| ⇦ | 
| ⇨ |
When a current of (2.5±0.5) A flows through a wire, it develops a potential difference of (20±1) V, then the resistance of wire is
Options
(a) (8±2)Ω
(b) (8±1.6)Ω
(c) (8±1.5)Ω
(d) (8±3)Ω
Correct Answer:
(8±2)Ω
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - If λ₁ and λ₂ are the wavelengths of the first members of the Lyman and Paschen
- A car moving with a speed of 50 kmh⁻¹ can be stopped by brakes after at least 6 m
- If a wire is stretched to four times its length, then the specific resistance
- A body undergoes no change in volume. Poisson’s ratio is
- The respective speeds of the molecules are 1,2,3,4 and 5 km/s.
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If λ₁ and λ₂ are the wavelengths of the first members of the Lyman and Paschen
- A car moving with a speed of 50 kmh⁻¹ can be stopped by brakes after at least 6 m
- If a wire is stretched to four times its length, then the specific resistance
- A body undergoes no change in volume. Poisson’s ratio is
- The respective speeds of the molecules are 1,2,3,4 and 5 km/s.
Topics: Current Electricity (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
R=V/I
R=20/2.5
R= 8 ohm
Now,
∆ R/R=∆ V/V+∆I/I
=1/20 +0.5/2.5
= 1/4
Therefore,
∆ R÷R=1÷4
∆ R=1÷4×R
∆R=1÷4×8
∆R=2
Therefore,
Resistance with error limits=R+ – ∆R
=(8+ – 2)ohm