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Chapter 8: Problem 61

If Young's modulus of iron is \(2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\) and the interatomic spacing between two molecules is \(3 \times 10^{-10} \mathrm{~m}\), the interatomic force constant is (A) \(60 \mathrm{~N} / \mathrm{m}\) (B) \(120 \mathrm{~N} / \mathrm{m}\) (C) \(3 \mathrm{~N} / \mathrm{m}\) (D) \(180 \mathrm{~N} / \mathrm{m}\)

Short Answer

Expert verified
The short answer to the problem based on the given step-by-step solution is: The interatomic force constant is \(60 \mathrm{~N} / \mathrm{m}\) (Option A).

Step by step solution

01

Identify given values and formula

In this problem, we have given: Young's modulus (Y): 2 脳 10^11 N / m^2 Interatomic spacing: 3 脳 10^-10 m We will use the formula: \(Y = \frac{F_{constant}}{A} \times \frac{L_0}{\delta L}\)
02

Simplify the formula

Since we are considering an arbitrary length and cross-sectional area, we can rewrite the formula as \(Y = F_{constant} \times \frac{L_0}{\delta L}\)
03

Solve for Force Constant

Now, we will solve for the interatomic force constant (F_constant) by rearranging the formula as follows: \(F_{constant} = Y \times \frac{\delta L}{L_0}\)
04

Insert given values into the formula

Now, we will insert the given values of Young's modulus and interatomic spacing into the equation: \(F_{constant} = (2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}) \times \frac{(3 \times 10^{-10} \mathrm{~m})}{(1 \mathrm{~m})}\)
05

Calculate the Force Constant

After inserting the given values, we will calculate the interatomic force constant: \(F_{constant} = 60 \mathrm{~N} / \mathrm{m}\)
06

Choose the correct option

From the calculations above, the correct option for the interatomic force constant is: (A) 60 N / m

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Most popular questions from this chapter

Air is pushed into a soap bubble of radius \(r\) to double its radius. If the surface tension of the soap solution is \(S\), the work done in the process is (A) \(8 \pi r^{2} S\) (B) \(12 \pi r^{2} S\) (C) \(16 \pi r^{2} S\) (D) \(24 \pi r^{2} S\)

Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by \(75 \times 10^{-4} \mathrm{~N}\), force due to the weight of the liquid. If the surface tension of water is \(6 \times 10^{-2} \mathrm{~N} / \mathrm{m}\), the inner circumference of the capillary must be (A) \(1.25 \times 10^{-2} \mathrm{~m}\) (B) \(0.50 \times 10^{-2} \mathrm{~m}\) (C) \(6.5 \times 10^{-2} \mathrm{~m}\) (D) \(12.5 \times 10^{-2} \mathrm{~m}\)

A gas having density \(\rho\) flows with a velocity \(v\) along a pipe of cross- sectional area \(s\) and bent at an angle of \(90^{\circ}\) at a point \(A\). The force exerted by the gas on the pipe at \(A\) is (A) \(\frac{\sqrt{2} s v}{\rho}\) (B) \(s v^{2} \rho\) (C) \(\frac{\sqrt{3} s v^{2} \rho}{2}\) (D) \(s v^{2} \rho\)

The surface tension of a liquid is \(5 \mathrm{~N} / \mathrm{m}\). If a film is held on a ring of area \(0.02 \mathrm{~m}^{2}\), its total surface energy is about (A) \(5 \times 10^{-2} \mathrm{~J}\) (B) \(2.5 \times 10^{-2} \mathrm{~J}\) (C) \(2 \times 10^{-1} \mathrm{~J}\) (D) \(3 \times 10^{-1} \mathrm{~J}\)

If \(\mathrm{S}\) is stress and \(Y\) is Young's modulus of material of a wire, the energy stored in the wire per unit volume is \([\mathbf{2 0 0 5}]\) (A) \(2 S^{2} Y\) (B) \(\frac{S^{2}}{2 Y}\) (C) \(\frac{2 Y}{S^{2}}\) (D) \(\frac{S}{2 Y}\)
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