The ratio of diameters of two wires of same materials is n: 1. The length of each wire is 4 m. On applying the same load, the increase in length of thin wire will be (n 1)

Question

The ratio of diameters of two wires of same materials is n: 1. The length of each wire is 4 m. On applying the same load, the increase in length of thin wire will be (n > 1) (A) n2 times (B) n times (C) 2 n times (D) (2 n+1) times

Tardigrade Admin · Accepted Answer

As, Y=(((l/A)/Δ l)/l)=(F l/A Δ l) or Y=(F l × 4/π D2 × Δ l) or Δ l propto (1/D2) or (Δ L2/Δ L1)=(D12/D22)=(n2/1)

Q. The ratio of diameters of two wires of same materials is $n : 1$ . The length of each wire is $4 m$ . On applying the same load, the increase in length of thin wire will be $(n > 1)$

$n^{2}$ times

$n$ times

$2 n$ times

$(2 n + 1)$ times

As, $Y = \frac{\frac{\frac{l}{A}}{Δ l}}{l} = \frac{Fl}{A Δ l}$
or $Y = \frac{Fl \times 4}{π D ^{2} \times Δ l}$
or $Δ l \propto \frac{1}{D ^{2}}$
or $\frac{Δ L _{2}}{Δ L _{1}} = \frac{D _{1}^{2}}{D _{2}^{2}} = \frac{n ^{2}}{1}$

Q. The ratio of diameters of two wires of same materials is n:1. The length of each wire is 4m. On applying the same load, the increase in length of thin wire will be (n>1)

n2 times

n times

2n times

(2n+1) times