Tag: elastic energy

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Questions Related to elastic energy

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

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?

  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
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
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.
Multiple choice physics mechanical properties of solids applications of elasticity elastic energy properties of matter

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 

  1. $3:\pi$

  2. $\pi :3$

  3. $1:\pi$

  4. $\pi :1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

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

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

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

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

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

Reason: $R = {{\rho l} \over a}$

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When a wire is stretched to 3 times its length, its volume remains constant, so the cross-sectional area becomes 1/3 of the original. Since R = rho * (L/A), the new resistance becomes rho * (3L) / (A/3) = 9 * (rho * L/A) = 9R.

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

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

  1. $1.33 m, 1m$

  2. $1m, 1.33 m$

  3. $1.5 m, 1.33 m$

  4. $1.33m, 1.5 m$
Reveal answer Fill a bubble to check yourself
A Correct answer
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
Multiple choice physics properties of material substances poisson ratio poisson's ratio elastic energy

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

  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 $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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:

  1. $0.1$

  2. $0.2$

  3. $0.4$

  4. $0.5$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $

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

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.

  1. $0.15$%

  2. $0.02$%

  3. $0.24$%

  4. $0.48$%
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $
$V=\pi { R }^{ 2 }l$
$\therefore \cfrac { \Delta V }{ V } \times 100=\left( 2\cfrac { \Delta R }{ R } +\cfrac { \Delta l }{ l }  \right) \times 100=\left[ 2\times \left( -0.8\times { 10 }^{ -3 } \right) +4.0\times { 10 }^{ -3 } \right] \times 100=2.4\times { 10 }^{ -3 }\times 100=0.24$%