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NEET-XII-Physics

14: Some Mechanical Properties of Matter

with Solutions - page 5

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  • Qstn #1
    A load of 10 kg is suspended by a metal wire 3 m long and having a cross-sectional area 4 mm2. Find
    Ans : Given:
    Mass of the load (m) = 10 kg
    Length of wire (L) = 3 m
    Area of cross-section of the wire (A) = 4 mm2 = 4.0 × 10-6 m2
    Young's modulus of the metal Y = 2.0 × 1011 N m-2
  • #1-a
    the stress
    Ans : Stress = F/A
    F = mg
    = `` 10\times 10`` = 100 N (g = 10 m/s2)
    `` \therefore \frac{F}{A}=\frac{100}{4\times {10}^{-6}}``
    `` =2.5\times {10}^{7}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
  • #1-b
    the strain and
    Ans : Strain = `` \frac{\Delta L}{L}``
    ​
    Or, `` \,\mathrm{\,Strain\,}=\frac{\,\mathrm{\,Stress\,}}{Y}``
    `` \,\mathrm{\,Strain\,}=\frac{2.5\times {10}^{7}}{2\times {10}^{11}}``
    `` =1.25\times {10}^{-4}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
  • #1-c
    the elongation. Young modulus of the metal is 2.0 × 1011 N m-2.
    Ans : Let the elongation in the wire be `` ∆L``.
    `` \,\mathrm{\,Strain\,}=\frac{\Delta L}{L}``
    `` \Rightarrow \,\mathrm{\,\Delta \,}L=\left(\,\mathrm{\,Strain\,}\right)\times L``
    `` =1.25\times {10}^{-4}\times 3``
    `` =3.75\times {10}^{-4}\,\mathrm{\,m\,}``
    Page No 300:
  • Qstn #2
    A vertical metal cylinder of radius 2 cm and length 2 m is fixed at the lower end and a load of 100 kg is put on it. Find
    Ans : Given:
    Radius of cylinder (r) = 2 cm = `` 2\times {10}^{-2}\,\mathrm{\,m\,}``
    Length of cylinder (L) = 2 m
    Mass of the load = 100 kg
    Young's modulus of the metal = `` 2\times {10}^{11}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
  • #2-a
    the stress
    Ans : Stress(ρ) is given by: `` \frac{F}{A}``
    Here, F is the force given by mg = `` 100\times 10=1000\,\mathrm{\,N\,}`` ( Taking g = 10 m/s2)
    A is the area of cross-section = πr2 = 4π`` \times {10}^{-4}{\,\mathrm{\,m\,}}^{2}``
    `` \Rightarrow S\,\mathrm{\,tress\,}\rho =\frac{mg}{A}``
    `` =\frac{\left(100\times 10\right)}{\left(4\,\mathrm{\,\pi \,}\times {10}^{-4}\right)}``
    `` =7.96\times {10}^{5}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
  • #2-b
    the strain and
    Ans : Strain is given by:
    `` \,\mathrm{\,Strain\,}=\frac{\rho }{Y}=\frac{\left(7.96\times {10}^{5}\right)}{\left(2\times {10}^{11}\right)}``
    `` =4\times {10}^{-6}``
  • #2-c
    the compression of the cylinder. Young modulus of the metal = 2 × 1011 N m-2.
    Ans : Compression of the cylinder:
    ΔL = strain × L
    = 4 × 10-6 × 2 = 8 × 10-6 m
    Page No 300:
  • Qstn #3
    The elastic limit of steel is 8 × 108 N m-2 and its Young modulus 2 × 1011 N m-2. Find the maximum elongation of a half-metre steel wire that can be given without exceeding the elastic limit.
    Ans : Given:
    `` \,\mathrm{\,Elastic\,}\,\mathrm{\,limit\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}\frac{F}{A}=8\times {10}^{5}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
    `` \,\mathrm{\,Young\,}\text{'}\,\mathrm{\,s\,}\,\mathrm{\,modulus\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}Y=2\times {10}^{11}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
    `` \,\mathrm{\,Length\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}\,\mathrm{\,wire\,}L=\frac{1}{2}\,\mathrm{\,m\,}=0.5\,\mathrm{\,m\,}``
    `` ``
    The elastic limit of steel indicates the maximum pressure that steel can bear.
    Let the maximum elongation of steel wire be `` ∆L``.
    `` Y=\frac{F}{A}\frac{L}{∆L}``
    `` \Rightarrow ∆L=\frac{FL}{AY}``
    `` \Rightarrow ∆L=\frac{8\times {10}^{5}\times \left(0.5\right)}{2\times {10}^{11}}``
    `` =2\times {10}^{-3}\,\mathrm{\,m\,}=2\,\mathrm{\,mm\,}``
    Hence, the required elongation of steel wire is 2 mm.
    Page No 300:
  • Qstn #4
    A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of
    Ans : Given:
    Young's modulus of steel = 2 × 1011 N m-2
    Young's modulus of copper = 1.3 × 10 11 N m-2
    Both wires are of equal length and equal cross-sectional area. Also, equal tension is applied on them.
    As per the question:
    `` {L}_{\,\mathrm{\,steel\,}}={L}_{\,\mathrm{\,Cu\,}}``
    `` {A}_{\,\mathrm{\,steel\,}}={A}_{\,\mathrm{\,Cu\,}}``
    `` {F}_{\,\mathrm{\,Cu\,}}={F}_{\,\mathrm{\,Steel\,}}``
    `` ``
    Here: Lsteel and LCu denote the lengths of steel and copper wires, respectively.
    Asteel and ACu denote the cross-sectional areas of steel and copper wires, respectively.
    Fsteel and FCu denote the tension of steel and cooper wires, respectively.
    `` \left(\,\mathrm{\,a\,}\right)\frac{\,\mathrm{\,Stress\,}\,\mathrm{\,of\,}\,\mathrm{\,Cu\,}}{\,\mathrm{\,Stress\,}\,\mathrm{\,of\,}\,\mathrm{\,Steel\,}}=\frac{{F}_{\,\mathrm{\,Cu\,}}}{{A}_{\,\mathrm{\,Cu\,}}}\frac{{A}_{\,\mathrm{\,Steel\,}}}{{F}_{\,\mathrm{\,Steel\,}}}=1``
    (b)
    `` \frac{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,Cu\,}}{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}}=\frac{{\displaystyle \frac{∆{L}_{\,\mathrm{\,Steel\,}}}{{L}_{\,\mathrm{\,Steel\,}}}}}{{\displaystyle \frac{∆{L}_{\,\mathrm{\,cu\,}}}{{L}_{\,\mathrm{\,cu\,}}}}}=\frac{{F}_{\,\mathrm{\,Steel\,}}{L}_{\,\mathrm{\,Steel\,}}{A}_{cu}{Y}_{cu}}{{A}_{\,\mathrm{\,Steel\,}}{Y}_{\,\mathrm{\,Steel\,}}{F}_{cu}{L}_{cu}}``
    `` \left(\,\mathrm{\,Using\,}\frac{∆L}{L}=\frac{F}{AY}\right)``
    `` \Rightarrow \frac{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,Cu\,}}{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}}=\frac{{\,\mathrm{\,Y\,}}_{\,\mathrm{\,cu\,}}}{{\,\mathrm{\,Y\,}}_{\,\mathrm{\,Steel\,}}}=\frac{1.3\times {10}^{11}}{2\times {10}^{11}}``
    `` \Rightarrow \frac{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,C\,}\,\mathrm{\,u\,}}{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}}=\frac{13}{20}``
    `` \Rightarrow \frac{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}}{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,Cu\,}}=\frac{20}{13}``
    Hence, the required ratio is 20 : 13.
    Page No 300:
  • Qstn #5
    In figure the upper wire is made of steel and the lower of copper. The wires have equal cross section. Find the ratio of the longitudinal strains developed in the two wires.

    Figure
    Ans : Given that both wires are of equal length and equal cross-sectional area,
    the block applies equal tension on both of them.
    ∴ `` {L}_{\,\mathrm{\,steel\,}}={L}_{\,\mathrm{\,Cu\,}}``
    `` {A}_{\,\mathrm{\,steel\,}}={A}_{\,\mathrm{\,Cu\,}}``
    `` {F}_{\,\mathrm{\,Cu\,}}={F}_{\,\mathrm{\,Steel\,}}``
    `` ``
    `` ``
    `` \frac{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}Cu}{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}}=\frac{{\displaystyle \frac{∆{L}_{\,\mathrm{\,Steel\,}}}{{L}_{\,\mathrm{\,Steel\,}}}}}{{\displaystyle \frac{∆{L}_{\,\mathrm{\,cu\,}}}{{L}_{\,\mathrm{\,cu\,}}}}}=\frac{{F}_{\,\mathrm{\,Steel\,}}{L}_{\,\mathrm{\,Steel\,}}{A}_{cu}{Y}_{cu}}{{A}_{\,\mathrm{\,Steel\,}}{Y}_{\,\mathrm{\,Steel\,}}{F}_{cu}{L}_{cu}}``
    `` \left(\,\mathrm{\,Using\,}\frac{∆L}{L}=\frac{F}{AY}\right)``
    `` \Rightarrow \frac{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}Cu}{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}}=\frac{{\,\mathrm{\,Y\,}}_{\,\mathrm{\,cu\,}}}{{\,\mathrm{\,Y\,}}_{\,\mathrm{\,Steel\,}}}=\frac{1.3\times {10}^{11}}{2\times {10}^{11}}``
    `` \Rightarrow \frac{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,steel\,}}{\,\mathrm{\,Strain\,}\,\mathrm{\,of\,}\,\mathrm{\,C\,}\,\mathrm{\,u\,}}=\frac{20}{13}=1.54``
    Hence, the required ratio of the longitudinal strains is 20 : 13.
    Page No 300:
  • Qstn #6
    The two wires shown in figure are made of the

    Figure
    same material which has a breaking stress of 8 × 108 N m-2. The area of cross section of the upper wire is 0.006 cm2 and that of the lower wire is 0.003 cm2. The mass m1 = 10 kg, m2 = 20 kg and the hanger is light.
  • #6-a
    Find the maximum load that can be put on the hanger without breaking a wire. Which wire will break first if the load is increased?
    Ans : Given:
    `` \,\mathrm{\,Breaking\,}\,\mathrm{\,stress\,}\,\mathrm{\,of\,}\,\mathrm{\,wire\,}=8\times {10}^{8}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
    `` \,\mathrm{\,Area\,}\,\mathrm{\,of\,}\,\mathrm{\,cross\,}-\,\mathrm{\,section\,}\,\mathrm{\,of\,}\,\mathrm{\,upper\,}\,\mathrm{\,wire\,}\left({A}_{\,\mathrm{\,u\,}}\right)=0.006{\,\mathrm{\,cm\,}}^{2}=6\times {10}^{-7}\,\mathrm{\,m\,}``
    `` \,\mathrm{\,Area\,}\,\mathrm{\,of\,}\,\mathrm{\,cross\,}-\,\mathrm{\,section\,}\,\mathrm{\,of\,}\,\mathrm{\,lower\,}\,\mathrm{\,wire\,}\left({A}_{\,\mathrm{\,l\,}}\right)=0.003{\,\mathrm{\,cm\,}}^{2}=3\times {10}^{-7}\,\mathrm{\,m\,}``
    `` {m}_{1}=10\,\mathrm{\,kg\,},{m}_{2}=20\,\mathrm{\,kg\,}``
    Tension in lower wire `` {T}_{l}={m}_{1}g+w``
    Here: g is the acceleration due to gravity
    w is the load
    ∴ Stress in lower wire`` =\frac{{\,\mathrm{\,T\,}}_{l}}{{\,\mathrm{\,A\,}}_{l}}=\frac{{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{l}}``
    `` \Rightarrow \frac{{m}_{1}g+w}{{A}_{l}}=8\times {10}^{8}``
    `` \Rightarrow w=\left[\left(8\times {10}^{8}\right)\times \left(3\times {10}^{-7}\right)\right]-100``
    `` ``
    `` \Rightarrow w=140\,\mathrm{\,N\,}\,\mathrm{\,or\,}14\,\mathrm{\,kg\,}``
    Now, tension in upper wire `` {T}_{2}={m}_{1}g+{m}_{2}g+w``
    ∴ Stress in upper wire`` =\frac{{\,\mathrm{\,T\,}}_{u}}{{\,\mathrm{\,A\,}}_{u}}=\frac{{m}_{\mathit{2}}g+{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{u}}``
    `` \Rightarrow \frac{{m}_{2}g+{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{u}}=8\times {10}^{8}``
    `` \Rightarrow w=180\,\mathrm{\,N\,}\,\mathrm{\,or\,}18\,\mathrm{\,kg\,}``
    For the same breaking stress, the maximum load that can be put is 140 N or 14 kg. The lower wire will break first if the load is increased.
  • #6-b
    Repeat the above part if m1 = 10 kg and m2 = 36 kg.
    Ans : `` \,\mathrm{\,If\,}{m}_{1}=10\,\mathrm{\,kg\,}\,\mathrm{\,and\,}{m}_{2}=36\,\mathrm{\,kg\,}``:
    Tension in lower wire `` {T}_{l}={m}_{1}g+w``
    Here: g is the acceleration due to gravity
    w is the load
    ∴ Stress in lower wire:
    `` \Rightarrow \frac{{\,\mathrm{\,T\,}}_{l}}{{\,\mathrm{\,A\,}}_{l}}=\frac{{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{l}}=8\times {10}^{5}``
    `` \Rightarrow w=140\,\mathrm{\,N\,}``
    Now, tension in upper wire `` {T}_{2}={m}_{1}g+{m}_{2}g+w``
    ∴ Stress in upper wire:
    `` \Rightarrow \frac{{\,\mathrm{\,T\,}}_{u}}{{\,\mathrm{\,A\,}}_{u}}=\frac{{m}_{2}g+{m}_{1}g+w}{{\,\mathrm{\,A\,}}_{u}}=8\times {10}^{5}``
    `` \Rightarrow w=20\,\mathrm{\,N\,}``
    For the same breaking stress, the maximum load that can be put is 20 N or 2 kg. The upper wire will break first if the load is increased.
    Page No 300:
  • Qstn #7
    Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope. Find the Young modulus of the material of the rope if it extends in length by 1 cm. Original length of the rope = 2 m and the area of cross section = 2 cm2.
    Ans : Given:
    Force (F) applied by two persons on the rope = 100 N
    `` \,\mathrm{\,Original\,}\,\mathrm{\,length\,}\,\mathrm{\,of\,}\,\mathrm{\,rope\,}L=2\,\mathrm{\,m\,}``
    `` \,\mathrm{\,Extension\,}\,\mathrm{\,in\,}\,\mathrm{\,the\,}\,\mathrm{\,rope\,}∆L=0.01\,\mathrm{\,m\,}``
    `` \,\mathrm{\,Area\,}\,\mathrm{\,of\,}\,\mathrm{\,cross\,}-section\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,rope\,}A=2\times {10}^{-4}``
    `` \,\mathrm{\,We\,}\,\mathrm{\,know\,}\,\mathrm{\,that\,}:``
    `` \,\mathrm{\,Young\,}\text{'}\,\mathrm{\,s\,}\,\mathrm{\,modulus\,}Y=\frac{F}{A}\times \frac{L}{∆L}``
    `` =\frac{100}{2\times {10}^{-4}}\times \frac{2}{0.01}``
    `` \Rightarrow Y=1\times {10}^{8}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
    Hence, the required Young's modulus for the rope is `` 1\times {10}^{8}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``.
    Page No 300:
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