Question

A uniform wire is cut into 10 segments increasing in length in equal steps, the resistance of the shortest segment is R and the resistance of the other segments increases in steps of 8 ohm. If the resistance of the longest segment is 2R , find the value of R and the resistance of the original wire?

Answer 1

Answer:R=72 ohm and resistance of original wire is 1090ohm

Explanation: Using arithmetic progression

First term is R with common difference = 8

R +(10-1)8=2R

R+9*8=2R

R+72=2R

R=72 ohm

Original resistance = 1090ohm

By using sum of terms of AP

Answer 2

Answer:

The value of R is equal to 72Ω and the resistance of the original wire is equal to 1080Ω.

Explanation:

We have given, the resistance of first segment = RΩ

Resistance of second segment will be = R + 8Ω

Resistance of 3rd segment will be = (R + 8 +8)Ω = R + 2(8)Ω

We can use arithmetic progression , $a_{n}=a+ (n-1)d$

Resistance of 4th segment = R + 4(8)Ω

Similarly keep going......

Resistance of 10th segment = R + (10-1)8Ω = R + 7(8)Ω = R + 72Ω

As we cut the wire into segments the length of segment increases,

Therefore 10th segment will be the longest segment having resistance equal to 2R.

R + 72Ω = 2R

R = 72Ω

Resistances of segments will be as: 72, 80, 88, ................,144

Resistance of the original wire will be sum of resistances of all segments, we can use the formula of sum for arithmetic progression:-

$S_{n}=\frac{n}{2}(a+L)$

Where n = 10 and a is first segment's resistance = 72,

L is last segment's resistance = 144

$S_{n}= \frac{10}{2}(72+144)$

$S_{n}=5(216)$

$S_{n}=1080$Ω

Therefore, the value of resistance of original wire is equal to 1080Ω.