The extension produced in a wire by the application of a load is $3.0$ mm. The extension produced in a wire of the same material and length but half the radius by the same load is:

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Solution

The correct option is C $12$ mm
Formulae,

$\frac{Δ l_{2}}{Δ l_{1}} = \frac{a_{1}}{a_{2}}$

$= \frac{a_{1}}{\frac{a_{1}}{4}}$

$\Rightarrow Δ l_{2} = 4 Δ l_{1}$

$∴ Δ l_{2} = 4 \times 3 = 12 m m$

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The extension produced in a wire by the application of a load is 3.0 mm. The extension produced in a wire of the same material and length but half the radius by the same load is:

The correct option is C 12 mmFormulae,Δl2Δl1=a1a2=a1a14⇒Δl2=4Δl1∴Δl2=4×3=12mm

The extension produced in a wire by the application of a load is $3.0$ mm. The extension produced in a wire of the same material and length but half the radius by the same load is:

The correct option is C $12$ mm
Formulae,

$\frac{Δ l_{2}}{Δ l_{1}} = \frac{a_{1}}{a_{2}}$

$= \frac{a_{1}}{\frac{a_{1}}{4}}$

$\Rightarrow Δ l_{2} = 4 Δ l_{1}$

$∴ Δ l_{2} = 4 \times 3 = 12 m m$