Equal weights are suspended from the wires of same material and same lengths but with radii in the ratio $1 : 2$ . The ratio of extensions produced in them will be

4 : 1

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1 : 4

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1 : 2

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2 : 1

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Solution

The correct option is C $4 : 1$
We know that $Δ l = \frac{F L}{A Y}$ . Thus, for same force, same material (Y) and same length, extension is inversely proportional to area, or $r^{2}$
Thus, for radii in ratio $1 : 2$ , extension will be in ratio $4 : 1$

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Equal weights are suspended from the wires of same material and same lengths but with radii in the ratio 1:2. The ratio of extensions produced in them will be

The correct option is C 4:1We know that Δl=FLAY. Thus, for same force, same material (Y) and same length, extension is inversely proportional to area, or r2Thus, for radii in ratio 1:2, extension will be in ratio 4:1

Equal weights are suspended from the wires of same material and same lengths but with radii in the ratio $1 : 2$ . The ratio of extensions produced in them will be

The correct option is C $4 : 1$
We know that $Δ l = \frac{F L}{A Y}$ . Thus, for same force, same material (Y) and same length, extension is inversely proportional to area, or $r^{2}$
Thus, for radii in ratio $1 : 2$ , extension will be in ratio $4 : 1$