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Yes, completing a BBA degree from Sasurie College of Arts and Science makes students eligible for various jobs in government departments and organisations. This happens because BBA equips students with many employability skills, such as communication, critical thinking, planning and organising. Tabulated below are some of the job profiles for students to pursue after completion of a BBA degree:

Note: The above-mentioned are industry-based average payouts. The actual salary may differ.

Lady Shri Ram College is ranked 10th under the Colleges category by NIRF 2024,  4th by India Today 2023 for BA,  4th by India Today 2023 for BCom, and 15th by India Today 2023 for BSc. Lady Shri Ram College was established in 1956 and is affiliated with Delhi University. The college is ranked one by Outlook 2023 under the Humanities & Social Sciences category. Lady Shri Ram College for Women–Delhi University (LSR–DU) is accredited with an A+ Grade by NAAC. Lady Sriram College is approved by the UGC.

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

To apply for the programmes offered at Principal K.M. Kundnani College of Pharmacy, students need to register themselves at the official website of the accepted entrance exam. Interested students need to secure a valid score in the exam and participate in the counselling process for seat allotment. The application process is mainly carried out online. Students need to fill out the application form duly by completing the application form with the right information. Uploading/ attaching the specified credentials is the next step. The filled out form then needs to be submitted.

ν = 2.19 x 10⁶ m s^-1

As per de Brogile’s equation,

λ= m/v = (6.626 x 10^-34 kg m² s^-1) / (9.1 x 10^-31 kg) x (2.19 x 10⁶ m s^-1)

6.626/91 * 2.19=  x 10^-34+25 m = 0.33243 x 10^-9 m = 332.43 pm

Sasurie College of Arts and Science provides great placements to its students, 10 companies participated in 2024 placement drive. Refer to the table below to know more:

Let $I={\displaystyle {\int }_1^3\frac{dx}{x^2(x+1)}}.$

The integrand is of the form.

1 = Ax (x + 1) + B (x + 1) + Cx²

= A (x² + x) + B (x + 1) + Cx²

Comparing the coefficients,

A + C = 0 ____ (1)

A + B = 0 ______ (2)

B = 1 ________ (3)

Putting Equation (3) in (2),

A + 1 = 0

A = -1.

and putting value of A in Equation (1),

-1 + C = 0

C = 1

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

$\therefore I={\displaystyle {\int }_1^3\frac{dx}{x^2(x+1)}}={\displaystyle {\int }_1^3\frac{-dx}{x}}+{\displaystyle {\int }_1^3\frac{dx}{x^2}}+{\displaystyle {\int }_1^3\frac{dx}{x+1}}$

$=-{[log|x|]}_1^3+{\left[\frac{x^{-2+1}}{-2+1}\right]}_1^3+{[log?x+1]}_1^3$

$=[-log3+log1]-{\left[\frac{1}{x}\right]}_1^3+\left[log|3+1|-log|1+1|\right]$

$=-log3+0-\left[\frac{1}{3}-1\right]+log4-log2$

$=-log3-\frac{(1-3)}{3}+log{2}^2-log2$

$=-log3-\frac{(-2)}{3}+2log2-log2$

$=log2-log3+\frac{2}{3}$

$=\frac{2}{3}+log\frac{2}{3}$

Hence proved.

The primary eligibility requirement to get admission into BTech courses is candidates must complete their Class 12 from the PCM stream with a minimum aggregate of 45%. Additionally, for the selection process, the college accepts TNEA counselling. This is the online counselling process conducted by Anna University for admission to BE/ BTech programmes in Tamil Nadu engineering colleges. It is based on Class 12 marks, and no entrance exam is conducted.

λ = 800 pm = 800 x 10^-12 m

m = 1.675 x 10^-27 kg

As per de Brogile’s equation,

ν =h/mλ  = (6.626 x 10^-34 kg m² s^-1) / (1.675 x 10^-27 kg) x (800 x 10^-12 m)

=6.626/1.675 * 8  x 10^-34+27+10

= 0.494 x 10³ ms^-1

= 494 ms^-1

Yes, it is worth pursuing MA at Dr. Shantilal K Somaiya School of Art as it is one of the schools of Somaiya Vidyavihar University and has been accredited by NAAC with an A grade in 2025. The college offers an MA programme to students for a duration of 2 years in full-time mode. Further, Dr. Shantilal K Somaiya School of Art has a panel of guest lecturers to inspire students in their respective careers. 

The placement rate for Sasurie College of Arts and Science has not been disclosed yet. However, the placement highlights recorded during Sasurie College of Arts and Science placements 2024 is given in the table below:

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

The benzene here would undergo Friedel-Crafts alkylation reaction where carbocation is formed as an intermediate, thus secondary carbocation is formed as an intermediate due to its greater stability than that of the primary carbocation.

The final product obtained is 

As per de Brogile’s equation,

λ = h / mv = (6.626 x 10^-34 kg m² s^-1) / (9.1 x 10^-31 kg) x (1.6 x 10⁶ ms^-1)

= 0.455 x 10^-9 m = 0.455 nm = 455 pm.

Let I = ${\displaystyle {\int }_1^4[|x-1|+|x+2|+|x-3|]}dx$

I = ${\displaystyle {\int }_1^4|}x-1|dx+{\displaystyle {\int }_1^4|}x+2|dx+{\displaystyle {\int }_1^4|}x-3|dx$

I = I₁ + I₂ + I₃______(1)

So, I₁ = ${{\displaystyle \int }}^4$ |x – 1| dx . $\{\begin{array}{l}(x-1)x-1>0\Rightarrow x>1\\ (x-1)x-1<0\Rightarrow x<1\end{array}$

= ${\displaystyle {\int }_1^4(}x-1)dx$

= ${\left[\frac{x^2}{2}-x\right]}_1^4=\left(\frac{4^2}{2}-\frac{1^2}{2}\right)-(4-1)$

= $\frac{15}{2}-3=\frac{15-6}{2}=\frac{9}{2}.$

I₂ = ${\displaystyle {\int }_1^4|}x-2|dx$ $\{\begin{array}{l}x-2x-2>0,x>2\\ -(x-2)x-2<0,x<2.\end{array}$

= ${\displaystyle {\int }_1^2-}(x-2)dx+{\displaystyle {\int }_2^4(}x-2)dx$

= $-{\left[\frac{x^2}{2}-2x\right]}_1^2+{\left[\frac{x^2}{2}-2x\right]}_2^4$

= $-\left[\left(\frac{2^2}{2}-\frac{1}{2}\right)-(2\times 2-2\times 1)\right]+\left[\left(\frac{4^2}{2}-\frac{2^2}{2}\right)-(2\times 4-2\times 2)\right]$

= $-\left[\frac{4-1}{2}-(4-2)\right]+[(8-2)-(8-4)]$

= $-\frac{3}{2}+2+6-4$

= $\frac{-3+4+12-8}{2}=\frac{5}{2}$

 I₃ = ${\displaystyle {\int }_1^4|}x-3|dx$ $\{\begin{array}{l}(x-3)x-3>0,x>3\\ -(x-3)x-3<0,x<3\end{array}$

= ${\displaystyle {\int }_1^3-}(x-3)dx+{\displaystyle {\int }_3^4(}x-3)dx$

= $-{\left[\frac{x^2}{2}-3x\right]}_1^3+{\left[\frac{x^2}{2}-3x\right]}_3^4$

= $-\left[\left(\frac{3^2}{2}-\frac{1^2}{2}\right)-(3·3-3·1)\right]+\left[\left(\frac{4^2}{2}-\frac{3^2}{2}\right)-(3\times 4-3\times 3)\right]$

= $-\left[\frac{8}{2}-6\right]+\left[\frac{7}{2}-3\right]$

= $-4+6+\frac{7}{2}-3=\frac{-8+12+7-6}{2}=\frac{5}{2}$

Hence Equation (1) becomes

I = $\frac{9}{2}+\frac{5}{2}+\frac{5}{2}$

I = $\frac{19}{2}$

The highest package offered during Sasurie College of Arts and Science placements 2024 stood at INR 3.50 LPA. The table below presents the same: