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The table below states the average course fees at the Nirmala BCA colleges in Thrissur.

Disclaimer: This information is sourced from the official website and may vary.

Let P be the point on hypotenuse of a triangle. ABC, t angle at B.

Which is at distance a& b from the sides of the triangle.

Let < BAC = < MPC = Ø.

Then, in… ΔANP,

$\Rightarrow ta{n}^3\theta =\frac{a}{b}$

$tan\theta ={\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right..}$

At tan Ø = ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3,$}\right.}$

$\frac{d^2\stackrel{?}{q}}{d{\theta }^2}=+ve.>0.$ { Øall trigonometric fxⁿ are + ve in I^st quadrant}.

So, z is least for tanØ = ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3.$}\right.}$

As, Sec²Ø = 1 + tan²Ø = 1 + ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3.$}\right.}$ = $\frac{b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3.$}\right.}}$

secØ = ${\left[\frac{b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ = $\frac{{\left(a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right..}}$

And tan²Ø = $\frac{1}{co{t}^2\theta }$

cot²Ø = ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

And cosec²Ø = 1 + cot²Ø = 1 + ${\left(\frac{a}{b}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ = $\frac{a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

cosecØ = $\frac{{\left(a^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+b^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right..}}{a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}$

The S.P. Group of Colleges provide good placements to its graduating students. The students are given excellent placement support every year placing them in top companies. Check out the table below to know the key highlights of the S.P. Group of Colleges placement 2021-2023:

Note- the placement report for 2024 is yet to be released.

Yes, BNM Institute of Technology is ranked by various ranking bodies. Over the past few years, BNM Institute of Technology has been ranked under multiple streams, namely Engineering, Management, and others. The below-mentioned are the BNM Institute of Technology rankings:

Let ‘x’ metre be the radius of the semi-circular opening mounded on the length ‘2x’ side of rectangle. Then, let ‘y’ be the breadth of the rectangle.

Then, perimeter of the window = 10m

$\Rightarrow \frac{2\pi x}{2}+2x+y+y=10$

x + 2x + 2y + = 10.

$\Rightarrow y=\frac{10-(\pi +2)\pi }{2.}$

Let the area of the window be A.

Then, A = $\frac{1}{2}\left(\pi x^2\right)+2xy.$

$=\frac{\pi x^2}{2}+2x·\left[\frac{10-(x+2)x}{2}\right].$

$=\frac{\pi x^2+20x-2\pi x^2-4x^2}{2}$

= $\frac{1}{2}$ [-πx²- 4x² + 20x].

So, $\frac{d}{dx}=\frac{1}{2}$ [ -2πx - 8x + 20]

And $\frac{d\stackrel{2}{}}{d{x}^2}=\frac{1}{2}$ [ -2π - 8] = -π -4 = -( π+ 4)

At $\frac{d}{dx}=0.$

$\frac{1}{2}$ [ -2πx - 8x + 20] = 0

2x + 8x = 20

x = $\frac{20}{2\pi +8}$ = $\frac{10}{\pi +4}.$

At x = $\frac{10}{\pi +4},\frac{d^{\mathrm{2A}}}{d{x}^2}$ = -( π+ 4) < 0

Øx = $\frac{10}{x+4}$ is a point of minima.

And y = $\frac{10-(x+2)\times \left(\frac{10}{x+4}\right)}{2}$

$=\frac{10(\pi +4)-(\pi +2)10}{2(\pi +4)}$

$=\frac{10\pi +40-10\pi -20}{2(x+4)}=\frac{10}{\pi +4.}$

Ø Dimensions of the window are

length = 2x = $\frac{20}{\pi +4}m$

breadth = y $\frac{10}{\pi +4}m.$

radius = y = $\frac{10}{\pi +4}m.$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i) the reaction is endothermic and requires very high temperature is correct since dinitrogen and dioxygen are the main constituents of air. These gases do not react with each other at a normal temperature. The dissociation energy of N₂is very high due to the presence of triple bond and it is very stable. 

The S.P. Group of Colleges have been providing successful placements over the years. The students are given excellent placement support every year placing them in top companies. Check out the table below to know the key highlights of the S.P. Group of Colleges placement in 2021 to 2023:

Note- the placement report for 2024 is yet to be released. 

After earning a BA in Psychology from MIT School of Vedic Sciences, students can pursue diverse career paths. Popular options include roles as Clinical Psychologists, Mental Health Counselors, School Psychologists, and Research Psychologists. Graduates can also specialize as Industrial-Organizational Psychologists, Neuropsychologists, Forensic Psychologists, Child Psychologists, Rehabilitation Counselors, or Substance Abuse Counselors, depending on further education and training.

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Yes, of course! Take a look at the table below for some key highlights about the BCA colleges in Thrissur.

Disclaimer: This information is sourced from the official website and may vary.

BNM Institute of Technology offers PhD, undergraduate, and postgraduate programmes to eligible students. These courses are offered across various streams such as Engineering, Management, and others. At the PG level, the college provides M.Tech and MBA programmes to students. The college also offers PhD programmes to students. At the UG level, the college offers B.E./BTech courses to students. 

Let r and s be the radius of the circle and length of side of square.

Then, sum of perimeter of circle and square = k

2πr +4s = k

s = $\frac{k-2\times r}{4}$

The area A be the total areas of the circle and square.

Then, A = πr² + s²

$=\pi r^2+{\left(\frac{k-2\pi r}{4}\right)}^2=\pi r^2+\frac{x^2+4{\pi }^2{r}^2-4\pi rk}{16}$

$=\frac{16\pi r^2+x^2+4{\pi }^2{r}^2-4\pi k\cdot r}{16}$

$=\frac{1}{16}\left[\left(16\pi +4{\pi }^2\right){\pi }^2-4\pi kr+x^2\right].$

So, $\frac{dA}{dr}=\frac{1}{16}\left[\left(16\pi +4x^2\right)\cdot 2n-4\pi k.\right].$

And $\frac{d^{\mathrm{2A}}}{d{r}^2}=\frac{1}{16}\left[\left(6\pi +4{\pi }^2\right)2\right]=\frac{16\pi +4{\pi }^2}{8}$

At $\frac{dA}{dr}=0$

$\frac{1}{16}\left[\left(16x+4x^2\right)2x-4\pi k\right]=0$

$\Rightarrow \left(16x+4x^2\right)2x-4\pi k=0$

$\Rightarrow 4\pi [4+\pi ]2r=4\pi k.$

$r=\frac{k}{2(4+\pi ).}$

At $r=\frac{x}{2(4+\pi )},\frac{d^{\mathrm{2A}}}{d{x}^2}=\frac{16\pi +4{\pi }^2}{8}>0.$

s = $\frac{2x}{2(4+x).}$

s = 2r.

Hence, proved.

Yes, Shri Guru Ram Rai University admissions for the academic year 2025-2025 are open and currently ongoing for various courses, including BSc, BCom, BA, MBA, etc. Interested candidates can visit the official website of the university to apply for the preferred course. Before applying, students must ensure that they fulfil the course-specific eligibility criteria set by the university.

Let x and y meters be the length and breath of the rectangular base of the tank respectively.

Then, volume V of the tank is

V = length × depth × breath.

V = 2xy = 8m³(given).

$\Rightarrow y=\frac{8}{2x}=\frac{4}{x}.$

Let ‘t’ be the total cost of building the tank.

Then, t = cost of base + cost of sides.

= 70xy + 45[4x+4y]     {there are four sides.

= 70xy + 180x+ 180y.

= $=70\cdot x\cdot \frac{4}{x}+180x+180\times \frac{4}{x}.$

$t=280+180x+\frac{720}{x}.$

So, $\frac{dz}{dx}=0+180-\frac{720}{x^2}.$

And $\frac{d^{2}t}{d{x}^2}=\frac{1440}{x^3}$

At $\frac{d^2}{dx}=0\Rightarrow 180-\frac{720}{x^2}=0$

$\Rightarrow x^2=\frac{720}{180}=4$

x = ± 2

x = 2, (x = length and it cannot be negative)

At x = 2, $\frac{d^{2}t}{d{x}^2}=\frac{1440}{2^3}=\frac{1440}{8}=180>0.$

x = 2 is point of maxima.

Hence, minimum cost = $280+180\times 2+\frac{720}{2}$ = 280 + 360 + 360 = 1000.

Candidates willing to get admission must take part in the BNMIT admission process before the last date of application. Candidates must know their BNMIT eligibility criteria and apply accordingly. Follow the BNM Institute of Technology application process mentioned below:

Step 1: Visit the official website of BNM Institute of Technology.

Step 2: Select the required course.

Step 3: Pay the application fee.

Step 4: Fill out and submit the application form with the required information.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (ii) Troposphere is correct since the lowest region of atmosphere in which the human beings along with other organisms live is called troposphere. It extends up to the height of  approximately 10 km from sea level. 

The given equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$  (1)

Let the major axis be along x-axis so, vertex is at $(\pm a,0)$

Let ΔABC be the isosceles triangle inscribed on the

ellipse with one vertex C at (a, 0).

Then, let A have Co-ordinate (x0, yo) from figure.

So, Co-ordinate of B = (x0, y0)

As A and B lies on the ellipse, from equation (i),

The application portal for Shri Guru Ram Rai University BSc courses is open. Candidates seeking admission must apply for the preferred course by submitting the application form available on the university's official website. The last date to apply is 31 Aug, 2025. Students must fill out the form before the prescribed last date.