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

The nitro group strongly deactivate the benzene towards the electrophilic substitution reaction due to the -R and -I effect

The nuclear charge experienced by electrons depends on the distance between the nucleus and orbital. The greater is this distance, the lesser is the effective nuclear charge. Among all the p orbitals, 4p orbital lies the farthest from the nucleus and thus experiences the lowest effective nuclear charge because of the maximum magnitude of screening or shielding effect.

Let I ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}·}2ta{n}^{3}xdx$

$=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}ta{n}^2}xtanxdx$

$=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\left(se{c}^{2}x-1\right)}tanxdx\left\{?se{c}^{2}x=ta{n}^{2}x+1\right\}$

$=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}se{c}^2}xtanxdx-2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}tan}xdx$

$=\mathrm{2I}{}_1+2[log]{(cosx)}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$

$=\mathrm{2I}{}_1+2\left[log\left(cos\frac{\pi }{4}\right)-log(cos0)\right]$

$=\mathrm{2I}{}_1+2(lo\mathrm{g1/\surd 2}$
$\frac{\mathrm{}}{}-log1)$

$=\mathrm{2I}{}_1+2(log·2^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-0)$

$I=\mathrm{2I}{}_1+2\times \left(\frac{-1}{2}\right)log2=\mathrm{2I}{}_1-log2.\_\_\_\_(1).$

Where I${}_1={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}se{c}^2}xtanxdx$

Let tan x = t =>sec²xdx = dt

When, x = 0, t = tan 0. = 0

$x=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.,t=tan\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.=1$

${}_1={\displaystyle {\int }_0^{1}t}dt={\left[\frac{t^2}{2}\right]}_0^1=\frac{1}{2}-0=\frac{1}{2}$

So, Equation (1) becomes,

$=2\times \frac{1}{2}-log2$

=1 - log 2.

The candidates applying to the IILET 2026 should have passed or appeared in Class 12th or an equivalent examination. Applicants belonging to General/OBC category must haveleast 45% marks or an equivalent CGPA and, 40% marks or an equivalent CGPA if belonging to SC/ ST category.  

The electrons may be assigned to the following orbitals :
(i) 4d
(ii) 3d
(iii) 4p
(iv) 3d
(v) 3p
(vi) 4p.
The increasing order of energy is :
(v) < (ii) = (iv) < (vi) = (iii) < (i)

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

Due to the +R effect halogens are ortho and para directing groups.

Sasurie College of Arts and Science has an impressive placement record. The placement statistics for 2024 are given in the table below:

LHS = I$={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}si{n}^3}xdx.\{sin\mathrm{3A}=3si\mathrm{nA}-4si{n}^{\mathrm{3\; A}}$

$={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{4}}(3sinx-sin3x)dx.si{n}^{\mathrm{3A}}=\frac{1}{4}(3si\mathrm{nA}-sin\mathrm{3A})$

$=\frac{1}{4}\left[3{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}sin}xdx-{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$q$}\right.}sin}3xdx\right]$

$=\frac{1}{4}\left\{3{[-cosx]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left[\frac{-cos3x}{3}\right]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right\}$

$=\frac{3}{4}(-cos\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+cos0)-\frac{1}{12}\left(-cos\frac{3\pi }{2}+cos3\times 0\right)$

$=\frac{3}{4}(-0+1)-\frac{1}{12}(-0+1)$

$=\frac{3}{4}-\frac{1}{12}=\frac{9-1}{12}=\frac{8}{12}=\frac{2}{3}$

Sasurie College of Arts and Science  has released its placement report for 2024. The key highlights of the 2024 placement statistics are given in the table below:

Since actual momentum is smaller than the uncertainty in measuring momentum, therefore, the momentum of electron cannot be defined

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

The electron donating group increases the reactivity in electrophilic substitution reaction whereas the electron withdrawing group decreases the reactivity in electrophilic substitution.

Thus, the order of reactivity

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$={\displaystyle {\int }_{-1}^1{x}^{17}}·co{s}^{4}xdx$

Here f (x) = x¹⁷ cos4x

f ( -x) = ( -x)¹⁷ cos⁴ ( -x)

= -x¹⁷ cos⁴x

= f (x)

i e, odd fxⁿ

As ${\displaystyle {\int }_a^{a}f}(x)dx=0$ for odd fxⁿ

therefore, I = 0.

Dr. Shantilal K Somaiya School of Art application process 2025 is conducted online. A step-by-step process for filling out the online application form is mentioned below:

Step 1: Visit the official website of Dr. Shantilal K Somaiya School of Art (application.somaiya.edu/widget/#/signup? groupId=2).

Step 2: Fill out the form.

Step 3: Complete and submit the same. 

For a correspondence MA program, you'll generally need to submit documents like your graduation degree and mark sheets, 10th and 12th mark sheets, identity proof, passport-sized photographs, and a category certificate ( if needed)

The number of students placed during Sasurie College of Arts and Science placements 2024 was 182. The table below states the same:

ν = 4.37 x 10⁵ m s^-1, m = 0.1 kg

As per de Brogile’s equation,

λ= m/v = (6.626 x 10^-34 kg m² s^-1) / (0.1 kg) x (4.37 x 10⁵ m s^-1)

=6.626/0.437  x 10^-34-5 m

= 1.516 x 10^-38 m

LHS= ${\displaystyle {\int }_0^{1}x}{e}^{x}dx=x{\displaystyle {\int }_0^1{e}^x}dx-{\displaystyle {\int }_0^1\frac{dx}{dx}}\int e^{x}dxdx$

$={\left[x{e}^x\right]}_0^1-{\displaystyle {\int }_0^1{e}^x}dx$

$=\left[1e^1-0\times e^0\right]-{\left[e^x\right]}_0^1$

$=\left(e^1-0\right)-\left(e^1-e^0\right)$

= e-e + e⁰

= e⁰ = 1=RHS

The highest package and the lowest package recorded at Sasurie College of Arts and Science placements 2024 is given in the table below: