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Physics MCQs for NEET — Practice Questions with Answers

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The colour sequence in a carbon resistor is red, brown, orange and silver. The resistance of the resistor is

21 × 10³ ± 10%

23 × 10¹ ± 10

21 × 10³ ± 5%

12 × 10³ ± 5%

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Explanation

Significant figures		Multiplier	Tolerance
Red	Brown	Orange	Silver
2	1	10³	± 10%

∴ R = $21 \times 10^{3} \pm 10 %$

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Two resistors of resistance R₁ and R₂ having R₁ > R₂ are connected in parallel. For equivalent resistance R, the correct statement is :

$R > R_{1} + R_{2}$

$R_{1} < R < R_{2}$

$R_{2} < R < (R_{1} + R_{2})$

R < R₁

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Explanation

Equivalent resistance of parallel resistors is always less than any of the member of the resistance system.

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A wire of resistance R is divided into 10 equal parts. These parts are connected in parallel, the equivalent resistance of such connection will be :

0.01 R

0.1 R

10 R

100 R

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Explanation

Each part will have a resistance $r = R / 10$

Let equivalent resistance be r_R, then

$\frac{1}{r_{R}} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} . ..........10 times$

$∴ \frac{1}{r_{R}} = \frac{10}{r} = \frac{10}{R / 10} = \frac{100}{R} \Rightarrow r_{R} = \frac{R}{100} = 0 .01 R$

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Three resistors each of 2 ohm are connected together in a triangular shape. The resistance between any two vertices will be

4/3 ohm

3/4 ohm

3 ohm

6 ohm

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Explanation

When three resistors of equal resistance R are connected in a triangular shape, the effective resistance between any two vertices is R/3 by applying the parallel resistance formula. Since each resistor is 2Î©, the effective resistance between any two vertices is 2/3 = 4/3Î©.

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There are n similar conductors each of resistance R. The resultant resistance comes out to be x when connected in parallel. If they are connected in series, the resistance comes out to be :

x/n²

n²x

x/n

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Explanation

In parallel, $x = \frac{R}{n}$ ∴ R = nx

In series, R + R + R .... n times = nR = n (nx) = n²x

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Two resistances are joined in parallel whose resultant is $\frac{6}{8}$ ohm. One of the resistance wire is broken and the effective resistance becomes 2Ω. Then the resistance in ohm of the wire that got broken was

3/5

6/5

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Explanation

If resistances are $R_{1}$ and $R_{2}$ then $\frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{6}{8}$ …..(i)

Suppose $R_{2}$ is broken then $R_{1} = 2 Ω$ …..(ii)

On solving equations (i) and (ii) we get $R_{2} = 6 / 5 Ω$

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The equivalent resistance of resistors connected in series is always :

Equal to the mean of component resistors

Less than the lowest of component resistors

In between the lowest and the highest of component resistors

Equal to the sum of component resistors

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Explanation

$R_{series} = R_{1} + R_{2} + R_{3} + . . . . .$

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A cell of negligible resistance and e.m.f. 2 volts is connected to a series combination of 2, 3, and 5 $Ω$ . The potential difference in volts between the terminals of 3 $Ω$ resistance will be :

0.6 $Ω$

2/3 $Ω$

3 $Ω$

6 $Ω$

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Four wires of equal length and of resistances 10 ohms each are connected in the form of a square. The equivalent resistance between two opposite corners of the square is :

10 ohm

40 ohm

20 ohm

10/4 ohm

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Two resistors are connected (a) in series (b) in parallel. The equivalent resistance in the two cases is 9 ohm and 2 ohms respectively. Then the resistances of the component resistors are :

2 ohm and 7 ohm

3 ohm and 6 ohm

3 ohm and 9 ohm

5 ohm and 4 ohm

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Explanation

$R_{1} + R_{2} = 9$ and $\frac{R_{1} R_{2}}{R_{1} + R_{2}} = 2$ ⇒ $R_{1} R_{2} = 18$

$R_{1} - R_{2} = \sqrt{{(R_{1} + R_{2})}^{2} - 4 R_{1} R_{2}} = \sqrt{81 - 72} = 3$

$R_{1} = 6 Ω, R_{2} = 3 Ω$

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Frequently Asked Questions

Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.