Problem : A uniform wire is cut into 10 segments increasing in length in equal steps, the resistance of the shortest segment is R and the resistances of the other segments increases in steps of 8 Ω. If the resistance of the longest segment is 2R, find the value of R and hence find the resistance of the original wire.
Doubt by Kulwinder
Solution :
Let the length of the shortest segment by a then length of the other segments would be 2a, 3a, 4a, 5a, 6a, 7a, 8a, 9a, 10a.
The corresponding resistance of these segments would be R, R+8, R+16, R+24, R+32, R+40, R+48, R+56, R+64, R+72.
Resistance of Longest Segment = 2R (Given)
R + 72 = 2R
72 = 2R - R
72 = R
R = 72 Ω
Resistance of original Wire = R1 + R2 + R3 + . . . + R10
= R + R+8 + R+16 + R+24 + R+32 + R+40 + R+48 + R+56 + R+64 + R+72
= 10R + 8 (1+2+3+4+5+6+7+8+9)
= 10R + 8 [9(9+1)]2
= 10R + 4 × 90
= 10R +360
= 10 (72) + 360 [R = 72 Ω]
= 720 + 360
= 1080 Ω
The corresponding resistance of these segments would be R, R+8, R+16, R+24, R+32, R+40, R+48, R+56, R+64, R+72.
Resistance of Longest Segment = 2R (Given)
R + 72 = 2R
72 = 2R - R
72 = R
R = 72 Ω
Resistance of original Wire = R1 + R2 + R3 + . . . + R10
= R + R+8 + R+16 + R+24 + R+32 + R+40 + R+48 + R+56 + R+64 + R+72
= 10R + 8 (1+2+3+4+5+6+7+8+9)
= 10R + 8 [9(9+1)]2
= 10R + 4 × 90
= 10R +360
= 10 (72) + 360 [R = 72 Ω]
= 720 + 360
= 1080 Ω
