Class 11th

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

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

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

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

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

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

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

Let a and b be the intercepts on x and y-axis

Then a + b = 9

$\Rightarrow b=9-a\left(1\right)$

Using intercept form equation of line with a and b intercepts are

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

$\Rightarrow \frac{x}{a}+\frac{y}{9-a}=1\left(2\right)$

As point (2, 2) passes the line having equation of form equation (2) we can write as

$\frac{2}{a}+\frac{2}{9-a}=1$

$\Rightarrow \frac{2\left(9-a\right)+2a}{a\left(9-a\right)}=1$

$\Rightarrow 18-2a+2a=\left(9-a\right)a$

$\Rightarrow 18=9a-a^2$

$\Rightarrow a^2-9a+18=0$

$\Rightarrow a^2-3a-6a+18=0$

$\Rightarrow a\left(a-3\right)-6\left(a-3\right)=0$

$\Rightarrow \left(a-3\right)\left(a-6\right)=0$

So, a = 3, 6

Case I

When a = 3, b = 9 – 3 = 6. Then the equation of line is

$\frac{x}{3}+\frac{y}{c}=1$

$\Rightarrow \frac{2x+y}{6}=1$

$\Rightarrow 2x+y=6$

$\Rightarrow 2x+y-6=0$

Case II

When a = 6, b = 9 – 6 = 3. Then equation of line is

$\frac{x}{6}+\frac{y}{3}=1$

$\Rightarrow \frac{x+2y}{6}=1$

$\Rightarrow x+2y=6$

$\Rightarrow x+2y-6=0$

Hence, equation of line is 2x + y – 6 or x + 2y – 6 = 0

26.

Let the line cuts x and y axis at intercept c. Then a = b = c

Then by intercept form the equation of line is

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

$\Rightarrow \frac{x}{c}+\frac{y}{c}=1$

$\Rightarrow x+y=c\left(1\right)$

Since equation (1) passes through point (2, 3).

Hence 2 + 3 = c

$\Rightarrow c=5$

So, substituting value of c in equation (1) we get

$\Rightarrow x+y=5$

$\Rightarrow x+y-5=0$