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The TOEFL exam is good for students who want to study abroad in English. It is mostly useful where universities teach in English. TOEFL also helps for getting a student visa sometimes. If you plan to study in US, UK, Canada, Australia, or Europe, the TOEFL test is widely accepted and many universities prefer it.

Let A = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right].$

We write, A = IA

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 5\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A. (R1 $\leftrightarrow$ R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2-2(1)\hfill & \hfill 5-2(3)\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1-2(0)\hfill & \hfill 0-2(1)\hfill \end{array}\right]$ A. (R2→R22R1)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -2\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$ A. (R2→ (1)R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1-3(0)\hfill & \hfill 3-3(1)\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0-3(-1)\hfill & \hfill 1-3(2)\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$ A. (R1→R13R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill -5\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$ A.

∴ A^-1 = $\left[\begin{array}{cc}\hfill 3\hfill & \hfill -5\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right].$

Let A = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 7\hfill & \hfill 4\hfill \end{array}\right].$

We write, A = IA,

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 7\hfill & \hfill 4\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 7\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A. (R1 $\leftrightarrow$ R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 7-3(2)\hfill & \hfill 4-3(1)\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0-3(1)\hfill & \hfill 1-3(0)\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A.(R1→R1 3R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -3\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2-2(1)\hfill & \hfill 1-2(1)\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -3\hfill & \hfill 1\hfill \\ \hfill 1-2(-3)\hfill & \hfill 0-2(1)\hfill \end{array}\right]$ A. (R2→R2 2R1)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -3\hfill & \hfill 1\hfill \\ \hfill 7\hfill & \hfill -2\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -3\hfill & \hfill 1\hfill \\ \hfill -7\hfill & \hfill 2\hfill \end{array}\right]$ A. (R2→(1)R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1-0\hfill & \hfill 1-1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -3-(-7)\hfill & \hfill 1-2\hfill \\ \hfill -7\hfill & \hfill 2\hfill \end{array}\right]$ A. (R1→R1R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 4\hfill & \hfill -1\hfill \\ \hfill -7\hfill & \hfill 2\hfill \end{array}\right]$ A.

∴A^-1 = $\left[\begin{array}{cc}\hfill 4\hfill & \hfill -1\hfill \\ \hfill -7\hfill & \hfill 2\hfill \end{array}\right]$

MBA in Business Analytics is a two-year postgraduate course, divided into four semesters, and offered in full-time mode. This course is offered through faculty of Management. The total number of seats offered for this specialisation is 180. Check out the number of seats allocated to other specialisations in the below table:

Let A = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 7\hfill \end{array}\right]$

We write, A = 1A.

$\to \left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 7\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{cc}\hfill 5\hfill & \hfill 7\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A((R₂ $\leftrightarrow$ R₁)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 5-2(2)\hfill & \hfill 7-2(3)\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 0-2(1)\hfill & \hfill 1-2(0)\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A. (R1→R1–2R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2-2(1)\hfill & \hfill 3-2(1)\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -2\hfill & \hfill 1\hfill \\ \hfill 1-2(-2)\hfill & \hfill 0-2(1)\hfill \end{array}\right]$ A.(R2→R2–2R1)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -2\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1-0\hfill & \hfill 1-1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -2-5\hfill & \hfill 1-(-2)\hfill \\ \hfill 5\hfill & \hfill -2\hfill \end{array}\right]$ A(R1→R1–R2)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill -7\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill -2\hfill \end{array}\right]$ A.

∴ A^-1 = $\left[\begin{array}{cc}\hfill -7\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill -2\hfill \end{array}\right]$

Let A = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 7\hfill \end{array}\right].$

We write, A = IA.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 7\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2-2(1)\hfill & \hfill 7-2(3)\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0-2(1)\hfill & \hfill 1-0\hfill \end{array}\right]$ A (R₂→R₂–2R₁)

$\Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow \left[\begin{array}{cc}\hfill 1-0\hfill & \hfill 3-3(1)\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1-3(-2)\hfill & \hfill 0-3(1)\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right]$ A(R₁→R₁–3R₂)

$\Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 7\hfill & \hfill -3\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right]$ A.

∴A^-1 = $\left[\begin{array}{cc}\hfill 7\hfill & \hfill -3\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right].$

S.P. Mandali's Ramnarain Ruia Autonomous College UG cutoff 2025 has been released for multiple courses, including BSc. As per the first merit list, the cutoff percentage ranged between 60 and 94.6 for the students belonging to the General AI quota. The most and least sought-after BSc specialisations during the round 1 admission were BSc in Biotechnology and B.Sc. (Physics, Chemistry, Mathematics), respectively. For the rest of the specialisations, refer to the table below: 

Candidates wanting to enrol in the BA LLB course of Motherhood University must meet the prescribed eligibility criteria. The aspirants who are Class 12 or equivalent exam pass from a recognised board with at least 45% aggregate can apply. Additionally, the minimum aggregate requirement for OBC category candidates is 42% and 40% for SC/ ST candidates. Students must also have the relevant documents at the time of admission or form filling. 

Let A = $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right].$

We write, A = IA.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2-1\hfill & \hfill 1-1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1-0\hfill & \hfill 0-1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A (R₁→R₁–R₂)

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ A.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1-1\hfill & \hfill 1-0\hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0-1\hfill & \hfill 1-(-1)\hfill \end{array}\right]$ = A (R₂→R₂–R₁).

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$ A.

∴ A^-1  = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right].$

The cost of studying an MMS at Mumbai colleges can be estimated by taking a look at the table below:

Disclaimer: This information is sourced from the official website and may vary.

Let A= $\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right].$

We write A = IA.

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right]={\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]}^{1}A.$

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 2-2(1)\hfill & \hfill 3-2(-1)\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0-2(1)\hfill & \hfill 1-0\hfill \end{array}\right]$ (R₂ → R₂ –2R₁)

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right]A.$

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill \raisebox{1ex}{$0$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]=[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}]A\left(R_2\to \frac{1}{5}{R}_2\right)$

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right].$

$\left[\begin{array}{cc}\hfill 1+0\hfill & \hfill -1+1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1-2/5\hfill & \hfill 0+1/5\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]A\left(R_1\to R_1+R_2\right)$

$\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=[\begin{array}{cc}\hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}]A.$

∴ A^-1  = $\left[\begin{array}{cc}\hfill \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \\ \hfill \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \end{array}\right]$

Given, A = $\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill -sin\alpha \hfill \\ \hfill sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]$ . Then, A’ = $\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill sin\alpha \hfill \\ \hfill -sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]$

and A + A’ = I.

$\Rightarrow$ $\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill -sin\alpha \hfill \\ \hfill sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]$ + $\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill sin\alpha \hfill \\ \hfill -sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right].$

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2cos\alpha +cos\alpha \hfill & \hfill -sin\alpha +sin\alpha \hfill \\ \hfill sin\alpha -sin\alpha \hfill & \hfill cos\alpha +cos\alpha \hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow$ $\left[\begin{array}{cc}\hfill 2cot\alpha \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2cot\alpha \hfill \end{array}\right]$ = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right].$

Equating the corresponding element of the matrix we get,

2 cos $\alpha$ = 1

$\Rightarrow$ cos $\alpha =\frac{1}{2}$

$\Rightarrow$ $\alpha$ = cos -$\frac{11}{2}$ = cos^-1$\left(cos\frac{}{3}\right)$$$

Option B is correct

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

In the solid state BeCl₂ has a polymeric chain like structure. Be atom is surrounded by four Cl atoms among which two of them are bonded through covalent bond and other two are through co-ordinate bond. Structure of BeCl₂ is given by structure A.

In gaseous state beryllium chloride exists as dimer (Be₂Cl₄) which dissociates to the monomer at about 1200 K temperature which is linear in structure.

Structure of BeCl₂ in gaseous state is given by the structure B.

Given A and B are symmetric matrices,

(E) Then, A’ = A and B’ = B.

Now, (AB - BA)’ = (AB)’-  (BA)’

= B’A’ - A’B’.

= BA - AB

(AB - BA)’ = - (AB - BA)

AB - BA is a skew symmetric matrix

∴ Option A is correct.

Yes, under its MPA program, Maya Somaiya School of Music and Performing Arts offers five specialisations. Out of these five, three specialisations deal with Music and the other two specialisations deal with different forms of Classical Dance. The below list showcases all the MPA courses offered at the institute:

• MPA in Vocal
• MPA in Swar Vadya
• MPA in Taal Vadya
• MPA in Bharata Natyam
• MPA in Kathak

The table given below lists the Mumbai MMS colleges with the highest placement salary. Students can also get an idea about the top recruiting companies that hire graduates from the MMS colleges in Mumbai:

Disclaimer: This information is sourced from the official website and may vary.

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

The flame is due to the excitation of the electron from its higher energy state to the lower energy states. Due to the small atomic and ionic size of beryllium and magnesium, electrons are tightly bound to the atom. The electrons of Be and Mg does not gain excitation from the energy provided by the flame. Hence they do not show any flame in the flame test.

When selecting Mohamed Sathak College of Arts and Science for a BSc, students should consider factors such as specialisations offered, curriculum relevance, faculty expertise, and placement support. It's also important to assess industry exposure, internship opportunities, tuition fees, infrastructure, and student feedback to ensure that the programme aligns with long-term career goals.