COMEDK 2024 Evening Shift | Current Electricity Question 2 | Physics

The resistance of a $$10 \mathrm{~m}$$ long wire is $$10 \Omega$$. Its length is increased by $$25 \%$$ by stretching the wire uniformly. The new resistance is

$$18.6 \Omega$$

$$15.6 \Omega$$

$$12.8 \Omega$$

$$14.9 \Omega$$

COMEDK 2024 Evening Shift

MCQ (Single Correct Answer)

-0

Four resistors, each of resistance R, are connected as shown in the figure below.

COMEDK 2024 Evening Shift Physics - Current Electricity Question 1 English

The total resistance between points 1 and 3 is 0.5 R.

The total resistance between points 1 and 6 is $$3.5 \mathrm{R}$$.

The total resistance between points 3 and 6 is 2 R.

The total resistance between points 2 and 4 is $$0.5 \mathrm{~R}$$.

COMEDK 2024 Evening Shift

MCQ (Single Correct Answer)

-0

A voltmeter of resistance $$1000 \Omega .0 .5 \mathrm{~V} /$$ div is to be converted into a voltmeter to make it to read $$=1 \mathrm{~V} / \mathrm{div}$$. The value of high resistance to be connected in series with it is

$$6000 \Omega$$

$$5000 \Omega$$

$$4000 \Omega$$

$$1000 \Omega$$

COMEDK 2024 Morning Shift

MCQ (Single Correct Answer)

-0

The resistance of the galvanometer and shunt of an ammeter are $$90 \mathrm{~ohm}$$ and $$10 \mathrm{~ohm}$$ respectively, then the fraction of the main current passing through the galvanometer and the shut respectively are:

$$\frac{1}{90}$$ and $$\frac{1}{10}$$

$$\frac{1}{10}$$ and $$\frac{1}{90}$$

$$\frac{1}{10}$$ and $$\frac{9}{10}$$

$$\frac{9}{10}$$ and $$\frac{1}{10}$$

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