Mechanical Properties Of Solids - Class 11 Medical Physics - Extra Questions

$$strain \propto stress$$

So, on applying stress it gets elongated.But when it placed horizontally,it does not get any stress to be elongated.

$$F={ \gamma  }_{ steel }\times { \alpha  }_{ steel }\times \Delta T\times A$$

   $$=20\times { 10 }^{ 10 }\times 12\times { 10 }^{ -6 }\times 25\times 5\times { 10 }^{ -6 }$$

    $$=30000\times { 10 }^{ 10 }\times { 10 }^{ -12 }$$

    $$=300N$$

Strain $$={ \alpha  }_{ steel }\times \Delta T$$

           $$=12\times { 10 }^{ -6 }\times 25$$

           $$=300\times { 10 }^{ -6 }$$

           $$=3\times { 10 }^{ -4 }$$.

Given $$d=1\times { 10 }^{ -3 }m$$, $$F=20N$$

$$Y =2\times { 10 }^{ 11 }{ N }/{ { m }^{ 2 } }$$, $$r=\dfrac { d }{ 2 } =0.5\times { 10 }^{ -3 }m=5\times { 10 }^{ -4 }m$$

$$\therefore$$ Cross sectional area $$ A=\pi { r }^{ 2 }=3.142\times 25\times { 10 }^{ -8 }{ m }^{ 2 }$$

Strain energy per unit volume of the wire $$=\dfrac { 1 }{ 2Y  } { \left( stress \right)  }^{ 2 }$$

                                                                  $$=\dfrac { { \left( { F }/{ A } \right)  }^{ 2 } }{ 2Y  } $$

                                                                  $$=\dfrac { { F }^{ 2 } }{ 2{ A }^{ 2 }Y  } $$

                                                                   $$=\dfrac { { \left( 20 \right)  }^{ 2 } }{ 2\times { \left( 3.142\times 25\times { 10 }^{ -8 } \right)  }^{ 2 }\times 2\times { 10 }^{ 11 } } $$

                                                                    $$=1622{ J }/{ { m }^{ 3 } }$$.

in order to increase the volume(v) and water sample by 0.01% increasing the pressure P

$$\dfrac { V\times 0.01 }{ 100 } =\triangle V$$

$$B=\dfrac { PV }{ \triangle V } $$

$$=2.1\times { 10 }^{ 9 }\times { 10 }^{ -4 }$$

$$=2.1\times { 10 }^{ 5 }\dfrac { N }{ m^2 } $$

We note that:  
                                             $$\dfrac{\Delta V}{V}=\dfrac{\Delta l^3}{l^3}=\dfrac{(\Delta l+l)^3-l^3}{l^3}=\dfrac{3\Delta l}{l}$$
Thus, the pressure required is 
                                              $$p=\dfrac{3B\Delta l}{l}=\dfrac{3(1.4 \times10^{11} N/m^2 )(85.5cm -85.0cm)}{85.5cm}=2.4\times10^9 \,N/m^2$$  

$$ 10 \times 1.013 \times 10^{5} Pa $$

$$ B =\dfrac{P.V}{\Delta V} $$ 

$$\dfrac{\Delta V}{V}$$ is the fractional change in volume $$\dfrac{\Delta V}{V} = \dfrac{P}{B} = \dfrac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}$$

$$ = 2,73 \times 10^{-5}$$ 

$$ A = \pi (0.015)^{2} m^{2}$$

$$ max stress = 10^{8} Nm^{2}$$

Let F be the maximum force .

$$Stress = \dfrac{F}{A} $$

$$ \therefore F = max stress \times area. $$ 

$$ = 10^{8} \times\pi \times(0.015)^{2} $$

$$ = 7.065 \times 10^{4} N $$

 = 70, 650 N

 $$ \therefore $$ The maximum load is 70,650 N 

Shearing Force $$ = Shearing stress \times Area $$

$$ 8.4 \times 10^{7} \times (3 \times 10^{-1})^{2}$$

$$ = 7.56 \times 10^{4}N $$

75 . 6 KN .

$$ = 3200 Nm^{-2}$$

$$ Shear Strain = \dfrac{\Delta l}{l}$$

But$$ \Delta I = 0.2 cm $$

I = 5 cm 

$$ Shear Strain = \dfrac{0.2 cm}{5 cm}$$

$$ = 0.04 $$ 

Shear Modulus $$ = \dfrac{Shear stress}{Shear strain} $$

$$ = \dfrac{3200}{0.04}$$

$$ = 80 \times 10^{3}Nm^{-2} or 80 kPa .$$ 

$$Y=\dfrac{\dfrac{T}{A}}{\dfrac{\Delta L}{L}}$$

$$\displaystyle U=\frac{1}{2}F.\Delta l=\frac{1}{2}F.(\frac{Fl}{AY})=\frac{F^2l}{2AY}$$  

$$\implies x=2$$

Cross -sectional area of wir B,
$$a_2=2.0mm^2=2.0×10^{-6}m^2$$

Young’s modulus for steel, 
$$Y_1​=2×10^11Nm^{−2}$$

Young’s modulus for aluminium, 
$$Y_2​=7.0×10^10Nm^{−2}$$

(a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
$$Stress in the wire =\dfrac{ Force}{Area}=\dfrac{F}{a}$$
If the two wires have equal stresses, then:
$$\dfrac{F_1}{a_1}=\dfrac{F_2​}{a_2​}$$
Where, 
$$F_1​$$= Force exerted on the steel wire
$$F_2​$$= Force exerted on the aluminum wire
$$\dfrac{F_1​}{F_2​}=\dfrac{a_1​}{a_2​}=\dfrac{1}{2}$$.... (i)
The situation is shown in the figure.

Taking torque about the point of suspension, we have:
$$F_1​y=F_2​(1.05−y)$$

$$\dfrac{F_1}{F_2​}=\dfrac{1.05−y}{y}$$  ..... (ii)
Using equations (i) and (ii), we can write:
$$\dfrac{1.05−y}{y}=\dfrac{1}{2}$$
$$2(1.05−y)=y$$
$$y=0.7m$$
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b) Young's modulus =$$\dfrac{ Stress}{Strain}$$

$$Strain = \dfrac{Stress}{Young's modulus} =\dfrac{\dfrac{F}{a}}{Y}$$
If the strain in the two wires is equal, then:
$$\dfrac{\dfrac{F_1​}{a_1​}}{Y_1}=\dfrac{\dfrac{F_2​}{a_2​}}{Y_2}$$
$$\dfrac{F_1}{F_2}=\dfrac{a_1Y_1}{a_2Y_2}$$

$$\dfrac{a_1}{a_2}=\dfrac{1}{2}$$

$$\dfrac{F_1}{F_2}=\dfrac{1×2×10^{11}}{2×7×10^{10}}=\dfrac{10}{7}$$ ..... (iii)

Taking torque about the point where mass m, is suspended at a distance
$$y_1$$ from the side where wire A attached, we get:
$$F_1​y_1​=F_2​(1.05–y_1​)$$

$$\dfrac{F_1}{F_2}=\dfrac{1.05−y_1​}{y_1}$$ ..... (iv)

Using equations (iii) and (iv), we get:
$$\dfrac{1.05−y_1​}{y_1}=\dfrac{10}{7}
$$7(1.05−y_1​)=10y_1​$$
$$y_1​=0.432m$$
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached.