Question

Q.2 a wire of radius 2 mm has resistance 2.5 . If it is stretched such that its new resistance becomes 40 , the new radius is

Answer 1

Answer:

A 4 Ω resistance wire is doubled on it. Calculate the new resistance of the wire. When the wire is doubled on itself, its length becomes halved (L/2) and area of cross-section becomes doubled (2 A). Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W

Answer 2

The new radius is 1 mm.

Explanation:

Given that,

Radius = 2 mm

Resistance = 2.5 ohm

New resistance = 40 ohm

The volume of wire remains constant after stretch

We need to calculate the resistance

Using formula of resistance

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

$R=\dfrac{\rho l}{\pi r^2}$....(I)

We need to calculate the new radius

Using formula of resistance

$R'=\dfrac{\rho 2\timesl}{\dfrac{A}{2}}$

$R'=\dfrac{\rho 2l}{\dfrac{\pi r'^2}{2}}$....(II)

From equation (I) and (II)

$\dfrac{R}{R'}=\dfrac{\dfrac{\rho l}{\pi r^2}}{\dfrac{\rho 2l}{\dfrac{\pi r'^2}{2}}}$

$\dfrac{R}{R'}=\dfrac{r'^2}{4r^2}$

Put the value into the formula

$\dfrac{2.5}{40}=\dfrac{r'^2}{4\times(2\times10^{-3})^2}$

$r'=\sqrt{\dfrac{2.5\times4\times(2\times10^{-3})^2}{40}}$

$r'=0.001\ m$

$r'=1\ mm$

Hence, The new radius is 1 mm.

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Topic : resistivity

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