Tag: elasticity

Questions Related to elasticity

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=$1/2\times Stress\times Strain\times Volume=1/2\times \dfrac{Mg}{A} \times \dfrac{l}{L}\times AL=\dfrac{Mgl}{2}$