| ⇦ |
| ⇨ |
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: - A L-C-R circuit with L=1.00 mH, C=10μF and R=50Ω, is driven with 5 V AC voltage.
- The electromagnetic waves detected using a thermopile and used in physical therapy
- The focal length of the objective of a terrestrial telescope is 80 cm
- A body covers 200 cm in the first 2 seconds and 220 cm in the next 4 seconds.
- The dimensional formula for Young’s modulus is
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A L-C-R circuit with L=1.00 mH, C=10μF and R=50Ω, is driven with 5 V AC voltage.
- The electromagnetic waves detected using a thermopile and used in physical therapy
- The focal length of the objective of a terrestrial telescope is 80 cm
- A body covers 200 cm in the first 2 seconds and 220 cm in the next 4 seconds.
- The dimensional formula for Young’s modulus is
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