Two rods of same material and volume having circular cross-section...

Two rods of same material and volume having circular cross-section are subjected to tension T. Within the elastic limit, same force is applied to both the rods. Diameter of the first rod is half of the second rod, then the extensions of first rod to second rod will be in the ratio

Options

A $4: 1$
B $16: 1$
C $32: 1$
D $2: 1$

Correct answer

B. $16: 1$

Step-by-step solution

The young's modulus of the material is given by

\begin{aligned} \mathrm{Y} &=\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{l} / \mathrm{l}} \\ \mathrm{Y} &=\frac{\mathrm{F} \times \mathrm{l}}{\mathrm{A} \times \Delta \mathrm{l}} \\ \mathrm{Y} &=\frac{\mathrm{F} \times \mathrm{l} \times \mathrm{A}}{\mathrm{A}^{2} \times \Delta \mathrm{l}} \quad(\mathrm{V}=\mathrm{lA}) \\ \mathrm{Y} &=\frac{\mathrm{FV}}{\mathrm{A}^{2} \times \Delta \mathrm{l}} \ldots . .(1) \end{aligned}

From the given question,

\mathrm{F}, \mathrm{V}, \mathrm{Y}

are same. From the equation (1),

\Delta \mathrm{l} \propto \frac{1}{\mathrm{~A}^{2}}\Delta \propto \frac{1}{\mathrmr^{4}}

where

\mathrm{r}=

radius

\Delta \propto \frac{1}{\mathrm{~d}^{4}}

where,

\mathrm{d}=

diameter The diameter of

\mathrm{A}

is half of the diameter of

\mathrm{B}\mathrm{d}_{\mathrm{A}}=\frac{\mathrm{d}_{\mathrm{B}}}{2}\mathrm{d}_{\mathrm{B}}=2 \mathrm{~d}_{\mathrm{A}}

Ratio,

\frac{\Delta \mathrm{l}_{\mathrm{A}}}{\Delta_{\mathrm{B}}}=\frac{\mathrm{d}_{\mathrm{B}}^{4}}{\mathrm{~d}_{\mathrm{A}}^{4}}=\left(\frac{2 \mathrm{~d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{A}}}\right)^{4}\frac{\Delta \mathrm{l}_{\mathrm{A}}}{\Delta \mathrm{l}_{\mathrm{B}}}=16

The correct option is A.

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