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Mechanical Properties of Solids Test - 15

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Mechanical Properties of Solids Test - 15
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  • Question 1
    1 / -0
    The ratio of lateral strain to the linear strain within elastic limit is known as:
    Solution
    The ratio of lateral strain to the linear strain within elastic limit is known as Poisson's ratio

    The correct option is (d)
  • Question 2
    1 / -0
    A spring is made of steel and not of copper because
    Solution
    Steel is more elastic than copper. Due to this reason that springs are made of steel not copper.
  • Question 3
    1 / -0
    Substances that break just after elastic limit is reached are known as 
    Solution
    Substances that break just after elastic limit is reached are known as brittle substances
  • Question 4
    1 / -0
    The force required to double the length of the steel wire of area of cross section $$5\times 10^{-5}m^{2}\quad (Y=20\times 10^{10}Pa)$$ in $$N$$ is:
    Solution
    $$Y=\cfrac{stress}{strain}$$
    $$Y=\cfrac{\cfrac{F}{A}}{\cfrac{\triangle l}{l}}$$
    $$\triangle l=l$$
    $$F=YA$$
    $$=20\times 10^{10}\times 5\times 10^{-5}$$
    $$=100\times 10^{10}\times 10^{-5}$$
    $$=10^7$$
  • Question 5
    1 / -0
    A wire is stretched to double its length. The strain is :
    Solution
    Longitudinal strain $$= \dfrac{change\,\, in\,\, length}{original\,\, length}$$  
    If $$l$$ be the original length and if stretched to double of its length , change in length $$= l$$  
    So , strain $$= 1$$
  • Question 6
    1 / -0
    Young's modulus of Perfectly elastic body is
    Solution
    Young's modulus for a perfectly elastic body is infinity. For an elastic body, there is no deformation on the application of load. Therefore, the strain is equal to zero. The denominator becomes zero which leads to the infinite value of Young's modulus.
  • Question 7
    1 / -0
    A wire whose cross-sectional area is $$4\ mm^{2}$$ is stretched by $$0.1\ mm$$ by a certain load. If a similar wire of double the area of cross-section is under the same load, then the elongation would be
    Solution
    $$0.05mm$$

    Formula,

    $$y=\dfrac{Fl}{Ae}$$

    $$\Rightarrow e\propto \dfrac{1}{A}$$

    $$\dfrac{e_2}{0.1}=\dfrac{4}{8}=\dfrac{1}{2}$$

    $$\therefore e_2=0.05mm$$
  • Question 8
    1 / -0
    Bulk modulus was first defined by

    Solution

  • Question 9
    1 / -0
    Minimum and maximum values of Poisson’s ratio for a metal lies between
    Solution

  • Question 10
    1 / -0
    The only elastic modulus that applies to fluids is

    Solution

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