question 17 of 30, Section A.

An electric bulb of 500 watt at 100 volt is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500 W.

(1)20 $\Omega$ Official answer
(2)10 $\Omega$
(3)5 $\Omega$
(4)30 $\Omega$

Official answer

Option 1

NTA final key.

Chapter: Current Electricity
Topic: Ohm's Law and Resistance
Idea tested: Joule Heating and Power

Same idea in other shifts

Asked 15× in all

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8 Apr 2024, Shift 2 · Q56A heater is designed to operate with a power of 1000 W in a 100 V line. It is connected in combination with a resistance of $10\ \Omega$ …MediumNumerical valueHas a figure
31 Jan 2024, Shift 2 · Q41By what percentage will the illumination of the lamp decrease if the current drops by 20%?EasySingle correct
13 Apr 2023, Shift 1 · Q50Different combination of 3 resistors of equal resistance R are shown in the figures. The increasing order for power dissipation is: (A)…MediumDiagram basedHas a figure
13 Apr 2023, Shift 1 · Q60A potential $V_o$ is applied across a uniform wire of resistance R. The power dissipation is $P_1$ . The wire is then cut into two equal…EasyNumerical value
28 Jul 2022, Shift 2 · Q26An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.