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Mechanical Properties of Solids Test - 61

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Mechanical Properties of Solids Test - 61
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  • Question 1
    1 / -0
    A metal rod of Young's modulus 2×1010Nm22\times { 10 }^{ 10 }N{ m }^{ -2 } undergoes an elastic strain of 0.02%. The energy per unit volume stored in the rod, in JoulesJoules is ?
    Solution
    Energy per unit voume = Energy density\text{Energy per unit voume = Energy density}
    =12× stress × strain =12×FA×Δll=12×(γl)(Δll)=12Y(Δll)2=12×2×1010×(0.02100)2\begin{array}{l}=\dfrac{1}{2} \times \text { stress } \times \text { strain } \\=\dfrac{1}{2}\times\dfrac{F}{A} \times \dfrac{\Delta l}{l}\\=\dfrac{1}{2} \times\left(\dfrac{\gamma}{l}\right)\left(\dfrac{\Delta l}{l}\right) \\=\dfrac{1}{2} Y\left(\dfrac{\Delta l}{l}\right)^{2} \\=\dfrac{1}{2} \times 2 \times 10^{10} \times\left(\dfrac{0.02}{100}\right)^{2}\end{array}
    =400Joule=400 \mathrm{Joule}
    Ans: (A)\text{Ans: (A)}
  • Question 2
    1 / -0
    A wire elongates by ll mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (In mm) -
    Solution

     Hence at equilibriun T=WWA=Yl1Ll1=WLAY\begin{array}{l}\text { Hence at equilibriun }\\\qquad \begin{array}{l}T=W \\\frac{W}{A}=Y\frac{l_{1}}{L} \\\Rightarrow l_{1}=\frac{W L}{A Y}\end{array}\end{array}

     At equilibnumu T=W for each  side of wire WA=Yl2Ll2=WLAY\begin{array}{l}\text { At equilibnumu } T=W \text{ for each } \\\text { side of wire }\rightarrow\\\frac{W}{A}=Y \frac{l_{2}}{L}\\\Rightarrow l_{2}=\frac{W L}{A Y}\end{array}



     hence elongation will be same  ans (A)\text { hence elongation will be same }\Rightarrow\text { ans } \rightarrow(A)

  • Question 3
    1 / -0
    A metal wire of length 2.5m2.5 m and area of cross section 1.5×106m21.5\times { 10 }^{ -6 }{ m }^{ -2 }, is stretched through 2mm2 mm. Calculate the work done during stretching. Y=1.2×1011Nm2Y=1.2\times { 10 }^{ 11 }N{ m }^{ -2 }
    Solution

  • Question 4
    1 / -0
    A brass wire of diameter 1mm1 mm and of length 2m2 m is stretched by applying a force of 20N20 N. If the increase in length is 0.51mm0.51 mm. Then the Young's modulus of the wire is 
    Solution
    Given,
    F=20N,l=2m,d=1mm,Δl=0.51mmF=20N,l=2m,d=1mm,\Delta l=0.51mm
    We know,

    Y=FlAΔlY=\dfrac{Fl}{A\Delta l}

    Y=20×2π52×108×0.51×103Y=\dfrac{20\times 2}{\pi 5^2 \times 10^{-8}\times 0.51\times 10^{-3}}

    Y=4025π×0.51×1011Y=\dfrac{40}{25\pi\times 0.51}\times 10^11

    Y=0.9984×1011=9.984×1010Y=0.9984\times 10^11=9.984\times 10^{10}

    Option D\textbf D is the correct answer
  • Question 5
    1 / -0
     A force of 15N15 N increases the length ofa by 1mm1 \mathrm { mm }. The additional force require increase the length by 2.5mm2.5 \mathrm { mm } in N is 
    Solution

    We know,

    Y=FLAΔLY=\dfrac{FL}{A\Delta L}

    For a wire, A,L,Y are constant.

    FΔL=\dfrac{F}{\Delta L}=constant

    Let extra Force needed be F F,

    151=F+152.5\dfrac{15}{1}=\dfrac{F+15}{2.5}

    F+15=2.5×15F+15=2.5\times 15

    F=1.5×15=22.5F=1.5\times 15=22.5

    Option B\textbf B is the correct answer
  • Question 6
    1 / -0
    Ball A strikes with velocity u elastically with identical ball B at rest, inclined at an angle of with line joining the centers of two balls. What will be the speed of ball B after the collision: 
    Solution
    It is not possible for ball AA to collide with ball BB at an angle to the line joining their centers because it will always consider with the line joining their centers  
    \rightarrow It will always collide along the line joining their centers. 
    As the collision is elastic the velocity of ball BB will be uu.

  • Question 7
    1 / -0
    A material has poisson's ratio 0.30.3. If a uniform rod of it suffers a longitudinal strain of 25×10325\times 10^{-3}, then the percentage increase in its volume is
    Solution
    μ=12Y6B310=12Y6BY6B=12310=15Y6B=15YB=65\begin{aligned}\mu &=\frac{1}{2}-\frac{Y}{6 B}\\\Rightarrow& \frac{3}{10}=\frac{1}{2}-\frac{Y}{6 B}\\\Rightarrow & \frac{Y}{6 B}=\frac{1}{2}-\frac{3}{10}=\frac{1}{5} \\\Rightarrow & \frac{Y}{6 B}=\frac{1}{5}\\\Rightarrow & \frac{Y}{B}=\frac{6}{5}\end{aligned}

    (ΔV/V)(Δl/l)=65(Δv/v)=25×103×65=30×103ΔVV%=30×103×100=3%Ans:(C)\begin{aligned}\Rightarrow \frac{(\Delta V / V)}{(\Delta l / l)} &=\frac{6}{5} \\\Rightarrow \quad(\Delta v / v)&=\frac{25 \times 10^{-3} \times 6}{5}=30\times10^{-3}\\\Rightarrow \quad \frac{\Delta V}{V}\%&=30 \times 10^{-3} \times 100=3 \% \\A n s: &(C)\end{aligned}
  • Question 8
    1 / -0
    When a weight of 10kg10 \,kg is suspended from a copper wire of length 3m3 \,m and diameter 0.4mm0.4 \,mm. Its length increases by 2.4cm2.4 \,cm. If the diameter of the wire is doubled, then the extension is its length will be :
    Solution
    Mass =10kg= 10 \,kg          D=0.4mmD = 0.4 \,mm
    f=mxg=100Nf = mxg = 100 N

    A=πr2=π×(0.4×103)24=πD24A = \pi r^2 = \dfrac{\pi \times (0.4 \times 10^{-3})^2}{4} = \dfrac{\pi D^2}{4}
    L=3L = 3

    T=fLAΔLT = \dfrac{fL}{A\Delta L}

    Y=100×3π×(0.4×103)24×ΔLY = \dfrac{100 \times 3}{\dfrac{\pi \times (0.4 \times 10^{-3})^2}{4} \times \Delta L}

    ΔL=300×4π(0.4×103)2×4\Delta L = \dfrac{300 \times 4}{\pi (0.4 \times 10^{-3})^2 \times 4}

    Now, YY is same for the material.
    Y=F×LA×ΔLY = \dfrac{F \times L'}{A' \times \Delta L'}

    ΔL=f×L44×A×Y\Delta L' = \dfrac{f \times \dfrac{L}{4}}{4 \times A \times Y}

    ΔL=100×344×1.25×107×1×1011\Delta L' = \dfrac{\dfrac{100 \times 3}{4}}{4 \times 1.25 \times 10^{-7} \times 1 \times 10^{11}}

    ΔL=754×1.25×104\Delta L' = \dfrac{75}{4 \times 1.25 \times 10^4}

    ΔL=1.5×103m\Delta L' = 1.5 \times 10^{-3}m
    ΔL=1.5cm\Delta L' = 1.5 \,cm
  • Question 9
    1 / -0
    The Young's modulus a rubber string 8 cm long and density 1.5 kg/ m3m^{3} is 5×108N/m25\times 10^{8}N/m^{2}, is suspended on the ceiling in a room. The increases in length due to its own weigth will be 
    Solution

    Given that,

    Length of rubber band,  l=8cml=8\,cm

    Density, d=1.5Kg/m3d=1.5\,Kg/{{m}^{3}}

    Young’s modulus, Y=5×108N/m2Y=5\times {{10}^{8}}\,N/{{m}^{2}}

    The relation for increase in length and young’s modulus is as follows:

    l=L2dg2Y l=\dfrac{{{L}^{2}}dg}{2Y}

    l=(8×102)21.5×102×5×108 l=\dfrac{{{(8\times {{10}^{-2}})}^{2}}1.5\times 10}{2\times 5\times {{10}^{8}}}

    l=64×104×1.5×10109 l=\dfrac{64\times {{10}^{-4}}\times 1.5\times 10}{{{10}^{9}}}

    l=96×1012 l=96\times {{10}^{-12}}

    l=9.6×1011m l=9.6\times {{10}^{-11}}\,m

  • Question 10
    1 / -0
    Write Copper, Steel, Glass and Rubber in order of increasing coefficient of elasticity
    Solution
    D.
    rubber, glass, copper, steel
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