Tag: elastic energy

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Questions Related to elastic energy

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

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} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The elastic potential energy stored in a twisted wire is given by the formula U = (1/2) * C * alpha^2, where C is the torsional rigidity (C = (pi * eta * r^4) / (2 * L)). Substituting C into the formula yields U = (pi * eta * r^4 * alpha^2) / (4 * L).

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

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. $7y + 5x$
Reveal answer Fill a bubble to check yourself
B Correct answer
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,  $Z=5y-4x$


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

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. $15\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The work done by a restoring force is negative when the string is stretched. The energy stored in the string is converted into heat when the string returns to its equilibrium position, equal in magnitude to the work done.

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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).

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

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 $\dfrac {Mgl}{3}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

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

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

State whether true or false :
The hollow shaft is much stronger than a solid shaft of same mass, same length and same material.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Modulus of elasticity = $\dfrac { stress }{ strain } $     and, stress = $\dfrac { Force }{ Area } $

$\therefore $  Elasticity $\alpha \dfrac { 1 }{ Area } $
As follow shaft, so, elasticity of the follow shaft is more than a solid one. As elasticity measures rigidity. So, hollow shaft is stronger.

Multiple choice physics mechanical properties of solids applications of elasticity elastic energy properties of matter
 A cable that can support a load of 1000 N is cut into equal parts. the maximum load that can be supported by the either part is:-

  1. 1000 N

  2. 2000 N

  3. 500 N

  4. 250 N
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Maximum load supported by the cable is directly proportional to the breaking stress.

Since,

Breaking stress = F / A

where F is the force and A is the cross-sectional area.

As we see that the breaking stress is independent of the length of the cable. So, if the cable is cut in two equal parts, the maximum load that can be supported by either parts of the cable remain the same as before.

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

State whether true or false :
The metal used in construction of a bridge should have high Young's modulus.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Young's modulus measures modulus of elasticity of a material. 

Elasticity defines rigidity of the material. 
So, for construction, we need metals which have high elasticity i.e. high young's modulus.

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

A silver wire of length $10 $ metre and cross-sectional area $10^{-8} m^{2}$ is suspended vertically and a weight of $10 N$ is attached to it. Young's modulus of silver and its resistivity are $7 \times 10^{10} N/m^{2}$ and $1.59 \times 10^{8} N/m^{2}$ \Omega - m$ respectively. The increase in its resistance is equal to:-

  1. $0.0455 \Omega$

  2. $0.455 \Omega$

  3. $0.91 \Omega$

  4. $0.091 \Omega$
Reveal answer Fill a bubble to check yourself
C Correct answer