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10 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Twelve Some Basic Principles and Techniques

64. Assertion (A): Sulphur present in an organic compound can be estimated quantitatively by Carious method.

Reason (R): Sulphur is separated easily from other atoms in the molecule and gets precipitated as light-yellow solid.

(i) Both A and R are correct, and R is the correct explanation of A.

(ii) Both A and R are correct, but R is not the correct explanation of A.

(iii) Both A and R are not correct.

(iv) A is not correct but R is correct.

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alok kumar singh

Contributor-Level 10

This is a Assertion Type Questions as classified in NCERT Exemplar

64. Option (iii) i.e., Both A and R are not correct is the answer since sulphur is estimated by Carius method in the form of white precipitate of BaSO₄ on heating with fuming and BaCl₂. If light yellow solid is obtained means impurities are present. It is filtered, washed and then dried to get pure BaSO₄ .

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10 months ago

Matrices Maths

42. Express the following matrices as the sum of a symmetric and a skew symmetric

matrix:

(i) $[\begin{matrix} 3 & 5 \\ 1 & - 1 \end{matrix}]$ (ii) $[\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$

(iii) $[\begin{matrix} 3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2 \end{matrix}]$ (iv) $[\begin{matrix} 1 & 5 \\ - 1 & 2 \end{matrix}]$

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Vishal Baghel

Contributor-Level 10

(i) Let A = $[\begin{matrix} 3 & 5 \\ 1 & - 1 \end{matrix}] .$

Then, A’ = $[\begin{matrix} 3 & 1 \\ 5 & - 1 \end{matrix}] .$

Let P = $\frac{1}{2}$ (A + A’) = $\frac{1}{2}$ ${[\begin{matrix} 3 & 5 \\ 1 & - 1 \end{matrix}] + [\begin{matrix} 3 & 1 \\ 5 & - 1 \end{matrix}]} = \frac{1}{2} [\begin{matrix} 3 + 3 & 5 + 1 \\ 1 + 5 & - 1 + \end{matrix}]$

= $\frac{1}{2} [\begin{matrix} 6 & 6 \\ 6 & - 2 \end{matrix}] = [\begin{matrix} 3 & 3 \\ 3 & - 1 \end{matrix}] .$

Then, P’ = $[\begin{matrix} 3 & 3 \\ 3 & - 1 \end{matrix}]$ = P.

∴ P = $\frac{1}{2}$ (A + A’) is symmetric matrix

Let Q = $\frac{1}{2}$ (A + A’) = $\frac{1}{2} {[\begin{matrix} 3 & 5 \\ 1 & - 1 \end{matrix}] - [\begin{matrix} 3 & 1 \\ 5 & - 1 \end{matrix}]} = \frac{1}{2} [\begin{matrix} 3 - 3 & 5 - 1 \\ 1 - 5 & - 1 - (- 1) \end{matrix}]$

= $\frac{1}{2} [\begin{matrix} 0 & 4 \\ - 4 & 0 \end{matrix}] = [\begin{matrix} 0 & 2 \\ - 2 & 0 \end{matrix}] .$

Then Q.’ = $[\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}]$ = (-1) $[\begin{matrix} 0 & 2 \\ - 2 & 0 \end{matrix}]$ = (-1) Q.

$\Rightarrow$ Q.’ = Q,

∴ Q = $\frac{1}{2}$ (A - A’) is a symmetric matrix

Now, P + Q = $\frac{1}{2}$ (A + A’) + $\frac{1}{2}$ (A - A’)

$\Rightarrow$ P + Q = $[\begin{matrix} 3 & 3 \\ 3 & - 1 \end{matrix}] + [\begin{matrix} 0 & 2 \\ - 2 & 0 \end{matrix}] = [\begin{matrix} 3 + 0 & 3 + 2 \\ 3 - 2 & - 1 + 0 \end{matrix}] = [\begin{matrix} 3 & 5 \\ 1 & - 1 \end{matrix}]$ = A.

This A is represented as a sun of symmetric and skew symmetric matrix

Let A = $[\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}] .$

Then A’ = $[\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}] .$

Now, A + A’ = $[\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}] + [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$

= $[\begin{matrix} 6 + 6 & - 2 + (- 2) & 2 + 2 \\ - 2 + (- 2) & 3 + 3 & - 1 + (- 1) \\ 2 + 2 & - 1 + (- 1) & 3 + 3 \end{matrix}]$ = $[\begin{matrix} 12 & - 4 & 4 \\ - 4 & 6 & - 2 \\ 4 & - 2 & 6 \end{matrix}]$

Let P = $\frac{1}{2}$ (A + A’) = $\frac{1}{2} [\begin{matrix} 12 & - 4 & 4 \\ - 4 & 6 & - 2 \\ 4 & - 2 & 6 \end{matrix}] = [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$

Then, P’ =&n

...more

(i) Let A = $[\begin{matrix} 3 & 5 \\ 1 & - 1 \end{matrix}] .$

Then, A’ = $[\begin{matrix} 3 & 1 \\ 5 & - 1 \end{matrix}] .$

= $\frac{1}{2} [\begin{matrix} 6 & 6 \\ 6 & - 2 \end{matrix}] = [\begin{matrix} 3 & 3 \\ 3 & - 1 \end{matrix}] .$

Then, P’ = $[\begin{matrix} 3 & 3 \\ 3 & - 1 \end{matrix}]$ = P.

∴ P = $\frac{1}{2}$ (A + A’) is symmetric matrix

= $\frac{1}{2} [\begin{matrix} 0 & 4 \\ - 4 & 0 \end{matrix}] = [\begin{matrix} 0 & 2 \\ - 2 & 0 \end{matrix}] .$

Then Q.’ = $[\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}]$ = (-1) $[\begin{matrix} 0 & 2 \\ - 2 & 0 \end{matrix}]$ = (-1) Q.

$\Rightarrow$ Q.’ = Q,

∴ Q = $\frac{1}{2}$ (A - A’) is a symmetric matrix

Now, P + Q = $\frac{1}{2}$ (A + A’) + $\frac{1}{2}$ (A - A’)

This A is represented as a sun of symmetric and skew symmetric matrix

Let A = $[\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}] .$

Then A’ = $[\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}] .$

Now, A + A’ = $[\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}] + [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$

Let P = $\frac{1}{2}$ (A + A’) = $\frac{1}{2} [\begin{matrix} 12 & - 4 & 4 \\ - 4 & 6 & - 2 \\ 4 & - 2 & 6 \end{matrix}] = [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$

Then, P’ = $[\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$ = P’

∴ P = $\frac{1}{2}$ (A + A’) is asyntri matrix.

A - A’ = $[\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}] - [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

Let Q = $\frac{1}{2}$ (A - A’) = $\frac{1}{2} [\begin{matrix} 0 & 0 & 0 \\ 0 & σ & 0 \\ σ & 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] .$

Q’ = (-1) $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$ = -Q.

$\Rightarrow$ Q’ = -Q.

∴ Q = $\frac{1}{2}$ (A - A’) is a skew symmetric matrix

Have P + Q = $[\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}] + [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = [\begin{matrix} 6 & - 2 & 2 \\ - 2 & 3 & - 1 \\ 2 & - 1 & 3 \end{matrix}]$ = A.

Then A is represented as a sum of symmetric & skew symmetric matrix

(iii) Let A = $[\begin{matrix} 3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2 \end{matrix}]$

Then, A’ = $[\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}]$

Let P = $\frac{1}{2}$ (A + A’) = $\frac{1}{2} {[\begin{matrix} 3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & - 5 & 2 \end{matrix}] + [\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}]}$

= $\frac{1}{2} [\begin{matrix} 3 + 3 & 3 - 2 & - 1 - 4 \\ - 2 + 3 & - 2 - 2 & 1 - 5 \\ - 4 - 1 & - 5 + 1 & 2 + 2 \end{matrix}] = \frac{1}{2} [\begin{matrix} 6 & 1 & - 5 \\ 1 & - 4 & - 4 \\ - 5 & - 4 & 4 \end{matrix}]$

= $[\begin{matrix} 3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}]$

Then P’ = $[\begin{matrix} 3 & \frac{1}{2} & \frac{- 5}{2} \\ \frac{1}{2} & - 2 & - 2 \\ \frac{- 5}{2} & - 2 & 2 \end{matrix}]$ = P.

∴ P = $\frac{1}{2}$ (A + A’ ) is symmetric matrix.

Let Q = $\frac{1}{2}$ (A - A’) = $\frac{1}{2} {[\begin{matrix} 3 & 3 & - 1 \\ - 2 & - 2 & 1 \\ - 4 & 45 & 2 \end{matrix}] - [\begin{matrix} 3 & - 2 & - 4 \\ 3 & - 2 & - 5 \\ - 1 & 1 & 2 \end{matrix}]}$

= $\frac{1}{2} [\begin{matrix} 3 - 3 & 3 - (- 2) & - 1 - (- 4) \\ - 2 - 3 & - 2 - (- 2) & 1 - (- 5) \\ - 4 - (- 1) & - 5 - 1 & 2 - 2 \end{matrix}] .$

= $\frac{1}{2} [\begin{matrix} 0 & 5 & 3 \\ - 5 & 0 & 6 \\ - 3 & - 6 & 0 \end{matrix}] = [\begin{matrix} 0 & \frac{5}{2} & \frac{3}{2} \\ \frac{- 5}{2} & 0 & 3 \\ \frac{- 3}{2} & - 3 & 0 \end{matrix}]$

Then Q’ = $[\begin{matrix} 0 & \frac{- 5}{2} & \frac{- 3}{2} \\ \frac{5}{2} & 0 & - 3 \\ \frac{3}{2} & 3 & 0 \end{matrix}]$ = (-1) $[\begin{matrix} 0 & \frac{5}{2} & \frac{3}{2} \\ \frac{5}{2} & 0 & 3 \\ \frac{3}{2} & 3 & 0 \end{matrix}]$ = (-1) Q.

$\Rightarrow$ Q’ = Q.

∴ Q = $\frac{1}{2}$ (A - A’) is a skew symmetric matrix

∴ P + Q = symmetric

= .

= $[\begin{matrix} 3 & \frac{6}{2} & \frac{- 2}{2} \\ \frac{:}{2} & - 2 & 1 \\ - \end{matrix}$

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New Question 10 months ago

Computers & IT (Non-Engg) Information Technology

What CUET score is required for BCA admission at Babasaheb Bhimrao Ambedkar University?

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Venkatesh Gorla Contributor-Level 8

For admission to BCA at Babasaheb Bhimrao Ambedkar University (via the Amethi satellite centre), CUET 2022–23 cut offs in the General/EWS category ranged roughly from 162 to 221 marks. Actual eligibility changes yearly based on competition and available seats, so final cut offs for BBAU’s main campus will reflect that variation.

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New Question 10 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Ten The S Block Elements

Write Lewis strucure of O₂^– ion and find out oxidation state of each oxygen atom? What is the average oxidation state of oxygen in this ion?

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Payal Gupta Contributor-Level 10

This is a short answer type question as classified in NCERT ExemplarThe lewis structure of O₂^– ion is,

Oxygen atom having no charge has 6 electrons, so its oxidation number is zero. Oxygen atoms containing −1 charge have 7 electrons, so its oxidation number is −1. The average oxidation state of oxygen in this ion is, =1/2

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New Question 10 months ago

Maya Somaiya School of Music and Performing Arts MPA Eligibility

Who is eligible to apply for MPA at Maya Somaiya School of Music and Performing Arts?

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Shailja Rawat Contributor-Level 10

Candidates seeking admission to Maya Somaiya School of Music and Performing Arts MPA courses must fulfil the eligibility criteria set by the institute. Check below to know the course-specific eligibility requirements:Vocal, Swar Vadya, Taal Vadya:

Completed a three-year BPA (Music) or equivalent examination with 45% or more, or
Passed an undergraduate degree and must have passed Music for not less than five years in a recognised institution, or
An undergraduate degree in any subject recognised by the UGC and having learnt Music for not less than five years from a well-known teacher/guru

Bharata Natyam & Kathak:

Completed

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Maharashtra Business Management Studies

Are MMS courses in Mumbai expensive?

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Muskan Chugh Contributor-Level 10

Some MMS courses in Mumbai are relatively more affordable than MBA programmes in the city. At Shiksha, you can find 10+ MMS colleges in Mumbai that cost INR 1 to 3 Lakh for tuition only. Around 2 institutes offer the same under INR 1 Lakh. See the most affordable options with good quality programmes.

Affordable MMS Courses in Mumbai	Tuition Fee
SIMSREE MMS course	INR 69,000
XIBMS MMS course	INR 45,000
VIVA Institute of Management and Research MMS Course	INR 1.62 Lakh
Shivajirao S. Jondhle Institute of Management Science and Research MMS Course	INR 1.24 Lakh
S.A.V Acharya Institute of Management Studies MMS Course	INR 1 Lakh

Note: All info. is from official sites and can change.

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New Question 10 months ago

Somaiya Vidyavihar Department of Library and Information Science

Can I get any scholarship while pursuing MLISc at Somaiya Vidyavihar University's DLIS?

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Himanshi Pandey Contributor-Level 10

Somaiya Vidyavihar University provides financial aid to candidates enrolled in the MLISc course at Department of Library and Information Science. The financial aid is provided in the form of tuition fee concessions. Candidates are required to fulfil the specified eligibility criteria to avail of the financial aid. Listed below are the criteria for various the available financial aid:

EWS/BPL: Applicants with family income up to INR 8 LPA are eligible for financial aid worth up to 100% on the tuition fees
PWD: Applicants with benchmark disabilities (40% or more) are eligible for a concession of 50% on the tuition fees
Defence Personnel: Wa

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Chemistry NCERT Exemplar Solutions Class 11th Chapter Twelve Some Basic Principles and Techniques

63. Assertion (A): All the carbon atoms in H₂C=C=CH₂are sp² hybridized Reason (R): In this molecule, all the carbon atoms are attached to each other by double bonds. (i) Both A and R are correct, and R is the correct explanation of A. (ii) Both A and R are correct, but R is not the correct explanation of A. (iii) Both A and R are not correct. (iv) A is not correct but R is correct.

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alok kumar singh Contributor-Level 10

This is a Assertion Type Questions as classified in NCERT Exemplar63. Option (iv) i.e., A is not correct, but R is correct is the answer since hybridization of C can be found out by counting σ bonds and π bonds present on C atom. This can be shown as-

If C has 3σ bonds then it is sp² hybridized. If C has 2σ bonds then it is sp hybridized.

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New Question 10 months ago

Matrices Maths

41. Find $\frac{1}{2} (A+A')$ and $\frac{1}{2} (A -A')$ , when A= $[\begin{matrix} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{matrix}]$

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Vishal Baghel

Contributor-Level 10

Given, A = $[\begin{matrix} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{matrix}]$

Then, A’ = $[\begin{matrix} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{matrix}]$

So, A + A’ = $[\begin{matrix} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{matrix}] + [\begin{matrix} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{matrix}] = [\begin{matrix} 0 & a - a & b - b \\ - a + a & 0 & c - c \\ - b + b & - c + c & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

$∴ \frac{1}{2}$ (A + A’) = $\frac{1}{2} [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] .$

And A - A’ = $[\begin{matrix} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{matrix}] - [\begin{matrix} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{matrix}] = [\begin{matrix} 0 & a - (- a) & b - (- b) \\ - a - a & 0 & c - (- c) \\ - b - b & - c - c & 0 \end{matrix}]$

= $[\begin{matrix} 0 & 2 a & 2 b \\ - 2 a & 0 & 2 c \\ - 2 b & - 2 c & 0 \end{matrix}]$

$∴ \frac{1}{2}$ (A - A’) = $\frac{1}{2} [\begin{matrix} 0 & 2 a & 2 b \\ - 2 a & 0 a & 2 c \\ - 2 b & - 2 c & 0 \end{matrix}]$ = $[\begin{matrix} 0 & a & b \\ - a & 0 & c \\ - b & - c & 0 \end{matrix}] .$

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10 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Ten The S Block Elements

In the Solvay process, can we obtain sodium carbonate directly by treating the solution containing (NH₄) 2CO₃ with sodium chloride? Explain.

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Payal Gupta

Contributor-Level 10

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

In the Solvay process, carbon dioxide is transferred through a concentrated solution of ammoniacontaining sodium chloride, which forms ammonium carbonate followed by ammonium hydrogen carbonate. The chemicals in ammonium hydrogen carbonate are different and are heated to form sodium carbonate. NH₃ is found in a solution containing NH₄Cl that is heated and treated with Ca (OH)₂? The reaction of (NH₄)2CO₃ with NaCl provides two products, Na₂CO₃ and NH₄Cl both soluble in water which do not shift to the right balance.

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10 months ago

Matrices Maths

40. For the matrix $[\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}]$ verify that

(i) (A + A') is a symmetric matrix

(ii) (A – A') is a skew symmetric matrix

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Vishal Baghel

Contributor-Level 10

Given, A = $[\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}]$

Then, A’ = $[\begin{matrix} 1 & 6 \\ 5 & 7 \end{matrix}]$

Let P = A + A’ = $[\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}] + [\begin{matrix} 1 & 6 \\ 5 & 7 \end{matrix}] = [\begin{matrix} 1 + 1 & 5 + 6 \\ 6 + 5 & 7 + 7 \end{matrix}] = [\begin{matrix} 2 & 11 \\ 11 & 14 \end{matrix}]$

So, P’ = $[\begin{matrix} 2 & 11 \\ 11 & 14 \end{matrix}]$ = P

i e, ( A + A’ )’ = A + A’.

Hence, A + A’ is symmetric matrix.

Let Q = A A’ = $[\begin{matrix} 1 & 5 \\ 6 & 7 \end{matrix}] - [\begin{matrix} 1 & 6 \\ 5 & 7 \end{matrix}] = [\begin{matrix} 1 - 1 & 5 - 6 \\ 6 - 5 & 7 - 7 \end{matrix}] = [\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}] .$

So,Q¹ = $[\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}]$ = (1) $[\begin{matrix} 0 & - 1 \\ - 1 & 0 \end{matrix}]$ = (1) Q.

$\Rightarrow$ Q¹ = Q.

i e, (A A’)’ = -(A - A’).

Have, A - A’ is a show symmetric matrix

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10 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Ten The S Block Elements

All compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents. Explain.

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Payal Gupta

Contributor-Level 10

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

Ionic compounds are formed from the alkali metals due to their large ionic size and low ionization enthalpy. Thus, they are soluble in water. But, due to the small ionic size, high ionization enthalpy, and high electronegativity of lithium, it forms compounds of covalent nature and thus, are soluble in organic solvents.

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10 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Twelve Some Basic Principles and Techniques

62. Assertion (A): Pent- 1- ene and pent- 2- ene are position isomers. Reason (R): Position isomers differ in the position of functional group or substituent.

(i) Both A and R are correct, and R is the correct explanation of A.

(ii) Both A and R are correct, but R is not the correct explanation of A.

(iii) Both A and R are not correct. (iv) A is not correct but R is correct.

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alok kumar singh

Contributor-Level 10

This is a Assertion Type Questions as classified in NCERT Exemplar

62. Option (i) i.e.,

Both A and R are correct, and R is the correct explanation of A is the answer as when two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton, they are known as position isomers and this phenomenon is termed as position isomerism. Pent-2-ene and pent-l-ene are position isomers because they differ in the position of the double bond.

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10 months ago

Matrices Maths

39. (i) Show that the matrix $[\begin{matrix} 1 & - 1 & 5 \\ - 1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix}]$ is a symmetric matrix.

(ii) Show that the matrix $[\begin{matrix} 0 & 1 & - 1 \\ - 1 & 0 & 1 \\ 1 & - 1 & 0 \end{matrix}]$ is a skew symmetric matrix.

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Vishal Baghel

Contributor-Level 10

(i) Given A = ${[\begin{matrix} 1 & - 1 & 5 \\ - 1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix}]}_{\cdot}$

Then, A’ = $[\begin{matrix} 1 \cdot & - \cdot 1 & \cdot 5 \\ - 1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix}]$

∴A’ = A.

Here, A is symmetric matrix

(ii) Given, A = $[\begin{matrix} 0 & 1 & - 1 \\ - 1 & 0 & 1 \\ 1 & - 1 & 0 \end{matrix}]$

Then, A’ = $[\begin{matrix} 0 & - 1 & 1 \\ 1 & 0 & - 1 \\ - 1 & 1 & 0 \end{matrix}] = (- 1) [\begin{matrix} 0 & 1 & - 1 \\ - 1 & 0 & 1 \\ 1 & - 1 & 0 \end{matrix}]$

$\Rightarrow$ A’ = (1) A.

$\Rightarrow$ A’ = A.

Hers A is a show symmetric matrix.

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10 months ago

B.Sc. Placement B.Sc.

Which companies have visited MSCAS for BSc placements in recent years?

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Himanshu Singh

Contributor-Level 10

In recent years, Mohamed Sathak College of Arts and Science has welcomed many well-known companies for BSc placements. During the 2024 placement season, firms like HCL, First Source Solution, Foxconn, and EOS Globe Solutions recruited students. The following year, 2025, companies such as Hitachi, India Filings, and Prudent took part, providing a variety of job opportunities across multiple sectors.

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10 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Twelve Some Basic Principles and Techniques

61. Assertion (A): Energy of resonance hybrid is equal to the average of energies of all canonical forms.

Reason (R) : Resonance hybrid cannot be presented by a single structure.

(i) Both A and R are correct, and R is the correct explanation of A.

(ii) Both A and R are correct, but R is not the correct explanation of A.

(iii) Both A and R are not correct.

(iv) A is not correct but R is correct

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alok kumar singh

Contributor-Level 10

This is a Assertion Type Questions as classified in NCERT Exemplar

61. Option (iv)

i.e., A is not correct, but R is correct is the answer. This is because canonical structures always have more energy than the resonance hybrid. Resonance hybrids are always found to be more stable than any of the canonical structures since the delocalization of electrons lowers the orbitals energy and gives stability.

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SRM School of Management, SRMIST - SRM Institute of Science and Technology, Chennai Colleges in Chennai

Which MBA specialisation at SRM School of Management has the highest number of seats?

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Atul Pruthi

Contributor-Level 9

SRM School of Management offers eleven specialsiations in the MBA course, from which the school does not offer the highest number of seats in any specialisation. However, it offers the highest number of seats in the general MBA course. Check out the below table to get an exact number of seats offered in the general MBA course:

Courses	Seats
Master of Business Administration (MBA)	450

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10 months ago

Chemistry NCERT Exemplar Solutions Class 11th Chapter Ten The S Block Elements

Why are BeSO₄ and MgSO₄ readily soluble in water while CaSO₄, SrSO₄ and BaSO₄ are insoluble?

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Payal Gupta

Contributor-Level 10

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

BeSO₄ and MgSO₄ readily soluble in water while CaSO₄, SrSO₄ and BaSO₄ are insoluble The greater hydration enthalpy of Be²⁺and Mg²⁺ ions overcome the lattice enthalpy factor and therefore, their sulfates are soluble.

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10 months ago

Matrices Maths

38. If (i) $[\begin{matrix} c o s α & s i n α \\ - s i n α & c o s α \end{matrix}]$ , then verify that A' A = I

(ii) If $[\begin{matrix} s i n α & c o s α \\ - c o s α & s i n α \end{matrix}]$ , then verify that A' A = I

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Vishal Baghel

Contributor-Level 10

(i) Given, A = $[\begin{matrix} c o s α & s i n α \\ - s i n α & c o s α \end{matrix}]$

Then, A’ = $[\begin{matrix} c o s α & - s i n α \\ s i n α & c o s α \end{matrix}]$

∴A’ A = $[\begin{matrix} c o s α & - s i n α \\ s i n α & c o s α \end{matrix}] [\begin{matrix} c o s α & s i n α \\ - s i n α & c o t α \end{matrix}]$

= $[\begin{matrix} c o s α \cdot c o s α + (- s i n α) (- s i n α) & c o s α \cdot s i n α + (- s i n α) c o s α \\ s i n α \cdot c o s α + c o s α (- s i n α) & s i n α \cdot s i n α + c o s α c o s α \end{matrix}]$

= $[\begin{matrix} c o s^{2} α + s i n^{2} α & c o s α s i n α - s i n α c o s α \\ s i n α c o s α - c o s α s i n α & s i n^{2} α + c o s^{2} α \end{matrix}]$

= $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ ${? c o s^{2} x + s i n^{2} x = 2}$

= A ’ A = 1.

Given,

(ii) 1 A = $[\begin{matrix} s i n α & c o s α \\ - c o t α & s i n α \end{matrix}]$

Then, A’ = $[\begin{matrix} s i n α & - c o s α \\ c o t α & s i n α \end{matrix}]$

∴A’ A = $[\begin{matrix} s i n α & - c o s α \\ c o t α & s i n α \end{matrix}] [\begin{matrix} s i n α & c o s α \\ - c o s α & s i n α \end{matrix}]$

= $[\begin{matrix} s i n \cdot α s i n α + (c o s α) (- c o s α) & s i n α \cdot c o s α + (- c o s α) c o s α \\ c o s α \cdot s i n α + s i n α (- c o s α) & c o s α \cdot c o s α + s i n α s i n α \end{matrix}]$

= $[\begin{matrix} s i n^{2} α + c o s^{2} α & s i n α c o t α - c o s α s i n α \\ c o s α s i n α - s i n α c o s α & c o s^{2} α + s i n^{2} α \end{matrix}]$

= $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ ${? c o s^{2} x + c o s c o s^{2} x = 1} .$

$\Rightarrow$ A’ A. = I

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10 months ago

Maharashtra Business Management Studies

Is it worth pursuing an MMS from Mumbai colleges?

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Muskan Chugh

Contributor-Level 10

A Master of Management Studies (MMS) degree can be quite beneficial for students who wish to build their careers in the fields of business administration and management. Students will develop theoretical knowledge and learn through practical internship experiences if they choose to pursue an MMS course from Mumbai colleges.

Students can evaluate the ROI of MMS colleges in Mumbai by comparing the fees and salary given in the table below:

College Name	Average Fees	Average Package
Jamnalal Bajaj Institute of Management Studies MMS	INR 6.01 Lakh- 7 Lakh	INR 26.4 LPA
Prin. L. N. Welingkar Institute of Management Development MMS	INR 6.7 Lakh	INR 10 LPA
SIES College of Management Studies MMS	INR 4.52 Lakh	INR 7.5 LPA
Lala Lajpatrai Institute of Management MMS	INR 2.79 Lakh	INR 5 LPA
TISS - Tata Institute of Social Sciences MMS	INR 80,000	INR 6 LPA

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

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