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Chapter 7: Problem 857

Two wires of equal lengths are made of the same material wire A has a diameter that is twice as that of wire \(B\). If identical weights are suspended from the ends of these wires the increase in length is......... (A) Four times for wire \(\mathrm{A}\) as for wire \(\mathrm{B}\). (B) Twice for wire \(\mathrm{A}\) as for wire \(\mathrm{B}\). (C) Half for wire \(\mathrm{A}\) as for wire \(\mathrm{B}\). (D) One-fourth for wire \(\mathrm{A}\) as for wire \(\mathrm{B}\).

Short Answer

Expert verified
The increase in length for wire A is one-fourth that of wire B, which corresponds to option (D).

Step by step solution

01

Identify the Young's modulus formula

Young's modulus (Y) is a measure of the stiffness of a material, and it is defined as the ratio of tensile stress to tensile strain. The formula for Young's modulus is given by: \[Y = \frac{F/A}{\Delta L / L}\] Where: - Y is Young's modulus - F is the applied force (weight) - A is the cross-sectional area - ΔL is the change in length - L is the original length
02

Relate the diameter and cross-sectional area of the two wires

First, let's find the relationship between the cross-sectional areas of both wires. The cross-sectional area of a wire is given as: \[A = \pi (\frac{d}{2})^{2}\] Where d is the diameter of the wire. Since wire A has a diameter twice the diameter of wire B: \[d_{A} = 2 d_{B}\] Now, we find the ratio of the cross-sectional areas of both wires. \[\frac{A_{A}}{A_{B}} = \frac{\pi (\frac{d_{A}}{2})^{2}}{\pi (\frac{d_{B}}{2})^{2}}\]
03

Simplify the ratio of cross-sectional areas

Now, simplify the equation by substituting the relationship between the diameters: \[\frac{A_{A}}{A_{B}} = \frac{\pi (\frac{2 d_{B}}{2})^{2}}{\pi (\frac{d_{B}}{2})^{2}}\] \[\frac{A_{A}}{A_{B}} = \frac{d_{A}^2}{d_{B}^2} = \frac{(2d_{B})^2}{d_{B}^2} = 4\] The cross-sectional area of wire A is 4 times larger than that of wire B.
04

Use the Young's modulus formula to compare the increase in length

We will use the Young's modulus formula to find the relationship between the increase in lengths for both wires. The wires are made of the same material, so their Young's modulus is the same. Since Y, F, and L are constant for both wires, we have: \[\frac{A_{A}}{\Delta L_{A}} = \frac{A_{B}}{\Delta L_{B}}\] Now, substitute the relationship of the cross-sectional areas and solve for the ratio of the increase in lengths: \[\frac{4A_{B}}{\Delta L_{A}} = \frac{A_{B}}{\Delta L_{B}}\] \[\frac{\Delta L_{A}}{\Delta L_{B}} = \frac{1}{4}\] So, when identical weights are suspended from the ends of both wires, the increase in length for wire A is one-fourth that of wire B, which corresponds to option (D).

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Most popular questions from this chapter

Which is the dimensional formula for modulus of rigidity? (A) \(\mathrm{M}_{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (B) \(\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\) (C) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{-1}\) (D) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{-2}\)

A vessel contains \(110 \mathrm{~g}\) of water the heat capacity of the vessel is equal to \(10 \mathrm{~g}\) of water. The initial temperature of water in vessel is \(10^{\circ} \mathrm{C}\) If \(220 \mathrm{~g}\) of hot water at \(70^{\circ} \mathrm{C}\) is poured in the vessel the Final temperature neglecting radiation loss will be (A) \(70^{\circ} \mathrm{C}\) (B) \(80^{\circ} \mathrm{C}\) (C) \(60^{\circ} \mathrm{C}\) (D) \(50^{\circ} \mathrm{C}\)

Two drops of the same radius are falling through air with a steady velocity for \(5 \mathrm{~cm}\) per sec. If the two drops coakesce the terminal velocity would be (A) \(10 \mathrm{~cm}\) per sec (B) \(2.5 \mathrm{~cm}\) per sec (C) \(5 \times(4)^{(1 / 3)} \mathrm{cm}\) per sec (D) \(5 \times \sqrt{2} \mathrm{~cm}\) per sec

If temperature of an object is \(140^{\circ} \mathrm{F}\) then its temperature in centigrade is (A) \(105^{\circ} \mathrm{C}\) (B) \(32^{\circ} \mathrm{C}\) (C) \(140^{\circ} \mathrm{C}\) (D) \(60^{\circ} \mathrm{C}\)

When two soap bubbles of radius \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\left(\mathrm{r}_{2}>\mathrm{r}_{1}\right)\) coalesce, the radius of curvature of common surface is........... (A) \(r_{2}-r_{1}\) (B) \(\left[\left(r_{2}-r_{1}\right) /\left(r_{1} r_{2}\right)\right]\) (C) $\left[\left(\mathrm{r}_{1} \mathrm{r}_{2}\right) /\left(\mathrm{r}_{2}-\mathrm{r}_{1}\right)\right]$ (D) \(\mathrm{r}_{2}+\mathrm{r}_{1}\)
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