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

The rate-determining step involved in the reaction is 

CH₃-CH=CH₂ + HX → CH₃-CH⁺-CH₃ + X^-

So the rate-determining step depends on the bond energy of HX i.e. higher the bond energy lesser would be reactivity.

Thus the order of reactivity of these halogen acids is

HI>HBr>HCl

The placements at Sasurie College of Arts and Science are good. The following table presents the placement statistics for Sasurie College of Arts and Science.

The IILET entrance exam application window is open. The aspirants will be able to fill the form through the Indore Institute of Law website. Use your registered email ID and mobile number to register for the exam.

Given, ν = (3.29 x 10¹⁵ Hz) (1/3² – 1/n²)

= (3.29 x 10¹⁵ Hz) (1/3² – 1/n²)

(3 x 10⁸ ms^-1) / (1.285 x 10^-6 m) = (3.29 x 10¹⁵ Hz) (1/3² – 1/n²)

2.3346 x 10¹⁴ = (3.29 x 10¹⁵ Hz) (1/3² – 1/n²)

2.3346 / 32.9 = 1/3² – 1/n²

0.071 = 1/9 – 1/n²

1/n² = 1/9 – 0.071= 0.111 – 0.071 = 0.04

n² = 1/ 0.04 = 25

=> n = 5

For n = 5, Paschen series lies in infrared region of the spectrum.

$\begin{array}{l}2I={\displaystyle {\int }_0^n\frac{\pi tanx}{secx+tanx}dx\Rightarrow 2I=\pi {\displaystyle {\int }_0^n\frac{\frac{sinx}{cosx}}{\frac{1}{cosx}+\frac{sinx}{cosx}}}}dx\\ \Rightarrow 2I=\pi {\displaystyle {\int }_0^{\pi }\frac{sinx+1-1}{1+sinx}dx}\\ \Rightarrow 2I=\pi {\displaystyle {\int }_0^{\pi }1.dx-\pi {\displaystyle {\int }_0^{\pi }\frac{1}{1+sinx}}dx}\\ \Rightarrow 2I=\pi {\displaystyle {\int }_0^{\pi }1.dx-\pi {\displaystyle {\int }_0^{\pi }\frac{\left(1-sinx\right)}{\left(1+sinx\right)\left(1-sinx\right)}}dx}\end{array}$

$\begin{array}{l}\Rightarrow 2I=\pi {\left[x\right]}_0^{\pi }-\pi {\displaystyle {\int }_0^{\pi }\frac{1-sinx}{co{s}^{2}x}dx}\\ \Rightarrow 2I={\pi }^2-\pi {\displaystyle {\int }_0^{\pi }\left(se{c}^{2}x-tanxsecx\right)}dx\\ \Rightarrow 2I={\pi }^2-\pi {\left[tanx-secx\right]}_0^{\pi }\end{array}$

$\begin{array}{l}\Rightarrow 2I={\pi }^2-\pi \left[tan\pi -sec\pi -tan0+sec0\right]\\ \Rightarrow 2I={\pi }^2-\pi \left[0-\left(-1\right)-0+1\right]\\ \Rightarrow 2I={\pi }^2-2\pi \\ \Rightarrow 2I=\pi \left(\pi -2\right)\\ I=\frac{\pi }{2}\left(\pi -2\right)\end{array}$

Lady Shri Ram College offers professional courses in Journalism and Mass Communication and Elementary Education. Some of the departments under LSR College for Women are Economics, Commerce, Statistics, Psychology, Sanskrit, Computer – Science, etc.

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

The rotation around C-C single bond is possible but it is not completely free due to the torsional strain which is about 1^-20 KJ mol^-1.

λ = 150 pm, v = 1.5 x 10⁷ m s^-1

Kinetic energy, K.E. = ½ mv2 = ½ x 9.1 x 10^-31 kg x (1.5 x 10⁷ ms^-1)²

= [ (9.1 x 1.5 x 1.5) / 2] x 10^{-31 +14}

= 10.2375 x 10^-17 J = 1.02375 x 10^-16 J

K.E = hc/ λ = (6.626 x 10^-34 kg m² s^-1) / (3 x 10⁸ ms^-1) / (1.5 x 10^-10 m)

= [ (6.626 x 3) x 10^-34+8+10] / 1.5  

= 13.252 x 10^-16 J

We know, E = W₀ + K.E.

W₀ = E – K.E.  = (13.252 – 1.024) x 10^-16 J

= 12.228 x 10^-16 J

= 12.228 10^-16 / 1.602 x 10^-19

= 7.63 x 10³ eV

Dr. Shantilal K Somaiya School of Art (SSA) is a private college and is located in Mumbai, Maharashtra. It is one of the schools of Somaiya Vidyavihar University and has been accredited by NAAC with an A grade in 2025. Apart from this, the college has a panel of guest lecturers to inspire students in their respective careers and to give them tips and invaluable insights in every course. 

Let I = ${\displaystyle {\int }_0^{\pi /2}sin}2xta{n}^{-1}(sinx)dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}2}sinxcosxta{n}^{-1}(sinx)dx$

Putting sin x = t =>cos xdx = dt.

whenx = 0, t = sin 0 = 0.

$x=\pi /2t=sin\pi /2=1$

? I = ${\displaystyle {\int }_0^{1}2}·tta{n}^{-1}(t)dt$

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

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

= $2\left\{\left[ta{n}^{-1}(1)\times \frac{1}{2}-ta{n}^{-1}(0)\times \frac{0}{2}\right]-\frac{1}{2}{\displaystyle {\int }_0^1\frac{\left(1+t^2\right)-1}{1+t^2}}dt\right\}$

= $2\left[\frac{\pi }{8}-0\right]-\frac{2}{2}\left\{{\displaystyle {\int }_0^1\frac{1+t^2}{1+t^2}}dt-{\displaystyle {\int }_0^1\frac{dt}{1+t^2}}\right\}$

= $\frac{\pi }{4}-{\displaystyle {\int }_0^{1}d}t+{\left[ta{n}^{-1}t\right]}_0^1$

= $-\frac{\pi }{4}-{[t]}_0^1+\left[ta{n}^{-1}(1)-ta{n}^{-1}(0)\right]$

= $\frac{\pi }{4}-1+\frac{\pi }{4}=2\times \frac{\pi }{4}-1=\frac{\pi }{2}-1$

Yes, Sasurie College of Engineering does offer BTech and BE courses for the duration of four years, divided into eight semesters. Moreover, the college offers these courses in several specialisations, such as Mechanical Engineering, Computer Science Engineering, Robotics & Automoation, etc. The college also offers lateral entry admission to some of its BTech courses. 

λ = 256.7 nm = 256.7 x 10^-9 m

K.E. = 0.35 eV

E = hc/ λ = (6.626 x 10^-34Js) / (3 x 10⁸ ms^-1) / (256.7 x 10^-9 m)

= (6.626 x 3 x 10^-17) J/ 256.7

= (6.626 x 3 x 10^-17) / (256.7 x 1.602 x 10^-19) eV

E = 4.83 eV

The potential applied to silver gets converted into kinetic energy of the photoelectron.

So, Kinetic energy, K.E= 0.35 V

=> K.E= 0.35 eV          

E = W₀ + K.E.

=> W₀ = E – K.E.

= 4.83 eV – 0.35 eV = 4.48 eV.

Yes, LSR is ranked by various ranking bodies. Over the past few years, Lady Shri Ram College has been ranked under multiple streams, namely Commerce, Political Science, Journalism, and others. A few of the important Lady Shri Ram College rankings are mentioned below:

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

Yes, it will execute geometrical isomerism as the product formed can be either cis-2-butene or trans-2-butene.

Let I = ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{sinx+cosx}{9+16si{n}^{2}x}}dx$

Let sin x – cos x = t. =>(cosx + sin x) dx = dt.

and (sin x – cos x)² = t²

sin²x + cos²x – 2 sin x cos x = t²

1 – sin2x = t².

sin2t = 1 - t².

When x = 0, t = sin 0 – cos 0 = –1$$

? I = ${\displaystyle {\int }_{-1}^0\frac{dt}{9+16\left(1-t^2\right)}}={\displaystyle {\int }_{-1}^0\frac{dt}{9+16-16t^2}}$

= ${\displaystyle {\int }_{-1}^0\frac{dt}{25-16t^2}}$

= ${\displaystyle {\int }_{-1}^0}\frac{dt}{16\left(\frac{25}{16}-t^2\right)}$

= $\frac{1}{16}{\displaystyle {\int }_{-1}^0\frac{dt}{{\left(\frac{5}{4}\right)}^2-t^2}}$

= $\frac{1}{16}{\left[\frac{1}{2\times \left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)}log\left|\frac{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+t}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.-t}\right|\right]}_{-1}^0\left\{{\displaystyle \int \frac{dx}{a^2-x^2}=\frac{1}{2a}log\left|\frac{a+x}{a-x}\right|}\right\}$

= $\frac{1}{16}\times \frac{4}{2\times 5}{\left[log\left|\frac{5+4t}{5-4t}\right|\right]}_{-1}^0$

= $\frac{1}{40}\left[log\frac{5+4\times 0}{5-4\times 0}-log\frac{5+4(-1)}{5-4(-1)}\right]$

= $\frac{1}{40}\left[log\frac{5}{5}-log\frac{1}{9}\right]$

= $\frac{-1}{40}\left[log1-log{9}^{(-1)}\right]$

= $\frac{1}{40}[0-(-1)log9]$

= $\frac{1}{40}log9$

= $\frac{1}{40}log{3}^2=\frac{2}{40}log3$

= $\frac{1}{20}log3$