Class 12th

(i) We have, $R=\{\left(x,y\right):3x-y=0\}$ a relation in set A= $\{1,2,3..........14\}$

For $x\in A,y=3x$ or $y\ne x$ i.e.,

$\left(x,x\right)$ does not exist in R

$\therefore$ R is not reflexive.

For $\left(x,y\right)\in R,y=3x$

Then $\left(y,x\right)x\ne 3y$

So $\left(y,x\right)\notin R$

$\therefore$ R is not symmetric

For $\left(x,y\right)\in R$ and $\left(y,z\right)\in R$ . We have

$y=3x$ and $z=3y$

Then $z=3\left(3x\right)=9x$

i.e., $\left(x,z\right)\notin R$

$\therefore$ R is not Transitive

(ii) We have,

R= $\{\left(x,y\right):y=x+5\&x<4\}$ is a relation in N

= $\{\left(1,1+5\right),\left(2,2+5\right),\left(3,3+5\right)\}$

= $\{\left(1,6\right),\left(2,7\right),\left(3,8\right)\}$

Clearly, R is not reflexive as $\left(x,x\right)\notin R$ and $x<4\&x\in N$

Also, R is not symmetric as $\left(1,6\right)\in R$ but $\left(6,1\right)\notin R$

And for $\left(x,y\right)\in R\left(y,z\right)\notin R$ . Hence, R is not Transitive.

(iii) R= $\{\left(x,y\right);y$ is divisible by x $\}$ is a relation in set

A= $\{1,2,3,4,5,6\}$

So, R= $\{\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,2\right),\left(2,4\right),\left(2,6\right),\left(3,3\right),\left(3,6\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\}$

Hence, R is reflexive because $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\in R$ i.e., $\left(x,x\right)\in R$

R is not symmetric as $\left(1,2\right)\in R$ but $\left(2,1\right)\notin R$

And for $x,y,z\in A$ and $\left(x,y\right)\in R\&\left(y,z\right)\in R$

$\frac{y}{z}=n$ and $\frac{z}{y}=m$ where $n\&m\in N$

Then, $z=my=m(nx)=mn.x$

$\frac{z}{x}=m.n,m,mn\in N$

Hence, $\left(x,z\right)\in R$

$\therefore$ R is transitive

(iv) R= $\{\left(x,y\right):x-y$ is an integer $\}$ is a relation in set Z

For $x\in Z$

$x-x=0$ is an integer

So, $\left(x,x\right)\in R$ i.e., R is reflexive

For $x,y\in Z$ and $\left(x,y\right)\in R$

$x-y$ is an integer

$-\left(y-x\right)$ is an integer

$\left(y-x\right)$ is an integer

So, $\left(y,x\right)\in R$ i.e., R is symmetric

For $x,y,z\in R$ and $\left(x,y\right)\in R\&\left(y,z\right)\in R$ We have,

$x-y$ is an integer

$y-z$ is an integer

So, $\left(x-y\right)+\left(y-z\right)$ is also an integer

$x-z$ is an integer

So, $\left(x,z\right)\in R$ i.e., R is transitive.

(v) (a) R= $\{\left(x,y\right):x$ and $y$ work at same place $\}$ in set A of human being.

For $x\in A$ we get

$x\&x$ work at same place

$\left(x,x\right)\in R$ So, R is reflexive.

For $x,y\in A$ and $\left(x,y\right)\in R$ We get,

$x\&y$ work at same place

$y\&x$ work at same place

$\left(y,x\right)\in R$ . So, R is symmetric.

For $x,y,z\in A$ and $\left(x,y\right)$ and $\left(y,z\right)\in R$ . We have,

$x\&y$ work at same place

$y\&z$ work at same place

$x\&z$ work at same place

i.e., $\left(x,z\right)\in R$ . So, R is Transitive.

(b) R= $\{\left(x,y\right):x\&y$ live in same locality $\}$

For $x\in A$ , we get,

$x\&x$ live in same locality

$\left(x,x\right)\in R$ So, R is reflexive.

For $x,y\in A$ and $\left(x,y\right)\in R$ we get,

$x\&y$ live in same locality

$y\&x$ live in same locality

i.e., $\left(y,x\right)\in R$ . So, R is symmetric.

For $x,y,z\in A$ and $\left(x,y\right)$ & $\left(y,z\right)\in R$ . Then

$x,y$ live in same locality

$y,z$ live in same locality

$x,z$ live in same locality

So, $\left(x,z\right)\in R$ i.e., R is transitive.

(c) R= $\{\left(x,y\right):x$ is exactly 7 cm taller than y $\}$

For $x\in A$ ,

Height $\left(x\right)\ne 7+$ height of $\left(x\right)$

So, $\left(x,x\right)\notin R$ . i.e., R is not reflexive.

For, $x,y\in A$ and $\left(x,y\right)\in R$ we have,

Height $\left(x\right)$ $=7+height\left(y\right)$

But $height\left(y\right)\ne 7+height\left(x\right)$

So, $\left(y,x\right)\in R$ i.e., R is not symmetric.

For $x,y,z\in A$ and $\left(x,y\right)\in R$ and $\left(y,z\right)\in R$ we have,

Height $\left(x\right)=7+height\left(y\right)$

And $height\left(y\right)=7+height\left(z\right)$

So, $height\left(x\right)=7+7+height\left(z\right)=14+height\left(z\right)$

i.e., $\left(x,z\right)\notin R$

So, R is not transitive.

(d) R= $\{\left(x,y\right):x$ is wife of y $\}$

For $x\in A,$

X is not wife of x.

So, $\left(x,x\right)\notin R$ i.e., R is not reflexive.

For $x,y\in A$ and $\left(x,y\right)\in R$ ,

X is wife of y but y is not wife of x

So, $\left(y,x\right)\notin R$ . i.e., R is not symmetric.

For $x,y,z\in A$ and $\left(x,y\right)$ and $\left(y,z\right)\in R$

X is wife of y

Y is wife of z

But y can never be husband & wife simultaneously

So, R is not transitive.

(e) R= $\{\left(x,y\right):x$ is father of y $\}$

For $x\in A$ ,

X is not a father of x

So, $\left(x,x\right)\notin R$ i.e., R is not reflexive.

For $x,y\in A$ and $\left(x,y\right)\in R$

X is father of y

Y is father of z

But x is not father of z

So, $\left(x,z\right)\notin R$ i.e., R is not transitive.

7.70

As Bond dissociation energy generally decreases on moving down the group as the atomic size of the element increases. However, among halogens, the bond dissociation energy of F₂ is lower than that of Cl₂ and F₂ due to the small atomic size of

Thus increasing order for bond dissociation energy among halogens is as follows:I₂2

As Bond dissociation energy of H-X molecules where X is the halogen decreases with increase in the atomic HI is the strongest acid as it loses H atom easily due to weak bonding between H and I.

So Increasing acid strength is as follows: HF

Basic strength decreases as we move from Nitrogen to Bismuth down the group as the size of the atom increases which is electron density of the atom decreases.

So Basic Strength is as follows: BiH3

9.32 Focal length of the convex lens, $f_1$ = 30 cm

The liquid acts as a mirror, focal length of the liquid = $f_2$

Focal length of the system (convex lens + liquid), $f$ = 45 cm

For a pair of optical systems placed in contact, the equivalent focal length is given as

$\frac{1}{f}$ = $\frac{1}{f_1}$ + $\frac{1}{f_2}$ or $\frac{1}{f_2}=\frac{1}{f}-\frac{1}{f_1}$ = $\frac{1}{45}$ - $\frac{1}{30}$

$f_2=$ - 90 cm

Let the refractive index of the lens be ${\mu }_1$ and the radius of curvature of one surface be R

Hence, the radius of curvature of the other surface is –R

R can be obtained by using the relation

$\frac{1}{f_1}$ = ( ${\mu }_1-1\left)\right(\frac{1}{R}$ + $\frac{1}{\left|R\right|}$ ) = (1.5 – 1)( $\frac{2}{R})$

$\frac{1}{30}$ = $\frac{1}{R}$ , so R = 30 cm

Let the refractive index of the liquid be ${\mu }_2$

The radius of curvature of the liquid on the side of the plane mirror = $\infty$

Radius of curvature of the liquid on the side of the lens, R = -30 cm

The value of ${\mu }_2$ can be calculated using the relation,

$\frac{1}{f_2}$ = ( ${\mu }_2-1\left)\right(\frac{1}{-R}$ - $\frac{1}{\infty }$ )

$\frac{-1}{90}$ = ( ${\mu }_2-1\left)\right(\frac{1}{30}$ - 0)

${\mu }_2$ - 1 = $\frac{1}{3}$

${\mu }_2=\frac{4}{3}$ = 1.33

Hence, the refractive index of the liquid is 1.33.

9.30 Distance between the objective mirror and the secondary mirror, d = 20 mm

Radius of curvature of objective mirror,  $R_1$ = 220 mm

Hence focal length of the objective mirror,  $f_1$ = $\frac{R_1}{2}$ = 110 mm

Radius of curvature of secondary mirror,  $R_2$ = 140 mm

Hence focal length of the objective mirror,  $f_2$ = $\frac{R_2}{2}$ = 70 mm

The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror. Hence, the virtual object distance for the secondary mirror,

u = $f_1-d$ = 110 – 20 = 90 mm

Applying the mirror formula for the secondary mirror, we can calculate the image distance ( $v$ ) as:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f_2}$ or $\frac{1}{v}$ = $\frac{1}{f_2}-\frac{1}{u}$ = $\frac{1}{70}$ - $\frac{1}{90}$

$v$ = 315 mm

Hence, the final image will be formed 315 mm away from the secondary mirror.

9.29 Focal length of the objective lens, $f_o$ = 140 cm

Focal length of the eyepiece, $f_e$ = 5 cm

Least distance of distinct vision, d = 25 cm

In normal adjustment, the separation between the objective lens and the eyepiece

= $f_o+f_e=140+5=145cm$

Height of the tower $h_1=100m$

Distance of the tower (object) from the telescope, u = 3 km = 3000 m

The angle subtended by the tower at the telescope is given as : $\theta =\frac{h}{u}$ = $\frac{100}{3000}$ = $\frac{1}{30}$ rad

The angle subtended by the image produced by the objective lens is given as $\theta =\frac{h_2}{f_o}$ , where $h_2$ = height of the image of the tower formed by the objective lens

So, $h_2$ = $\theta *f_o$ = $\frac{1}{30}*140$ =4.7 cm

Therefore, the objective lens forms a 4.7 cm tall image of the tower.

Image is formed at a distance, d = 25 cm

The magnification of the eyepiece is given by the relation

m = 1 + $\frac{d}{f_e}$ = 1 + $\frac{25}{5}$ = 6

Height of the final image = m $*h_2$ = 6 $*4.7=28.2cm$