Question

A fuse wire melts at 5 A. It is desired that the fuse wire of the same material melt at 10 A. The new

the radius of the wire is

(a) 4 times (b) 2 times

(c) 1⁄2 times (d) 1⁄4 times

pls answer

Answer 1

Answer:

b 2 times

Explanation:

because when radius increase the melting point increase s due to more volume which takes more time to heat or melt.

Answer 2

Answer:

The radius of the new wire will be 2 times the original wire i.e. option (b).

Explanation:

For a given wire volume remains constant so,

$V_1=V_2$

$A_1\times l_1=A_2 \times l_2$

$\frac{A_1}{A_2}=\frac{l_2}{l_1}$             (1)

V=volume of the wire

A= area of the wire

l=length of the wire

r=radius of the wire

For resistance we have;

$R=\rho \frac{l}{A}$

$\frac{R_1}{R_2}=\frac{l_1}{l_2} \times \frac{A_2}{A_1}$      (2)

l₁=2πr₁

l₂=2πr₂

A₁=πr₁²

A₂=πr₂²

By putting equation (1), and, l₁,l₂, A₁, A₂ in equation (2) we get;

$\frac{R_1}{R_2}=\frac{r_2}{r_1}$      (3)

And,

$\frac{I_1}{I_2}=\frac{V}{R_1}\times \frac{R_2}{V}$

$\frac{I_1}{I_2}=\frac{R_2}{R_1}$            (4)

From the questions we have,

I₁=5A

I₂=10A

From equations (3) and (4) we have;

$\frac{I_1}{I_2}=\frac{r_1}{r_2}$           (5)

By placing I₁ and I₂ in equation (5) we get;

$\frac{5}{10}=\frac{r_1}{r_2}$

$r_2=2r_1$

Hence, The radius of the new wire will be 2 times the original wire i.e. option (b).