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It is better not to wear jeans. If you do, make sure they are plain, clean, and not torn. Formal pants are a better choice for looking professional.

Yes, students willing to apply for Shri Guru Ram Rai University BSc courses can apply offline. In order to do so, students are required to visit the admissions office located on the university's campus. After reaching the office, students must obtain and fill out the application form. After that, they must submit INR 1,000 as the application fee.

Note: The application fees mentioned above is as per the official sources. However, it is indicative and subject to change.

Let r and h be the radius and height of the one in scribed in the sphere of radius R.

Then, is ΔOBC, rt angle at B (h-r)² + r² = R²

$\Rightarrow$ h² + R²- 2hR + h² = R²

$\Rightarrow$ r² = 2hR -h²

Then the volume v of the cone is, $V=\frac{1}{3}\times r^{2}h$ $=\frac{1}{3}\pi \left(2hR-h^2\right)h$

$=\frac{1}{3}\pi \left(2{\mathrm{Rh}}^2-h^3\right).$

$\Rightarrow \frac{\mathrm{dV}}{dh}=\frac{\pi }{3}\left(4\mathrm{Rh}-3h^2\right).$

$\frac{d^{\mathrm{2V}}}{d{h}^2}=\frac{\pi }{3}(\mathrm{4R}-6h).$

At $\frac{\mathrm{dV}}{dh}=0$

$-\frac{\pi }{3}\left(\mathrm{4R}h-3h^2\right)=0.$

4Rh – 3h² = 0.

h(4R – 3h) = 0.

h = 0 and $h=\frac{4R}{3}$

As h> 0, $h=\frac{4}{3}.$

At $h=\frac{\mathrm{4R}}{3},\frac{d^{\mathrm{2v}}}{d{h}^2}=\frac{\pi }{3}\left[4R-6\times \frac{\mathrm{4rR}}{3}\right]$

$=\frac{1}{3}[\mathrm{4R}-8R]=-\frac{4\mathrm{R\pi }}{3}<0$

$h=\frac{\mathrm{4R}}{3}$ is a point of maxima.

and $r^2=2\mathrm{hR}-h^2=2\times \frac{\mathrm{4R}}{3}R-{\left(\frac{\mathrm{4R}}{3}\right)}^2$

$=\frac{3\times \mathrm{8R}{}^2}{9}-\frac{1\mathrm{6R}{}^2}{9}$

$r^2=\frac{\mathrm{8R}{}^2}{9}$

Hence, Volume of Cone, $=\frac{1}{3}\pi r^{2}h$

$=\frac{1}{3}\pi \times \frac{\mathrm{8R}{}^2}{9}\times \frac{\mathrm{4R}}{3}$

$=\frac{8}{27}\times \frac{4}{3}\mathrm{\pi R}{}^3=\frac{8}{27}\times$ Volume of sphere.

Be polite, speak clearly, and answer honestly. Dress neatly, bring all your documents, and smile. Confidence and good manners make a good impression.

Yes, you can wear a kurti. Just make sure it is clean, simple, and not too flashy. Pair it with neat leggings or pants.

There is no strict dress code, but you should dress nicely. Wearing formal or semi-formal clothes helps you look serious and respectful.

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

(i) For a binary solution having both components as volatile liquids (viz. CHCl₃ and CH₂Cl₂), the total pressure will be

p= $\frac{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{v}\mathrm{a}\mathrm{p}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{m}\mathrm{i}\mathrm{x}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}}{\mathrm{b}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{y}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{d}\mathrm{s}\mathrm{ }}$

p₁= partial vapour pressure of component 1 (ie. CHCl₃)

p₂= partial vapour pressure of component 2 (ie. CH₂Cl₂)

(ii) For a solution containing non-volatile solute ie. NaCl (s) and H₂O (l), the Raoult’s law is applicable only to vaporisable component (1) ie. H₂O (l) and total vapour pressure is written as 

P= P₁= x₁P₁⁰+ x₂P₂⁰

= x₁P₁⁰+ (1-x₁) P₂⁰

(P₁⁰-P₂⁰)X¹ + P₂⁰

p= $\frac{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{v}\mathrm{a}\mathrm{p}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{m}\mathrm{i}\mathrm{x}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{}}{\mathrm{b}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{r}\mathrm{y}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{d}\mathrm{s}\mathrm{ }}$

p₁= partial vapour pressure of component 1 (CHCl₃ ) 

Let x and y in ‘m’ be the length of side of the square the radius of the circle respectily

Then, length of wire = perimeter of square + circumference of circle

$\Rightarrow$ 28 = 4x + 2πy

$\Rightarrow$ 2x + πy = 14

$\Rightarrow y=\frac{14-2x}{x}$

The combine area A of the square and the circle is

A = x² + πy²

$=x^2+\pi {\left[\frac{14-2x}{\pi }\right]}^2$

${=x^2+\frac{[2(7-x)}{x}]}^2$

$=x^2+\frac{4}{\pi }{(7-x)}^2.$

So, $\frac{\mathrm{dA}}{dx}=2x+\frac{4\cdot 2}{\pi }(7-x)+(-1)=2x-\frac{8}{\pi }(7-x)$

$\frac{d^{\mathrm{2A}}}{d{x}^2}=2+\frac{8}{\pi }$

At, $\frac{\mathrm{dA}}{dx}=0$

$\Rightarrow 2x-\frac{8(7-x)}{\pi }=0$

$\Rightarrow 2x=\frac{8}{\pi }(7-x)$

$\Rightarrow 2\pi x=56-8x$

$\Rightarrow (2x+8)x=56$

$\Rightarrow x=\frac{56}{2\pi +8}=\frac{28}{x+4}.$

At, $x=\frac{28}{\pi +4},\frac{d^{\mathrm{2A}}}{d{x}^2}=2+\frac{\gamma }{\pi }>0$

$\therefore x=\frac{28}{\pi +4}$ isa point of minima

Hence, length of square = $4x=4\left(\frac{28}{1+4}\right)=\frac{112}{\pi +4}m$

and length of circle = 2πy

$=2\pi \left(\frac{14-2x}{\pi }\right)$

$=2\pi \left[\frac{14}{\pi }-\frac{2}{1}\left[\frac{28}{\pi +4}\right]\right]$

$=28-4\left[\frac{28}{\pi +4}\right]=28-\frac{112}{1+4}$

$=\frac{28x+112-112}{x+4}$

$=\frac{28\pi }{\pi +4}$

Light colors like white, light blue, or pastel shades are good. They look clean and professional. Avoid shirts with bright colors or loud patterns.

Yes, Shri Guru Ram Rai University BSc applications can be submitted online via the institute's online admission portal. The steps to apply online are as follows:

1. Go to the official website of the university.

2. Click on 'Admissions 2025' on the Home page.

3. Register by providing the required details.

4. Login and fill out the application form.

5. Upload the documents and pay the application fee of INR 1,000.

6. Click on the submit button to apply.

DES Pune University offers a wide range of programmes to students at the UG and PG levels. The college is mainly concerned with Engineering and Management streams. Some of the popular courses offered at DES University Pune are BBA, BSc, BA, BDes, MBA, MCA, MSW, MSc, MA, MTech, etc. Further, DES Pune University courses are AICTE-approved and are offered to students in full-time mode. 

You should wear neat and clean clothes that look professional. A nice shirt, blouse, or kurti with simple pants is a good choice. Try to avoid very casual clothes.

The volume v of a cylinder of height h and radius r is

V = πr²h = 100

$\Rightarrow h=\frac{100}{\pi r^2}.$

Let, s be the surface area then

S = 2r²hr(r + h) = $2x\left(r^2+\frac{100r}{\pi r^2}\right)$

$=2\pi \left(n^2+\frac{100}{\pi n}\right).$

$\therefore \frac{dS}{dr}=2\pi \left[2x-\frac{100}{1r^2}\right]$

$\Rightarrow \frac{d^{2}S}{d{r}^2}=2\pi \left[2+\frac{200}{\pi r^3}\right]$

At, $\frac{ds}{dr}=2\pi \left[2r-\frac{100}{\pi r^2}\right]=0$

$\Rightarrow 2r=\frac{100}{\pi r^2}$

$\Rightarrow M{}^3=\frac{50}{\pi }\Rightarrow M={\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.},>0.$

At, $x={\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.},\frac{d^{\mathrm{2S}}}{d{r}^2}=2\pi \left[2+\frac{200}{\pi {\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\times 3\right]$

$=2\pi [2+4]=12\pi >0$

$\therefore r={\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ isa point of minimum

And $h=\frac{100}{\pi {\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=2{\left[\frac{50}{\pi }\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}.$

This is a  Assertion Type Questions as classified in NCERT Exemplar

Option (i)

 i.e., Both A and R are correct, and R is the correct explanation of A is the answer since pesticides are the synthetic toxic chemicals with ecological repercussions and their repeated use gives rise to pests that are resistant to that group of pesticides and hence making the pesticides ineffective. Therefore, as insect resistance of DDT increased, other organic toxins such as Aldrin and Dieldrin got introduced in the market by the pesticide industry. Most of the organic toxins are water insoluble and are non-biodegradable. 

This university can be considered a cost-effective solution to international students as the tuition fee at this university ranges between EUR 980 - 8,000 (INR 92,826 - 7,57,760) at PG level. International students are required to pay for the entire degree cost includes components like accommodation, transportation, health insurance etc.

To crack Srusti Academy of Management placement interview students can follow the steps given below:

·       Review your resume

·       Practice common interview questions

·       Showcase your strengths

·       Be confident and articulate

Let r and h be the radius and height of the cylinder

So, r, h > 0

The total surface area s is given by

S = 2πr(h + r) = content .

$\Rightarrow \frac{S}{2\pi }=r(h+r)$ = content = x (say)

$\Rightarrow h+r=\frac{u}{r}$

$\Rightarrow h=\frac{k}{r}-r=\frac{k-r^2}{r.}$

Then, the volume v of the cylinder

$V=\pi r^{2}h=\pi r^2\left(\frac{k-r^2}{r}\right)=\pi r\left(x-x^2\right)=\pi (kx-{\pi }^3)$

So, $\frac{dv}{dr}=\pi \left(x-3r^2\right)$

$\frac{d^2}{d{r}^2}=-6\pi r.$

For maximum, $\frac{dv}{dr}=0$

$\Rightarrow \pi \left[x-3x^2\right)=0$

$\Rightarrow r^2=\frac{x}{3}$

DES Pune University application process is conducted in an online mode. Candidates can visit the official website to apply for UG or PG courses. Find below the step-by-step guide to filling out the online application form for admission to various courses at DES Pune University:

Step 1: Visit the official website and click on "Apply Now".

Step 2: Fill all the required information such as personal and qualification details in the given slots.

Step 3: Upload all the documents in the prescribed format.

Step 4: Complete the application process by paying an application fee of INR 1,080.

The students can choose from a variety of Catholic University of Leuven courses which are available at both the levels: undergraduate and post-graduate in areas like IT and Software, Science, Business and Management Studies and Architecture and Planning. Some of the popular areas of specializations that international students can study here are Food and Nutrition, Business Economics, Physics and Cyber Security, among others.

This is a  Assertion Type Questions as classified in NCERT Exemplar

Option (iv)

i.e., A is not correct, but R is correct since ozone in the stratosphere is a product of ultraviolet radiations falling on oxygen molecules. The main cause of ozone layer depletion is the release of CFCs i.e., freons. With the depletion of the ozone layer, more ultraviolet radiation filters into the troposphere. These radiations lead to ageing of skin, cataract, sunburn, skin cancer etc.