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.