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The MBA course at Sasmira's Institute of Management Studies and Research is considered one of the best. The MBA programme recorded decent placement offers. Check out the table below to know about the highest package and companies at Sasmira's Institute of Management Studies and Research placement for MBA in 2025: 

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

V₁=2L ,V₂=3L

µ₁ = 4.0and µ₂ =5.2

p₁= 1.00 atm and p₂ = 2.00 atm

p₁V₁= µ₁RT₁

p₂V₂= µ₂RT₂

when the partition is removed the gases get mixed without any loss of energy . the mixture now attains a common equilibrium pressure and total volume of the system is sum of the volume of individual chambers V₁ and V₂

$\mu ={\mu }_1+{\mu }_2$ , V =V₁+V₂

From the kinetic theory of gases pV=2/3 E

For mole 1  ,P₁V₁= 2/3 ${\mu }_1{E}_1$

For mole 2  , P₂V₂= 2/3 ${\mu }_2{E}_2$

Total energy is ( ${\mu }_1{E}_1+{\mu }_2{E}_2$ )= 3/2 ( $P_1{V}_1+P_2{V}_2$ )

PV==2/3E_total = 2/3 $\mu E_{permole}$

P( $V_1+V_2$ )= $\frac{2}{3}\times \frac{3}{2}(p_1{V}_1+p_2{V}_2)$

P= $\frac{P_1{V}_1+P_2{V}_2}{V_1+V_2}$ = $\frac{1\times 2+2\times 3}{2+3}atm$

P=8/5 =1.6atm

Yes. In UPSC IES 2025, there is a negative marking. As per the official notification, there are four alternatives for the answers to every question. For each question for which a wrong answer has been given by the candidate, one-third of the marks assigned to that question will be deducted as a penalty.

The companies that come for recruitment represent a variety of sectors – including FMCG, BFSI, Consumer Durables, Consultancy, IT, and Media – offering appealing profiles and packages to students.

Some the top recruiters for the Sasmira's Institute of Management Studies and Research are listed below:

The Sasmira's Institute of Management Studies and Research provides good placements to its graduating students. Check out the table below to know the key highlights of the Sasmira's Institute of Management Studies and Research placement for MBA in 2025: :

$LetI={{\displaystyle \int }}^{\text{?}}{\left(si{n}^{-1}x\right)}^{2}dx$

Putting sin^-1x =θ=> x = sinθ, dx = cosθdθ.

$So,I={{\displaystyle \int }}^{\text{?}}{\theta }^2\cdot cos\theta d\theta ={\theta }^2{{\displaystyle \int }}^{\text{?}}cos\theta d\theta -{{\displaystyle \int }}^{\text{?}}\frac{d}{d\theta }{\theta }^2{{\displaystyle \int }}^{\text{?}}cos\theta d\theta d\theta .$

$={\theta }^{2}sin\theta -{{\displaystyle \int }}^{\text{?}}2\theta sin\theta d\theta .\\ ={\theta }^{2}sin\theta -2{{\displaystyle \int }}^{\text{?}}\theta sin\theta d\theta .$

$\begin{array}{cc}\hfill \hfill & \hfill ={\theta }^{2}sin\theta -2\left[\theta {{\displaystyle \int }}^{\text{?}}sin\theta d\theta -{{\displaystyle \int }}^{\text{?}}\frac{d\theta }{d\theta }{{\displaystyle \int }}^{\text{?}}sin\theta d\theta d\theta \right]\hfill \end{array}\\$

$={\theta }^{2}sin\theta -2\left[\theta (-cos\theta )-{{\displaystyle \int }}^{\text{?}}(-cos\theta )d\theta \right]\\$

$={\theta }^{2}sin\theta +2\theta cos\theta -2sin\theta +C$

$Let,I={\displaystyle \int x}co{s}^{-1}xdx$

Putting cos^-1 x =θ=> x = cosθ=>dx = - sinθdθ.

$So,I={{\displaystyle \int }}^{\text{?}}cos\theta \times \theta \cdot (-sin\theta )d\theta .$

$=-\frac{1}{2}{{\displaystyle \int }}^{\text{?}}\theta (2sin\theta cos\theta )d\theta$ {θsin 2θ = 2 sinθ cosθ}

$\begin{array}{l}=-\frac{1}{2}{{\displaystyle \int }}^{\text{?}}\theta sin2\theta d\theta .\\ =-\frac{1}{2}\left[\theta {{\displaystyle \int }}^{\text{?}}sin2\theta d\theta -{{\displaystyle \int }}^{\text{?}}\frac{d\theta }{d\theta }{{\displaystyle \int }}^{\text{?}}sin2\theta d\theta d\theta \right]\end{array}$

$\begin{array}{l}=-\frac{1}{2}\left[\frac{\theta (-cos2\theta )}{\theta }-{{\displaystyle \int }}^{\text{?}}\left(\frac{(-cos2\theta )}{2}\right)d\theta \right]\\ =\frac{\theta }{4}cos2\theta -\frac{1}{4}\cdot \frac{sin2\theta }{2}+C\\ =\frac{\theta }{4}cos2\theta -\frac{1}{8}(2sin\theta cos\theta )+C.\\ =\frac{\theta }{4}\left[2co{s}^2\theta -1\right]-\frac{1}{4}sin\theta cos\theta +C. \left\{?cos2\theta =2co{s}^2\theta -1\right\}.\\ \end{array}$$\begin{array}{l}\end{array}$

Both these institutes are decent options for BBA in all the aspects, such as its placements and fee structure. But most students choose DTU over NSUT because of the huge campus and big crowd. However, selecting a college depends completely on the choice of the students and their requirements. For more reviews on BBA from DTU, check out the verified students' reviews on Shiksha.

To identify gerund in a sentence, look for the word ending with '-ing', which is acting as a noun (not a verb or adjective).

Example:

• Painting is fun.
• I am trying to learn German language.
• She is practicing for tomorrow's cricket match.

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

Mean free path l=1/ $\sqrt[]{2}{d}^{2}n$

So n= number of molecules /volume

d = diameter of the molecule

l $\propto \frac{1}{d^2}$

d₁=1A^o, d₂=2A^o

l $\propto \frac{1}{d^2}$

$\frac{l_1}{l_2}={\left(\frac{d_2}{d_1}\right)}^2={\left(\frac{2}{1}\right)}^2=\frac{4}{1}$

l₁:l₂=4:1

$\begin{array}{l}{\displaystyle \int x}ta{n}^{-1}xdx=ta{n}^{-1}x{\displaystyle \int x}dx-\int \frac{d}{dx}ta{n}^{-1}x·{\displaystyle \int xdx.dx}\\ =ta{n}^{-1}x·\frac{x^2}{2}-\int \frac{1}{1+x^2}·\frac{x^2}{2}dx.\\ =\frac{x^2}{2}ta{n}^{-1}x-\frac{1}{2}\int \frac{x^2}{1+x^2}dx\\ =\frac{x^2}{2}ta{n}^{-1}x-\frac{1}{2}\int \frac{\left(1+x^2\right)-1}{1+x^2}dx\\ =\frac{x^2}{2}·ta{n}^{-1}x-\frac{1}{2}\left[{\displaystyle \int \frac{\left(1+x^2\right)}{1+x^2}}dx-{\displaystyle \int \frac{dx}{1+x^2}}\right]\\ =\frac{x^2}{2}·ta{n}^{-1}x-\frac{1}{2}\left[{\displaystyle \int d}x-ta{n}^{-1}x\right]\\ =\frac{x^2}{2}ta{n}^{-1}x-\frac{1}{2}\left[x-ta{n}^{-1}x\right]+C.\\ =\frac{1}{2}\left[x^{2}ta{n}^{-1}x-x+ta{n}^{-1}x\right]+C.\\ =\frac{1}{2}\left[\left(x^2+1\right)ta{n}^{-1}x·-x\right]+C\end{array}$

The UPSC (ESE) IES 2025 prelims exam carry a total of 500 marks. 
Paper 1: 200 marks
Paper 2: 300 Marks

In the field of Management,  DTU offers three year BBA via the University School of Management and Entrepreneurship (USME). Divided into six semesters of six months each, admission to BBA programme offered by DTU is based on merit scores obtained by the students in the best of four subjects in Class 12. Students, however, must have completed their Class 12 with 50% aggregate (with English as a subject) to be eligible for the course.

The Sasmira's Institute of Management Studies and Research provides decent placement packages to its graduating MBA students. The highest package offered at Sasmira's Institute of Management Studies and Research placement for MBA in 2025 stood at INR 24 LPA. Refer to the table to know the same:

Kindly go through the solution

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

Oxygen gas having 5 degrees of freedom

Energy per mole of the gas =5/2RT

For 2 moles of the gas total internal energy =2 $*$ 5/2RT=5RT

Neon is a monoatomic gas having 3 degrees of freedom

Energy per mole =3/2RT

We have 4 moles of Ne

Energy = 4 $*$ 3/2RT=6RT

Total energy =5RT+6RT=11RT

$\begin{array}{l}\int x^2·logx·dx=logx·\int x^{2}dx-\int \frac{d}{dx}logx·\int x^{2}dxdx\\ =logx·\frac{x^3}{3}-\int \frac{1}{x}·\frac{x^3}{3}dx\\ =\frac{x^3}{3}·logx-\int \frac{x^2}{3}dx\\ =\frac{x^3}{3}·logx-\frac{1}{3}\times \frac{x^3}{3}+c\\ =\frac{x^3}{3}logx-\frac{x^3}{9}+c.\end{array}$