Class 12th

Given,  $?=\left|\begin{array}{lll}5\hfill & 3\hfill & 8\hfill \\ 2\hfill & 0\hfill & 1\hfill \\ 1\hfill & 2\hfill & 3\hfill \end{array}\right|$

Co-factors of elements of second row,

? ? = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃

= 2 * 7 + 0 * 7 + 1 * (-7) = 14 + 0 - 7 = 7.

(i) Given, A = $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|$

So,

(ii)Given,  $\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|$

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

Total distance = 5km and loss is 2 dB/km

So total loss = 5 (2)= 10 dB

Total gain in amplifier  10+20= 30dB and gain in signal is 20dB

So by the formula 20= 10log_{10 $\frac{p_o}{p_i}$}

log_{10 $\frac{p_o}{p_i}$} = 2

so po/pi= 10²

so Po= Pi (100)= 101mW

(i) We know that,

Minor of element a_ij is m_ij and its co-factor is A_ij = (–1)^i+j M_ij

So,

M₁₁ = 3 and A₁₁ = (- 1)^{1 + 1} M₁₁ = 1 * 3 = 3

M₁₂ = 0 and A₁₂ = (-1)¹⁺² M₁₂ = -1 * 0 = 0

M₂₁ = -4 and A₂₁ = (-1)²⁺¹ M₂₁ = (-1) * (-4) = 4

M₂₂ = 2 and A₂₂ = (-1)²⁺² M₂₂ = 1 * 2 = 2

(ii) Given A = $\left|\begin{array}{cc}\hfill a\hfill & \hfill c\hfill \\ \hfill b\hfill & \hfill d\hfill \end{array}\right|$

So,

M₁₁ = d and A₁₁ = (-1)¹⁺¹ M₁₁ = 1 * d = d

M₁₂ = b and A₁₂ = (-1)¹⁺² M₁₂ = (-1) * b = -b

M₂₁ = c and A₂₁ = (-1)^{2+ 1} M₂₁ = (–1) * c = –c

M₂₂ = a and A₂₂ = (-1)²⁺² M₂₂ = 1 * a = a

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

In modulating signal we add career wave or noise signal to send it to receiver end and this wave is varied from time to time that is why more noise appear.

But in frequency modulation frequency is not varied so less noise appear.

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

Frequency tuned amplifier is $\frac{1}{2\pi \sqrt[]{LC}}=1MHz$

$\sqrt[]{LC}$ =1/2 $\pi *{10}^6$

$LC=2.54*{10}^{-14}s$

Given,

Area of triangle = 35 sq. Units

$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}\hfill 2\hfill & \hfill -6\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 4\hfill & \hfill 1\hfill \\ \hfill k\hfill & \hfill 4\hfill & \hfill 1\hfill \end{array}\right|=35.$

∴ Option D is correct.

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

Maximum amplitude = Am+Ac =12 while minimum amplitude is Ac-Am=3

Using elimination method = 2Ac= 18

Ac=9 and Am= 6V

So modulating index m = 6/9= 2/3

(i) Let P (x, y) be any point on line joining A (1, 2) & B (3, 6)

Then, area of triangle (ABP) = 0 {the point are collinear

$\frac{1}{2}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 6\hfill & \hfill 1\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill 1\hfill \end{array}\right|=0$

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

As frequency of B is more than A so it has more refractive index also and if a wave have higher refractive index then it has less angle of refraction. So wave B travel more in ionosphere.