A wire has resistance of 10 Ω. If it is stretched by 1 / 10 th of its length, then its resistance is nearly

Question

A wire has resistance of 10 Ω. If it is stretched by 1 / 10 th of its length, then its resistance is nearly (A) 9 Ω (B) 10 Ω (C) 11 Ω (D) 12 Ω

Tardigrade Admin · Accepted Answer

R=ρ l / A=10 Ω New length, l1=l+l / 10=11 l / 10 ∴ New area, A1=A l / l1=(10 A/11) ∴ New resistance, R1=ρ l1 / A1=ρ(11 l / 10) /(10 / 11) A =(121 ρ l/100 A)=(121/100) × 10=12.1 Ω

Q. A wire has resistance of $10Ω$ . If it is stretched by $1/10$ th of its length, then its resistance is nearly

$9 Ω$

$10 Ω$

$11 Ω$

$12 Ω$

$R = ρl / A = 10Ω$
New length, $l_{1} = l + l /10 = 11 l /10$
$∴$ New area, $A_{1} = A l / l_{1} = \frac{10 A}{11}$
$∴$ New resistance, $R_{1} = ρ l_{1} / A_{1} = ρ (11 l /10) / (10/11) A$
$= \frac{121 ρl}{100 A} = \frac{121}{100} \times 10 = 12.1Ω$

Q. A wire has resistance of 10Ω. If it is stretched by 1/10 th of its length, then its resistance is nearly

9Ω

10Ω

11Ω

12Ω