Tag: elastic energy

Questions Related to elastic energy

Two wires of different materials, each $2$m long and of diameter $2\,$mm, are joined in series to form a composite wire. What force will produce a total extension of $0.9$mm. $(Y _1=2\times 10^{11}\ Pa$ & $Y _2=6\times 10^{11}\ Pa)$.

elastic energy properties of material substances elasticity properties of matter physics

  1. $282.6$ N

  2. $212$ N

  3. $319.8$ N

  4. $382.6$ N

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Correct Option: B
Explanation:

Given.

Length of both wires, $L=2\,m$

Radius of both wires, $d=2mm=2\times {{10}^{-3}}m$

Total extension of joined wire, $\Delta {{L} _{net}}=0.9\,mm=9\times {{10}^{-4}}\,m$

Both wires join in series, so tension in both is equal

$T=\dfrac{{{Y} _{1}}A\Delta {{L} _{1}}}{L}=\dfrac{{{Y} _{2}}A\Delta {{L} _{2}}}{L}$  

Net extension in joined wire.

$\Rightarrow \Delta {{L} _{net}}=\Delta {{L} _{1}}+\Delta {{L} _{2}}=\dfrac{TL}{{{Y} _{1}}A}+\dfrac{TL}{{{Y} _{2}}A}=\dfrac{TL}{A}\left( \dfrac{1}{{{Y} _{1}}}+\dfrac{1}{{{Y} _{2}}} \right)$

$\Rightarrow \Delta {{L} _{net}}=\dfrac{TL}{A}\left( \dfrac{1}{{{Y} _{1}}}+\dfrac{1}{{{Y} _{2}}} \right)$

$\Rightarrow T=\dfrac{A\Delta {{L} _{net}}}{L\left( \dfrac{1}{{{Y} _{1}}}+\dfrac{1}{{{Y} _{2}}} \right)}=\dfrac{\dfrac{\pi }{4}{{\left( 2\times {{10}^{-3}} \right)}^{2}}\times 9\times {{10}^{-4}}}{2\left( \dfrac{1}{2\times {{10}^{11}}}+\dfrac{1}{6\times {{10}^{11}}} \right)}=212.05\ N$

Total force produced in joined wire is $212\ N$ 

Four identical hollow cylindrical columns of steel support a big structure of mass $50,000kg$. The inner and outer radii of each column are $30\ cm$ and $60\ cm$ respectively, Assuming the load distribution to be uniform. Calculate the compressional strain of each column,

elastic energy properties of material substances elasticity properties of matter physics

  1. $7.2\times 10^{-7}$

  2. $3.78\times 10^{-6}$

  3. $2.78\times 10^{-4}$

  4. $3.78\times 10^{-4}$

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Correct Option: A

To break a wire of 1 m length, minimum 40 kg weight is required. Then the wire of the same material of double radius and 6 m length will require breaking weight 

elastic energy properties of material substances elasticity properties of matter physics

  1. 80 kg weight

  2. 240 kg weight

  3. 200 kg weight

  4. 160 kg weight

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Correct Option: B

two wires of different material, each $2m$ long and of diameter $2mm$ are joined in series to form a composite wire.What force will produce a total extension of $0.9mm$ $\left( { Y } _{ 1 }=2\times { 10 }^{ 11 }N/{ m }^{ 2 },{ Y } _{ 2 }=7\times { 10 }^{ 11 }N/{ m }^{ 2 } \right) $

elastic energy properties of material substances elasticity properties of matter physics

  1. $22 N$

  2. $220 N$

  3. $120 N$

  4. 159 N$

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Correct Option: B

Which of the following shows greater increment in length when subjected to same load to wires made of same material:

elastic energy properties of material substances elasticity properties of matter physics

  1. $L = 1 m$ and $r = 1 mm$

  2. $L = 1 m$ and $r = 2 mm$

  3. $L = 2 m$ and $r = 1 mm$

  4. $L = 2 m$ and $r = 2 mm$

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Correct Option: A

A composite wire consists of a steel Wire of length 1 5 and a co uniform cross-sectional area of ${ 2.5\times  }10^{ -5 }{ m }^{ -5 }$.It is loaded with a mass of 200kg. Find the extension produced. Young's modulus of copper is ${ 2.5\times }10^{ 11 }{ Nm }^{ -2 }$ and that of steel ${ 2.0\times  }10^{ 11 }{ Nm }^{ -2 }$

elastic energy properties of material substances elasticity properties of matter physics

  1. 4.156 mm.

  2. 2.156 mm.

  3. 2.256 mm.

  4. 3.156 mm.

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Correct Option: B

A uniform rod of length L , area of cross-section A , mass m and Young 's modulus Y is pulled on  horizontal surface by a force f , such that the friction acting on it is F/2 . What if the elongation in the rod? 

elastic energy properties of material substances elasticity properties of matter physics

  1. $\frac { FL }{ 2AY } $

  2. $\frac { FL }{ AY }$

  3. $\frac { 3FL }{ 2AY }$

  4. $\frac { 3FL }{ 4AY }$

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Correct Option: A

A load of 2 kg produces an extension of 1 mm in a wire of 3 m in length and 1 mm In diameter. The Young's modulus of wire will be 

elastic energy properties of material substances elasticity properties of matter physics

  1. $3.25 \times 10 ^ { 10 } \mathrm { Nm } ^ { - 2 }$

  2. $7.48 \times 10 ^ { 12 } \mathrm { Nm } ^ { 2 }$

  3. $7.48 \times 10 ^ { 10 } \mathrm { Nm } ^ { - 2 }$

  4. $7.48 \times 10 ^ { - 10 } \mathrm { Nm } ^ { - 2 }$

Show answer
Correct Option: C
Explanation:
The applied load $=F = mg = 2 \times 9.8 = 19.6 N$
diameter $= 1 mm =1 \times {10^{ - 3}} m$
=> radius$ = 0.5 \times {10^{ - 7}} m$
Area , $A  = 3.14{r^2} = 3.14 \times 0.25 \times {10^{ - 6}}$
$=0.785 \times {10^{ - 6}}$
We know that$,$
$stress = Force/area$
$= 19.6/0.785 \times {10^{ - 6}}$
$24.96\times { 10^{ -6 } }$
change in length$, ΔL = 1 mm = {10^{ - 3}} m$
Length$, L  = 3 m$
$hence strain = ΔL/L $
$={10^{ - 3}}/3 $
we know that young's modulus is given by the ratio of stress and strain$,$
Hence$,$
$E  = 24.96 \times {10^{ - 6}}/{10^{ - 3}}/3$
$= 74.8 \times {10^9}$
$= 7.48 \times {10^{10}}N{m^{ - 2}}$
Hence, 
option $(C)$ is correct answer.

A wire is suspended by one end. At the other end, a weight equivalent to 20 N force is applied. If the increase in length is I mm, then increase in the f the wire will be

elastic energy properties of material substances elasticity properties of matter physics

  1. 0.01 J

  2. $0.02 \mathrm { J }$

  3. $0.04 J$

  4. $1.00 \mathrm { J }$

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Correct Option: A
Explanation:

$P.E = \frac{1}{2} \times 20 \times 0.001 = 0.01J$

Two wires of same length and same radius one of copper and another of steel are welded to form a long wire. An extension of $3cm$ is produced in it on applying a load at one of its ends. If the Young's modulus of steel is twice that of copper, then the extension in the steel wire will be

elastic energy properties of material substances elasticity properties of matter physics

  1. 1 cm

  2. 2 cm

  3. 1.5 cm

  4. 2.5 cm

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Correct Option: A
Explanation:

Let the length of copper wire and steel wires be L
Total extension in the joined wire of length 2L= 3cm
Let extension in copper wire be ${ l } _{ 1 }$ and extension in steel wire be ${ l } _{ 2 }$.
Let Young's Modulus of copper wire be Y. So, the Young's Modulus of steel wire is 2Y.(Given)
Since, Stress applied on the long wire is same we can write that,

$\dfrac { { l } _{ 1 } }{ L } \times Y=\dfrac { { l } _{ 2 } }{ L } \times 2Y$
Hence, we get
${ l } _{ 1 }={ 2l } _{ 2 }$                              .....(1)
We also know,
${ l } _{ 1 }{ +l } _{ 2 }=3 cm$                               .....(2)
Solving (1) and (2),
we get
${ l } _{ 2 }=1 cm$