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Mechanical Properties of Solids Test - 25

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Mechanical Properties of Solids Test - 25
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  • Question 1
    1 / -0
    A brass rod has a length of 0.2m0.2m, area of cross section 1.0cm2^{2} and young's modulus 1011^{11} Nm2^{-2}. If it is compressed by 5kg5kg weight along its length, then the change in its energy will be :
    Solution
    Young's Modulus, Y=F/AΔl/lY=\dfrac{F/A}{\Delta l/l}

    Δl=F/AY/l\Delta l=\dfrac{F/A}{Y/l}

    =5×9.8×0.21×104×1011=\dfrac{5\times 9.8\times 0.2}{1\times 10^{-4}\times 10^{11}}

    energy=12×F×Δlenergy=\dfrac{1}{2}\times F\times \Delta l

    =12×5×9.8×5×9.8×0.2104×1011=\dfrac{1}{2}\times 5\times 9.8\times \dfrac{5\times 9.8\times 0.2}{10^{-4}\times 10^{11}}

    =2.4×105J.=2.4\times 10^{-5}J.

    energy increases because the rod is compressed.
  • Question 2
    1 / -0
    A solid sphere of radius RR made of material of bulk modulus KK, is surrounded by a liquid in a cylindrical container. A massless piston of area AA floats on the surface of liquid. When a mass mm is placed on the piston to compress the liquid, the fractional change in the radius of the sphere δR/R\delta R/R is :
    Solution
    ΔVV=PK\dfrac{\Delta V}{V}=\dfrac{P}{K}

    Now, ΔVV=3δRR\dfrac{\Delta V}{V}=\dfrac{3\delta R}{R}

    and P=mgAP=\dfrac{mg}{A}

    So, 3δRR=mgAK\dfrac{3\delta R}{R}=\dfrac{mg}{AK}

    δRR=mg3AK\therefore \dfrac{\delta R}{R}=\dfrac{mg}{3AK}
  • Question 3
    1 / -0
    A spring of spring constant 5 x 103^{3}Nm1^{-1} is stretched initially by 5cm5 cm from the unstretched position. Then the work required to stretch it further by another 5cm5 cm is       
    Solution
    Work done =12k(x22x12)=\dfrac{1}{2}k(x{_{2}}^{2}-x{_{1}}^{2})

    =12×5×103((10)2(5)2)=\dfrac{1}{2}\times 5\times 10^{3}\left ( (10)^{2}-(5)^{2} \right )

    =12×5×103(75)10000=\dfrac{1}{2}\times 5\times 10^{3}\dfrac{(75)}{10000}

    =18.75=18.75 N-m
  • Question 4
    1 / -0
    A 20kg20 kg load is suspended from the lower end of a wire 10cm10 cm long and 1mm1 mm2^2 in cross sectional area. The upper half of the wire is made of iron and the lower half with aluminium. The total elongation in the wire is
    (Yiron=Y_{iron} = 20 x 1010^{10} N/m2^{2}, YAlY_{Al} == 7 x 1010^{10} N/m2^{2})
    Solution
    Y=F/AΔL/LΔL=FLAYY=\dfrac{F/A}{\Delta L/L} \Rightarrow \Delta L=\dfrac{FL}{AY}

    Therefore, ΔLi=FLiYiA\Delta L_i=\dfrac{FL_i}{Y_i A} and ΔLA=FLAYAA\Delta L_A =\dfrac{FL_A}{Y_A A}

    Total elongation =ΔLi+ΔLA=\Delta L_i+\Delta L_A
    =FLA(1Yi+1YA)=\dfrac{FL}{A}\left ( \dfrac{1}{Y_i}+\dfrac{1}{Y_A} \right )

    =200×5106×100(120×1010+17×1010)=\dfrac{200\times 5}{10^{-6}\times 100}\left ( \dfrac{1}{20\times 10^{10}}+\dfrac{1}{7\times 10^{10}} \right )

    =1.92×104m=1.92\times 10^{-4}m
  • Question 5
    1 / -0
    The length of a metal wire is l1l_{1} when the tension in it is F1F_{1} and l2l_{2} when the tension in it is F2F_{2}. The natural length of the wire is
    Solution
    We have
    l1l=F1lAYl_1-l=\dfrac{F_1l}{AY}
    and
    l2l=F2lAYl_2-l=\dfrac{F_2l}{AY}
    Thus we get
    l1lF1=l2lF2\dfrac{l_1-l}{F_1}=\dfrac{l_2-l}{F_2}
    or
    F2l1F2l=F1l2F1lF_2l_1-F_2l=F_1l_2-F_1l
    or
    l=F2l1F1l2F2F1l=\dfrac{F_2l_1-F_1l_2}{F_2-F_1}
  • Question 6
    1 / -0
    A uniform metal rod of 2mm2mm2^{2} cross section is heated from 0o^{o}C to 20o^{o}C. The coefficient of linear expansion of the rod is 12 x 106/0^{-6/0}C. Its Young's modulus of elasticity is 1011^{11} N/m2^{2}. The energy stored per unit volume of the rod is :
    Solution
    energy=12×stress×strain×volumeenergy=\dfrac{1}{2}\times stress\times strain\times volume

    energyvolume=12×stress×strain\dfrac{energy}{volume}=\dfrac{1}{2}\times stress\times strain

    =12×Y×strain×strain=\dfrac{1}{2}\times Y\times strain\times strain

    =12×Y×(strain)2=\dfrac{1}{2}\times Y\times (strain)^2

    =12×Y×(αΔT)2=\dfrac{1}{2}\times Y\times (\alpha \Delta T)^2

    =12×1011×(12×106×20)2=\dfrac{1}{2}\times 10^{11}\times (12\times 10^{-6}\times 20)^2

    =2880J/m3=2880J/m^3

  • Question 7
    1 / -0
    A piece of copper wire has twice the radius of steel wire. One end of the copper wire is joined to one end of steel wire so that both of them can be subjected to the same longitudinal force. YY for steel is twice that of copper. When the length of copper wire is increased by 11%, the steel wire will be stretched by
    Solution
    Y=F/AΔl/lY=\dfrac{F/A}{\Delta l/l}

    Δll=FYA\dfrac{\Delta l}{l}=\dfrac{F}{YA}

    Δlclc=FYcAc=FYcπ(rc)2\dfrac{\Delta l_c}{l_c}=\dfrac{F}{Y_cA_c}=\dfrac{F}{Y_c \pi (r_c)^2}------(1)

    Δlsls=FYsAs=FYsπ(rs)2\dfrac{\Delta l_s}{l_s}=\dfrac{F}{Y_sA_s}=\dfrac{F}{Y_s \pi (r_s)^2}------(2)

    Dividing (1) by (2)

    Δlc/lcΔls/ls=Ys(rs)2Yc(rc)2=YsYc×(rsrc)2=(2)×(12)2\dfrac{\Delta l_c/l_c}{\Delta l_s/l_s}=\dfrac{Y_s(r_s)^2}{Y_c(r_c)^2}=\dfrac{Y_s}{Y_c}\times \left ( \dfrac{r_s}{r_c} \right )^2=(2)\times (\dfrac{1}{2})^2

    Δlc/lcΔls/ls=12\Rightarrow \dfrac{\Delta l_c/l_c}{\Delta l_s/l_s}=\dfrac{1}{2}

    Δls/ls=2Δlc/lc\therefore \Delta l_s/l_s=2\Delta l_c/l_c

    =2=2%
  • Question 8
    1 / -0
    One end of uniform wire of length LL and of weight WW is attached rigidly to a point in the roof and a weight W1W_{1} is suspended from the lower end. If AA is the area of cross-section of the wire, the stress in the wire at a height 3L4\dfrac{3L}{4} from its lower end is:
    Solution
    stress=TensionAreastress=\dfrac{Tension}{Area}

    Tension at height 3L4\dfrac{3L}{4} from lower end

    is 34w+w1\dfrac{3}{4}w+w_1

    So, stress=34w+w1S=\dfrac{\dfrac{3}{4}w+w_1}{S}

  • Question 9
    1 / -0
    The length of an elastic spring is 'aa' when tension is 4N4 N and 'bb' when the tension is 5N5N. The length of the spring when tensionis 9N is
    Solution
    Y=F/AΔl/lY=\dfrac{F/A}{\Delta l/l}

    Δl=FlYA\Delta l=\dfrac{Fl}{YA}

    (al)=4lYA(a-l)=\dfrac{4l}{YA}-------(1)

    (bl)=5lYA(b-l)=\dfrac{5l}{YA}-------(2)

    Dividing (1) by (2)

    albl=45\dfrac{a-l}{b-l}=\dfrac{4}{5}

    5a5l=4bbl5a-5l=4b-bl

    l=5a4bl=5a-4b------(3)

    Δl=9lYA\Delta l=\dfrac{9l}{YA}             (when force =9N=9N)

    xl=9(al)4x-l=9\dfrac{(a-l)}{4}                ( BY (1))

    xl=9[a(5a4b)]/4x-l=9\left [ a-(5a-4b) \right ]/4  ( BY (3))

    x=9(ba)+lx=9(b-a)+l

    =5b4a=5b-4a

  • Question 10
    1 / -0
    A steel wire of length 5m is pulled to have an extension of 1mm. Its Y is 1.9 x 104^{4} N/m2^{2}. The energy per unit volume stored in it is                 
    Solution
    energy=12×stress×strain×volumeenergy=\dfrac{1}{2}\times stress\times strain\times volume

    energyvolume=12×stress×strain\dfrac{energy}{volume}=\dfrac{1}{2}\times stress\times strain

                   =12×γ×strain×strain=\dfrac{1}{2}\times \gamma \times strain\times strain

                   =12×1.9×104×15000×15000=\dfrac{1}{2}\times 1.9\times 10^{4}\times \dfrac{1}{5000}\times \dfrac{1}{5000}

                   =3.8×104J/m3=3.8\times 10^{-4} J/m^{3}
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