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To be eligible for admission to Padala Rama Reddi Law College for the LLB, candidates must have passed the qualifying examination of Intermediate level (10+2) or equivalent in a regular course with a minimum of 45% marks and also clear the LLB common entrance examination (LAWCET). For the 3-year LL.B program, a graduate with any discipline who has passed the qualifying examination (10+2+3) or equivalent in a regular course is eligible.

The components a candidate needs to pay for in the LLB course at Padala Rama Reddi Law College are:

• Tuition fees


• Examination fees


• Library fees


• Laboratory fees


• Sports and games fees


• Cultural activities fees


• Student welfare fund


• University fees

Note: The exact components and their respective fees may vary, and it's always best to check with the college administration or official website for the most up-to-date information.

To get admission to the LLB programme at Padala Rama Reddi Law College, you need to have the following documents:
1. Qualifying examination certificate (10+2 or equivalent)
2. LAW CET scorecard
3. Transfer certificate
4. Migration certificate
5. Date of birth certificate
6. Caste certificate (if applicable)
7. Income certificate (if applicable)

Shri Ram College of Commerce accepts CUET PG scores for admission to MCom. The cutoff has been released for the second round. The cutoff list is released category-wise for All India quota candidates. The cutoff for the General All India category candidates was closed at 190 in the second round. The cutoff for the OBC AI category candidates was closed at 175. For the SC AI category candidates, the closing score for MCom was closed at 125. For ST AI and EWS AI category candidates, the cutoff score was closed at 139 and  157, respectively. 

SSSS has an impressive placement record. The placement statistics for 2025 are given in the table below:

SSSS has announced its 2025 placement report. Have a look at the data tabulated below to know more:

$U_B=\frac{1}{2}L{l}^2............\left(i\right)$

P_R = l²R

=> 640 = (8)² R

R = $10\Omega$

from (i) 64 = $\frac{1}{2}\times L\times {\left(8\right)}^2$

L = 2H

$l=\frac{L}{R}=\frac{2}{10}=0.2second$        

The number of students placed during SSSS placements was 39. The table below presents the same:

$\frac{Q_H}{Q_L}=\frac{T_H}{T_L}$

$\Rightarrow Q_L=\frac{T_L}{T_H-T_L}\times \left(Q_H-Q_L\right)$

$\Rightarrow \frac{d{Q}_L}{dt}=\frac{T_L}{T_H-T_L}.\frac{d\left(Q_H-Q_L\right)}{dt}$

$=\frac{273-10}{25-\left(-10\right)}\times 35$

$=263J{s}^{-1}$

SSSS provides good  packages to its students. The placement package recorded in SSSS placements 2025 is given in the table below:

SSSS provides great placements to its students. Check out the table below to know about the SSSS placement statistics 2025:

$R_{eq}=0.6\Omega +\frac{\frac{4\times 12}{4+12}\times \frac{6\times 8}{6+8}}{\frac{4\times 12}{4+12}+\frac{6\times 8}{6+8}}\Omega$

$=0.6\Omega +\frac{3\times \frac{48}{14}}{3+\frac{48}{14}}\Omega$

$=0.6\Omega +\frac{\left(144/14\right)\Omega }{\left(\frac{90}{14}\right)}$

$\begin{array}{l}=0.6\Omega +1.6\Omega \\ =2.2\Omega \end{array}$

p = $\frac{v^2}{R_{eq}}=\frac{{\left(2.2V\right)}^2}{2.2\Omega }=2.2W$

The placement rate recorded during SSSS placements 2025 has not been released yet. Have a look at the table belwo to know more about SSSS placement statistics: 

The overall highest package offered during SSSS placements 2025 stood at INR 8 LPA. Check out the table below to know more about the highest package recorded at SSSS:

Passive voice is a type of sentence in which the subject receives the action instead of becoming the doer. Here, the focus will be on the object, and there will be a little focus on the subject or doer. I.e. A mango was eaten by Ram. In this sentence, the emphasis is on Mango instead of Ram. 

The placements at SSSS are good. The following table presents the placement statistics for SSSS:

$B_{axis}=\frac{{\mu }_{0}Nl{a}^2}{2{\left(a^2+r^2\right)}^{3/2}}$

$B_{cent}=\frac{{\mu }_{0}NI}{2a}$

Fractional change = $\frac{B_{centre}-B_{axis}}{B_{centre}}$

$=1-\frac{B_{axis}}{B_{centre}}$

$=1-\left(1-\frac{3r^2}{2a^2}\right)\left[r<<a\right]$

$=\frac{3r^2}{2a^2}$

F. B. D of Beaker

By NLM2         

$\begin{array}{l}f=ma=m{\omega }^{2}R\\ \Rightarrow m{\omega }^{2}R\le \mu N\\ R\le \mu g/{\omega }^2\end{array}$

Let’s say $\left|\overrightarrow{OA}\right|=\left|\overrightarrow{OB}\right|=\left|\overrightarrow{OC}\right|=\mathcal{l}$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}$

$\overrightarrow{OA}=\mathcal{l}cos30°\widehat{i}+\mathcal{l}sin30°\widehat{j}$

$\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}$

= $\left[\mathcal{l}cos30°\widehat{i}+\mathcal{l}sin30°\widehat{j}\right]+\left[\mathcal{l}cos60°\widehat{i}+\mathcal{l}sin60°\left(-\widehat{j}\right)\right]-\left[\mathcal{l}cos45°\left(-\widehat{i}\right)+\mathcal{l}sin45°\left(+\widehat{j}\right)\right]$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$

$=\mathcal{l}\left(\frac{\sqrt[]{3}}{2}+\frac{1}{2}+\frac{1}{\sqrt[]{2}}\right)\widehat{i}+\mathcal{l}\left[\frac{1}{2}-\frac{\sqrt[]{3}}{2}-\frac{1}{\sqrt[]{2}}\right]\widehat{j}$

$tan\theta =\frac{\mathcal{l}\left[\frac{1}{2}-\frac{\sqrt[]{3}}{2}-\frac{\sqrt[]{2}}{2}\right]}{\mathcal{l}\left[\frac{\sqrt[]{3}}{2}+\frac{1}{2}+\frac{\sqrt[]{2}}{2}\right]}=\frac{1-\sqrt[]{3}-\sqrt[]{2}}{1+\sqrt[]{3}+\sqrt[]{2}}$