Tag: applications of elasticity

Questions Related to applications of elasticity

Modulus of the wire then the energy density stored in the wire is

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. $\frac{1}{2}\gamma ^{2}T^{2}Y$

  2. $\frac{1}{3}\gamma ^{2}T^{2}Y^{3}$

  3. $\frac{1}{3}\frac{\gamma ^{2}T^{2}}{Y}$

  4. $\frac{1}{18}\gamma ^{2}T^{2}Y$

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Correct Option: D

An Indian rubber cord $L$ metre long and area of cross-section $A$ meter$^2$ is suspended vertically. Density of rubber is $\rho \ kg/$ meter$^3$ and Young's modulus of rubber is $Y$ Newton/metre$^2$. If the cord extends by $l$ metre under its own weight, then extension $l$ is:

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. $\dfrac{L^2 \rho g}{Y}$

  2. $\dfrac{L^2 \rho g}{2Y}$

  3. $\dfrac{L^2 \rho g}{4Y}$

  4. $\dfrac{Y}{L^2 \rho g}$

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Correct Option: B
Explanation:

A small differential element $dx$ at distance $x$ from the bottom of chord

Force acting on this element $ = \dfrac{M}{L} \times x \times g$
If extension  in this element is $dl$
Then
$\begin{array}{l} \dfrac { { dl } }{ { dx } } =\dfrac { f }{ { Ay } } =\dfrac { { Mg } }{ { LAy } } x \ \int  _{ 0 }^{ l }{ dl }=\dfrac { { Mg } }{ { LAy } } \int  _{ 0 }^{ l }{ xdx } \ l=\dfrac { { Mg{ L^{ 2 } } } }{ { 2LAy } } =\dfrac { { Mgl } }{ { 2Ay } }  \ If\, density\, is\, f\, then, \ \rho =\dfrac { m }{ { AL } }  \ So,\, l=\dfrac { { \rho g{ L^{ 2 } } } }{ { 2y } }  \ Hence\, option\, B\, is\, the\, correct\, answer. \end{array}$

A breaking stress of a material is ${ 10 }^{ 6 }N/{ m }^{ 2 }$ If density of material is $3\times 10^{ 3 }kg/{ m }^{ 3 }$, what should be the length of the material so that its breaks by it own weight?

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. 43.3 m

  2. 23.3 m

  3. 13.3 m

  4. 33.3 m

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Correct Option: D

Two wire of same radius and length are subjected to the same load, One wire is of steel and the other is copper. If Young's modulus of steel is twice that of copper, then the ratio of elastic energy stored per unit volume of steel to that of copper wire is

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. $2:1$

  2. $1:2$

  3. $1:4$

  4. $4:1$

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Correct Option: B
Explanation:
${ Y } _{ S }=2{ Y } _{ C }$, then the ratio of elastic energy stored per unit volume.
$E=\dfrac { 1 }{ 2 } \dfrac { { YL }^{ 2 }\alpha  }{ L } $
or,  $\dfrac { E }{ { l }^{ 2 }\alpha  } =\dfrac { 1 }{ 2 } \dfrac { Y }{ L } $
Since same length and same radius.
$\dfrac { { E }^{ 1 } }{ { l }^{ 2 }\alpha  } =\dfrac { 1 }{ 2 } \dfrac { { Y } _{ C } }{ L } $
$\dfrac { { E }^{ 11 } }{ { l }^{ 2 }\alpha  } =\dfrac { 1 }{ 2 } \times \dfrac { 2{ Y } _{ C } }{ L } $
$\dfrac { { E }^{ 1 } }{ { E }^{ 11 } } =\dfrac { 1 }{ 2 } $.

The number of independent elastic constant of a solid is=

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. 1

  2. 2

  3. 3

  4. 4

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Correct Option: C

The depression produced at the end of a $50 cm$ long cantilever on applying a load is $15 mm$. The depression produced at a distance of $30 cm $ from the rigid end will be

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. 3.24 mm

  2. 1.62 mm

  3. 6.48 mm

  4. 12.96 mm

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Correct Option: C
Explanation:

Length of cantilever, $L= 50cm=0.5m$
Deflection at the end $w(L)= 15mm$
Find: Deflection $w(x)$ at 30cm from rigid end.
Depression at a point in a cantilever beam with load at one end is given by,$ w(x)=\dfrac { P{ x }^{ 2 }(3L-x) }{ 6EI } $
We have,
$\dfrac { w(x) }{ w(L) } =\dfrac { \dfrac { P{ x }^{ 2 }(3L-x) }{ 6EI }  }{ \dfrac { PL^{ 3 } }{ 3EI }  } $

$\dfrac { w(0.3m) }{ w(0.5m) } =\dfrac { \dfrac { P{ (0.3) }^{ 2 }(3(0.5)-0.3) }{ 6EI }  }{ \dfrac { P(0.5)^{ 3 } }{ 3EI }  } $

This gives, $w(0.3m)=6.48mm$

A solid cylindrical rod of radius $3 mm$ gets depressed under the influence of a load through $8 mm$. The depression produced in an identical hollow rod with outer and inner radii of $4 mm$ and $2 mm$ respectively, will be

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. 2.7mm

  2. 1.9mm

  3. 3.2mm

  4. 7.7mm

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Correct Option: A
Explanation:

Depression in Solid cylinder $\delta _{1}= 8mm$
Its radius $r _{1}= 3mm$
Outer radius of Hollow cylinder $R _{2}= 4mm$
Inner radius of Hollow cylinder $r _{2}= 2mm$
Let depression in this cylinder be $\delta _{2}$.
Depression $\delta =\dfrac { W{ l }^{ 3 } }{ 12\pi{r }^{ 4 }Y } $
From the above equation we know that $\delta$ is proportional to $\dfrac{1}{r^{4}}$
Hence, we have
$\dfrac{\delta _{1}}{\delta _{2}}= \dfrac{{R _{2}}^{4}-{r _{2}}^{4}}{r _{1}^{4}}$
Substituting the values in above equation, we get
$\delta _{2}=2.7mm$

A beam of cross section area A is made of a material of Young modulus Y. The beam is bent into the arc of a circle of radius R. The bending moment is proportional to

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. $\displaystyle \frac{Y}{R}$

  2. $\displaystyle \frac{Y}{RA}$

  3. $\displaystyle \frac{R}{Y}$

  4. $YR$

Show answer
Correct Option: A
Explanation:

Bending moment $C=\dfrac{Y{I} _{G}}{R}$
Therefore, $C$ is proportional to $\dfrac{Y}{R}$

A steel wire of length $L$ and area of cross-section A shrinks by $\Delta l$ during night. Find the tension developed at night if Young's modulus is $Y$ and wire is clamped at both ends

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. $\displaystyle \frac{AYL}{\Delta l}$

  2. $AYL$

  3. $AY\Delta l$

  4. $\displaystyle \frac{AY\Delta l}{L}$

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Correct Option: D
Explanation:

$\displaystyle \frac{\Delta l}{L}=\frac{F}{AY}:or:F=\frac{AY\Delta l}{L}$ (using standard equation)

A wire of radius $1 mm$ is bent in the form of a circle of radius $10 cm$. The bending moment will be $(Y = 2\times10^{11}N/m^{2})$

physics mechanical properties of solids applications of elasticity elastic energy properties of matter

  1. 3.14 N/m

  2. 6.28 N/m

  3. 1.57 N/m

  4. 15.7 N/m

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Correct Option: C
Explanation:

Bending Moment $C=\dfrac { Y{ I } _{ G } }{ R } $
where $Y= 2\times { 10 }^{ 11 }N/{m}^{2}$
$R= 10cm= 0.1m$
${ I } _{ G }=\dfrac{\pi{R}^{4}}{4} = \dfrac{\pi{0.1}^{4}}{4} =7.85\times{10}^{-5} m^{4}$
Put these in the expression for $C$,
$C=1.57N/m$