Tag: elastic energy

Questions Related to elastic energy

A steel wire is stretched by 5 kg wt, If the radius of the wire is doubled its Young's modulus

  1. Remains unchanged

  2. Becomes double

  3. Becomes half

  4. Becomes 1/4 times


Correct Option: A

In the system shown in figure pulley is smooth. String is massless and inextensible. The acceleration of the system a, tensions ${T} _{1}\ and {T} _{1}\left (g=10{m/s}^{2}\right)$ are 

  1. $\dfrac { 20 }{ 3 } { m/s }^{ 2 },\dfrac { 50 }{ 3 } N,\dfrac { 60 }{ 3 } N$

  2. $\dfrac { 10 }{ 3 } { m/s }^{ 2 },\dfrac { 100 }{ 3 } N,\dfrac { 60 }{ 3 } N$

  3. $\dfrac { 20 }{ 3 } { m/s }^{ 2 },\dfrac { 100 }{ 3 } N,\dfrac { 60 }{ 3 } N$

  4. $\dfrac { 20 }{ 3 } { m/s }^{ 2 },\dfrac { 100 }{ 3 } N,\dfrac { 50 }{ 3 } N$


Correct Option: C

A stone of mass 'm' s projected from a rubber catapult of length 'l' and cross-sectional area A stretched by an amount 'e'. If Y be the young's modulus of rubber then the velocity of projection of stone?

  1. $Y \sqrt {\dfrac{Ae^2}{lm}}$

  2. $ \sqrt {\dfrac{Ae^2}{lm}}$

  3. $Y \sqrt {\dfrac{YAe^2}{lm}}$

  4. $Y \sqrt {\dfrac{YAe^4}{lm}}$


Correct Option: C

Two wires of equal length and cross section are suspended. their young's modulus are $Y _1$ and $Y _2$ respectively. their equivalent young's modulus of elasticity is

  1. $Y _1+Y _2$

  2. $Y _1Y _2$

  3. $Y _1-Y _2$

  4. $\dfrac{Y _1+Y _2}2$


Correct Option: C

In the Young's double slit experiment the intestines at two points $P _{1}$ and $P _{2}$ on the screen are respectively $I _{1}$ and $I _{2}$. If $P _{1}$ is located at the centre of bright fringe and $P _{2}$ is located at a distance equal to a quarter of fringe width from $P _{1}$, then $I _{1}/I _{2}$ is 

  1. $2$

  2. $1/2$

  3. $4$

  4. $16$


Correct Option: A

Modulus of the wire then the energy density stored in the wire is

  1. $\frac{1}{2}\gamma ^{2}T^{2}Y$

  2. $\frac{1}{3}\gamma ^{2}T^{2}Y^{3}$

  3. $\frac{1}{3}\frac{\gamma ^{2}T^{2}}{Y}$

  4. $\frac{1}{18}\gamma ^{2}T^{2}Y$


Correct Option: D

An Indian rubber cord $L$ metre long and area of cross-section $A$ meter$^2$ is suspended vertically. Density of rubber is $\rho \ kg/$ meter$^3$ and Young's modulus of rubber is $Y$ Newton/metre$^2$. If the cord extends by $l$ metre under its own weight, then extension $l$ is:

  1. $\dfrac{L^2 \rho g}{Y}$

  2. $\dfrac{L^2 \rho g}{2Y}$

  3. $\dfrac{L^2 \rho g}{4Y}$

  4. $\dfrac{Y}{L^2 \rho g}$


Correct Option: B
Explanation:

A small differential element $dx$ at distance $x$ from the bottom of chord

Force acting on this element $ = \dfrac{M}{L} \times x \times g$
If extension  in this element is $dl$
Then
$\begin{array}{l} \dfrac { { dl } }{ { dx } } =\dfrac { f }{ { Ay } } =\dfrac { { Mg } }{ { LAy } } x \ \int  _{ 0 }^{ l }{ dl }=\dfrac { { Mg } }{ { LAy } } \int  _{ 0 }^{ l }{ xdx } \ l=\dfrac { { Mg{ L^{ 2 } } } }{ { 2LAy } } =\dfrac { { Mgl } }{ { 2Ay } }  \ If\, density\, is\, f\, then, \ \rho =\dfrac { m }{ { AL } }  \ So,\, l=\dfrac { { \rho g{ L^{ 2 } } } }{ { 2y } }  \ Hence\, option\, B\, is\, the\, correct\, answer. \end{array}$

A breaking stress of a material is ${ 10 }^{ 6 }N/{ m }^{ 2 }$ If density of material is $3\times 10^{ 3 }kg/{ m }^{ 3 }$, what should be the length of the material so that its breaks by it own weight?

  1. 43.3 m

  2. 23.3 m

  3. 13.3 m

  4. 33.3 m


Correct Option: D

Two wire of same radius and length are subjected to the same load, One wire is of steel and the other is copper. If Young's modulus of steel is twice that of copper, then the ratio of elastic energy stored per unit volume of steel to that of copper wire is

  1. $2:1$

  2. $1:2$

  3. $1:4$

  4. $4:1$


Correct Option: B
Explanation:
${ Y } _{ S }=2{ Y } _{ C }$, then the ratio of elastic energy stored per unit volume.
$E=\dfrac { 1 }{ 2 } \dfrac { { YL }^{ 2 }\alpha  }{ L } $
or,  $\dfrac { E }{ { l }^{ 2 }\alpha  } =\dfrac { 1 }{ 2 } \dfrac { Y }{ L } $
Since same length and same radius.
$\dfrac { { E }^{ 1 } }{ { l }^{ 2 }\alpha  } =\dfrac { 1 }{ 2 } \dfrac { { Y } _{ C } }{ L } $
$\dfrac { { E }^{ 11 } }{ { l }^{ 2 }\alpha  } =\dfrac { 1 }{ 2 } \times \dfrac { 2{ Y } _{ C } }{ L } $
$\dfrac { { E }^{ 1 } }{ { E }^{ 11 } } =\dfrac { 1 }{ 2 } $.

The number of independent elastic constant of a solid is=

  1. 1

  2. 2

  3. 3

  4. 4


Correct Option: C