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Elasticity Test - 6

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Elasticity Test - 6
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  • Question 1
    1 / -0

    If the Bulk Modulus of lead is 8.0×109 N/m2 and the initial density of the lead is 11.4 g/cc, then under the pressure of 2.0×108 N/m2, the density of the lead is

    Solution

    Using the definition of Bulk modulus we have

     

  • Question 2
    1 / -0

    When the temperature of a gas is constant at 20°C and pressure is changed from P1=1.01×105 Pa to P2=1.165×105 Pa, then the volume changes by 10%. The Bulk modulus of the gas is

    Solution

    Bulk modulus = B

    change in pressure = ΔP

    change in volume = ΔV

     

  • Question 3
    1 / -0

    When a rubber ball is taken to the bottom of a sea of depth 1400 m, its volume decreases by 2%. The Bulk Modulus of rubber ball is [density of water is 1 g cc and g = 10 m/s2]

    Solution

    Bulk modulus is given by

     

  • Question 4
    1 / -0

    A spherical ball contracts in volume by 0.02%, when subjected to a normal uniform pressure of 50 atmosphere. The Bulk Modulus of its material is

    Solution

     

  • Question 5
    1 / -0

    For an elastic material

    Solution

    For an elastic material

    γ > η

    Young's modulus is greater than the shearing strain

     

  • Question 6
    1 / -0

    A wire is 2 m in length suspended vertically stretches by 10 mm when mass of 10 kg is attached to the lower end. The elastic potential energy gain by the wire is (take g = 10 m/s2)

    Solution

    Spring force by wire = weight

    K Δ x = mg

    K(0.01) = 10 × 9.8

    K = 9800 Nm−1

    Elastic potential energy is given by

    U = (0.5)K Δ x2   

    = (0.5) × 9800 × (0.01)2

    = 0.5 J

     

  • Question 7
    1 / -0

    A wire of length L and cross-sectional area A is made of material of Young's Modulus Y. The work done in stretching the wire by an amount x is

    Solution

    By conservation of energy, energy stored will be work done by the least required force to stretch the wire.

    Work = Favg x displacement

    ⇒ W = F/2 × x

    Now,

    So, work done,

    Alternate Method:

    Work done = ΔU = 1/2 × stress × strain × volume

    or, W = 1/2 × Y × (strain)2 × volume........(i)

    strain = x/L and Volume = A × L

     

    Putting the values of strain and volume in equation (i):

     

  • Question 8
    1 / -0

    Two exactly similar wires of steel and copper are stretched by equal forces. If the total elongation is 2 cm, then how much is the elongation in steel and copper wire respectively? Given, Ysteel = 20 × 1011 dyne/cm2, Ycopper = 12 × 1011 dyne/cm2.

    Solution

    Equal force is applied,

     

  • Question 9
    1 / -0

    Energy stored per unit volume in a stretched wire having Young's Modulus Y and stress 'S' is

    Solution

     

  • Question 10
    1 / -0

    A wire suspended vertically from one end and is stretched by attaching a weight 200 N to the lower end. The weight stretches the wire by 1 mm. The elastic potential energy gained by the wire is

    Solution

     

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