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Chandigarh University is best university ever. The university is really growing fast and it has always been the first choice of students. Almost students dream taking admission here, not because the university is good in curriculum, but also because the university offers a top-notch quality education, a wide range of specialisations and students get to learn a lot practical experience, even as per the change in demand and trends of industry, several workshops, seminars and guest lectures are also organised by the university so that students can interact with expert personalities of market and can learn from them. 

Positive degree is the most basic form of adjective that we use for a noun or pronoun, let's see how it is different from Comparative and Superlative degrees. The comparative degree compares two things by adding 'er' to the adjective and the superlative degree compares three or more things by adding 'est' to the adjective. For example:

The Positive Degree in English grammar is the most basic and simple form of an adjective or adverb used to describe a noun or a pronoun. It is also called the normal form of an adjective or adverb that describes a quality without comparing it to anything else.

Aniline has higher viscosity due to intermolecular H -bonding.

R shunt Resistance

Given

CD = 3m

CF = 2m

Current through potentiometer wire

I = $\frac{v}{R_0}=\frac{VA}{\rho L}$ $\frac{}{{}_{}}\frac{}{}$ (1)

A Area of potentiometer wire

$$  $\rho \to$ Resistivity of the wire

L Length of the potentiometer wire

Case 1 When 8 $\Omega$ $$ shunt is bal aced at 3m length

$V_{CD}=I{R}_{CD}=\frac{V}{\rho L}A\times \frac{\rho CF}{A}$

$V_{CD}=\frac{V\mathcal{l}}{L}=\frac{V\times 3}{L}$

VCD = VAB = E – I1r

= $-\frac{E}{R+r}r=\frac{V}{L}\times 3$ $\frac{}{}\frac{}{}$

$$ $\Rightarrow E\left[1-\frac{r}{8+r}\right]=\frac{V}{L}\times 3$  (2)

Case 2 When 4 $$ $\Omega$ shunt is balanced at 2m length

$V_{CF}=I{R}_{CF}=\frac{V}{\rho L}A\times \frac{\rho CF}{A}=\frac{V\times 2}{L}$

$V_{CF}=V_{AB}=E-I_{2}r$

= $E-\frac{E}{R+r}r$ $\frac{}{}$

= $E\left[1-\frac{r}{4+r}\right]$ $\frac{}{}$

$\frac{}{}\frac{}{}$  $\Rightarrow E\left(1-\frac{r}{4+r}\right)=\frac{V\times 2}{L}$ (3)

$\left(2\right)÷\left(3\right)$

$\frac{1-\frac{r}{8+r}}{1-\frac{r}{4+r}}=\frac{3}{2}$

$\Rightarrow \left(\frac{8}{8+r}\right)\times \left(\frac{4+8}{4}\right)=\frac{3}{2}$

$\Rightarrow 4\left(4+r\right)=3\left(8+r\right)$

$\Rightarrow 16+4r=24+3r$

r = 24 – 16

r = 8 $\Omega$

We know

Energy on any orbit is given by,

$E_n=\frac{-13.6}{n^2}{z}^2$

${}_{}\frac{}{}$  $\Rightarrow E_1=\frac{-13.6}{12}\times 9$ [n1 = 1] (1)

For 3rd orbit

$$ $E_3=\frac{-13.6}{9}\times$   [n2 = 3]

E3 = 13.6ev

$\Delta E=E_3-E_1$

= 13.6 – (13.6 × 9)

$\Delta E=8\times 13.6ev=photonenergy$

$8\times 13.6ev=\frac{hc}{\lambda }$

$\Rightarrow 8\times 13.6ev=\frac{1242ev}{\lambda }nm$

$\lambda =11.415\times 10^9$

= 114.15 × 1010m

3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

$\frac{}{}$ $\Rightarrow \frac{dy}{D}=\left(2n+1\right)\frac{\lambda }{2}$   [for Dark fringe at P]

n = 0, for first dark fringe

$\frac{dy}{D}=\frac{\lambda }{2}$

$\lambda =\frac{2d}{D}y$

$$ $=\frac{2d}{D}\times \frac{d}{2}[y=\frac{d}{2},Given$  first dark fringe is observed on the screen directly opposite to one of the slits]

$\lambda =\frac{2\times 0.6\times 0.6}{800\times 2}$

$\lambda =450mm$

In strong ligand field  ${\mathrm{C}\mathrm{o}}^{3+}$ will have ${\mathrm{t}}_{2\mathrm{g}}^6{\mathrm{e}}_{\mathrm{g}}^0$ of configuration and $\mathrm{\Delta }\mathrm{t}=\frac{4}{9}{\mathrm{\Delta }}_0$

for lamp,   $P=\frac{v^2}{R}$

$\Rightarrow R=\frac{25\times 25}{5}=125\Omega$              

$l_{max}=\frac{v}{R}=\frac{25}{125}=\frac{1}{5}A$    

$\Rightarrow I_{max}=\frac{220}{125+R}=\frac{1}{5}$

$\Rightarrow 1100=125+R$        

R = 975  $\Omega$

T₂ = 200 k

$T_1=527°C+273=800K$      

w = 12000 kJ = 12 × 10⁶ J

Q₁ =?

  $x=1-\frac{T_2}{T_1}=\frac{w}{Q_1}=1-\frac{200}{800}=\frac{12\times 10^6}{Q_1}$              

$\Rightarrow \frac{3}{4}=\frac{12\times 10^6}{Q_1}$      

$\Rightarrow Q_1=16\times 10^{6}J$        

$V_T=2000c{m}^3=2000\times 10^{-6}{m}^3\left(Totalvolumeofmixture\right)$

T = 300 K

${}_{}{}^{}{}^{}$  $P_T=100kPa=100\times 10^{3}Pa=10^5$ Pa

mT = 8.3 J/K1 mol1

Now, using Ideal gas equation

$P_{T}V=\left(x_1+x_2\right)RT$

$\Rightarrow 10^5\times 200\times 10^{-6}=\left(x_1+x_2\right)\times 8.3\times 300$

$\frac{x_1}{x_2}=\frac{0.06}{0.02}=3:1$

Is bulky base, so elimination is dominating.

Truth table for Input & Output

Now, Truth Table for option B.

After completing a BTech in Agricultural Engineering course from Central Agricultural University, candidates have various career opportunities in both the government and private sectors. The names of some of the government and private career opportunities after completion of the course are as follows:

• Krishi Bhavan roles
• Agricultural Development Officer (ADO) 
• Field Officer/ Block Technology Manager (BTM)
• Roles in Research Farms and Krishi Vigyan Kendras
• Agri-input companies
• Food processing units
• Agri-tech startups
• Seed production and nurseries, etc.

  $\mathrm{}{\mathrm{C}\mathrm{N}}^{-}\Rightarrow 14$ electrons, so $\mathrm{B}.\mathrm{O}=3$ and diamagnetic.

Given absorb energy = 10.2 eV

$E_f=\frac{-13.6}{n^2}=-3.4$

$n^2=4$

n = 2

While the university offers more than 400 UG and PG courses, its MBA, MSc, and MA are some of the courses which are more popular among students than others. An MBA from Warwick University is a good investment since the University of Warwick employment rate for MBA is 89%,  with an average salary of INR 1.45 Cr for graduates. Thus, pursuing MBA from this university is definitely worth it.