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

Explanation- R= $\rho l$ /A

Resistance of first conductor, RA= $\frac{\rho l}{\pi {(10}^{-3}\times 0.5)}$ 2

Resistance of second conductor, RB= $\frac{\rho l}{\pi \left. {{(10}^{-6}\right)-(0.5\times 0.5\times {10}^{-6})\}}$

Now ratio $\frac{R_A}{R_B}=\frac{\frac{\rho l}{\pi {(10}^{-3}\times 0.5)}2}{\frac{\rho l}{\pi \left. {{(10}^{-6}\right)-(0.5\times 0.5\times {10}^{-6})\}}}$ = 3:1

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

Time require to avoid the collision T= l/v where l = mean free path =1/ $\sqrt[]{2}\pi d^{2}n$

Where n = N/V

n=number of aeroplanes/volume

= $\frac{10}{20\times 20\times 1.5}=0.0167km$ ^-3

T= $\frac{1V}{\sqrt[]{2}\pi d^{2}N}\times \frac{1}{v}$

T= $\frac{1}{\sqrt[]{2}\times 3.14\times {20}^2\times 0.0167\times {10}^{-6}\times 150}$ = $\frac{{10}^6}{1776.25\times 2.505}$

$=\frac{{10}^6}{4449.5}=224.74h\approx 225h$

$\frac{x}{{(x-1)}^2(x-2)}=\frac{A}{(x-1)}+\frac{B}{{(x-1)}^2}+\frac{C}{x+2}$

= A(x- 1)(x + 2) + B(x + 2) + C(x- 1)²

= A(x² + 2x-x- 2) + B(x + 2) + C(x² + 1 - 2x)

= A(x² + x- 2) + B(x + 2) + C(x²- 2x + 1)

Comparing the co-efficient we get, A + C = 0 ¾ (1)

A + B - 2C = 1 ¾ (2)

- 2A + 2B + C = 0 ¾ (3)

EQⁿ (3) - 2 ´ eⁿQ. (2),

- 2A + 2B + C (2A + 2B - 4C) = 0 - 2 ´ 1.

=> - 4A + 5C = - 2 ¾ (4)

Eqⁿ (4) + 4 ´ Eqⁿ (1) we get,

- 4A + 5C + 4A + 4C = - 2 + 4 ´ 0

=> 9C = - 2

$\begin{array}{l}\Rightarrow C=-\frac{2}{9}\\ PuttingC=-\frac{2}{9}, in----- \left(1\right)\\ Putting value of A and C in \left(3\right) we get,\\ -2\times \frac{2}{9}+2B-\frac{2}{9}=0\\ \Rightarrow 2B=\frac{2}{9}+\frac{4}{9}\\ \Rightarrow 2B=\frac{6}{9}:\frac{2}{3}\\ \Rightarrow B=\frac{1}{3}.\end{array}$

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

Explanation- according to ohms law

I= $\frac{E+E}{R+r1+r2}$

The potential difference across terminal is

V=e-Ir= E- $\frac{2E}{R+r1+r2}$ r1=0

E= $\frac{2Er1}{R+r1+r2}$

1= $\frac{2r1}{R+r1+r2}$

R+r1+r2 = 2r1

R= r1-r2

$\frac{x}{\left(x^2+1\right)(x-1)}=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}$

=> x = (Ax + B)(x- 1) + C(x² + 1)

= Ax²- Ax + Bx- B + Cx² + C.

Comparing the co-efficients we get,

A + C = 0 ---- (1)

- A + B = 1 ---- (2)

- B + C = 0 ---- (3)

Adding (1) and (2) we get,

A + C - A + B = 0 + 1

=> B + C = 1 --- (4)

Adding (4) + (3) we get,

- B + C + B + C = 0 + 1

=> 2C = 1

And C = B from (3)

So, from Eqⁿ (1),

A = - C

$\begin{array}{l}A=-\frac{1}{2}.\\ \therefore {\displaystyle \int \frac{x}{\left(x^2+1\right)(x-1)}}dx={\displaystyle \int \frac{\left(-\frac{1}{2}x+\frac{1}{2}\right)}{x^2+1}}dx+{\displaystyle \int \frac{1}{2}}dx.\\ =\frac{1}{2}{\displaystyle \int \frac{1-x}{x^2+1}}dx+\frac{1}{2}{\displaystyle \int \frac{1}{x-1}}dx.\\ =\frac{1}{2}{\displaystyle \int \frac{1}{x^2+1}}dx-\frac{1}{4}{\displaystyle \int \frac{2xdx}{x^2+1}}+\frac{1}{2}log|x-1|+c\\ =\frac{1}{2}ta{n}^{-1}x-\frac{1}{4}log\left(x^2+1\right)+\frac{1}{2}log|x-1|+c.\end{array}$

The expected IISER 2026 application fee for various categories is mentioned below.

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

$\frac{1-x^2}{x(1-2x)}$ which is not a proper fraction

So, division is missing

$\begin{array}{l}So,\frac{1-x^2}{x(1-2x)}=\frac{1}{2}+\frac{(\raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1)}{x(1-2x)}\\ =\frac{1}{2}+\frac{1}{2}\frac{(2-x)}{x(1-2x)}\\ Let\frac{2-x}{x(1-2x)}=\frac{A}{x}+\frac{B}{(1-2x)}\end{array}$

=> 2 -x = A(1 - 2x) + B x.

   So, - 2A + B = - 1

   A = 2.

   So, B = - 1 + 2A = - 1 + 2 ´ 2 = - 1 + 4 = 3

$\begin{array}{l}\therefore {\displaystyle \int \frac{1-x^{2}dx}{x(1-2x)}}={\displaystyle \int \frac{1}{2}}dx+{\displaystyle \int \frac{1}{2}}\frac{(2-x)}{x(1-2x)}dx\\ =\frac{x}{2}+\frac{1}{2}\left\{{\displaystyle \int \frac{A}{x}}dx+{\displaystyle \int \frac{B}{1-2x}}dx\right\}\\ =\frac{x}{2}+\frac{1}{2}\{\left\{\frac{2}{x}dx+{\displaystyle \int \frac{3}{(1-2x)}}dx\right\}\\ =\frac{x}{2}+log\left|x\right|+\frac{3}{2}log\frac{|1-2x|}{(-2)}+c\\ =\frac{x}{2}+log\left|x\right|-\frac{3}{4}log|1-2x|+c\end{array}$

The Sasmira's Institute of Management Studies and Research has had decent placement opportunities over the years. The students are given excellent placement support every year by placing them in leading MNCs. 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: 

Yes, there are scholarships for BSc at IISER Tirupati. IISER Tirupati offers two scholarships, which are,  KVPY scholarship and the INSPIRE scholarship, wherein students are offered an amount of INR 5,000 per month. Candidates are advised to use the National Scholarship Portal 2.0 for all the information regarding the scholarships on offer.

the eligibility typically requires a minimum of 45% marks in Physics, Chemistry, and Mathematics at the 10+2 level

Nazareth College of Arts and Science has a diverse range of recruiters participating in its placement drives. Some key names include Larsen & Toubro Infotech, TCS BPS, Infosys BPO, Sundaram Finance, Tech Mahindra, Sutherland Global Services, Star Health Insurance, Mphasis, CAMS Outsourcing, and Qspiders. These recruiters highlight the college's strong industry connections and career support for BCom students.

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

Explanation- when all resistance are in parallel

$\frac{1}{R_p}=\frac{1}{R_1}+\dots \dots \dots \dots ..+\frac{1}{R_n}$ by multiplying this equation by R_min we have

$\frac{R_{min}}{R_p}=\frac{R_{min}}{R_1}+\dots \dots \dots \dots ..+\frac{R_{min}}{R_n}$

But there exist a term in RHS $\frac{R_{min}}{R_{min}}=1$ and other terms are positive so we have

$\frac{R_{min}}{R_p}=\frac{R_{min}}{R_1}+\dots \dots \dots \dots ..+\frac{R_{min}}{R_n}$ >1

This shows that equivalent resistance is less than maximum resistance available.

But when all resistance are in series

R_s =R₁+R₂………R_n

here must be R_max value in RHS

R_s= R₁+……R_max+….R_n

And R_s> R_max

So equivalent resistance is less than R_max

$\frac{2x}{x^2+3x+2}=\frac{2x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}$

2x = A(x + 2) + B(x + 1).

= (A + B) x + (2A + B).

Comparing the coefficients we get,

A + B = 2 ---(1)

2A + B = 0 ---- (2).

So, Eqⁿ (2) - (1),

2A + B - (A + B) = 0 - 2

Þ A = - 2

And B = 2 - A = 2 - (-2) = 2 + 2 = 4

B = 4

$\begin{array}{l}So,\frac{2x}{(x+1)(x+2)}=\frac{-2}{x+1}+\frac{4}{x+2}.\\ \therefore {\displaystyle \int \frac{2xdx}{(x+1)(x+2)}}=-2{\displaystyle \int \frac{dx}{x+1}}+4{\displaystyle \int \frac{dx}{x+2}}\\ =-2log|x+1|+4log|x+2|+c\\ =4log|x+2|-2log|x+1|+c\end{array}$

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 theSasmira's Institute of Management Studies and Research placement for MBA in 2025: 

No, IISER application forms had to be filled in online mode on the official website of the exam only. Candidates had to visit the official website of IISER 2025 and register first, which generated the credentials, or the username and password. After that, they had to log in and complete the application process by filling out the mandatory fields and paying the requisite application fee online. 

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