Two wires of same material and same extension $Δ l$ have lengths l and $3 l$ and diameters $3 d$ and d. What will be the ratio of the forces applied on the two?

1 / 27

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27

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Solution

The correct option is B $27$
$y = \frac{F L}{A Δ 1}$
where, y= youngs modulus
F=Applied force
A=Area of cross section

Now, $\frac{F_{1} L_{1}}{A_{1} Δ L_{1}} = \frac{F_{2} L_{2}}{A_{2} Δ L_{2}}$

=> $\frac{F_{1}}{F_{2}} = \frac{D_{1}^{2}}{D_{2}^{2}} \times \frac{L_{2}}{L_{1}} \times \frac{Δ l_{1}}{Δ l_{2}}$

=> $\frac{F_{1}}{F_{2}} = \frac{g D^{2}}{D^{2}} \times \frac{2 L}{L} \times \frac{l}{l}$

=> $\frac{F_{1}}{F_{2}} = 27$

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Two wires of same material and same extension Δl have lengths l and 3l and diameters 3d and d. What will be the ratio of the forces applied on the two?

The correct option is B 27y=FLAΔ1where, y= youngs modulusF=Applied forceA=Area of cross sectionNow, F1L1A1ΔL1=F2L2A2ΔL2=> F1F2=D21D22×L2L1×Δl1Δl2=> F1F2=gD2D2×2LL×ll=>F1F2=27

Two wires of same material and same extension $Δ l$ have lengths l and $3 l$ and diameters $3 d$ and d. What will be the ratio of the forces applied on the two?