| ⇦ | 
| ⇨ |
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 the rise in height of capillary of two tubes is 6.6 cm and 2.2 cm
- Two particles A and B having equal charges +6C, after being accelerated
- In P-N junction having depletion layer of thickness 10⁻⁶ m, the potential across
- The equation of state for 5 g of oxygen at a pressure P and temperature T,
- The potential difference that must be applied to stop the fastest photoelectrons
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If the rise in height of capillary of two tubes is 6.6 cm and 2.2 cm
- Two particles A and B having equal charges +6C, after being accelerated
- In P-N junction having depletion layer of thickness 10⁻⁶ m, the potential across
- The equation of state for 5 g of oxygen at a pressure P and temperature T,
- The potential difference that must be applied to stop the fastest photoelectrons
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