back to directory
NEET PHYSICSCURRENT ELECTRICITYEasy

Question

The colour code of resistance is given below: The values of resistance and tolerance, respectively are:

A

47 k\Omega , 10%

B

4.7 k\Omega , 5%

C

470 \Omega , 5%

D

470 k\Omega , 5%

Step-by-Step Solution

According to the NCERT text on carbon resistors, the value of resistance is determined by a colour code. The first two bands indicate the first two significant figures, the third band denotes the decimal multiplier, and the fourth band represents the tolerance. For the correct answer of 470 \Omega with 5% tolerance, the colour sequence would be Yellow (4), Violet (7), Brown (10¹), and Gold (5%).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from CURRENT ELECTRICITY. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSCURRENT ELECTRICITYcolourresistancevaluesresistancetolerance

More CURRENT ELECTRICITY Questions

View all

Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. The value of X is:

A.12.1 Ω
B.12.4 Ω
C.10.3 Ω
D.11.7 Ω
MediumSolve

The potential difference across the 100 Ω resistance in the following circuit is measured by a voltmeter of 900 Ω resistance. The percentage error made in reading the potential difference is:

A.10.9
B.0.1
C.1
D.10
MediumSolve

The equivalent resistance of the infinite network given below is:

A.2 Ω
B.(1 + √2) Ω
C.(1 + √3) Ω
D.(1 + √5) Ω
HardSolve

The sliding contact C is at one fourth of the length of the potentiometer wire (AB) from A as shown in the circuit diagram. If the resistance of the wire AB is R₀, then the potential drop (V) across the resistor R is:

A.4V₀R / (3R₀ + 16R)
B.4V₀R / (3R₀ + R)
C.2V₀R / (4R₀ + R)
D.2V₀R / (2R₀ + 3R)
HardSolve

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, then the emf of the second cell is:

A.1.27 V
B.2.25 V
C.3.27 V
D.3.25 V
EasySolve

The V-I graph for a conductor at temperatures $T_1$ and $T_2$ are as shown in the figure. The term $(T_2 - T_1)$ is proportional to:

A.$\cos 2\theta$
B.$\sin \theta$
C.$\cot 2\theta$
D.$\tan \theta$
HardSolve

In a meter bridge experiment, the null point is at a distance of 30 cm from A. If a resistance of 16 Ω is connected in parallel with resistance Y, the null point occurs at 50 cm from A. The value of the resistance Y is:

A.112/3 Ω
B.40/3 Ω
C.64/3 Ω
D.48/3 Ω
MediumSolve

The figure shows a 2.0 V potentiometer used for the determination of the internal resistance of a 1.5 V cell. The balance point of the cell in the open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. The internal resistance of the cell is:

A.1.68 Ω
B.0.13 Ω
C.0.31 Ω
D.1.12 Ω
MediumSolve

This neet physics practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice physics sets →