If the ratio of radii of two wires of same material is 3: 1 and ratio of their lengths is 5: 1, then the ratio of the normal forces that will produce the same extension in the length of two wires is

Question

If the ratio of radii of two wires of same material is 3: 1 and ratio of their lengths is 5: 1, then the ratio of the normal forces that will produce the same extension in the length of two wires is (A) 2: 1 (B) 4: 1 (C) 1: 4 (D) 1: 1

Tardigrade Admin · Accepted Answer

Y =( F ℓ/ A Δ ℓ) ; ⇒ F =( YA Δ ℓ/ℓ) (F1/F2)=(A1/A2) × (l2/l1) =(r12/r22) × (l2/l1)=(4/1) × (1/4)=1: 1 .

Q. If the ratio of radii of two wires of same material is $3 : 1$ and ratio of their lengths is $5 : 1$ , then the ratio of the normal forces that will produce the same extension in the length of two wires is

$2 : 1$

$4 : 1$

$1 : 4$

$1 : 1$

$Y = \frac{F ℓ}{A Δ ℓ}; \Rightarrow F = \frac{Y A Δ ℓ}{ℓ}$
$\frac{F _{1}}{F _{2}} = \frac{A _{1}}{A _{2}} \times \frac{l _{2}}{l _{1}}$
$= \frac{r _{1}^{2}}{r _{2}^{2}} \times \frac{l _{2}}{l _{1}} = \frac{4}{1} \times \frac{1}{4} = 1 : 1.$

Q. If the ratio of radii of two wires of same material is 3:1 and ratio of their lengths is 5:1, then the ratio of the normal forces that will produce the same extension in the length of two wires is

2:1

4:1

1:4

1:1