Self Studies

Mechanical Properties of Solids Test - 36

Result Self Studies

Mechanical Properties of Solids Test - 36
  • 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
    Which of the following affects the elasticity of a substance?
    Solution
    Elasticity depends on temperature.
    Also, elasticity changes due to impurities because they increase binding of crystal grains., hence enhancing elasticity.
    Hammering also changes the structure of material as crystal grains break up into smaller units which plays a large role in determining elasticity (increases elasticity). Also, annealing tends to form a uniform orientation of crystal grains, hence producing larger crystal and changing elasticity.
  • Question 2
    1 / -0
    A wire is stretched by a force $$F$$. If $$s$$ is the strain developed and $$Y$$ is Young's modulus of material of wire, then work done per unit volume is 
    Solution
    Work done per unit volume is same as energy per unit volume which is given by:
    $$\dfrac{1}{2}\times Stress \times Strain=1/2\times (Y\times Strain)\times Strain=\dfrac{1}{2}\times Y\times {(Strain)}^{2}=\dfrac{Y{s}^{2}}{2}$$
  • Question 3
    1 / -0
    Two wires of same material and length but cross-sections in the ratio $$1:2$$ are used to suspend the same loads. The extensions in them will be in the ratio
    Solution
    Let the cross section of wire be A , 2A
    $$\Delta L_1=\dfrac{Fl}{AY}$$
    $$\Delta L_2=\dfrac{Fl}{2AY}$$
    so, $$\dfrac{\Delta L_1}{\Delta _2}=2/1$$
  • Question 4
    1 / -0
    A spherical ball contracts in volume by $$0.02\%$$ when subjected to a pressure of $$100$$ atmosphere. Assuming one atmosphere $$=10^8\ Nm^{-2}$$, the bulk modulus of the material of the ball is 
    Solution
    $$B=-V\dfrac{dP}{dV}$$
    Given $$dP=100atm, \dfrac{dV}{V}=-0.0002$$
    Thus, $$B=-\dfrac{100}{-0.0002}=5\times{10}^{5}$$ in atm units.
    $$1\ atm = {10}^{5}Pa$$, thus in SI units, $$B=5\times {10}^{10}=50\times{10}^{9}\ N/{m}^{2}$$
  • Question 5
    1 / -0
    The bulk modulus of a perfectly rigid body is 
    Solution
    $$B=-V\dfrac{dP}{dV}$$
    Perfectly rigid body means there can be no change in volume, or $$dV$$ tends to zero. Thus, B tends to infinity.
  • Question 6
    1 / -0
    The amount of work done in increasing the length of a wire through $$1cm$$ will be
    Solution
    Work done = change in elastic potential energy=$$1/2\times Stress\times Strain\times Volume=1/2\times Y\times (Strain)^2\times AL$$
    =$$1/2\times Y\times (1cm/L)^2\times AL=\dfrac{AY}{2L}$$
  • Question 7
    1 / -0
    On stretching a wire, the elastic energy stored per unit volume is 
    Solution
    Energy stored per unit volume is given as $$\dfrac{1}{2}Stress\times Strain$$=$$\dfrac { 1 }{ 2 } \left[ \dfrac { F }{ A }  \right] \left[ \dfrac { l }{ L }  \right] =\dfrac { Fl }{ 2AL } $$
  • Question 8
    1 / -0
    A $$5$$ metre long wire is fixed to the ceiling. A weight of $$10kg$$ is hung at the lower end and is $$1$$metre above the floor. The wire was elongated by $$1mm$$. The energy stored in the wire due to stretched is 
    Solution
    $$W=\displaystyle\frac{1}{2}\times F\times l=\frac{1}{2}mgl$$
    $$=\displaystyle\frac{1}{2}\times 10\times 10\times 1\times 10^{-3}=0.05J$$
  • Question 9
    1 / -0
    A steel ring of radius $$r$$ and cross sectional area $$A$$ is fitted onto a wooden disc of radius $$R(R>r)$$. If the Young's modulus of steel is $$Y$$, then the force with which the steel ring is expanded is 
    Solution
    Let $$T$$ be the tension in the ring, then 
    $$Y=\displaystyle\frac{T.2\pi r}{A.2\pi(R-r)}=\frac{Tr}{A(R-r)}$$
    $$\therefore T=\displaystyle\frac{YA(R-r)}{r}$$
  • Question 10
    1 / -0
    A metallic rod of length $$l$$ and cross-sectional area $$A$$ is made of a material of Young modules $$Y$$. If the rod is elongated by an amount $$y$$, then the work done is proportional to 
    Solution
    $$ Volume  =A\times L  $$ or $$V=Al$$
    $$strain=\displaystyle \frac{Elongation}{\text{Original  length}}=\frac{Y}{l}$$
    Young's modulus $$=\displaystyle \frac{stress}{strain}$$
    Work done, $$W=\displaystyle \frac{1}{2}\times stress \times strain \times volume$$
    $$W=\displaystyle \frac{1}{2}\times Y\times (strain)^2\times Al$$
    $$=\displaystyle \frac{1}{2}\times Y\left[\frac{y}{l}\right]^2\times Al=\displaystyle\frac{1}{2}\left[\frac{YA}{l}\right]y^2\Rightarrow W\propto y^2$$
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