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6 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Three

If $A$ is a square matrix such that $A^{2} = I$ , then ${(A - I)}^{3} + {(A + I)}^{3} - 7 A$ is equal to

(A) A

(B) I - A

(C) I + A

(D) 3A

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} {(A - I)}^{3} + {(A + I)}^{3} - 7 A = A^{3} - I^{3} - 3 A^{2} I + 3 A I^{2} + A^{3} + I^{3} + 3 A^{2} I + 3 A I^{2} - 7 A \\ = 2 A^{3} + 6 A I^{2} - 7 A \\ = 2 A . A^{2} + 6 A I - 7 A \\ = 2 A I + 6 A I - 7 A [A^{2} = I] \\ = 8 A I - 7 A = 8 A - 7 A = A \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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6 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Three

$250 m L$ of a waste solution obtained from the workshop of a goldsmith contains $0.1 M$ ${A g N O}_{3}$ and $0.1 M A u C l$ . The solution was electrolyzed at $2 V$ by passing a current of $1 A$ for 15 minutes. The metal/metals electrodeposited will be : $(E_{{A g}^{+} / A g}^{o} = 0.8 V, E_{{A u}^{+} / A u}^{o} = 1.69 V)$

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6 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Three

If $A$ and $B$ are matrices of the same order, then $A B^{'} - B A^{'}$ is a

(A) Skew symmetric matrix

(B) Null matrix

(C) Symmetric matrix

(D) Unit matrix

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t P = (A B^{'} - B A^{'}) \\ P^{'} = {(A B^{'} - B A^{'})}^{'} \\ = {(A B^{'})}^{'} - {(B A^{'})}^{'} \\ = {(B^{'})}^{'} A^{'} - {(A^{'})}^{'} B^{'} [? {(A B)}^{'} = B^{'} A^{'}] \\ = B A^{'} - A B^{'} \\ = - (A B^{'} - B A^{'}) = - P \\ P^{'} = - P, s o i t i s a s k e w s y m m e t r i c m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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6 months ago

Class 12th physics ncert exemplar solutions class 12th chapter two

In an experiment of find out diameter of wire using screw gauge, the following observations were noted:

(A) Screw moves 0.5mm on main scale in one complete rotation

(B) Total divisions on circular scale = 50

(C) Main scale reading is 2.5 mm

(D) 45th division of circular scale is in the pitch line

(E) Instrument has 0.03 mm negative error

Then the diameter of wire is:

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Vishal Baghel

Contributor-Level 10

MS â€“ Reading = 2.5 mm.

50 division on CS = 0.5 mm on MS.

$∴ 4 5^{t h} d i v i s i o n o n C S = \frac{0.5}{50} * 45 m m M . S .$

= .45 mm on M.S.

So, diameter = 2.5 mm

+0.45 mm

+0.03 mm

= 2.98 mm

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6 months ago

Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Six

The processes of calcination and roasting in metallurgical industries, respectively, can lead to:

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alok kumar singh

Contributor-Level 10

Sol. Reactivity towards ${A g N O}_{3}$ depend on stability of carbocation formed

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6 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Three

10. If $A$ is a matrix of order $m * n$ and $B$ is a matrix such that $A B^{'}$ and $B^{'} A$ are both defined, then the order of matrix $B$ is

(A) $m * m$

(B) $n * n$

(C) $n * m$

(D) $m * n$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} O r d e r o f m a t r i x A = m * n \\ L e t o r d e r o f m a t r i x B b e K * P \\ I f A B^{'} i s d e f i n e d t h e n t h e o r d e r o f A B^{'} i s m * K i f n = P \\ I f B^{'} A i s d e f i n e d t h e n t h e o r d e r o f B^{'} A i s P * n w h e n K = m \\ N o w, o r d e r o f B^{'} = P * K \\ ∴ o r d e r o f B^{'} = K * P \\ = m * n [? K = m, P = n] \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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6 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Three

The matrix $[\begin{matrix} 0 & - 5 & 8 \\ 5 & 0 & 12 \\ - 8 & - 12 & 0 \end{matrix}]$ is a

(A) Diagonal matrix

(B) Symmetric matrix

(C) Skew symmetric matrix

(D) Scalar matrix

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t A = [\begin{matrix} 0 & - 5 & 8 \\ 5 & 0 & 12 \\ - 8 & - 12 & 0 \end{matrix}] \\ A^{'} = [\begin{matrix} 0 & 5 & - 8 \\ - 5 & 0 & - 12 \\ 8 & 12 & 0 \end{matrix}] \\ \Rightarrow A^{'} = - [\begin{matrix} 0 & - 5 & 8 \\ 5 & 0 & 12 \\ - 8 & - 12 & 0 \end{matrix}] = - A \\ A^{'} = - A, s o A i s a s k e w s y m m e t r i c m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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6 months ago

Class 12th physics ncert exemplar solutions class 12th chapter two

A traveling microscope has 20 divisions per cm on the main scale while its vernier scale has total 50 divisions and 25 vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope?

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Vishal Baghel

Contributor-Level 10

MSD = 20 divisions per cm 1 MSD = $\frac{1}{20} c m .$

VSD = 50 divisions

As, 25 VSD = 24 MSD

$∴ 1 V S D = \frac{24}{25} M S D$

L.C. = 1 MSD â€“ 1 VSD

= $= \frac{1}{500} c m = 0.002 c m$

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6 months ago

Class 12th Exemplar Solution

The matrix $[\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix}]$ is a

(A) Identity matrix

(B) Symmetric matrix

(C) Skew symmetric matrix

(D) None of these

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix}] \\ A^{'} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix}] = A \\ A^{'} = A, s o A i s a s y m m e t r i c m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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6 months ago

Class 12th Exemplar Solution

If matrix $A = [a_{i j}]_{2 * 2}$ , where $a_{i j} = 1$ if $i \neq j$ if $i = j$ ,then $A^{2}$ is equal to

(A) I

(B) A

(C) 0

(D) None of these

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A = {[a_{i j}]}_{2 * 2} \\ L e t A = {[\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]}_{2 * 2} \\ a_{11} = 0 [? i = j] \\ a_{12} = 1 [? i \neq j] \\ a_{21} = 1 [? i \neq j] \\ a_{22} = 0 [? i = j] \\ ∴ A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] \\ N o w, A^{2} = A . A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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