Class 11th

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

30.

Assuming L along y-axis and C along x-axis, we have two points (124.942, 20) and (125.134, 110) in xy-plane. By two-point form, the point L and C satisfies the equation.

y-124.942= $\frac{(125.134-124.942)}{(110-20)}$  (x-20)

$\Rightarrow$ y-124.942= $\frac{0.192}{90}$  (x - 20)

$\Rightarrow$ y-124.942= $\frac{0.032}{15}$  (x - 20)

$\Rightarrow$ 15y – 1874.13=0.032x -0.64

$\Rightarrow$ 15y= 0.032x +1873.49

$\Rightarrow$ y = 0.0021x+124.8993

$\Rightarrow$ L = 0.0021C + 124.8993

29.

The slope of line passing through (0, 0) and (–2, 9) is

$m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{9-0}{-2-0}=\frac{9}{-2}$

The line perpendicular to the line having slope m1 will have the slope

$m_2=\frac{-1}{m_1}=\frac{-1}{-9/2}=\frac{2}{9}$

So, the equation of line with slope m2 and passing $\left(x_0,y_0\right)=\left(-2,9\right)$ is

$m_2=\frac{y-y_0}{x-x_0}$

$\Rightarrow \frac{2}{9}=\frac{y-9}{x-\left(-2\right)}=\frac{y-9}{x+2}$

$\Rightarrow 2\left(x+2\right)=9\left(y-9\right)$

$\Rightarrow 2x+4=9y-81$

$\Rightarrow 2x-9y+4+81=0$

$\Rightarrow 2x-9y+85=0$