Wire stretched to double length (volume constant). New resistance: MCQ with Answer and Explanation

Wire stretched to double length (volume constant). New resistance:
A. R/2
B. R
C. 2R
D. 4R
Answer: Option D
Solution (By JKSSB Mock Tests)
Volume constant: A' = A/2 when L' = 2L. R' = ρ(2L)/(A/2) = 4ρL/A = 4R. Memory tip: 'Stretch wire: R ∝ L² when volume fixed'. Resistance proportionality frequently tested in competitive exams.

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Practice More Physics Questions

Question #1
When a body falls freely under gravity, which statement is correct?
A. Kinetic energy decreases
B. Total energy is not conserved
C. Mechanical energy remains constant
D. Potential energy increases

Correct Answer: Option C


Explanation:
In free fall under gravity alone, mechanical energy (kinetic + potential) is conserved, assuming no air resistance. Potential energy decreases while kinetic energy increases, sum constant. Law of conservation of mechanical energy. Non-conservative forces absent.

This question belongs to: Science Physics
Question #2
A stone tied to a string is whirled in a circle. The work done by tension in string is
A. Positive
B. Zero
C. Negative
D. Variable

Correct Answer: Option B


Explanation:
Tension is towards centre, displacement is tangential (velocity). Force perpendicular to displacement, cos90°=0, work zero. Centripetal force does no work, doesn't change speed.

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Question #3
Work done by centripetal force in circular motion is:
A. Positive
B. Depends on radius
C. Zero
D. Negative

Correct Answer: Option C


Explanation:
Centripetal force acts radially inward, perpendicular to tangential displacement. Work W = F·d·cosθ; θ=90° ⇒ cos90°=0 ⇒ W=0. No energy transfer occurs; speed remains constant in uniform circular motion. Memory tip: 'Perpendicular force ⇒ zero work'. Conceptual question testing work-energy understanding, common in competitive exams to identify misconceptions about circular motion.

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