Self Studies

Mechanical Properties of Solids Test - 28

Result Self Studies

Mechanical Properties of Solids Test - 28
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    A uniform pressure P is exerted by an external agent on all sides of a solid cube at a temperature t 0C^{0}C.By what amount should the temperature of the cube be raised in order to bring its volume back to its original volume before the pressure was applied if the bulk modulus is B and coefficient of volumetric expansion is γ\gamma
    Solution
    B=p[dvv]     dv=pvB     vγΔT=pvB       ΔT=pγB B = - \dfrac{p}{[\dfrac{dv}{v}]}  \implies dv = - \dfrac{pv}{B} \implies v\gamma \Delta T = \dfrac{pv}{B}  \implies \Delta T = \dfrac{p}{\gamma B}
  • Question 2
    1 / -0
    In Youngs expt., the distance between two slits is d/3 and the distance between the screen and the slits is 3D. The number of fringes in 1/3 m on the screen, formed by monochromatic light of wavelength 3λ3\lambda , will be? 
    Solution
    Triangle width = λDd\frac{\lambda D}{d}
    3λ(3D)d3\frac{3\lambda (3D)}{\frac{d}{3}}
    27λDd\frac{27\lambda D}{d}
    no. of fringes  = 13×27λDd\frac{1}{3} \times \frac{27\lambda D}{d}
    d81λD\frac{d}{81\lambda D}
  • Question 3
    1 / -0
    The Poissons ratio for inert gases is:
    Solution
    For mono atomic molecule
    Cv=3R2C_v = \dfrac {3R}{2}      (as can be derived from standard results)
    Cp=5R2C_p = \dfrac {5R}{2}
    Poissons ratio for any gas is given by:
    poissons ratio=CpCv\text{poissons ratio} = \dfrac {C_p}{C_v}
    =5R23R2= \dfrac {\dfrac {5R}{2}}{\dfrac {3R}{2}}
    =53= \dfrac {5}{3}
     1.66\simeq 1.66
  • Question 4
    1 / -0
    If SS is stress and YY is the Young's modulus of the material of a wire, the energy stored in the wire per unit volume is :
    Solution
    Energy stored per unit volume can be given as: 
     E =12×stress×strainE  =\dfrac{1}{2} \times stress \times strain -----------(1)      
    From Hooke's law :
    Young's modulus, Y=StressStrainY = \dfrac{Stress}{Strain}

        \implies    Strain  =StressY  =SYStrain   = \dfrac{Stress}{Y}   = \dfrac{S}{Y} -----------(2)  
      
    From equation (1) and (2): 
    \therefore    E=12×S×SYE = \dfrac{1}{2} \times S \times \dfrac{S}{Y}

    E= 12S2Y\Rightarrow E=  \dfrac{1}{2} \dfrac{S^2}{Y}
    Hence, the correct option is (B)(B)
  • Question 5
    1 / -0
    The load versus strain graph for four wires of the same material is shown in the figure. The thickest wire is represented by the line

    Solution
    According to the given diagram,
    we will get one important factor that " even after applying load to a maximum extent in all wire, extension of "OD" is less as compared to other wires. Therefore we can conclude that thickest wire is represented by "OD"only
  • Question 6
    1 / -0
    Four uniform wires of the same material are stretched by the same force. The dimensions of wire are as given below. The one which has the minimum elongation has :
    Solution
    According to Young's modulus of elasticity:

     Δl= Flπ r2Y  \Delta l =  \dfrac{Fl}{ \pi  r^{2}Y }

    Δl  1/r2 \Delta l  \propto  1/ r^{2} 
    Hence only option "D" is satisfying the relation.
  • Question 7
    1 / -0
    A rubber ball is brought into 200 m deep water, its volume is decreased by 0.1% then volume  elasticity coefficient of the material of ball will be:
    (Given ρ=103kg/m3(Given\ \rho = 10^3 kg/m^3 and g=9.8ms2) g = 9.8 ms^{-2})
    Solution
    According to bulk modulus of elasticity,
    B= ΔPΔV/VB =  -\dfrac{ \Delta P}{ \Delta V/V}

    According to pressure depth relation,
    P=hρg P = h \rho g

    B= hρgΔV/V B =  -\dfrac{h \rho g}{\Delta V/V}

    or, B= 200 × 103×9.811000 B =  \dfrac{200  \times  10^{3} \times 9.8}{ \dfrac{1}{1000} }

    or, B=19.6 × 108 N/m2 B = 19.6  \times  10^{8} N / m^{2}
  • Question 8
    1 / -0
    (i) For a Searle's experiment, in the graph shown, there are two readings a and b that are not lying on the straight line
    (ii) Experiment is not performed precisely

    Solution
    Both statements are correct because for Searle's experiment (determination of Young's modulus), the load versus deflection curve formed by the experimental points should be a straight line. Since the given data doesn't lie on straight line, it is only because the experiment was not performed precisely.
  • Question 9
    1 / -0
    The velocity of sound in glass in 50. 0m/sec, it's density is 2,927 gm c.c. Calculate Youngs modules of glass in dynes/cm2^{2}
    Solution
    Given
    Velocity of sound in glass= 50 m/s
    Density of glass = 2927g/cc = 2927000 kg/m3kg/m^{3}
    Young modulus of glass =?
    Solution
    We know that

    v=(YD)12v=(\dfrac{Y}{D})^{\dfrac{1}{2}}

    Y=v2 ×DY=v^{2}  \times D

    Y=502 ×2927000Y=50^{2}  \times 2927000

    Y=73.17×109dyne/cm2Y=73.17 \times 10^{9} dyne /cm^{2}


  • Question 10
    1 / -0
    A sample of a liquid has an initial volume of 1.5L1.5 L. The volume is reduced by 0.2mL,0.2 mL, when the pressure increases by 140kPa140 kPa. What is the bulk modulus of the liquid.

    Solution
    bulk modulus of elasticity,
    B= ΔPΔV/V B =  \dfrac{- \Delta P}{\Delta V/V} 
    B= 1.5×140×1030.2 × 103 B =  \dfrac{-1.5 \times140 \times 10^{3} }{-0.2  \times  10^{-3} }
    B=1.05 × 109Pa B = 1.05  \times  10^{9}Pa
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now