Two wires of the same material and length, having diameters in the ratio 2:1, are stretched by the same force. The potential energy per unit volume stored in the two wires will be in the ratio:-

Question

Two wires of the same material and length, having diameters in the ratio 2:1, are stretched by the same force. The potential energy per unit volume stored in the two wires will be in the ratio:- (A) 1:4 (B) 4:1 (C) 16:1 (D) 1:16

Tardigrade Admin · Accepted Answer

U=(1/2)(A Y/L)(Δ L)2 (U/2 L)=u=(1/2)Y((Δ L/L))2 (u1/u2)([((Δ L)1/(Δ L)2)])2=((A2/ A1))2=([((π /4) (d)2/(π )4 (2 d)2)])2=(1/16)

Q. Two wires of the same material and length, having diameters in the ratio $2 : 1,$ are stretched by the same force. The potential energy per unit volume stored in the two wires will be in the ratio:-

$1 : 4$

$4 : 1$

$16 : 1$

$1 : 16$

$U = \frac{1}{2} \frac{A Y}{L} (Δ L)^{2}$
$\frac{U}{2 L} = u = \frac{1}{2} Y (\frac{Δ L}{L})^{2}$
$\frac{u _{1}}{u _{2}} ([\frac{( Δ L ) _{1}}{( Δ L ) _{2}}])^{2} = (\frac{A _{2}}{A _{1}})^{2} = ([\frac{\frac{π}{4} ( d ) ^{2}}{\frac{π}{4} ( 2 d ) ^{2}}])^{2} = \frac{1}{16}$

Q. Two wires of the same material and length, having diameters in the ratio 2:1, are stretched by the same force. The potential energy per unit volume stored in the two wires will be in the ratio:-

1:4

4:1

16:1

1:16