| ⇦ |
| ⇨ |
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 bullet loses 1/20 of its velocity after penetrating a plank.How many planks
- The e.m.f. of a battery is 2 V and its internal resistance is 0.5 Ω. The maximum
- For the action of a common Base (CB) transistor, (E=emitter, B= base, C= collector)
- A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel.The lift starts moving upwards wi
- Two particles A and B, move with constant velocities V₁ and V₂. At the initial
Topics: Current Electricity
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A bullet loses 1/20 of its velocity after penetrating a plank.How many planks
- The e.m.f. of a battery is 2 V and its internal resistance is 0.5 Ω. The maximum
- For the action of a common Base (CB) transistor, (E=emitter, B= base, C= collector)
- A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel.The lift starts moving upwards wi
- Two particles A and B, move with constant velocities V₁ and V₂. At the initial
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