When a certain weight is suspended from a long uniform wire, its length increases by one cm. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one then the increase in length will be -

0.5 cm

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2 cm

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4 cm

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8 cm

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Solution

The correct option is C 4 cm
$c) : I = \frac{F L}{A Y} \Rightarrow l \propto \frac{1}{r^{2}}$ (F,L and Y are constant)
$\frac{l_{2}}{l_{1}} = (\frac{r_{1}}{r_{2}})^{2} = (2)^{2} = 4$
$\Rightarrow l_{2} = 4 l_{1}$ =4cm

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When a certain weight is suspended from a long uniform wire, its length increases by one cm. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one then the increase in length will be -

The correct option is C 4 cmc): I=FLAY⇒ l∝1r2 (F,L and Y are constant) l2l1=(r1r2)2=(2)2=4 ⇒ l2=4l1=4cm

The correct option is C 4 cm
$c) : I = \frac{F L}{A Y} \Rightarrow l \propto \frac{1}{r^{2}}$ (F,L and Y are constant)
$\frac{l_{2}}{l_{1}} = (\frac{r_{1}}{r_{2}})^{2} = (2)^{2} = 4$
$\Rightarrow l_{2} = 4 l_{1}$ =4cm