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Mechanical Properties of Solids Test - 11

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Mechanical Properties of Solids Test - 11
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  • Question 1
    1 / -0
    Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3 A. If the length of wire 1 increased by $$\displaystyle \Delta x$$ on applying force F, how much force is needed to stretch wire 2 by the same amount?
    Solution
    $$\displaystyle Y=\frac { FL }{ A\Delta L } $$
    or $$\displaystyle F=\frac { YA\Delta L }{ L } =\frac { Y{ A }^{ 2 }\Delta L }{ AL } $$
    $$\displaystyle $$
    $$\displaystyle =\frac { Y{ A }^{ 2 }\Delta L }{ V } =\frac { Y{ A }^{ 2 }\Delta x }{ V } $$

    $$ F' \displaystyle =\frac { Y{ A' }^{ 2 }\Delta L }{ V } =\frac { Y{ A' }^{ 2 }\Delta x }{ V } $$
    where, $$AL= V = $$ Volume of wire, Young's modulus is the same as both the wires are made of same material. It is given that, both the wires have same volume and same extension in length. Let's assume $$F'$$ force is needed to stretch wire 2 and its area of cross-section is A'.
    $$\displaystyle \therefore \quad \frac { F' }{ F } =\frac { { A' }^{ 2 } }{ { A }^{ 2 } } =\frac { { \left( 3A \right)  }^{ 2 } }{ { A }^{ 2 } } =9$$
    $$\displaystyle F'=9F$$
  • Question 2
    1 / -0
    How much force is required to produce an increase of $$0.2\%$$ in the length of a brass wire of diameter $$0.6\ mm$$? (Young's modulus for brass $$= 0.9 \times 10^{11} N / m^2$$) 
    Solution
    From Young's modulus relation,
    $$Y=\dfrac{F/A}{\Delta L/L}$$
    $$\implies F=YA(\dfrac{\Delta L}{L})$$
    $$=0.9\times 10^{11}\times \pi (0.0003)^2\times 0.002=50.8\ N\approx 51\ N$$
  • Question 3
    1 / -0
    The approximate depth of an ocean is $$2700\ m$$. The compressibility of water is $$45.4\times 10^{-11}Pa^{-1}$$ and density of water is $$10^{3} kg/m^{3}$$. What fractional compression of water will be obtained at the bottom of the ocean?
    Solution
    Compressibility of water,
    $$K = 45.4\times 10^{-11}Pa^{-1}$$
    density of water $$P = 10^{3} kg/ m^{3}$$
    depth of ocean, $$h = 2700\ m$$
    We have to find $$\dfrac {\triangle V}{V} = ?$$
    As we know, compressibility,
    $$K = \dfrac {1}{B} = \dfrac {(\triangle V/ V)}{P} (P = Pgh)$$
    So, $$(\triangle V/ V) = KPgh$$
    $$= 45.4 \times 10^{-11}\times 10^{3} \times 10\times 2700$$
    $$= 1.2258 \times 10^{-2}$$.
  • Question 4
    1 / -0
    When a body undergoes a linear tensile strain if experience a lateral contraction also. The ratio of lateral contraction to longitudinal strain is known as
    Solution
    Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force
  • Question 5
    1 / -0
    Which of the following is not dimension less
    Solution
    Strains are dimensionless, while stresses are not.
  • Question 6
    1 / -0
    A wire suspended vertically from one of its ends is stretched by attaching a weight of $$200N$$ to the lower end. The weight stretches the wire by $$1mm$$. Then the elastic energy stored in the wire is 
    Solution
    Elastic energy $$=\dfrac{1}{2}\times F\times x$$

    $$F=200N, x=1mm=10^{-3}m$$

    $$\therefore  E=\dfrac{1}{2}\times 200\times 1\times 10^{-3}=0.1J$$
  • Question 7
    1 / -0
    Energy per unit volume in a stretched wire is equal to 
    Solution
    Factual question. $$u=\dfrac{1}{2}stress\times strain$$
  • Question 8
    1 / -0
    Longitudinal strain can be produced in :
    Solution
    Longitudinal strain can only be generated in solids, not in liquids and gases because of inability to transfer the deformations
  • Question 9
    1 / -0
    Out of the following whose elasticity is independent of temperature
    Solution
    For most of the materials, elasticity decreases with temperature. One exception is invar steel.
  • Question 10
    1 / -0
    A wire ($$Y=2\times {10}^{11}N/m$$) has length $$1m$$ and area $$1m{m}^{2}$$. The work required to increased its length by $$2mm$$ is
    Solution
    Young's modulus, $$Y=\cfrac{F}{A}\times {l}{x}$$
    $$\Rightarrow$$ $$F=\cfrac{YAx}{l}$$
    Work done$$=\cfrac{1}{2}F\times x=\cfrac{1}{2}\cfrac{YA{x}^{2}}{l}$$
    $$=\cfrac{2\times {10}^{11}\times {10}^{-6}\times {(2\times {10}^{-3})}^{2}}{2\times 1}$$
    $$=0.4J$$
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