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Mechanical Properties of Solids Test - 63

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Mechanical Properties of Solids Test - 63
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  • Question 1
    1 / -0
    If the interatomic spacing in a steel wire is $$2.8\times 10^{-10}m$$ and $$Y_{real}=2\times 10^{11}N/m^{2}$$, then force constant in N/m is -
    Solution
    $$\begin{aligned}\text { Force constant } K, &=Y \times\text{ normal distance between atoms}\\&=2.8\times10^{-10}\times2\times10^{11}\\&=5.6\times10 \\&=56 \mathrm{~N} / \mathrm{m} \\\text{Hence answer is B}\end{aligned}$$
  • Question 2
    1 / -0
    One end of uniform wire of length $$L$$ and of weight $$W$$ is attached rigidly to a point in the roof and a weight $$W_{1}$$ is suspended from its lower end. If $$s$$ is the area of cross section of the wire, the stress in the wire at a height $$(3L/4)$$ from its lower end is
    Solution
    The total weight at $$(3L/4)$$ from the lower end:
    $$Weight = W_{1}+\dfrac{3L \times W}{4 \times L}$$
    So, the $$stress=weight/area$$
    i.e,$$stress=\dfrac{W_{1}+ \dfrac{3W}{4}}{s}$$
    hence Option C is correct
  • Question 3
    1 / -0
    A material has Poisson's ratio $$0.5$$. If a uniform rod of it suffers a longitudinal strain of $$3\times 10^{-3}$$, what will be percentage increase in volume?
    Solution
    $$\begin{array}{l}\mu \text { , poisson's ratio }=0.5\\\dfrac{\Delta l}{l}=3 \times 10^{-3} \\\text {To find: } \quad \dfrac{\Delta V}{V} \times 100=\text { percentage increase in volume. } \\\mu=\dfrac{1}{2}-\dfrac{Y}{6 B}\end{array}$$

    $$\begin{array}{l}\Rightarrow \dfrac{1}{2}=\dfrac{1}{2}-\dfrac{Y}{6 B} \\\Rightarrow \quad \dfrac{Y}{6 B}=0\\B\rightarrow \infty \\\left|\dfrac{F}{A}\right|=\left|B- \dfrac{\Delta V}{V}\right|\\\Rightarrow\quad\left|\dfrac{\Delta V}{V}\right|=\left|\dfrac{F}{B A}\right| \\\quad a s \quad B\longrightarrow \infty \\\dfrac{\Delta V}{V} \longrightarrow 0 .\end{array}$$

    $$\begin{array}{l}\text { Hence : } \dfrac{\Delta V}{V}\times100=0 \% \\\text { Ans : } D\end{array}$$
  • Question 4
    1 / -0
    A rod of length $$3$$ in and uniform cross-section area $$1mm^{2}$$ is subjected by four forces at different cross section as shown in the figure. Yungs modulus of the rod is

  • Question 5
    1 / -0
    A rod of uniform cross sectional area $$A$$ and length $$L$$ has a weight $$W$$. It is suspended vertically from a fixed support vertically from a fixed support. If Young's modulus for rod is $$Y$$, then elongation produced in rod is

    Solution
    $$\text { Each cross-section of rod is experiencing different amoun of stress. Hence,}$$

    $$\begin{array}{l}\text { Let us consider a small element of differential length } d x \\\text { at a distance } x \text { from free end of the rod.The stress }\end{array}$$

    $$\begin{array}{l}\text { at the position of this element is produced by the weight } \\\text { of the rod of length }x\text { lying below it i.e }(W / L) x\end{array}$$


    $$\sigma=\frac{(W / L) x}{A}=\dfrac{W x}{A L}$$

    $$\text { elongaluon } d \delta \text { is produced } \text { in differential}$$$$\text { element } d x=\dfrac{\sigma d x}{Y}=\dfrac{Ww x d x}{A Y L}$$

    $$\begin{aligned}\delta &=\int d\delta\\&=\int_{0}^{L}\frac{W x d x}{A Y L}=\frac{W}{Y A L}\left[\frac{x^{2}}{2}\right]_{0}^{L}=\frac{(W ) L}{2A Y}\end{aligned}$$
  • Question 6
    1 / -0
    Length of wire becomes 16 cm when certain tension is applied. If the tension is now doubled then wire is further elongated by 1 cm then original length of wire is
    Solution
    $$\text { Let the original length of the wire be }x\mathrm{~cm} \text { . }$$

    $$\dfrac{F}{A}=Y \dfrac{16-x}{x} \quad \longrightarrow(1)$$

    $$\text { if Tension doubles i.e, } 2 \mathrm{~F} \text { , final length of wire becomes } 17 \mathrm{~cm} \text { . }$$

    $$\begin{array}{l}\dfrac{2 F}{A}=Y \dfrac{17-x}{x}\rightarrow\text { (2) } \\\text { Using (2) }\&\mathbb{(1) } \\\dfrac{2(16-x)}{ x}=\dfrac{17-x}{x}\\\Rightarrow 32-2 x=17-x\\\Rightarrow\quad15=x\end{array}$$
  • Question 7
    1 / -0
    To apply a maximum stress of $$1.2\times10^8N/m^2$$ under the action of force $$40N$$ the minimum radius of wire required is
    Solution
    $$\begin{array}{l}\text { Max stress }=1.2 \times 10^{8}\mathrm{~N} / \mathrm{m}^{2} \\\text { Force}=40\mathrm{~N} \\\Rightarrow\text {  } A=\dfrac{F}{\text { Stress }}=\dfrac{40}{1.2 \times 10^{8}}=\dfrac{400}{12\times10^{8}}\end{array}$$

    $$\begin{aligned}&\Rightarrow \quad \pi r^{2}=\dfrac{400}{12\times 10^{8}}\\&\Rightarrow \quad r=\sqrt{\dfrac{400}{12\times 10^{8} \times \pi}}\\&\Rightarrow\text { } \quad r=0.32 \mathrm{~mm}\\&\text { Ans :(A) }\end{aligned}$$
  • Question 8
    1 / -0
    A brass bar, having cross sectional area 10 $$cm^2$$ is subjected to axial forces as shown in figure. Total elongation of the bar is (Take Y = $$8 \times 10^2 \ t/m^2$$).

    Solution
    $$\text { Since the system is in equilibrium } \sum F_{\text {right }}=\sum F_{\text {left }} $$

    $$\begin{array}{l}\text { Note: at section } B \text { net force } 3 t \text { is right is present.} \\\text { Similarly for section } C \text { too. }\end{array}$$

    $$\begin{aligned}\Delta l_{1} &=\dfrac{P_{1} L_{1}}{A Y}=\dfrac{(5 t)(0.6)}{A \times 8 \times 10^{2} \times  t}=\dfrac{3}{800 A} \rightarrow \mathbb{(1)} \\\Delta l_{2}&=\dfrac{(2 t)(1)}{A \times 8 \times 100 t}=\dfrac{1}{4004} \rightarrow(2) \\\Delta l_{3} &=\dfrac{(1 t)(1.2)}{A\times 800 t}=\dfrac{1.2}{800 A}\quad\rightarrow(3)\end{aligned}$$

    $$\begin{array}{l}\text { Since each cases is elongation, we add } \rightarrow \\\qquad(1+2+3)=\dfrac{3+2+1.2}{800A}=\dfrac{6 \cdot 2}{800 \times 10 \times 10^{-4}}=0.0775 cm\end{array}$$
  • Question 9
    1 / -0
    In a Young's double slit experiment, the intensity at the cetral maximum is $$l_{0-}$$. The intensity ata distance $$\beta/4$$ from the central maximum is ($$\beta $$is frige width)
    Solution
    In a Young 's double slit experimental the intensity at the central maximum is $$l_0$$. The intensity at a distance $$=\beta/4$$
    From the central maximum is 
    ($$\beta$$ is fringe width)
    Solution
    The intensity of central maxima is $$l_0$$, let $$l_1$$ and $$I_2$$ be the intensity emitted by the two slits $$S_1$$ and $$S_2$$ repectively. The expression for resultant intensity is
    $$l=l_1l_2+2\sqrt{l_1l_2\cos\phi}$$
    For central maxima
    $$l=l_0$$ and $$\phi=0$$

  • Question 10
    1 / -0
    In Searle's apparatus we have two wires. During experiment we study the extension in one wire. The use of second wire is-
    Solution
    Frist wire : extension due to applied load.
    Second wire : to study the effect of temperature and atmospheric conditions.

    Therefore, Answer is (B)
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