Class 11th

Area of cross-section, A = 15.2 mm $*19.1\mathrm{m}\mathrm{m}$ = 15.2 $*19.1*{10}^{-6}{\mathrm{m}}^2=2.9\mathrm{}*{10}^{-4}{\mathrm{m}}^2$

Force, F = 44500 N

Stress, F/A = (44500/$2.9\mathrm{}*{10}^{-4})$ N/ ${\mathrm{m}}^2$

Modulus of elasticity,  $\eta$ = Stress / Strains, Strains = Stress / $\eta$

For copper,  $\eta$ = 42$*{10}^9$ $\mathrm{N}/{\mathrm{m}}^2$

Strains = (44500/$2.9\mathrm{}*{10}^{-4})/($ 42 $*{10}^9)$ = 3.65 $*{10}^{-3}$

Mass of the big structure, M = 50000 kg = 50000 $*9.8N$ = 4.9 $*{10}^5$ N

Inner radius of the column, r = 30 cm = 0.3 m

Outer radius of the column, R = 60 cm = 0.6 m

Young's modulus of steel, Y = 2 $*{10}^{11}$ Pa

Total force exerted on 4 columns, F = 4.9 $*{10}^5$ N

Force exerted on single column, f = F/4 = 1.225 $*{10}^5$ N

Cross sectional area of each column, A = $\pi (R^2-r^2)$ = $\pi ({0.6}^2-{0.3}^2)$ = 0.848 $m^2$

Stress in each column = f/A

Young's modulus, Y = Stress / Strain, Strain = Stress / Y = f/ (A $*Y)$ = $\frac{1.225\mathrm{}*{10}^5}{0.848*2*{10}^{11}}=7.22*{10}^{-7\mathrm{}}$ m

Edge of the aluminium cube, L = 10 cm = 0.1 m, Area A = 0.01 $m^2$

Mass attached, m = 100 kg = 100 $*$ 9.8 = 980 N = Applied force F

Shear modulus $\eta$ = 25 GPa = 25 $*{10}^{9}Pa$

Shear modulus $\eta$ = Shear stress / Shear strain = $\frac{\frac{F}{A}}{\frac{?L}{L}}$ ,  $?L=\frac{F*L}{\eta *A}$ = $\frac{980*0.1}{25\mathrm{}*{10}^9*0.01}$ = 3.92 $*{10}^{-7}m$

Diameter of wires, d = 0.25 cm, radius, r = 0.125 cm

Cross-sectional area, $A_1$ = $A_2$ = $\pi *r^2=0.049{cm}^2$ = 4.908 $*{10}^{-6}$ $m^2$

Length of the steel wire, $L_1=1.5\mathrm{}\mathrm{m}$ , length of the brass wire, $L_2=1.0\mathrm{}\mathrm{m}$

Change in length of the steel wire $=?L_1,$ Change in length of the copper wire $=?L_2$

Total force exerted on the steel wire, $F_1$ = ( 4+6) kg = 10 kg = 98 N

Young's modulus of steel , $Y_1$ = $\frac{\frac{F_1}{A_1}}{\frac{?L_1}{L_1}}$ = 2.0 $*{10}^{11}$ Pa

$?L_1$ = $\frac{F_1}{A_1}*\frac{L_1}{Y_1}=\mathrm{}\frac{98*1.5}{4.908*{10}^{-6}*2.0*{10}^{11}}$ = 1.497 $*{10}^{-4}$ m

Similarly for brass wire, $F_2$ = 6 kg = 58.8 N, $Y_2$ = $0.91*{10}^{11}\mathrm{}\mathrm{P}\mathrm{a}$

$?L_2$ = $\frac{F_2}{A_2}*\frac{L_2}{Y_2}=\mathrm{}\frac{58.8*1.0}{4.908*{10}^{-6}*0.91*{10}^{11}}$ = 1.316 $*{10}^{-4}$ m

For a given stress, the strain in rubber is more than it is in steel, hence the Young's modulus of rubber is lesser than in steel. So the statement is False.

Shear modulus is the ratio of the applied stress to the change in the shape of a body. The stretching of a coil changes its shape. Hence, shear modulus of elasticity is involved in this process.

 = 2.2 $*{10}^{-4}{m}^3$

Material A has greater Young's modulus.

Material A is the strongest as it can withstand more strain than material B without fracture.

From the given graph, for the value stress 150 $*{10}^6$ N/ $m^2$ , the strain is 0.002

Young's modulus = $\frac{150\mathrm{}*{10}^6}{0.002}$ = 7.5 $*{10}^{10}$ N/ $m^2$

Yield strength is the maxium strength the material can withstand in elastic limit. From the graph, the yield strength is 300 $*{10}^6$ or 3 $*{10}^8\mathrm{N}/m^2$

Length of the steel wire, $L_1$ = 4.7 m

Area of cross-section of the steel wire, $A_1$ = 3.0 $*{10}^{-5}$ m2

Length of the copper wire, $L_2$ = 3.5 m

Area of cross-section of the copper wire, $A_2$ = 4.0 $*{10}^{-5}$ m2

Change in length, $\Delta {L=}_{}{L}_1-L_2$

Let the force applied = F

Young's modulus in steel wire,

$Y_1$ = $\frac{F_1}{A_1}*\frac{L_1}{\mathrm{\Delta }{L}_{}}$ ….(1)

Young's modulus in copper wire,

$Y_2$ = $\frac{F_2}{A_2}*\frac{L_2}{\mathrm{\Delta }{L}_{}}$ …….(2)

The ratio of Young's modulus

$\frac{\mathrm{}{Y}_1}{Y_2}$ = $\frac{F_1}{A_1}*\frac{L_1}{\mathrm{\Delta }{L}_{}}*\frac{A_2}{F_2}*\frac{\mathrm{\Delta }{L}_{}}{{\mathrm{L}}_2}$ = $\frac{L_1}{A_1}*\frac{A_2}{L_2}$ = $\frac{4.7*4*{10}^{-5}}{3*{10}^{-5}*3.5}$ = $\frac{18.8}{10.5}=1.79$

32.

Since P (a, b) is the mid-point of the line segment say AB with points A (0, y) and B (x, 0) we can write,

$\left(a,b\right)=\left(\frac{x+0}{2},\frac{y+0}{2}\right)$

$\Rightarrow a=\frac{x}{2}$

$b=\frac{y}{2}$

$\Rightarrow x=2a$
$y=2b$

So, the equation of line with x and y intercept 2a and 2b using intercept form is

$\frac{x}{\mathrm{2a}}+\frac{y}{\mathrm{2b}}=1$

$\Rightarrow \frac{x}{a}+\frac{y}{b}=2$

Hence, proved

31. 

Assuming the price per litre say P in x-axis and the corresponding demand say D in y-axis, we have two point (14, 980) and (16, 1220) in xy plane. Then the points (P, D) will satisfy the equation.

$\Rightarrow D-980=\frac{1220-980}{16-14}(P-14)$

$\Rightarrow D-980=\frac{240}{2}(P-14)$

$\Rightarrow D-120(P-14)+980$

$\Rightarrow D-12\mathrm{0P}-1680+980$

$\Rightarrow D=12\mathrm{0P}-700$

Which is the required relation

Where P = 17, we have

D = 120 * 17 – 700

D = 1340

Hence, the owner can sell 1340 litres of milk weekly at 17/litre