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7 months ago

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

Kindly Consider the following

$[\begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix}]$

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Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} | \\ C_{1} \to C_{1} - C_{2} \\ = | \begin{matrix} x^{2} - 2 x + 2 & x - 1 \\ 0 & x + 1 \end{matrix} | \\ = (x + 1) (x^{2} - 2 x + 2) - 0 \\ = x^{3} - 2 x^{2} + 2 x + x^{2} - 2 x + 2 = x^{3} - x^{2} + 2 \end{array}$

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7 months ago

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

If $x + y + z = 0$ , prove that $[\begin{matrix} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{matrix}] = x y z [\begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix}] .$

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L . H . S . \\ L e t Δ = | \begin{matrix} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow x a | \begin{matrix} z a & x b \\ x c & y a \end{matrix} | - y b | \begin{matrix} y c & x b \\ z b & y a \end{matrix} | + z c | \begin{matrix} y c & z a \\ z b & x c \end{matrix} | \\ \Rightarrow x a (y z a^{2} - x^{2} b c) - y b (y^{2} a c - x z b^{2}) + z c (x y c^{2} - z^{2} a b) \\ \Rightarrow x y z a^{3} - x^{3} a b c - y^{3} a b c + x y z b^{3} + x y z c^{3} - z^{3} a b c \\ \Rightarrow x y z (a^{3} + b^{3} + c^{3}) - a b c (x^{3} + y^{3} + z^{3}) \\ \Rightarrow x y z (a^{3} + b^{3} + c^{3}) - a b c (3 x y z) [? (x + y + z = 0) (∴ x^{3} + y^{3} + z^{3} = 3 x y z)] \\ \Rightarrow x y z (a^{3} + b^{3} + c^{3} - 3 a b c) \\ R . H . S . x y z | \begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix} | \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ \Rightarrow x y z | \begin{matrix} a + b + c & a + b + c & a + b + c \\ c & a & b \\ b & c & a \end{matrix} | \\ \Rightarrow x y z (a + b + c) | \begin{matrix} 1 & 1 & 1 \\ c & a & b \\ b & c & a \end{matrix} | (a + b + c c o m m o n f r o m R_{1}) \\ C_{1} \to C_{1} - C_{2} a n d C_{2} \to C_{2} - C_{3} \\ \Rightarrow x y z (a + b + c) | \begin{matrix} 0 & 0 & 1 \\ c - a & a - b & b \\ b - c & c - a & a \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow x y z (a + b + c) [1 | \begin{matrix} c - a & a - b \\ b - c & c - a \end{matrix} |] \\ \Rightarrow x y z (a + b + c) [{(c - a)}^{2} - (b - c) (a - b)] \\ \Rightarrow x y z (a + b + c) (c^{2} + a^{2} - 2 c a - a b + b^{2} + a c - b c) \\ \Rightarrow x y z (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) \\ \Rightarrow x y z (a^{3} + b^{3} + c^{3} - 3 a b c) [a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)] \\ L . H . S . = R . H . S . \\ H e n c e, p r o v e d . \end{array}$

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New answer posted

7 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

A proton and an a-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λ_pand λ_αrelated to each other?

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- $λ = h / \sqrt[]{2 m q v}$

$λ \propto \frac{1}{\sqrt[]{m q}}$

$\frac{λ_{p}}{λ_{a}} = \frac{\sqrt[]{m_{a} q_{a}}}{\sqrt[]{m_{p} q_{p}}} = \frac{\sqrt[]{4 m_{p} * 2 e}}{\sqrt[]{m_{p} * e}}$ = $\sqrt[]{8}$

So wavelength of proton is $\sqrt[]{8}$ times of alpha particle

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7 months ago

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

Prove that $[\begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} \\ c a - b^{2} & a b - c^{2} & b c - a^{2} \\ a b - c^{2} & b c - a^{2} & c a - b^{2} \end{matrix}]$ is divisible by $a + b + c$ and find the quotient.

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} \\ c a - b^{2} & a b - c^{2} & b c - a^{2} \\ a b - c^{2} & b c - a^{2} & a c - b^{2} \end{matrix} | \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ \Rightarrow [\begin{matrix} a b + b c + a c - a^{2} - b^{2} - c^{2} & c a - b^{2} & a b - c^{2} \\ a b + b c + a c - a^{2} - b^{2} - c^{2} & a b - c^{2} & b c - a^{2} \\ a b + b c + a c - a^{2} - b^{2} - c^{2} & b c - a^{2} & a c - b^{2} \end{matrix}] \\ \Rightarrow (a b + b c + a c - a^{2} - b^{2} - c^{2}) | \begin{matrix} 1 & c a - b^{2} & a b - c^{2} \\ 1 & a b - c^{2} & b c - a^{2} \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | (T a k i n g a b + b c + a c - a^{2} - b^{2} - c^{2} c o m m o n f r o m C_{1}) \\ R_{1} \to R_{1} - R_{2} a n d R_{2} \to R_{2} - R_{3} \\ \Rightarrow (a b + b c + a c - a^{2} - b^{2} - c^{2}) | \begin{matrix} 0 & c a - b^{2} - a b + c^{2} & a b - c^{2} - b c + a^{2} \\ 0 & a b - c^{2} - b c + a^{2} & b c - a^{2} - a c + b^{2} \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | \\ \Rightarrow (a b + b c + a c - a^{2} - b^{2} - c^{2}) | \begin{matrix} 0 & a (c - b) + (c + b) (c - b) & b (a - c) + (a + c) (a - c) \\ 0 & b (a - c) + (a + c) (a - c) & c (b - a) + (b + a) (b - a) \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | \\ \Rightarrow (a b + b c + a c - a^{2} - b^{2} - c^{2}) | \begin{matrix} 0 & (c - b) (a + b + c) & (a - c) (a + b + c) \\ 0 & (a - c) (a + b + c) & (b - a) (a + b + c) \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | \\ \Rightarrow (a b + b c + a c - a^{2} - b^{2} - c^{2}) (a + b + c) (a + b + c) | \begin{matrix} 0 & c - b & a - c \\ 0 & a - c & b - a \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | \\ \Rightarrow {(a + b + c)}^{2} (a b + b c + a c - a^{2} - b^{2} - c^{2}) | \begin{matrix} 0 & c - b & a - c \\ 0 & a - c & b - a \\ 1 & b c - a^{2} & a c - b^{2} \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow {(a + b + c)}^{2} (a b + b c + a c - a^{2} - b^{2} - c^{2}) [1 | \begin{matrix} c - b & a - c \\ a - c & b - a \end{matrix} |] \\ \Rightarrow {(a + b + c)}^{2} (a b + b c + a c - a^{2} - b^{2} - c^{2}) [(c - b) (b - a) - {(a - c)}^{2}] \\ \Rightarrow {(a + b + c)}^{2} (a b + b c + a c - a^{2} - b^{2} - c^{2}) (b c - c a - b^{2} + a b - a^{2} - c^{2} + 2 a c) \\ \Rightarrow {(a + b + c)}^{2} (a b + b c + a c - a^{2} - b^{2} - \end{array}$

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New answer posted

7 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Two particles A and B of de-Broglie wavelengths A, and combine to form a particle C. The process conserves momentum. Find the de-Broglie wavelength of the particle C. (The motion is one-dimensional)

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- from conservation of momentum

P_c=P_A+P_B

= $\frac{h}{λ_{c}} = \frac{h}{λ_{A}} + \frac{h}{λ_{B}}$

$\frac{h}{λ_{c}} = \frac{h λ_{A} + h λ_{B}}{λ_{A} λ_{B}}$

$λ_{c}$ = $\frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

Case 1 when both momentum are positive

$λ_{c}$ = $\frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

Case 2 when both momentum are negative

$λ_{c}$ = $\frac{λ_{A} λ_{B}}{λ_{A} + λ_{B}}$

Case 3 when 1 is negative and 2 is positive

$λ_{c}$ = $\frac{λ_{A} λ_{B}}{λ_{B} - λ_{A}}$

Case4 when 1 is positive and 2 is negative

$λ_{c}$ = $\frac{λ_{A} λ_{B}}{λ_{A} - λ_{B}}$

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7 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Nine

(i) Consider a thin lens placed between a source (S) and an observer (O) (Figure). Let the thickness of the lens vary as w(b) =w₀– b²/α , where b is the vertical distance from the pole, w₀is a constant. Using Fermat's principle, i.e., the time of transit fora ray between the source and observer is an extremum find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length.

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Payal Gupta

Contributor-Level 10

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

Time from S to P₁ is

t₁= $\frac{S P_{1}}{c} = \frac{\sqrt[]{u^{2} + b^{2}}}{c}$

$\frac{u}{c} (1 + \frac{1}{2} \frac{b^{2}}{u^{2}})$

Time from P₁ to O is

t₂= $\frac{P_{1} O}{c} = \frac{\sqrt[]{v^{2} + b^{2}}}{c}$ ; $\frac{v}{c} (1 + \frac{1}{2} \frac{b^{2}}{v^{2}})$

the time required travel through lens is t₁= $\frac{(n - 1) w b}{c}$

so total time is t=1/c(u+v+b²/2D+(n-1)(wo+b²/ $\propto$ ))

after solving we get $\propto = 2 (n - 1) D$

differentiating with respect to time

t=1/c(u+v+b²/2D+(n-1)K1In(K₂b))

dt/db=0=b/D-(n-1)K₁/b

b²= (n-1)K1D

b= $\sqrt[]{(n - 1) K_{1} D}$

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7 months ago

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

If $a + b + c \neq 0$ and $[\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}] = 0,$ then prove that $a = b = c$ .

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} G i v e n t h a t : a + b + c \neq 0 a n d [\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}] = 0 \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ \Rightarrow [\begin{matrix} a + b + c & b & c \\ a + b + c & c & a \\ a + b + c & a & b \end{matrix}] = 0 \\ \Rightarrow (a + b + c) | \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{matrix} | = 0 (T a k i n g a + b + c c o m m o n f r o m C_{1}) \\ \Rightarrow a + b + c \neq 0 ∴ | \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{matrix} | = 0 \\ R_{1} \to R_{1} - R_{2} a n d R_{2} \to R_{2} - R_{3} \\ \Rightarrow | \begin{matrix} 0 & b - c & c - a \\ 0 & c - a & a - b \\ 1 & a & b \end{matrix} | = 0 \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow 1 | \begin{matrix} b - c & c - a \\ c - a & a - b \end{matrix} | = 0 \\ \Rightarrow (b - c) (c - a) - {(c - a)}^{2} = 0 \\ \Rightarrow a b - b^{2} - a c + b c - c^{2} - a^{2} + 2 a c = 0 \\ \Rightarrow - a^{2} - b^{2} - c^{2} + a b + b c + a c = 0 \\ \Rightarrow a^{2} + b^{2} + c^{2} - a b - b c - a c = 0 \\ \Rightarrow &thin \end{array}$

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7 months ago

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

Given $A = [- \begin{matrix} 2 & 2 & - 4 \\ 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}], a n d B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}],$ find $B A$ and use this to solve the system of equations:

$y + 2 z = 7, x - y = 3, 2 x + 3 y + 4 z = 17 .$

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted

7 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then, what inference can you make about their frequencies?

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