Home
Class 10
PHYSICS
A wire of resistance R is stretched to t...

A wire of resistance R is stretched to thrice its original length keeping the volume constant. Calculate its new resistance.

A

R

B

3R

C

9R

D

27R

Text Solution

Verified by Experts

Since, volume is constant, i.e., IA= constant and
`RproptoI/Atherefore, RproptoI (I^2)/(AI) or R proptolI^2`
So, if length is stretched 3 times then resistance will increase to 9 times.
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    CENGAGE PHYSICS|Exercise MANDATORY EXERCISE(EXERCISE SET III)|5 Videos

  • CURRENT ELECTRICITY

    CENGAGE PHYSICS|Exercise MANDATORY EXERCISE(EXERCISE SET IV)|3 Videos

  • CURRENT ELECTRICITY

    CENGAGE PHYSICS|Exercise OLYMPIAD AND NTSE LEVEL EXERCISES|10 Videos

  • ATOMIC AND NUCLEAR PHYSICS

    CENGAGE PHYSICS|Exercise OLYMPIAD AND NTSE LEVEL EXERCISES |8 Videos

  • ELECTROMAGNETIC INDUCTION

    CENGAGE PHYSICS|Exercise OLYMPIAD AND NTSE LEVEL EXERCISES|10 Videos

Similar Questions

Explore conceptually related problems

A conductor of resistance R and conductance G is stretched to twice its original length keeping its volume constant. Calculate the new resistance and the conductance of the conductor.

A wire of 10 Omega resistance is stretched to thrice its original length. What will be its (i) new resistivity and (ii) new resistance?

A wire of resistance 4 Omega is stretched to twice its original length. The resistance of stretched wire would be

A wire of resistance 10 Omega is drawn out so that its length is thrice its original length. Calculate its new resistance (resistivity and density of the material remain uncharged). " " Or A wire of resistance 10 Omega is drawn out so that its length is increased by twice its original length. Calculate its new resistance.

A wire of resistance 10 Omega is drawn out so that its length is thrice its original length. Calculate its new resistance (resistivity and density of the material remain uncharged).

A wire of resistance 4 Omega is stretched to twice its original length. In the process of stretching, its area of cross-section gets halved. Now, the resistance of the wire is

A wire with 15 Q resistance is stretched by one tenth of its original length and volume of wire is kept constant. Then its resistance will be

A given copper wire of resistance R is stretched to reduce its diameter to half its previous value. What sould be its new resistance ?

A wire of resistance 4 Omega is stretched to four times of its original length resistance of wire now becomes