A wire is stretched through $1 m m$ by certain load. The extension produced in the wire of same material with double the length and radius will be

4 m m

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3 m m

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1 m m

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0.5 m m

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Solution

The correct option is D $0.5 m m$
$Δ l = \frac{F L}{A E} D o u b l e l e n g t h a n d r a d i u s m e a n s, L^{'} = 2 L a n d A^{'} = 2^{2} A = 4 A T h u s Δ l^{'} = \frac{F .2 L}{4 A . E} = \frac{Δ l}{2} = \frac{1}{2} = 0.5 m m$

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A wire is stretched through 1mm by certain load. The extension produced in the wire of same material with double the length and radius will be

The correct option is D 0.5mmΔl=FLAEDoublelengthandradiusmeans,L′=2LandA′=22A=4AThusΔl′=F.2L4A.E=Δl2=12=0.5mm

A wire is stretched through $1 m m$ by certain load. The extension produced in the wire of same material with double the length and radius will be

The correct option is D $0.5 m m$
$Δ l = \frac{F L}{A E} D o u b l e l e n g t h a n d r a d i u s m e a n s, L^{'} = 2 L a n d A^{'} = 2^{2} A = 4 A T h u s Δ l^{'} = \frac{F .2 L}{4 A . E} = \frac{Δ l}{2} = \frac{1}{2} = 0.5 m m$