JEE Main 2024 (Online) 1st February Morning Shift | Current Electricity Question 27 | Physics

The reading in the ideal voltmeter $(\mathrm{V})$ shown in the given circuit diagram is :

JEE Main 2024 (Online) 1st February Morning Shift Physics - Current Electricity Question 47 English

JEE Main 2024 (Online) 1st February Morning Shift Physics - Current Electricity Question 47 English

$3 \mathrm{~V}$

$10 \mathrm{~V}$

$5 \mathrm{~V}$

$0 \mathrm{~V}$

JEE Main 2024 (Online) 1st February Morning Shift

MCQ (Single Correct Answer)

-1

A galvanometer has a resistance of $50 ~\Omega$ and it allows maximum current of $5 \mathrm{~mA}$. It can be converted into voltmeter to measure upto $100 \mathrm{~V}$ by connecting in series a resistor of resistance :

$19500 \Omega$

$5975 \Omega$

$20050 \Omega$

$19950 \Omega$

JEE Main 2024 (Online) 31st January Evening Shift

MCQ (Single Correct Answer)

-1

By what percentage will the illumination of the lamp decrease if the current drops by 20%?

26%

36%

46%

56%

JEE Main 2024 (Online) 31st January Evening Shift

MCQ (Single Correct Answer)

-1

The resistance per centimeter of a meter bridge wire is $$r$$, with $$X \Omega$$ resistance in left gap. Balancing length from left end is at $$40 \mathrm{~cm}$$ with $$25 \Omega$$ resistance in right gap. Now the wire is replaced by another wire of $$2 r$$ resistance per centimeter. The new balancing length for same settings will be at

10 cm

80 cm

40 cm

20 cm

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