Two wires of the same material have lengths in the ratio 1- 2 and their radii are in the ratio 1- -2- If they are stretched by applying equal forces- the increase in their lengths will be in the ratio

The correct option is D 1 : 1Y=Fl/π r^2 lor l=F/π r^2y l ∝ 1/r^2 l' ∝ 21/(√(2)r)^2or l' ∝1/r^2 Therefore l/ l'=1 ...

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Byju's Answer

Standard XII

Physics

Young's Modulus of Elasticity

Two wires of ...

Question

Two wires of the same material have lengths in the ratio 1 : 2 and their radii are in the ratio 1 : $\sqrt{2}$ . If they are stretched by applying equal forces, the increase in their lengths will be in the ratio

√2 : 2

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2 : √2

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1 : 1

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1 : 2

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Solution

The correct option is C 1 : 1
$Y = \frac{F l}{π r^{2} △ l} o r △ l = \frac{F}{π r^{2} y}$
$△ l \propto \frac{1}{r^{2}}$
$△ l^{'} \propto \frac{21}{(\sqrt{2} r)^{2}} o r △ l^{'} \propto \frac{1}{r^{2}}$
Therefore $\frac{△ l}{△ l^{'}}$ =1

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Two wires of the same material have lengths in the ratio 1 : 2 and their radii are in the ratio 1 : √2. If they are stretched by applying equal forces, the increase in their lengths will be in the ratio

The correct option is C 1 : 1Y=Flπ r2△ lor △ l=Fπ r2y △ l ∝ 1r2 △ l′ ∝ 21(√2r)2or △ l′ ∝1r2 Therefore △ l△ l′=1

Two wires of the same material have lengths in the ratio 1 : 2 and their radii are in the ratio 1 : $\sqrt{2}$ . If they are stretched by applying equal forces, the increase in their lengths will be in the ratio

The correct option is C 1 : 1
$Y = \frac{F l}{π r^{2} △ l} o r △ l = \frac{F}{π r^{2} y}$
$△ l \propto \frac{1}{r^{2}}$
$△ l^{'} \propto \frac{21}{(\sqrt{2} r)^{2}} o r △ l^{'} \propto \frac{1}{r^{2}}$
Therefore $\frac{△ l}{△ l^{'}}$ =1