Tag: elastic energy

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Multiple choice elastic energy properties of material substances elasticity properties of matter physics

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. $(Y _1=2\times 10^{11}\ Pa$ & $Y _2=6\times 10^{11}\ Pa)$.

  1. $282.6$ N

  2. $212$ N

  3. $319.8$ N

  4. $382.6$ N
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $212\ N$ 

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

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,

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

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

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

  4. $3.78\times 10^{-4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Strain = Stress / Y. Stress = Force / Area. Force = (50000 * 9.8) / 4 columns. Area = pi * (R^2 - r^2). Using Y = 2e11 Pa for steel, the calculation yields approximately 7.2e-7.

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

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) $

  1. $22 N$

  2. $220 N$

  3. $120 N$

  4. 159 N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In series, total extension = deltaL1 + deltaL2 = F*L1/(A*Y1) + F*L2/(A*Y2). Given L1=L2=2m, A=pi*(0.001)^2, solve for F. F = 0.9e-3 / (L/A * (1/Y1 + 1/Y2)). Calculation leads to 220 N.

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

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 ${ 2.0\times  }10^{ 11 }{ Nm }^{ -2 }$

  1. <span>4.156 mm.</span>

  2. <span>2.156 mm.</span>

  3. <span>2.256 mm.</span>

  4. <span>3.156 mm.</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice elastic energy properties of material substances elasticity properties of matter physics

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? 

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

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

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

  4. $\frac { 3FL }{ 4AY }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The tension varies along the rod due to friction. Integrating the strain along the length L with a force F and friction F/2 results in the total elongation being FL / (2AY).

Multiple choice elastic energy properties of material substances elasticity properties of matter physics

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 

  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. $7.48 \times 10 ^ { - 10 } \mathrm { Nm } ^ { - 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
Multiple choice elastic energy properties of material substances elasticity properties of matter physics

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

  1. 1 cm

  2. 2 cm

  3. 1.5 cm

  4. 2.5 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
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
${ l } _{ 2 }=1 cm$