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The highest package offered at Sasmira's Institute of Management Studies and Research placement for MBA in 2025 stood at INR 24 LPA. Check out the table below to know more about company-wise packages offered during the Sasmira's Institute of Management Studies and Research placement in 2025:

$\begin{array}{l}{\displaystyle \int x}log2xdx=log2x·{\displaystyle \int x}dx-\int \frac{d}{dx}log2x{\displaystyle \int x}dxdx\\ =\frac{x^2}{2}·log2x-\int \frac{1}{2x}\times \frac{d(2x)}{dx}\times \frac{x^2}{2}dx+C\\ =\frac{x^2}{2}·log2x-\int \frac{1}{2x}\times 2\times \frac{x^2}{2}{1}^{dx}+C\\ =\frac{x^2}{2}log2x-\frac{1}{2}{\displaystyle \int x}dx+c\\ =\frac{x^2}{2}log2x-\frac{1}{2}·\frac{x^2}{2}+c\\ =\frac{x^2}{2}log2x-\frac{x^2}{4}+c\end{array}$

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:

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

Mumbai University released its first UG merit list for the 2025 session on May?27,?2025.

For N.R.?Bhagat College (Arts, Science & Commerce), the closing rank in the first round was approximately 833. This means candidates within the top ~833 in the merit list secured a B.A. seat in this round .

$\begin{array}{l}{\displaystyle \int x}logxdx=logx{\displaystyle \int x}dx-\int \frac{d}{dx}logx·{\displaystyle \int x}dxdx\\ =logx\times \frac{x^2}{2}-\int \frac{1}{x}\times \frac{x^2}{2}dx\\ =\frac{x^2}{2}·logx-\frac{1}{2}{\displaystyle \int x}dx\\ =\frac{x^2}{2}logx-\frac{1}{2}\times \frac{x^2}{2}+c\\ =\frac{x^2}{2}logx-\frac{x^2}{4}+c.\end{array}$

Yes, a gerund phrase (gerund + modifiers/objects) can act as the subject of a sentence.

Example:

• Eating too must sugar is unhealthy.
• Running every morning keeps me healthy.
• Learning new languages opens up many opportunities.

Answer – (b, c)

Explanation – According to the relation that p/q=r/s if galvanometer shows no deflection. In the above equation $\frac{R1}{R2}=\frac{R3}{R4}$ the value of R4 depends upon R1 and R2 . if the value of R1 and R2 will be feeble the value of R4 would be affected.

$\begin{array}{l}\int x^2{e}^{x}dx=x^2\int e^{x}dx-\int \frac{d{x}^2}{dx}\int e^{x}dxdx\\ =x^2·e^x-{\displaystyle \int 2}x{e}^{x}dx\\ =x^2{e}^x-2[x\int e^{x}dx-\int \frac{dx}{dx}\int e^{x}dxdx]\\ =x^2{e}^x-2\left[x{e}^x-\int e^{x}dx\right]\\ =x^2{e}^x-2x{e}^x+2e^x+c\\ =e^x\left[x^2-2x+2\right]+c\end{array}$

Yes, Siva Sivani Degree College does offer management quota seats for the BBA course. To get the management quota seats. the institute follows the rules and regulations according to Osmania University. Moreover, the number of seats assigned or percentage is not mentioned on the official website of the college. 

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: 

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

V_rms= $\sqrt[]{\frac{3RT}{M}}$

V_rms= $\sqrt[]{T}$

$\frac{V_{rms1}}{V_{rms2}}=\sqrt[]{\frac{T_1}{T_2}}$

T₁=27^oC= 27+273=300K

T₂=127^oC= 127+273= 400K

$\frac{100}{V_{rms2}}=\sqrt[]{\frac{300}{400}}=\frac{\sqrt[]{3}}{2}$

V_rms²= $\frac{2\times 100}{\sqrt[]{3}}=\frac{200}{\sqrt[]{3}}m/s$

$\begin{array}{l}{\displaystyle \int x}sin3xdx=x{\displaystyle \int sin}3xdx-\int \frac{dx}{dx}{\displaystyle \int sin}3xdxdx.\\ =-x\frac{cos3x}{3}+\int \frac{cos3x}{3}\\ =\frac{-xcos3x}{3}+\frac{sin3x}{3\times 3}+c\\ =-\frac{x}{3}cos3x+\frac{1}{9}sin3x+c\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- according to the relation $\rho$ = $\frac{m}{ne2t}$ also T $\propto \frac{1}{t}$

$\begin{array}{l}{\displaystyle \int x}sinxdx=x{\displaystyle \int sin}dx-{\displaystyle \int \frac{d}{dx}x{\displaystyle \int sinx}dxdx}\\ =x(-cosx)-{\displaystyle \int (}-cosx)dx\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (a, d)

Explanation- potential drop across AB is independent to R'  also when we decrease the value of R the current will  also very large. According to the relation I=e/R+r

$\begin{array}{l}\frac{1}{x(x^2+1)}=\frac{x}{x^2(x^2+1)}\\ Putting,x^2=t\Rightarrow 2xdx=dt\end{array}$

$\begin{array}{l}{\displaystyle \int \frac{dx}{x(x^2+1)}}={\displaystyle \int \frac{xdx}{x^2(x^2+1)}=\frac{1}{2}}{\displaystyle \int \frac{dt}{t(t+1)}}\\ =\frac{1}{2}{\displaystyle \int \frac{(t+1)-t}{t(t+1)}}dt\\ =\frac{1}{2}\left\{{\displaystyle \int \frac{dt}{t}}-{\displaystyle \int \frac{dt}{t+1}}\right\}\\ =\frac{1}{2}\left[log\left|t\right|-log\left|t+1\right|\right]+\; C\\ =\frac{1}{2}log\left|x^2\right|-\frac{1}{2}log\left|x^2+1\right|+\; C\\ =\frac{2}{2}log\left|x\right|-\frac{1}{2}log\left|x^2+1\right|+c\\ =log\left|x\right|-\frac{1}{2}log\left|x^2+1\right|+c\\ So,option\left(A\right)iscorrect.\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (b, d)

Explanation-Algebraic sum of the currents flowing towards any point in an electric network is zero. It also tell us about the conservation of charge.