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Elastic energy - class-XI

Description: elastic energy
Number of Questions: 35
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Tags: elasticity properties of matter properties of material substances physics
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The elastic energy stored per unit volume in a stretched wire is

  1. $\cfrac { 1 }{ 2 } \cfrac { { \left( stress \right) }^{ } }{ Y } $

  2. $\cfrac { 1 }{ 2 } \cfrac { { \left( stress \right) }^{ 2 } }{ Y } $

  3. $\cfrac { 1 }{ 2 } \cfrac { { \left( stress \right) }^{ 2 } }{ { Y }^{ 2 } } $

  4. $\cfrac { 1 }{ 2 } \cfrac { { \left( stress \right) }^{ } }{ { Y }^{ 2 } } $


Correct Option: B
Explanation:

The elastic energy stored per unit volume in a stretched wire is
$u=\cfrac { 1 }{ 2 } \times stress\times strain=\dfrac { { (stress) }^{ 2 } }{ 2Y } \left( \because Y=\cfrac { stress }{ strain }  \right) $ 

If S is stress and Y is Young's modulus of the material of a wire, the energy stored in the wire per unit volume is:

  1. $\frac{S}{2Y}$

  2. $\frac{2Y}{S^2}$

  3. $\frac{S^2}{2Y}$

  4. $2S^2Y$


Correct Option: C
Explanation:
Energy stored per unit volume can be given as: 
 $E  =\dfrac{1}{2} \times stress \times strain $ -----------(1)      
From Hooke's law :
Young's modulus, $Y = \dfrac{Stress}{Strain}$

$\implies$    $Strain   = \dfrac{Stress}{Y}   = \dfrac{S}{Y}$ -----------(2)  
  
From equation (1) and (2): 
$\therefore$    $E = \dfrac{1}{2} \times S \times \dfrac{S}{Y} $

$\Rightarrow E=  \dfrac{1}{2} \dfrac{S^2}{Y}$
Hence, the correct option is $(C)$

One end of an aluminium wire whose diameter is 2.5 mm is welded to one end of a copper diameter is 1.8 mm. The composite wire carries a steady current i of I .3 A. What is the current density in each wire ?

  1. $j _{Cu}= 26 A/cm^2, j _{Al}= 51 A/cm^2$

  2. $j _{Cu}=51A/cm^2 , j _{Al}= 26A/cm^2$

  3. $j _{Cu}=40A/cm^2, j _{Al}= 60 A/cm^2$

  4. $j _{Cu}= 60 A/cm^2, j _{Al}= 40 A/cm^2$


Correct Option: C

Two wires are of same material. Wire 1 is of 4 times longer than wire 2 and area of wire 1 is 4 times less than wire 2. Compare the stresses if they are elongated by the same load

  1. 1/2

  2. 4

  3. 1/4

  4. 2


Correct Option: C
Explanation:

The stress is given by Force/Area . 
Comparing the stresses, we get, $(stress _1/stress _2)=(L _2/L _1)=1/4$

The correct option is (c)

Two wires of different material but of same radius and length are stretched by the same load, the ratio of the stresses in the material will be same

  1. True

  2. False


Correct Option: A
Explanation:

The stress depends only on the load and the area of cross section. Thus, the stress produces will also be same

Two wires of different material but of same radius and different length are stretched by the loads in the ratio 1:3, the ratio of the stresses in the material will be same

  1. 3:1

  2. 2:3

  3. 3:2

  4. 1:3


Correct Option: D
Explanation:

The stress is directly proportional to the force applied and the inversely proportional to area of cross section. Thus as force ratio is 1:3, the stress ratio will also be 1:3

The correct option is (d)

The total strain energy stored in a body is known as 

  1. Resilience

  2. Toughness

  3. Modulus of resilience

  4. None of the above


Correct Option: A
Explanation:

The strain energy is released when the object is unloaded, which is nothing but resilience

Resilience is the ability of a material to absorb energy when deformed, and release that energy upon unloading

The correct option is (a)

A material capable of absorbing large amount of energy before fracture is known as 

  1. Ductility

  2. Toughness

  3. Resilience

  4. Plasticity


Correct Option: B
Explanation:

Toughness is the ability of a material to absorb energy and plastically deform without fracturing. 

It is defined as the amount of energy per unit volume that a material can absorb before rupturing.

The correct option is option(b)

A copper wire $1.0$ m and a steel wire of length $0.5$ m having equal cross-sectional areas are joining end to end. The composite wire is stretched by a certain load which stretches the copper wire by $1$ mm. If the Young's modulus  of copper steel are respectively $1.0\times 11^{11}Nm^{-1}$ and $ 2.0 \times 10^{11} Nm^{-2}$, the total extension of the composite wire is

  1. $1.75mm$

  2. $2.0 mm$

  3. $1.50 mm$

  4. $1.25 mm$


Correct Option: D

Give the MKS units for the following quantities.
Young's modulus.

  1. $2N/m^2$.

  2. $3N/m^2$.

  3. $4N/m^2$.

  4. $N/m^2$.


Correct Option: D

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


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,

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

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

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

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


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 

  1. 80 kg weight

  2. 240 kg weight

  3. 200 kg weight

  4. 160 kg weight


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) $

  1. $22 N$

  2. $220 N$

  3. $120 N$

  4. 159 N$


Correct Option: B

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

  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 $r = 2 mm$


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

  1. 4.156 mm.

  2. 2.156 mm.

  3. 2.256 mm.

  4. 3.156 mm.


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? 

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

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

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

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


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 

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


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

  1. 0.01 J

  2. $0.02 \mathrm { J }$

  3. $0.04 J$

  4. $1.00 \mathrm { J }$


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

  1. 1 cm

  2. 2 cm

  3. 1.5 cm

  4. 2.5 cm


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
${ l } _{ 2 }=1 cm$

Wire of length $L$ is stretched by length l when a force $F$ is applied at one end. If elastic limit is not exceeded, the amount of energy stored in wire is 

  1. $Fl$

  2. $\dfrac{1}{2}Fl$

  3. $\dfrac{Fl^2}{L}$

  4. $\dfrac{1}{2}\dfrac{El^2}{L}$


Correct Option: C

A composite rodd consists of a steel rod of length $25cm$ and area $2A$ and a copper rod of length $50cm$ and area $A$. The composite rod is subjected to an axial load $F$. If the Young's modulii of steel and copper are in the ration $2:1$, then

  1. The extension produced in copper rod will be more

  2. The extension in copper and steel parts will be in the ratio $1 : 2$

  3. The stress applied to copper rod will be more

  4. No extension will be produced in the steel rod


Correct Option: A,B,C
Explanation:
$Stress=\dfrac{Force}{Area}$. Since force applied on both materials is same, but area of steel is more, so stress for copper will be more.
$\Delta l=\dfrac { FL }{ AY } $
Force is same for both.
Thus, $Extension,\Delta l \propto \dfrac{l}{AY}$
Thus$\dfrac{{\Delta l} _{Steel}}{{\Delta l} _{Copper}} \propto \dfrac{{(\dfrac{l}{AY})} _{Steel}}{{(\dfrac{l}{AY})} _{Copper}}$
=$\dfrac{{(\dfrac{25}{2A.2Y})}}{{(\dfrac{50}{AY})}}=1:8$

Which of the following are correct?

  1. For a small deformation of a material, the ratio (stress/strain)decreases.

  2. For a large deformation of a material, the ratio (stress/strain) decreases

  3. Two wires mad of different materials, having the same diameter and length are connected end to end. A force is applied. This stretches their combined length by $2mm$. Now, the strain is same in both the wire but stress is different.

  4. None of these is correct.


Correct Option: D
Explanation:

(A)and (B) The ratio(stress/strain) remains constant for a material and is called the modulus of elasticity of the material.
(C)Stress is same on both wires as equal amount of force gets transmitted to each wire. Since, each wire is made up of different material, they will have different strains.

Work done on stretching a rubber will be stored in it as :

  1. chemical energy

  2. heat energy

  3. muscular energy

  4. potential energy


Correct Option: D
Explanation:

When the rubber is stretched , the work done in stretching the rubber band is converted into elastic strain energy within the rubber.

Elastic strain energy is a form of potential energy.
Hence the correct option is (D).

A brass rod of length 2 m and cross-sectional area 2.0 $\displaystyle cm^{2}$ is attached end to end to a steel rod of length L and cross-sectional area 1.0 $\displaystyle cm^{2}.$ The compound rod is subjected to equal and opposite pulls of magnitude $\displaystyle 5\times 10^{4}N$ at its ends. If the elongations of the two rods are equal the length of the steel rod (L) is
($\displaystyle Y _{Brass}=1.0\times 10^{11}N/m^{2}: : and: : Y _{Steel}=2.0\times 10^{11}N/m^{2}$)

  1. 1.5 m

  2. 1.8 m

  3. 1 m

  4. 2 m


Correct Option: D
Explanation:

$k=\dfrac{YA}{L}$

$k _S=\dfrac{Y _SA _S}{L},k _B=\dfrac{Y _BA _B}{L _B}$
$F=5\times 10^4 N$
$\Delta l _S=\Delta l _b=\Delta l$
$F=k _B\Delta l$
$\dfrac{Y _SA _S}{L}=\dfrac{Y _BA _B}{L _B}$
$(\dfrac{Y _S}{Y _B}).(\dfrac{L _B}{L})=\dfrac{A _B}{A _S}$
$2\times \dfrac{2}{L}=2$
$L=2m$

If in a wire of Young's modulus $Y$, longitudinal strain $X$ is produced then the potential energy stored in its unit volume will be :

  1. $0.5Y{X}^{2}$

  2. $0.5{Y}^{2}X$

  3. $2Y{X}^{2}$

  4. $Y{X}^{2}$


Correct Option: A
Explanation:

We know that, the potential energy stored per unit volume is,

$W=\cfrac{1}{2} \times strss \times strain$
And, 
$Y=\cfrac{stress}{Strain}$
$\therefore Stress=Y\times strain$
$\therefore W=\cfrac{1}{2} \times Y \times {(strain)}^{2}$
$\Rightarrow\cfrac{1}{2}Y{X}^{2}=0.5 Y{X}^{2}$ 

A composite wire of a uniform cross-section $5.5\times 10^{-5}m^{2}$ consists of a steel wire of length $1.5\ m$ and a copper wire of length with a mass of $200\ kg$ is [Young's modulus of steel is $2\times 10^{11} N\ m^{-2}$ and that of copper is $1\times 10^{11}Nm^{-2}$. Take $g = 10\ ms^{-2}]$

  1. $1\ mm$

  2. $2\ mm$

  3. $3\ mm$

  4. $4\ mm$


Correct Option: A
Explanation:
$\Delta Ps=F\cdot \dfrac{P}{A}\cdot Ys$

$=2000\times \dfrac {1.5}{5.5\times10^{-5}}\times 2\times 10^{11}$

$=\dfrac{3\times 10^{-2}}{55\times 2}$

$=0.00027m$

$\Delta Pc=2000\times \dfrac {20}{5.5\times10^{-5}}\times 10\times 10^{11}$

$=0.00072m$

$\Delta P=\Delta Ps+\Delta Pc=1mm$

In an experiment on the determination of Young's Modulus of a wire by Searle's method, following data is available:
Normal length of the wire (L) = $110$cm
Diameter of the wire (d) = $0.01cm$
Elongation in the wire(l) = $0.125cm$
This elongation is for a tension of $50$N. The least counts for corresponding quantities are $0.01cm, 0.00005 cm, $ and $0.001cm$, respectively. Calculate the maximum error in calculating the value of Young's modulus(Y).

  1. $8\%$

  2. $1.809\%$

  3. $1.09\%$

  4. cant say


Correct Option: B

When a weight of 5 kg is suspended from a copper wire of length 30 m and diameter 0.5 mm, the length of the wire increases by 2.4 cm. If the diameter is doubled, the extension produced is :

  1. 1.2 cm

  2. 0.6

  3. 0.3 cm

  4. 0.15 cm


Correct Option: B
Explanation:

$Y \, = \, \dfrac{Mg \, \times \, 4 \, \times \, 1}{\pi D^2 \, \times \, \Delta l} \, or \, \Delta l \, \propto \, \dfrac{1}{D^2}$
(i) when D is double, $\Delta l$ becomes one-fourth. i.e.,
$\dfrac{1}{4} \, \times \, 2.4 \, cm \, i.e., 0.6 \, cm \, \times \, 2.4 \, cm$ i.e. 0.6 cm.

The maximum load a wire can with stand without breaking, when it is stretched to twice of its original length, will:

  1. be half

  2. be four time decreased

  3. be double

  4. remain same


Correct Option: D
Explanation:

the maximum load a wire can with stand without breaking it is stretched to twice of its original length is remain same$.$

Hence,
option $(D)$ is correct answer.

A uniform wire of length L and radius r is twisted by a angle $ \angle \alpha$. If modulus of rigidity of the wire is $ \eta  $, then the elastic potential energy stored in wire, is

  1. $ \frac{\pi \eta r^{4}\alpha }{2L^{2}} $

  2. $ \frac{\pi \eta r^{4}\alpha^{2} }{4L} $

  3. $ \frac{\pi \eta r^{4}\alpha }{4L^{2}} $

  4. $ \frac{\pi \eta r^{4}\alpha^{2} }{2L} $


Correct Option: B

The length of an elastic string is $x$ metre when the tension is $8\ N$. Its length is $y$ metre when the tension is $10\ N$. What will be its length, when the tension is $18\ N$?

  1. $2x + y$

  2. $5y - 4x$

  3. $7y - 5x$

  4. $7y + 5x$


Correct Option: B
Explanation:

Let, original length of the spring is L metre and, Y = $\dfrac { F.L }{ A.l } $

Now, when F = 8N, and l = (x - l)m then, $Y=\dfrac { 8.L }{ A.\left( x-L \right)  } \quad \longrightarrow (I)$
and when F=10N, and l = (y - l)m then, $y=\dfrac { 10.L }{ A.\left( y-L \right)  } m\quad \longrightarrow (II)$
From equation (I) and (II) we get,
$\dfrac { 8L }{ A\left( x-L \right)  } =\dfrac { 10L }{ A\left( y-L \right)  } $
or,  $8\left( y-L \right) =10\left( x-L \right) $
or,    $4y-4L=5x-5L$
or,                $L=5x-4y$
When, F=18N,
Let, length of the wire will be Z metre.
$\therefore \quad Y=\dfrac { 18.L }{ A.\left( Z-L \right)  } \quad \longrightarrow (III)$
From equation (I) and (III) we get,
$\dfrac { 8L }{ A\left( x-L \right)  } =\dfrac { 18L }{ A\left( Z-L \right)  } $
or,  $9\left( x-L \right) =4\left( Z-L \right) $
or,  $4Z=9x-9L+4L$
            $=9x-5L$
            $=9x-25x+20y$    [putting value of L]
or,  $Z=5y-4x$


Work done by restoring force in a string within elastic limit is $-10\ J$. The maximum amount of heat produced in the string is :

  1. $10\ J$

  2. $20\ J$

  3. $5\ J$

  4. $15\ J$


Correct Option: A

If work done in stretching a wire by 1 mm is 2J. Then the work necessary for stretching another wire of same material but with double the radius and half the length by 1 mm in joule is

  1. 1/4

  2. 4

  3. 8

  4. 16


Correct Option: D
Explanation:

The stretching force $F=\dfrac{YA\Delta l}{l}$

where $Y=$Young's modulus, $A=$Area of cross-section of wire, $l=$actual length of wire, $\Delta l=$increase in length.
$F=\dfrac{Y\pi r^{2}\Delta l}{l}$
As the material is same $Y$ does not change.
$\dfrac{F _1}{F _2}=\dfrac{\dfrac{r _1^{2}\Delta l _1}{l _1}}{\dfrac{r _2^{2}\Delta l _2}{l _2}}$
Here $\Delta l _1=1mm$
$\Delta l _2=1mm$
$l _2=\dfrac{1}{2}l _1$
$r _2=2r _1$
$\dfrac{F _1}{F _2}=\dfrac{\dfrac{r _1^{2}\times 1mm}{l _1}}{\dfrac{4r _1^{2}\times 1mm}{\dfrac{1}{2}l _1}}$
$\dfrac{F _1}{F _2}=\dfrac{1}{8}$
The work done in stretching wire by amount $\Delta l$ is $W=\dfrac{1}{2} F\Delta l$
Hence $\dfrac{W _1}{W _2}=\dfrac{F _1}{F _2}=\dfrac{1}{8}$
As $F _1=2$
$F _2=2\times 8=16$
Hence the correct option is (D).

When a body mass $M$ is attached to power end of a wire (of length $L$) whose upper end is fixed, then the elongation of the wire is $l$. In this situation mark out the correct statement(s).

  1. Loss in gravitational potential energy of $M$ is $Mgl$.

  2. Elastic potential energy stored in the wire is $\dfrac {Mgl}{2}$

  3. Elastic potential energy stored in the wire is $Mgl$

  4. Elastic potential energy stored in the wire is $\dfrac {Mgl}{3}$


Correct Option: A,B
Explanation:

Since it moves $l$ distance against gravity, so gravitational potential energy=Mgl
Elastic potential energy=$1/2\times Stress\times Strain\times Volume=1/2\times \dfrac{Mg}{A} \times \dfrac{l}{L}\times AL=\dfrac{Mgl}{2}$

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