A conductor of resistance 3 Ω is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohms is

Question

A conductor of resistance 3 Ω is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohms is (A) (9/2) (B) (8/3) (C) 2 (D) 1

Tardigrade Admin · Accepted Answer

R=3Ω When stretched Râ²=4× 3=12 Ω Therefore each side of the triangle has resistance 4 Ω <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-mdacytddmlvkdysa.jpg" /> RA B=(8 × 4/8 + 4)=(32/12)=(8/3)Ω

Q. A conductor of resistance $3 Ω$ is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohms is

$\frac{9}{2}$

$\frac{8}{3}$

$2$

$1$

$R = 3Ω$
When stretched $R^{'} = 4 \times 3 = 12 Ω$
Therefore each side of the triangle has resistance $4 Ω$

$R_{A B} = \frac{8 \times 4}{8 + 4} = \frac{32}{12} = \frac{8}{3} Ω$

Q. A conductor of resistance 3Ω is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohms is

29​

38​

2

1

R=3Ω When stretched R′=4×3=12Ω Therefore each side of the triangle has resistance 4Ω RAB​=8+48×4​=1232​=38​Ω

Q. A conductor of resistance $3 Ω$ is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohms is

$\frac{9}{2}$

$\frac{8}{3}$

$2$

$1$

$R = 3Ω$
When stretched $R^{'} = 4 \times 3 = 12 Ω$
Therefore each side of the triangle has resistance $4 Ω$

$R_{A B} = \frac{8 \times 4}{8 + 4} = \frac{32}{12} = \frac{8}{3} Ω$