Tag: elastic energy

Questions Related to elastic energy

A light rod of length $2\ m$ is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross section $0.1\ cm^{2}$. A weight is suspended from a certain point of the rod such that equal stress are produced in both the wires. Which of the following are correct?

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

  1. The ratio of tension in the steel and brass wires is $0.5$

  2. The load is suspended at a distance of $400/3cm$ from the steel wire

  3. Both (a) and (b) are correct

  4. Neither (a) nor (b) are correct

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Correct Option: A,B,C
Explanation:
As $Stress=Force(Tension\ here)/Area$
As, $stress _{steel}=stress _{brass}$
$\Rightarrow Tension _{steel}/Tension _{brass}=Area _{steel}/Area _{brass}=0.1/0.2=0.5$
option A is correct.

As tension is inversely proportional to the distance of suspension of load. So distance of suspension for brass=0.5 x distance of suspension for steel.
As distance of suspension for steel+distance of suspension for brass$=2m=200cm$
So distance of suspension for steel+0.5xdistance of suspension for steel$=2m=200cm$
distance of suspension for steel$=\dfrac{2}{1.5}m=\dfrac{400}{3}cm$
option B is correct.

Both option A and B is correct.

For the same cross-section area and for a given load, the ratio of depression for the beam of a square cross-section and circular cross-section is 

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

  1. $3:\pi$

  2. $\pi :3$

  3. $1:\pi$

  4. $\pi :1$

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Correct Option: A
Explanation:

$\displaystyle \delta=\displaystyle\frac{Wl^3}{3YI}$, where $W=$load$,\ l=$length of beam$,\ I=$moment of inertia$=\dfrac{b{d}^{3}}{12}$ for rectangular beam, and for square beam$,\ b=d.$ Thus, ${I} _{1}=\dfrac{{b}^{4}}{12}$

Now, for circular cross section, $\displaystyle I _2=\left[\dfrac{\pi r^4}{4}\right]$

$\therefore \delta _1=\dfrac{Wl^3\times 12}{3Yb^4}=\dfrac{4Wl^3}{Yb^4}$

and $\delta _2=\dfrac{Wl^3}{3Y(\pi r^$/4)}=\displaystyle\frac{4Wl^3}{3Y(\pi r^4)}$

Thus, $\dfrac{\delta_1}{\delta_2}=\dfrac{3\pi r^4}{b^4}=\dfrac{3\pi r^4}{(\pi r^2)^2}=\dfrac{3}{\pi}$
$(\because b^2=\pi r^2$ as they have same cross sectional area)

A beam of metal supported at the two edges is loaded at the centre. The depression at the centre is proportional to 

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

  1. $Y^2$

  2. $Y$

  3. $1/Y$

  4. $1/Y^2$

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Correct Option: C
Explanation:

Young's modulus: $ Y=\dfrac { \dfrac { F }{ a }  }{ \dfrac { \Delta l }{ l }  } =\dfrac { Fl }{ a(\Delta l) } $

$ \Delta l\rightarrow \dfrac { 1 }{ Y } $

The buckling of a beam is found to be more if __________.

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

  1. The breadth of the beam is large

  2. The beam material has large value of Young's modulus

  3. The length of the beam is small

  4. The depth of the beam is small

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Correct Option: D
Explanation:

Critical buckling stress of a column formula is given by 

$\sigma=\dfrac{F}{A}=\dfrac{{\pi}^2 r^2 E}{L^2}$
where $\sigma$ = critical stress
$L$= unsupported length of the column
$r=$ least radius 
So if the depth of the beam i small, buckling of a beam will be more.

Assertion: When a wire is stretched to three times its length, its resistance becomes 9 times

Reason: $R = {{\rho l} \over a}$

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

  1. both, Assertion and Reason are true and the reason is correct explanation of the Assertion

  2. both, Assertion and Reason are true and the reason is not correct explanation of the Assertion

  3. Assertion is true, but the reason is false.

  4. Both, Assertion and reason and false

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Correct Option: A

A light rod of length $2.00 m$ is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross section $10^{-3}m^{2}$ and the other is of brass of cross-section $2\times10^{-3}m^{2}$ . Find out the position along the rod at which a weight may be hung to produce.(Youngs modulus for steel is 2x10$^{11}$N /m$^{2}$ and for brass is 10$^{11}$N / m$^{2}$ )
a) equal stress in both wires
b) equal strains on both wires

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

  1. $1.33 m, 1m$

  2. $1m, 1.33 m$

  3. $1.5 m, 1.33 m$

  4. $1.33m, 1.5 m$

Show answer
Correct Option: A
Explanation:
For  equal  stress
$ \dfrac{F _{1}}{A _{1}} = \dfrac{F _{2}}{A _{2}}$
$ \dfrac{F _{1}}{F _{2}} = \dfrac{A _{1}}{A _{2}} = \dfrac{10^{-3} m^{2}}{2 \times 10^{-3}m^{2}} = \dfrac{1}{2}$
$ 2F _{1} = F _{2}$
For  balance  of  rod
$ W = F _{1} + F _{2}$
$W = \dfrac{3 F _{2}}{2}$
$ F _{2} = \dfrac{2}{3} W$
Now equating torque
$Wx = F _{2} \times 2$
$x = \dfrac{2}{3} \times 2 = \dfrac{4}{3} = 1.33m$
For equal strain
$ \dfrac{\triangle l _{1}}{l} = \dfrac{\triangle l _{2}}{l}$
or
$\dfrac{\sigma _1}{Y _1}=\dfrac{\sigma _2}{Y _2}$
or
$\dfrac{F _1}{10^{-3}\times 2\times 10^{11}}=\dfrac{F _2}{2\times 10^{-3}\times 10^{11}}$
Thus we get $F _1=F _2$.
So, weight  will  be  hanging  mid - way 1m

A material has Poisson's ratio $0.5$. if a uniform rod of it suffers a longitudinal strain of $2\times {10}^{3}$, then the percentage change in volume is

physics properties of material substances poisson ratio poisson's ratio elastic energy

  1. $0.6$

  2. $0.4$

  3. $0.2$

  4. zero

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Correct Option: D
Explanation:

Poisson's ratio is a material property. Therefore it doesn't change with dimensions.  

Option D

Which of the following statements is correct regarding Poisson's ratio?

physics properties of material substances poisson ratio poisson's ratio elastic energy

  1. It is the ratio of the longitudinal strain to the lateral strain

  2. Its value is independent of the nature of the material

  3. It is unitless and dimensionless quantity

  4. The practical value of Poisson's ratio lies between $0$ and $1$

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Correct Option: C
Explanation:

The ratio of the lateral strain to longitudinal strain is called Poisson's ratio.
Hence, option (a) is an incorrect statement.
Its value depends only on the nature of the material. 
Hence, option (b) is an incorrect statement.
It is the ratio of two like physical quantities.
Therefore, it is unitless and dimensionless quantity.
Hence option (c) is a correct statement
The practical value of Poisson's ratio lies between $0$ and $0.5$
hence option (d) is an incorrect statement.

If the volume of a wire remains constant when subjected to tensile stress, the value of Poisson's ratio of the material of the wire is:

physics properties of material substances poisson ratio poisson's ratio elastic energy

  1. $0.1$

  2. $0.2$

  3. $0.4$

  4. $0.5$

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Correct Option: D
Explanation:

Let $L$ be the length, $r$ be the radius of the wire.
Volume of the wire is
$V=\pi { r }^{ 2 }L$
Differentiating both sides, we get
$\Delta V=\pi (2r\Delta r)L+\pi { r }^{ 2 }\Delta L\quad $
As the volume of the wire remains unchanged when it gets stretched, so $\Delta V=0$. Hence
$0=2\pi rL\Delta r+\pi { r }^{ 2 }\Delta L$
$\therefore \cfrac { \Delta r/r }{ \Delta L/L } =-\cfrac { 1 }{ 2 } $
$Poisson's\quad ratio=\cfrac { Lateral\quad strain }{ Longitudinal\quad strain } =-\cfrac { \Delta r/r }{ \Delta L/L } =\cfrac { 1 }{ 2 } =0.5\quad $

A material has Poisson's ratio $0.2$. If a uniform rod of its suffers longitudinal strain $4.0\times {10}^{-3}$, calculate the percentage change in its volume.

physics properties of material substances poisson ratio poisson's ratio elastic energy

  1. $0.15$%

  2. $0.02$%

  3. $0.24$%

  4. $0.48$%

Show answer
Correct Option: C
Explanation:

Given:
 Poisson's ratio $0.2$. 
 longitudinal strain $4.0\times 10^{−3}$
As $\sigma =-\cfrac { \Delta R/R }{ \Delta l/l } $
$\therefore \cfrac { \Delta R }{ R } =-\sigma \cfrac { \Delta l }{ l } =-0.2\times 4.0\times { 10 }^{ -3 }=-0.8\times { 10 }^{ -3 }\quad $
$V=\pi { R }^{ 2 }l$
$\therefore \cfrac { \Delta V }{ V } \times 100=\left( 2\cfrac { \Delta R }{ R } +\cfrac { \Delta l }{ l }  \right) \times 100=\left[ 2\times \left( -0.8\times { 10 }^{ -3 } \right) +4.0\times { 10 }^{ -3 } \right] \times 100=2.4\times { 10 }^{ -3 }\times 100=0.24$%