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The recent cutoff for the BPharm course at Sabarmati University has not been released. However, it is expected that the latest cutoff will soon be released. To get an idea about how much a competition can be increased and decreased for admission to the BPharm course in the upcoming years, one can check out the previous year's cutoff:

Hi.

The admissions counsellor / admission office at Sabarmati University for B.Pharm admissions:

Phone Numbers

You can call these admission enquiry numbers:

• +91 95952 56677 – Admission queries 

• +91 90819 11124 – General contact (also useful for admissions) 

Try calling during office hours (typically 10:00 AM – 5:00 PM IST) for faster response.

Sabarmati University, Ahmedabad, requires several documents to verify the eligibility and identity for BPharm courses. The list of the documents that are required at the time of filling out the application form is as follows:

• Class 10 and 12 Mark Sheets 
• Transfer Certificate
• Migration Certificate
• Character Certificate
• Caste or Category Certificate
• Entrance exam Scorecard
• Domicile Certificate
• Passport-sized Photographs
• Identification Proof

$\mathrm{}a\mathrm{c}\mathrm{o}\mathrm{s}?\theta =b\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)=c\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)=k$

$a=\frac{k}{\mathrm{c}\mathrm{o}\mathrm{s}?\theta },b=\frac{k}{\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)},c=\frac{k}{\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)}$

$ab+bc+ca=k^2\frac{\left[\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)+\mathrm{c}\mathrm{o}\mathrm{s}?\theta +\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)\right]}{\mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)\mathrm{c}\mathrm{o}\mathrm{s}?\theta \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)}$

$={\mathrm{k}}^2\left[\frac{\mathrm{c}\mathrm{o}\mathrm{s}?\theta +2\mathrm{c}\mathrm{o}\mathrm{s}?(\theta +\pi )\cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\frac{\pi }{3}\right)}{\mathrm{c}\mathrm{o}\mathrm{s}?\theta \cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)\cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)}\right]$

$={\mathrm{k}}^2\left[\frac{\mathrm{c}\mathrm{o}\mathrm{s}?\theta -2\mathrm{c}\mathrm{o}\mathrm{s}?\theta \cdot \frac{1}{2}}{\mathrm{c}\mathrm{o}\mathrm{s}?\theta \cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{2\pi }{3}\right)\cdot \mathrm{c}\mathrm{o}\mathrm{s}?\left(\theta +\frac{4\pi }{3}\right)}\right]=0$

$\mathrm{c}\mathrm{o}\mathrm{s}??=\frac{(a\hat{i}+b\hat{j}+c\hat{k})\cdot (b\hat{i}+c\hat{j}+a\hat{k})}{\sqrt[]{a^2+b^2+c^2}\cdot \sqrt[]{b^2+c^2+a^2}}=ab+bc+ca=0$

$?=\frac{\pi }{2}$

The selection process for the BPharm course involves applicants securing a valid sscorein the GUJCET entrance exam. Moreover, to get selected, applicants must complete their basic eligibility criteria first. The minimum eligibility criteria is applicants must complete their Class 12 from a Science stream PCB/ PCM with decent scores.

The table below provides some details about the top-ranked private MBA colleges in Tamil Nadu.

Disclaimer: This information is sourced from the official website and may vary.

$\left|z_1-1\right|=\mathrm{R}\mathrm{e}?\left(z_1\right)$ Let $z_1{x}_1+i{y}_1$and $z_2=x_2+i{y}_2$

${\left({\mathrm{x}}_1-1\right)}^2+{\mathrm{y}}_1^2={\mathrm{x}}_1^2$

${\mathrm{y}}_1^2-2{\mathrm{x}}_1+1=0$

$\left|z_2-1\right|=\mathrm{R}\mathrm{e}?\left(z_2\right)$

${\left(x_2-1\right)}^2+y_2^2=x_2^2$

${\mathrm{y}}_2^2-2{\mathrm{x}}_2+1=0$

$\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\left({\mathrm{y}}_1+{\mathrm{y}}_2\right)=2\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)$

${\mathrm{y}}_1+{\mathrm{y}}_2=2\left(\frac{{\mathrm{x}}_1-{\mathrm{x}}_2}{{\mathrm{y}}_1-{\mathrm{y}}_2}\right)$

$\mathrm{a}\mathrm{r}\mathrm{g}?\left(z_1-z_2\right)=\frac{\pi }{6}$

${\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}?\left(\frac{{\mathrm{y}}_1-{\mathrm{y}}_2}{{\mathrm{x}}_1-{\mathrm{x}}_2}\right)=\frac{\pi }{6}$

$\frac{{\mathrm{y}}_1-{\mathrm{y}}_2}{{\mathrm{x}}_1-{\mathrm{x}}_2}=\frac{1}{\sqrt[]{3}}$

$\therefore {\mathrm{y}}_1+{\mathrm{y}}_2=2\sqrt[]{3}$

$\Rightarrow \mathrm{}\mathrm{I}\mathrm{m}?\left(z_1+z_2\right)=2\sqrt[]{3}$

The table below provides some details about the top-ranked private MBA colleges in Tamil Nadu.

Disclaimer: This information is sourced from the official website and may vary.

$\mathrm{}488=\frac{n}{2}\left. [2\left(\frac{100}{5}\right)+(n-1)\left(\frac{2}{5}\right)\right]$  

$488=\frac{n}{2}(101-n)$

$\Rightarrow \mathrm{}{\mathrm{n}}^2-101\mathrm{n}+2440=0$

$\Rightarrow \mathrm{n}=61$ or 40

For $n=40\mathrm{}\Rightarrow \mathrm{}{T}_n>0$  

For $\mathrm{n}=61\mathrm{}\Rightarrow \mathrm{}{\mathrm{T}}_{\mathrm{n}}<0$

${\mathrm{T}}_{\mathrm{n}}=\frac{100}{5}+(61-1)\left(-\frac{2}{5}\right)=-4$

Number of mole of $\mathrm{X}=\frac{6.022\times {10}^{22}}{6.022\times {10}^{23}}=\frac{10}{\text{Molar mass of}\mathrm{X}}$

So molar mass of $X=100\mathrm{g}$

Molarity $=\frac{5}{100\times 2}=0.025\mathrm{M}$

Ans. $=0.025\mathrm{M}$

$\mathrm{M}=25\times {10}^{-3}$

So $P=25$

According to Faraday law

  $\mathrm{W}=\mathrm{Z}\mathrm{I}\mathrm{t}\times \eta \mathrm{}$ Where $\eta =$ efficiency

or $W=\frac{E}{96500}\times 1\times t\times \eta$

Putting values

$.104=\frac{(52/3)\times 2\times 8\times 60\times \eta }{96500}\Rightarrow \eta =0.6$

So percentage efficiency $=60\mathrm{\%}$

Check the comparison listed below between Manipal University Dubai and IMT Dubai. This will help students in gaining more clarity on these universities respectively: