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Mechanical Properties of Solids Test - 16

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Mechanical Properties of Solids Test - 16
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  • Question 1
    1 / -0
    Ratio of transverse to axial strain is 
    Solution
    Hookes law states that stress is proportional to strain up to elastic limit. If p is the stress induced in material and e the corresponding strain, then according to Hooke's law, 
    PE\dfrac{P}{E} = E, a constant.
  • Question 2
    1 / -0
    The modulus of elasticity is dimensionally equivalent to
    Solution
    Hookes law establishes the relationship between stress and strain
    Stress: The force per unit area
    Strain: The elongation or contraction per unit length (dimensionless)
    The ratio of stress to strain is known as the elastic modulus of the material
    Elastic Modulus = stressstrain= \displaystyle \frac{stress}{strain} Hence, the modulus of elasticity is dimensionally equivalent to the stress
  • Question 3
    1 / -0
    A wire elongates by 1mm1 mm when a load WW is hung from it. If the wire goes over a pulley and the two weights WW, each are hung at the two ends, then the elongation of the wire will be:
    Solution
    One of the important concepts in theory regarding stress and strain is that an object which undergoes longitudinal strain under lateral stress is actually under the action of the force from both ends.
    The given condition is similar to hanging the load on one end, since the ceiling applies the required reaction force to make the situations similar.
  • Question 4
    1 / -0
    An iron bar of length LL, cross-section AA and Young's modulus YY is pulled by a force FF from ends so as to produce an elongation ll. Which of the following statements is correct ?
    Solution
    We know l=FLAY\rightarrow l =\dfrac{FL}{AY}
    l 1Al  \propto \dfrac{1}{A}
  • Question 5
    1 / -0
    A wire of length L and radius r fixed at one end and a force F applied to the other end produces and extension ll . The extension produced in another wire of the same material of length 2Land radius 2r by a force 2 F is:
    Solution
    Y=FLAlY=\dfrac{FL}{Al}
    Y - Young's Modulus
    F - Applied Force
    A - Cross Section Area
    l - Elongation

    For 1st Wire

    Y=FL(π)(r2)lY = \dfrac{FL}{(\pi)(r^2)l} ----   (M)

    For 2nd Wire
    z - elongation in 2nd wire
    Y=2F2L(π)(22)(r2)zY = \dfrac{2F2L}{(\pi)(2^2)(r^2) z}  ------  (N)

    Dividing M by M
     1=zl\Rightarrow  1 = \dfrac{z}{l}
    Therefore z=lz = l
  • Question 6
    1 / -0
    Elongation of a wire under its own weight is independent of :
    Solution
    Mass == M

    Force =ρ= \rho  ALg

    Let weight of the rod is W.

    Consider a small length dx of the rod at distance x from the fixed end. The tension T in element equals the weight of rod below it.

    T=(Lx)WLT=(L-x)\dfrac{W}{L}

    Elongation =L×Stressy=(Lx)WdxLAy=\dfrac{L\times Stress}{y}=\dfrac{(L-x)Wdx}{LAy}

    Total elongation =OL(Lx)WdxLAy= \int_{O}^{L}\dfrac{(L-x)Wdx}{LAy}=WLAy(Lxx22)0L=WL2Ay=\dfrac{W}{LAy}(Lx-\dfrac{x^{2}}{2})_{0}^{L}=\dfrac{WL}{2Ay}

    Now,  W=ρALgW=\rho ALg

    Putting it in there:ΔL=ρAL2g2Ay\Delta L=\dfrac{\rho AL^2g}{2Ay}=ρL2g2y=\dfrac{\rho L^2g}{2y}
  • Question 7
    1 / -0
    Assertion (A): Stress is restoring force per unit area.
    Reason (R) : Interatomic forces in solids are responsible for the property of elasticity
    Solution
    Both assertion and reason are true statements but first is just a definition of how we consider stress and second being the reason behind elastcity
  • Question 8
    1 / -0
    If stress is numerically equal to young's modulus,the elongation will be
    Solution
    Hint\textbf{Hint}: Young's modulus of elasticity
    Step 1\textbf{Step 1}:
    Young's modulus of elasticity is the ratio of stress to strain.
    Youngsmodulus=stressstrainYoung's \quad modulus= \dfrac{stress}{strain}
    Stress is defined as force per unit area.
    Stress=FAStress=\dfrac{F}{A} where F is force and A is  area.
    Strain=ΔLLStrain=\dfrac{\Delta L}{L} where 'L' is the length of the wire and ΔL\Delta L is the change in the length of the wire.
    Step 2\textbf{Step 2}:
    When the change in length to the original length is minimum, then modulus of elasticity is dimensionally equivalent to the stress since it is given stress is numerically equivalent to young's modulus ie.., Stress=YStress=Y
    Thus option C is correct.







  • Question 9
    1 / -0
    A metal string is fixed between rigid supports. It is initially at negligible tension. Its Young's modulus is Y, density is ρ\rho  and coefficient of linear expansion is α\alpha. It is now cooled through a temperature t, transverse waves will move along it with a speed of :
    Solution
    By the definition of Young's modulus, we have 
    y=(FA)/(Ll)  F=YAl(αt)ly=\left ( \dfrac{F}{A} \right )/\left ( \dfrac{\bigtriangleup L}{l} \right )\ \ \Rightarrow F=\dfrac{YAl(\alpha t)}{l}
    l=l α t\bigtriangleup l=l\ \alpha\ t
    \therefore F = YA α\alpha t
    The velocity V=Tμ=YA αtρ(A)V=\sqrt{\dfrac{T}{\mu}}=\sqrt{\dfrac{YA\ \alpha t}{\rho(A)}}
    μ=ρ(A)\mu = \rho(A)
    V=Y αtρ\Rightarrow V = \sqrt{\dfrac{Y\ \alpha t}{\rho}}
  • Question 10
    1 / -0
    Three wires A, B, C made of different materials elongated by 1.5, 2.5, 3.5 mm, under a load of 5kg. If the diameters of the wires are the same,the most elastic material is that of
    Solution
    Elasticity = (force x length of wire)/(area x elongation of wire)
    Here, force is same for all the three wires as load is same for all.
    So, the measurement of elasticity is inversly proportional to elongation of wire and wire A has least elongation soits elasticity is maximum.
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