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Course wise placements for Sasurie College of Arts and Science have not been released yet. However, students can refer to the overall number of students placed and the highest salary given in the table below:

Let $=\int \frac{cos2xdx}{{(sinx+cosx)}^2}$

$=\int \frac{co{s}^{2}x-si{n}^{2}x}{{(sinx+cosx)}^2}dx\{?cos2x=co{s}^{2}x-si{n}^{2}x$

$=\int \frac{(cosx+sinx)(cosx-sinx)}{{(sinx+cosx)}^2}dx\{?a^2-b^2=(x+b)(x-b)$

$=\int \frac{(cosx-sinx)}{sinx+cosx.}dx$

$=log|sinx+cosx|+c\{{\displaystyle \int \frac{f^{\prime }(x)}{f(x)}dx=log|f(x)|+c}$

So, option B is correct.

(a) For n = 4, No. of sub-shells = (l = 0, l = 1, l = 2, l = 3) = 4.

(b) Total number of orbitals which can be present = n² = 4² = 16.
Each orbital can have an electron with m_s = – 1/2. Total no. of electrons with m, = – 1/2 is 16.

The BBA course offered by Sasurie College of Arts and Science BBA runs for a duration of three years. During this course duration, students get to study various subjects related to the field of Management. Some of these subjects are Human Resources Management, Financial Management,  Cost Accounting, Basics of Marketing, and more. Those who wish to pursue this course can submit their application on the university's official website.

Telangana NEET counselling 2025 process comprises various steps. It is being conducted by Kaloji Narayana Rao University of Health Sciences, Telangana. The steps by steps process is given below: 

Registration

Choice Filling & Locking

Provisional Allotment

Final Allotment 

Merit List

Reporting to the college

Let $=\int \frac{dx}{e^x+e^{-x}}$

$=\int \frac{dx}{e^x+\frac{1}{e^x.}}=\int \frac{e^{x}dx}{e^x·e^x+1}$

$=\int \frac{e^{x}dx}{e^{2x}+1}$

Putting e^x = t

e^xdx = dt.

$\therefore =\int \frac{dt}{t^2+1.}$

= tan^{- 1} t + c

= tan^{- 1} (e^x) + c

therefore, Option A is correct.

(a) P (Z=15) : [Ne]¹⁰3s²3p³  No. of unpaired electrons = 3
(b) Si (Z=14) : [Ne]¹⁰3s²3p²  No. of unpaired electrons = 2
(c) Cr (Z=24): [Ar]¹⁸4s¹3d⁵  No. of unpaired electrons = 6
(d) Fe (Z=26): [Ar]¹⁸4s²3d⁶  No. of unpaired electrons = 4
(e) Kr (Z=36) : [Ar]¹⁸4s²3d¹⁰4p⁶  No. of unpaired electrons = Nil.

This is a short answer type question as classified in NCERT Exemplar

In presence of (Ph-CO-O)₂it lead to CH₃-CH₂-CH₂Br as the reaction undergo by free radical mechanism

However in absence of (Ph-CO-O)₂ the reaction undergo by carbocation intermediate thus lead to give CH₃-CHBr-CH₃

To be able to enroll in the Sasurie College of Arts and Science BBA program, students must follow the below-mentioned steps:

1. Ensure that all the eligibility conditions are met.

2. Students who are eligible can apply by visiting the institute's official website.

3. If allocated a seat, report for the document verification process.

4. Pay the seat acceptance fees.

At Sasurie College of Arts and Science, there are a total of 50 seats available for the BBA course. These seats are allocated based on merit obtained by students in the qualifying examination. Students must note that this seat intake is taken from official sources. However, it is subject to change on a yearly basis at the discretion of concerned authorities.

Configuration of the two elements are:

Al (Z = 13): [Ne]¹⁰3s²3p¹

Si (Z = 14): [Ne]¹⁰3s²3p²
The unpaired electrons in silicon (Si) will experience more effective nuclear charge because the atomic number of the element Si is more than that of Al.

Various companies visit Sasurie College of Arts and Science for placements some of them are given below:

Let $I={\displaystyle {\int }_0^1{e}^{2-3x}dx}$

We know that,

${\displaystyle {\int }_a^{b}f\left(x\right)}dx=\left(b-a\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\left(a\right)+f\left(a+h\right)+...+f\left(a\left(n-1\right)h\right)\right]$

Where, $h=\frac{b-a}{n}$

Here, $a=0,b=1$ and $f\left(x\right)e^{2-3x}$

$\Rightarrow h=\frac{1-0}{n}=\frac{1}{n}$

$\therefore {\displaystyle {\int }_0^1{e}^{2-3x}dx}=\left(1-0\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\left(0\right)+f\left(0+h\right)+...+f\left(0+\left(n-1\right)h\right)\right]$

$\begin{array}{l}=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2+e^{2-3x}+...+e^{2-3\left(n-1\right)h}\right]=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{1+e^{-3h}+e^{-6h}+e^{-9h}+...+e^{-3\left(n-1\right)h}\right\}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-{\left(e^{-3h}\right)}^n}{1-\left(e^{-3h}\right)}\right\}\right]=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-e^{\frac{-3}{n}n}}{1-e^{\frac{-3}{n}}}\right\}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2\left(1-e^{-3}\right)}{1-e^{\frac{-3}{n}}}\right]=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{1}{e^{\frac{-3}{n}}-1}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2\left(1-e^{-3}\right)}{1-e^{\frac{-3}{n}}}\right]=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{1}{e^{\frac{-3}{n}}-1}\right]\end{array}$

$\begin{array}{l}=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\left(-\frac{1}{3}\right)\left[\frac{-\frac{3}{n}}{e^{\frac{-3}{n}}-1}\right]=\frac{e^2\left(e^{-3}-1\right)}{3}\underset{n\to \infty }{lim}\left[\frac{-\frac{3}{n}}{e^{\frac{-3}{n}}-1}\right]\\ =\frac{-e^2\left(e^{-3}-1\right)}{3}(1)\\ =\frac{-e^{-1}+e^2}{3}\\ =\frac{1}{3}\left(e^2-\frac{1}{e}\right)\end{array}$

This is a short answer type question as classified in NCERT Exemplar

The total fee for BTech courses at Sasurie College of Engineering includes multiple components, such as tuition fees, admission fees, and various other charges. The total tuition fee for a B.Tech. course ranges from INR 1.6 Lacs to INR 2 lakh. The fee amount mentioned is sourced from the official website/sanctioning body and is subject to change. Therefore, it is indicative. 

The average package for Sasurie College of Arts and Science is not out yet. However students can refer to the highest and average package provided in the table below:

Greater the penetration of the electron present in a particular orbital towards the nucleus more will be the magnitude of the effective nuclear charge. Based upon this

(i) 2s orbital is closer to the nucleus than 3s orbital and hence it will experience more effective nuclear charge.

(ii) 4d orbital will experience more effective nuclear charge due to its closer proximity to 4f orbital.

(iii) 3p orbital will experience more effective nuclear charge as it is closer to the nucleus

Kindly go through the solution