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$\begin{array}{l}I={\displaystyle \int si{n}^{3}x}co{s}^{3}xdx\\ ={\displaystyle \int co{s}^3}xsi{n}^{2}xsinx.dx\\ ={\displaystyle \int co{s}^3}x(1-co{s}^{2}x).sinxdx\\ Putcosx=t\\ \Rightarrow -sin.x.dx=dt\\ I=-{\displaystyle \int t^3}(1-t^2)dt\end{array}$

$\begin{array}{l}=-{\displaystyle \int \left(t^3-t^5\right)}dt=-\left\{\frac{t^4}{4}-\frac{t^6}{6}\right\}+\; C\\ =-\left\{\frac{co{s}^{4}x}{4}-\frac{co{s}^{6}x}{6}\right\}+\; C\\ =\frac{-co{s}^{4}x}{4}+\frac{co{s}^{6}x}{6}+\; C\\ =\frac{1}{6}co{s}^{6}x-\frac{1}{4}co{s}^{4}x+\; C\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(a) F= V $\rho G$

F $\propto \rho$

$F_{4^{0}C}>F_{0^{0}C}$

Hence buoyancy will be less in water at 0⁰C than that in water at 4⁰C.

Prepositions are the words which are used before the objects (nouns or pronouns). Preposition words connect nouns and pronouns with other parts of the sentence. Some of the examples of Prepositions are:

• He cleaned the shelf with a piece of cloth.
• I rushed into the room after seeing lizard.
• He sacrificed his life for his country.

In the above examples, with, into and for words are prepositions.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- (i)The cesium atoms, are situated at the corners of a cube and Cl atom is situated at the centre of the cube. From the given figure, we can analyse that the chlorine atom is at equal distance from all the eight comers of cube where cesium atoms are placed. Thus, due to symmetry the electric field due to all Cs atoms, on Cl atom will cancel out. Hence net electric field at the centre of cube is zero.

(ii) we know, f=qE  E= Kq/r²= Ke/r²

F= e (E)=e (ke/r²)=ke²/r²

Distance= $\sqrt[]{r^2+r^2+r^2}$ = $\sqrt[]{{0.2}^2+{0.2}^2+{0.2}^2}*$ 10^-9m

F=8.99 $*$ 10⁹ (1.6 $*$ 10^-16)²/ (0.346 $*$ 10^-9)²=1.92 $*$ 10^-9N

$\begin{array}{l}I={\displaystyle \int si{n}^3\left(2x+1\right)dx}\\ ={\displaystyle \int si{n}^2\left(2x+1\right).sin}\left(2x+1\right)dx\\ ={\displaystyle \int \left\{1-co{s}^2\left(2x+1\right)\right\}}sin\left(2x+1\right)dx\\ Puttingcos\left(2x+1\right)=t\\ \Rightarrow -2sin\left(2x+1\right)dx=dt\\ \Rightarrow sin\left(2x+1\right)dx=\frac{-dt}{2}\\ I=\frac{-1}{2}{\displaystyle \int \left(1-t^2\right)}dt\\ =\frac{-1}{2}\left\{t-\frac{t^3}{3}\right\}+\; C\\ =\frac{-1}{2}\left\{cos(2x+1)-\frac{co{s}^3(2x+1){}^3}{3}\right\}+\; C\\ =\frac{-cos(2x+1)}{2}+\frac{co{s}^3(2x+1){}^6}{6}+\; C\\ =\frac{-1}{2}cos(2x+1)+\frac{1}{6}co{s}^3(2x+1)+\; C\end{array}$

Though the campus placement packages vary depending on the colleges, the average placement packages for BFTech go from INR 5.5 LPA to 9 LPA. The salary packages increase eventually when you gain relevant experience in the industry and polish your skills.  

$\begin{array}{l}Here,cosAcosB=\frac{1}{2}\{cos(A+B)+cos(A-B)\}\\ I={\displaystyle \int cos2x\left(cos4xcos6x\right)dx}\\ ={\displaystyle \int cos}2x\left[\frac{1}{2}cos(4x+6x)+cos(4x-6x)\right]dx\\ ={\displaystyle \int cos2x\left[\frac{1}{2}(cos10x+cos(-2x))\right]dx}\\ =\frac{1}{2}{\displaystyle \int cos}2xcos10x+cos2xcos(-2x)dx\\ =\frac{1}{2}{\displaystyle \int cos}2xcos10x+cos2xdx\left[\therefore cos\left(-x\right)=cosx\right]\\ =\frac{1}{2}\left[\frac{1}{2}\{cos2x+10x+cos2x-10\}+\left\{\frac{1+cos4x}{2}\right\}\right]dx\\ =\frac{1}{4}{\displaystyle \int cos}12x+cos8x+1+cos4x\cdot dx\\ =\frac{1}{4}[\frac{sin12x}{12}+\frac{sin8x}{8}+x+\frac{cos4x}{4x}]+\; C\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(b) Lowest point for scale A is 30⁰ and lowest point for scale B is 0⁰. Highest point for the scale A is 180⁰ and for scale B is 100⁰

$\frac{{\mathrm{t}}_{\mathrm{A}}-{\mathrm{L}\mathrm{E}\mathrm{P}}_{\mathrm{A}}}{{\mathrm{U}\mathrm{F}\mathrm{P}}_{\mathrm{A}}-{\mathrm{L}\mathrm{F}\mathrm{P}}_{\mathrm{A}}}=\frac{{\mathrm{t}}_{\mathrm{B}}-{\mathrm{L}\mathrm{F}\mathrm{P}}_{\mathrm{B}}}{{\mathrm{U}\mathrm{F}\mathrm{P}}_{\mathrm{B}}-{\mathrm{L}\mathrm{F}\mathrm{P}}_{\mathrm{B}}}$

$\frac{{\mathrm{t}}_{\mathrm{A}}-\mathrm{}30}{180-30}=\frac{{\mathrm{t}}_{\mathrm{B}}-0}{100-0}\mathrm{}\mathrm{}$

$\frac{t_A-30}{150}=\frac{t_B}{100}$ ,LFP= lower fixed point , UFP= upper fixed point $\propto \rho$

Explanation- (i) F= $\frac{Kq2}{r2}$ = 9 $\times$ 10^{9 $\times$} (34.8 $\times$ 10³)²/ (10^-2)²=1.09 $\times$ 10²³N

                      (ii) F= $\frac{Kq2}{r2}$ = 9 $\times$ 10^{9 $\times$} (34.8 $\times$ 10³)²/ (100)²=1.09 $\times$ 10¹⁵N

                      (iii) F= $\frac{Kq2}{r2}$ = 9 $\times$ 10⁹ $\times$ (34.8 $\times$ 10³)²/ (10⁶)²=1.09 $\times$ 10⁷N

$\begin{array}{l}Here,\\ sinAcosB=\frac{1}{2}\left\{sin(A+B)+sin(A-B)\right\}\\ sin3xcos4x=\frac{1}{2}(sin(3x+4x)+sin(3x-4x))\\ Then,\\ {\displaystyle \int sin}3xcos4xdx={\displaystyle \int \frac{1}{2}}[sin(3x+4x)+sin\left(3x-4x\right)dx\\ =\frac{1}{2}{\displaystyle \int sin}(3x+4x)+sin\left(3x-4x\right)dx\\ =\frac{1}{2}{\displaystyle \int sin}7x+sin(-x)dx\\ =\frac{1}{2}\left[-\frac{cos7x}{7}+cosx\right]+c\\ =\frac{-cos7x}{14}+\frac{cosx}{2}+c\end{array}$

$\begin{array}{l}Here,\\ si{n}^2\left(2x+5\right)=\frac{1-cos2(2x+5)}{2}\\ =\frac{1-cos(4x+10)}{2}\\ Then,\\ ={\displaystyle \int si{n}^2}(2x+5)dx={\displaystyle \int \frac{1-cos(4x+10)}{2}}dx\\ =\frac{1}{2}{\displaystyle \int 1}\cdot dx-\frac{1}{2}{\displaystyle \int cos}(4x+10)dx\\ =\frac{1}{2}\cdot x-\frac{1}{2}\frac{sin(4x+10)}{4}+\; C\\ =\frac{x}{2}-\frac{1}{8}sin(4x+10)+\; C\end{array}$

Yes, Sewa Community College Andheri does provide placement support to students officially enrolled in the BBA courses. The placement cell plays a vital role in on-campus placements. Moreover, students must note that the college may have a placement policy that students may be required to adhere to. For more information, they are advised to connect with the college.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- 1 molar mass of Al has N_A (Avogadro number)= 6.023 $*$ 10²³

M= $\frac{NA}{M}*$ m= 6.023 $*$ 10²³/27 $*$ 0.75 = 1.6 $*$ 10²²

Magnitude of positive and negative charges in one paisa coin = Nze

As aluminium has 13 electron and 13 protons so

=  1.6 $*$ 10²² $*$ 13 $*$ 1.6 $*$ 10^-19=33.28 $*$ 10³C.

The steps involved in the admission process of NIMS - Nehru Institute of Management Studies are: 

Step 1: Filling out the application form for the course.

Step 2: Appear for TANCET. 

Step 3: Candidates who clear the cutoff set by the institute can pay the course fee and secure their seat. 

Candidates pursuing BTech at Bharati Vidyapeeth College of Engineering must meet the eligibility criteria by securing minimum aggregate in merit. Furthermore, candidates must secure a minimum aggregate of 45% in Class 12. In addition to this, candidates must secure a valid score in accepted entrance exams such as JEE Main, MHT CET, GATE or other equivalent examinations. For final admission, candidates must qualify the counselling round for final seat allocation.

This is a multiple choice answer as classified in NCERT Exemplar

L = angular momentum = Iw = constant

I₁w₁=I₂w₂

Due to expansion of the rod I₂>I₁

$\frac{w_2}{w_1}=\frac{I_1}{I_2}1$

w₂1 so angular velocity decreases.

The last date to apply for BCom (Hons) and BCom courses at Amity University Noida was June 15, 2025. Applicants were advised to complete the application process well before the deadline to avoid last-minute issues.