Question

The extension of a wire by application of load is 0.3cm the extension in a wire of same material but of double the length and half the radius of cross section by the same load will be in (CM)

Answer 1

Answer:

2.4cm

Explanation:

Explanation in figure

Answer 2

The extension of the wire is 2.4 cm.

Explanation:

Given that,

Extension of wire e= 0.3 cm

Length l'= 2l

Radius $r'= \dfrac{r}{2}$

We need to calculate the extension of the wire

Using formula of young modulus

$Y=\dfrac{Fl}{\pi r^2 e}.....(I)$

For new wire,

$Y=\dfrac{Fl'}{\pi r'^2 e'}....(II)$

From equation (I) and (II)

$\dfrac{Fl}{\pi r^2 e}=\dfrac{Fl'}{\pi r'^2 e'}$

Here, load is same

Put the value into the formula

$\dfrac{Fl}{\pi\times r^2\times0.3}=\dfrac{F2l}{\pi\times(\dfrac{r^2}{4})\times e'}$

$e=2\times4\times0.3$

$e=2.4\ cm$

Hence, The extension of the wire is 2.4 cm.

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Topic : extension of wire

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