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

If arenes would undergo electrophilic addition reaction then they would lose their aromaticity which would lead to comparatively less stability. However, alkenes undergo electrophilic addition reaction via carbocation intermediate.

Power of the laser E = Nhv = Nh c/λ, where N is the number of photos emitted

= [ (5.6 x 10²⁴) x (6.626 x 10^-34Js) x (3 x 10⁸ ms^-1)] / (337.1 x 10^-9 m)

= 3.3 x 10⁶ J

Let I = ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{sinxcosx}{co{s}^{4}x+si{n}^{4}x}}dx.$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{sinxcosx}{co{s}^{4}x\left(1+\frac{si{n}^{4}x}{co{s}^{4}x}\right)}}dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{\frac{sinx}{cosx}·\frac{cosx}{cosx}\frac{1}{co{s}^{2}x}}{\left(1+ta{n}^{4}x\right)}}dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{tanx·se{c}^{2}x}{1+ta{n}^{4}x}}dx$

= $\frac{1}{2}{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{2tanx·se{c}^{2}x}{1+ta{n}^{4}x}}dx.$

Putting tan²x = t =>2 tan x?sec²xdx = dt

When x = 0, t = tan²x = tan² 0 = 0

x = $\frac{\pi }{4}$ t = tan² $\frac{\pi }{4}$ = 1² = 1.

? I = $\frac{1}{2}{\displaystyle {\int }_0^1\frac{dt}{1+t^2}}=\frac{1}{2}{\left[ta{n}^{-1}t\right]}_0^1$

= $\frac{1}{2}\left[ta{n}^{-1}(1)-ta{n}^{-1}(0)\right]$

= $\frac{1}{2}\left[\frac{\pi }{4}-0\right]=\frac{\pi }{8}$

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The given radiations in increasing order of wavelength are:

Cosmic rays < X-rays < radiation from microwave oven < amber light from traffic signal < radiation from FM radio

IILET 2026 application forms window is closed. Candidates will have to submit an online application form along with the required documents including qualifying exam details, category, residential address, contact details, etc.

Let I = ${\displaystyle {\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }{e}^x}\left(\frac{1-sinx}{1-cosx}\right)dx.$

= ${\displaystyle {\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }{e}^x}\left[\frac{1-2sin\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2si{n}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]dx.$

= ${\displaystyle {\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }{e}^x}\left[\frac{1}{2}cose{c}^2\frac{x}{2}-cot\frac{x}{2}\right]dx$

= – ${\displaystyle {\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }{e}^x}\left[cot\frac{x}{2}-\frac{1}{2}cose{c}^2\frac{x}{2}\right]dx$

= $-{\left[e^x·cot\frac{x}{2}\right]}_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\pi }$ $\{?\int e^x\left[f(x)+f^{\prime }(x)\right]dx=e^{n}f(x)$

= $-\left[e^{\pi }cot\frac{x}{2}-\frac{\pi }{e^2}cot\frac{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2}\right]$

= $-\left[e^{\pi }\times 0-e^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}·cot\frac{1}{4}\right]$

= $0+e^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times 1.$

= $e^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Let the no. of electrons in the ion= x
∴ the no. of the protons= x + 3 (as the ion has three units positive charge)
and the no. of neutrons= x +  (x×30.4 / 100) = x+ 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons = (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x)

=>2.304x = 56 – 3 = 53
=>x = 53 / 2.304 = 23
Atomic no. of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

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

The bond energy of HCl is higher than that of HBr thus it is not cleaved by free radical mechanism to exhibit peroxide effect. However in case of HI the bond energy is so low that the iodine radical forms readily and after formation it combines to form an iodine molecule.

My PwBD NEET score 65 mujhe MBBS colleg mil jayega 

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Kindly go through the solution

Let the no. of electron in the ion = x
∴ The no. of protons = x – 1 (as the ion has one unit negative charge)
and the no. of neutrons = x +  (x×11.1 / 100) = 1.111 x
Mass of the ion = No. of protons + No. of neutrons
(x – 1) + (1.111 x)

Given mass of the ion = 37
∴ (x – 1) + (1.111 x) = 37

=> 2.111 x = 37 + 1 = 38
x = 38 / 2.111 = 18
No. of electrons = 18; No. of protons = 18 – 1 = 17
Atomic no. of the ion = 17; atom corresponding to ion = Cl
Symbol of the ion = ³⁷₁₇Cl^–

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

The structure of A is

The reaction involved are as follows:

An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
Number of neutrons = x + (x×31.7/100) = 1.371x
Now, Mass no. of element = no. of protons + no. neutrons

⇒ x + 1.317x = 81
⇒ x = 34.958

    x ?  35  

∴ No. of protons = 35, No. of neutrons = 81 – 35 =46
Atomic number of element (Z) = No. of protons = 35
The element with atomic number (Z) 35 is bromine (₃₅⁷⁹Br)

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