Elasticity - class-XI

The elastic energy stored per unit volume in a stretched wire is
$u=\cfrac { 1 }{ 2 } \times stress\times strain=\dfrac { { (stress) }^{ 2 } }{ 2Y } \left( \because Y=\cfrac { stress }{ strain }  \right) $ 

The stress is given by Force/Area . 
Comparing the stresses, we get, $(stress _1/stress _2)=(L _2/L _1)=1/4$

The correct option is (c)

The stress depends only on the load and the area of cross section. Thus, the stress produces will also be same

The stress is directly proportional to the force applied and the inversely proportional to area of cross section. Thus as force ratio is 1:3, the stress ratio will also be 1:3

The correct option is (d)

The strain energy is released when the object is unloaded, which is nothing but resilience

Resilience is the ability of a material to absorb energy when deformed, and release that energy upon unloading

The correct option is (a)

Toughness is the ability of a material to absorb energy and plastically deform without fracturing. 

It is defined as the amount of energy per unit volume that a material can absorb before rupturing.

The correct option is option(b)

Extension deltaL = FL / AY. Since F and A are the same, deltaL is proportional to L/Y. For copper: 1mm = F1 / (A1e11). For steel: deltaL_s = F0.5 / (A2e11) = 0.25mm. Total extension = 1 + 0.25 = 1.25mm.

Young's modulus is defined as stress divided by strain. Stress is force per unit area (N/m^2) and strain is dimensionless, so the unit of Young's modulus is N/m^2.

Given.

Length of both wires, $L=2\,m$

Radius of both wires, $d=2mm=2\times {{10}^{-3}}m$

Total extension of joined wire, $\Delta {{L} _{net}}=0.9\,mm=9\times {{10}^{-4}}\,m$

Both wires join in series, so tension in both is equal

$T=\dfrac{{{Y} _{1}}A\Delta {{L} _{1}}}{L}=\dfrac{{{Y} _{2}}A\Delta {{L} _{2}}}{L}$  

Net extension in joined wire.

$\Rightarrow \Delta {{L} _{net}}=\Delta {{L} _{1}}+\Delta {{L} _{2}}=\dfrac{TL}{{{Y} _{1}}A}+\dfrac{TL}{{{Y} _{2}}A}=\dfrac{TL}{A}\left( \dfrac{1}{{{Y} _{1}}}+\dfrac{1}{{{Y} _{2}}} \right)$

$\Rightarrow \Delta {{L} _{net}}=\dfrac{TL}{A}\left( \dfrac{1}{{{Y} _{1}}}+\dfrac{1}{{{Y} _{2}}} \right)$

$\Rightarrow T=\dfrac{A\Delta {{L} _{net}}}{L\left( \dfrac{1}{{{Y} _{1}}}+\dfrac{1}{{{Y} _{2}}} \right)}=\dfrac{\dfrac{\pi }{4}{{\left( 2\times {{10}^{-3}} \right)}^{2}}\times 9\times {{10}^{-4}}}{2\left( \dfrac{1}{2\times {{10}^{11}}}+\dfrac{1}{6\times {{10}^{11}}} \right)}=212.05\ N$

Total force produced in joined wire is $212\ N$ 

Strain = Stress / Y. Stress = Force / Area. Force = (50000 * 9.8) / 4 columns. Area = pi * (R^2 - r^2). Using Y = 2e11 Pa for steel, the calculation yields approximately 7.2e-7.

Breaking weight is proportional to the cross-sectional area of the wire, meaning it scales with the square of the radius (r^2), and is independent of length. Doubling the radius increases the breaking weight by a factor of 2^2 = 4. Thus, 40 kg * 4 = 160 kg.

In series, total extension = deltaL1 + deltaL2 = FL1/(AY1) + FL2/(AY2). Given L1=L2=2m, A=pi*(0.001)^2, solve for F. F = 0.9e-3 / (L/A * (1/Y1 + 1/Y2)). Calculation leads to 220 N.

From Young's modulus relation, delta L = (F * L) / (Y * A) = (F * L) / (Y * pi * r^2). The extension is directly proportional to length L and inversely proportional to the square of the radius r^2. Maximizing L/r^2 gives the greatest increment; for option C, L = 2 and r = 1, giving a ratio of 2/1 = 2, which is the highest among the choices.

The tension varies along the rod due to friction. Integrating the strain along the length L with a force F and friction F/2 results in the total elongation being FL / (2AY).

$P.E = \frac{1}{2} \times 20 \times 0.001 = 0.01J$

Let the length of copper wire and steel wires be L
Total extension in the joined wire of length 2L= 3cm
Let extension in copper wire be ${ l } _{ 1 }$ and extension in steel wire be ${ l } _{ 2 }$.
Let Young's Modulus of copper wire be Y. So, the Young's Modulus of steel wire is 2Y.(Given)
Since, Stress applied on the long wire is same we can write that,

$\dfrac { { l } _{ 1 } }{ L } \times Y=\dfrac { { l } _{ 2 } }{ L } \times 2Y$
Hence, we get
${ l } _{ 1 }={ 2l } _{ 2 }$                              .....(1)
We also know,
${ l } _{ 1 }{ +l } _{ 2 }=3 cm$                               .....(2)
Solving (1) and (2),
we get
${ l } _{ 2 }=1 cm$

The elastic potential energy stored in a stretched wire is given by the work done in stretching it, which equals the average force multiplied by the displacement, 1/2 * F * l.

(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.

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

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

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

Y = (FL) / (Al) = (4FL) / (pid^2l). The relative error is deltaY/Y = deltaF/F + deltaL/L + 2deltad/d + deltal/l. Plugging in the values: 0.01/50 (negligible) + 0.01/110 + 2(0.00005/0.01) + 0.001/0.125. Calculation yields approximately 1.809%.

$Y , = , \dfrac{Mg , \times , 4 , \times , 1}{\pi D^2 , \times , \Delta l} , or , \Delta l , \propto , \dfrac{1}{D^2}$
(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 it is stretched to twice of its original length is remain same$.$

The elastic potential energy stored in a twisted wire is given by the formula U = (1/2) * C * alpha^2, where C is the torsional rigidity (C = (pi * eta * r^4) / (2 * L)). Substituting C into the formula yields U = (pi * eta * r^4 * alpha^2) / (4 * L).

Let, original length of the spring is L metre and, Y = $\dfrac { F.L }{ A.l } $