Class 12th

Candidates seeking admission to the certificate and BVoc programme can enrol for admission with Class 12 marks. Candidates must complete Class 12 to enrol for various courses. The college offers more than 10 certificate courses. Zee Institute of Creative Art admissions are based on merit.

$\left\{\left(x+2\right)e^{\frac{y+1}{x+2}}+\left(y+1\right)\right\}dx=\left(x+2\right)dy.$              

  $\Rightarrow \frac{dy}{dx}=e^{\left(\frac{y+1}{x+2}\right)}+\left(\frac{y+1}{x+2}\right)-\left(i\right)$

(1) -> $v+\left(x+2\right)\frac{dv}{dx}=e^v+v.$  

$\Rightarrow \frac{dv}{e^v}=\frac{dx}{x+2}$ Integrating both side

Let $e^{e-2/3}=a\Rightarrow 0<\left|\frac{x+2}{3}\right|<a$  

$\therefore a=-3a-2\&\beta =3a-2$              

$\Rightarrow \left|\alpha +\beta \right|=4$               

As we all know that like charges repel each other. This means that there must be a repulsive electromagnetic force between positively charged protons. However, the nuclear force is said to be 100 times stronger than electromagnetic force which makes it strong enough to hold protons together within a nucleus. This nuclear force is much stronger than coulomb force at short ranges.

$f\left(x\right)=[\begin{array}{l}3\left(1-\frac{\left|x\right|}{2}\right);\left|x\right|\le 2\\ 0;\left|x\right|>2\end{array}$

$\therefore g\left(x\right)=f\left(x+2\right)-f\left(x-2\right)[\begin{array}{l}0x<-4\\ 3\left(1-\frac{\left|x-2\right|}{2}\right)-4\le x\le 0\\ -3\left(1-\frac{\left|x-2\right|}{2}\right)0<x\le 4\\ 0x>4.\end{array}$

g(x) is continuous every where but not differentiable

at x = -4, -2, 2 and 4

$\therefore n=0\&m=4\Rightarrow n+m=4$

For 90 mA in zener diode, current in R should be greater than 90 mA.

$\frac{45}{R}=90*10^{-3}\Rightarrow R=\frac{45}{90*10^{-3}}=500\Omega$

$i=\frac{\epsilon }{R+r}$

$v_R=\frac{\epsilon R}{R+r}$

$\frac{1.25=\frac{\epsilon \left(5\right)}{5+r}}{1=\frac{\epsilon \left(2\right)}{2+r}}$

r = 1

Using above equ : $\epsilon =\frac{3}{2}=\frac{15}{10}V$

The bag has 25-paise, 50-paise and 1 -rupee coins in the ratio 8 : 2 : 7. Now this is the ratio of the number of coins while the total value of the coins has been given. So we first need to convert coins into value.

Let the coins be 8x, 2x and 7x respectively.

8x coins of 1/4 rupee each = value is Rs. 2x.

2x coins of 1/2 rupee each = value is Rs. x.

7x coins of 1 rupee each = value is Rs. 7x.

Now

2x + x + 7x = 770

10x = 770 x = 77

Number of 25 paise and 50 coins = 10 x = 770 coins

from newtan's law of cooling :

$\frac{dT}{dt}=-k\left(T-T_0\right)$

Using average value :

$\frac{75-65}{5}=-k\left[\frac{\left(75+65\right)2}{2}-25\right]$

$\frac{\left(65-T\right)}{5}=-k\left[\frac{\left(65+T\right)}{2}-25\right]$

$\Rightarrow \frac{10}{65-T}=\frac{45}{\frac{65+T}{2}-25}$

$\Rightarrow T=57°C$