Tag: elastic energy

Questions Related to elastic energy

Wire of length $L$ is stretched by length l when a force $F$ is applied at one end. If elastic limit is not exceeded, the amount of energy stored in wire is 

elastic energy properties of material substances elasticity properties of matter physics

  1. $Fl$

  2. $\dfrac{1}{2}Fl$

  3. $\dfrac{Fl^2}{L}$

  4. $\dfrac{1}{2}\dfrac{El^2}{L}$

Show answer
Correct Option: C

A composite rodd consists of a steel rod of length $25cm$ and area $2A$ and a copper rod of length $50cm$ and area $A$. The composite rod is subjected to an axial load $F$. If the Young's modulii of steel and copper are in the ration $2:1$, then

elastic energy properties of material substances elasticity properties of matter physics

  1. The extension produced in copper rod will be more

  2. The extension in copper and steel parts will be in the ratio $1 : 2$

  3. The stress applied to copper rod will be more

  4. No extension will be produced in the steel rod

Show answer
Correct Option: A,B,C
Explanation:
$Stress=\dfrac{Force}{Area}$. Since force applied on both materials is same, but area of steel is more, so stress for copper will be more.
$\Delta l=\dfrac { FL }{ AY } $
Force is same for both.
Thus, $Extension,\Delta l \propto \dfrac{l}{AY}$
Thus$\dfrac{{\Delta l} _{Steel}}{{\Delta l} _{Copper}} \propto \dfrac{{(\dfrac{l}{AY})} _{Steel}}{{(\dfrac{l}{AY})} _{Copper}}$
=$\dfrac{{(\dfrac{25}{2A.2Y})}}{{(\dfrac{50}{AY})}}=1:8$

Which of the following are correct?

elastic energy properties of material substances elasticity properties of matter physics

  1. For a small deformation of a material, the ratio (stress/strain)decreases.

  2. For a large deformation of a material, the ratio (stress/strain) decreases

  3. Two wires mad of different materials, having the same diameter and length are connected end to end. A force is applied. This stretches their combined length by $2mm$. Now, the strain is same in both the wire but stress is different.

  4. None of these is correct.

Show answer
Correct Option: D
Explanation:

(A)and (B) The ratio(stress/strain) remains constant for a material and is called the modulus of elasticity of the material.
(C)Stress is same on both wires as equal amount of force gets transmitted to each wire. Since, each wire is made up of different material, they will have different strains.

Work done on stretching a rubber will be stored in it as :

elastic energy properties of material substances elasticity properties of matter physics

  1. chemical energy

  2. heat energy

  3. muscular energy

  4. potential energy

Show answer
Correct Option: D
Explanation:

When the rubber is stretched , the work done in stretching the rubber band is converted into elastic strain energy within the rubber.

Elastic strain energy is a form of potential energy.
Hence the correct option is (D).

A brass rod of length 2 m and cross-sectional area 2.0 $\displaystyle cm^{2}$ is attached end to end to a steel rod of length L and cross-sectional area 1.0 $\displaystyle cm^{2}.$ The compound rod is subjected to equal and opposite pulls of magnitude $\displaystyle 5\times 10^{4}N$ at its ends. If the elongations of the two rods are equal the length of the steel rod (L) is
($\displaystyle Y _{Brass}=1.0\times 10^{11}N/m^{2}: : and: : Y _{Steel}=2.0\times 10^{11}N/m^{2}$)

elastic energy properties of material substances elasticity properties of matter physics

  1. 1.5 m

  2. 1.8 m

  3. 1 m

  4. 2 m

Show answer
Correct Option: D
Explanation:

$k=\dfrac{YA}{L}$

$k _S=\dfrac{Y _SA _S}{L},k _B=\dfrac{Y _BA _B}{L _B}$
$F=5\times 10^4 N$
$\Delta l _S=\Delta l _b=\Delta l$
$F=k _B\Delta l$
$\dfrac{Y _SA _S}{L}=\dfrac{Y _BA _B}{L _B}$
$(\dfrac{Y _S}{Y _B}).(\dfrac{L _B}{L})=\dfrac{A _B}{A _S}$
$2\times \dfrac{2}{L}=2$
$L=2m$

If in a wire of Young's modulus $Y$, longitudinal strain $X$ is produced then the potential energy stored in its unit volume will be :

elastic energy properties of material substances elasticity properties of matter physics

  1. $0.5Y{X}^{2}$

  2. $0.5{Y}^{2}X$

  3. $2Y{X}^{2}$

  4. $Y{X}^{2}$

Show answer
Correct Option: A
Explanation:

We know that, the potential energy stored per unit volume is,

$W=\cfrac{1}{2} \times strss \times strain$
And, 
$Y=\cfrac{stress}{Strain}$
$\therefore Stress=Y\times strain$
$\therefore W=\cfrac{1}{2} \times Y \times {(strain)}^{2}$
$\Rightarrow\cfrac{1}{2}Y{X}^{2}=0.5 Y{X}^{2}$ 

A composite wire of a uniform cross-section $5.5\times 10^{-5}m^{2}$ consists of a steel wire of length $1.5\ m$ and a copper wire of length with a mass of $200\ kg$ is [Young's modulus of steel is $2\times 10^{11} N\ m^{-2}$ and that of copper is $1\times 10^{11}Nm^{-2}$. Take $g = 10\ ms^{-2}]$

elastic energy properties of material substances elasticity properties of matter physics

  1. $1\ mm$

  2. $2\ mm$

  3. $3\ mm$

  4. $4\ mm$

Show answer
Correct Option: A
Explanation:
$\Delta Ps=F\cdot \dfrac{P}{A}\cdot Ys$

$=2000\times \dfrac {1.5}{5.5\times10^{-5}}\times 2\times 10^{11}$

$=\dfrac{3\times 10^{-2}}{55\times 2}$

$=0.00027m$

$\Delta Pc=2000\times \dfrac {20}{5.5\times10^{-5}}\times 10\times 10^{11}$

$=0.00072m$

$\Delta P=\Delta Ps+\Delta Pc=1mm$

In an experiment on the determination of Young's Modulus of a wire by Searle's method, following data is available:
Normal length of the wire (L) = $110$cm
Diameter of the wire (d) = $0.01cm$
Elongation in the wire(l) = $0.125cm$
This elongation is for a tension of $50$N. The least counts for corresponding quantities are $0.01cm, 0.00005 cm, $ and $0.001cm$, respectively. Calculate the maximum error in calculating the value of Young's modulus(Y).

elastic energy properties of material substances elasticity properties of matter physics

  1. $8\%$

  2. $1.809\%$

  3. $1.09\%$

  4. cant say

Show answer
Correct Option: B

When a weight of 5 kg is suspended from a copper wire of length 30 m and diameter 0.5 mm, the length of the wire increases by 2.4 cm. If the diameter is doubled, the extension produced is :

elastic energy properties of material substances elasticity properties of matter physics

  1. 1.2 cm

  2. 0.6

  3. 0.3 cm

  4. 0.15 cm

Show answer
Correct Option: B
Explanation:

$Y \, = \, \dfrac{Mg \, \times \, 4 \, \times \, 1}{\pi D^2 \, \times \, \Delta l} \, or \, \Delta l \, \propto \, \dfrac{1}{D^2}$
(i) when D is double, $\Delta l$ becomes one-fourth. i.e.,
$\dfrac{1}{4} \, \times \, 2.4 \, cm \, i.e., 0.6 \, cm \, \times \, 2.4 \, cm$ i.e. 0.6 cm.

The maximum load a wire can with stand without breaking, when it is stretched to twice of its original length, will:

elastic energy properties of material substances elasticity properties of matter physics

  1. be half

  2. be four time decreased

  3. be double

  4. remain same

Show answer
Correct Option: D
Explanation:

the maximum load a wire can with stand without breaking it is stretched to twice of its original length is remain same$.$

Hence,
option $(D)$ is correct answer.