Class 12th

(i) Area of the triangle = 4 sq. units (given)

$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}\hfill k\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 4\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=4$

(ii) Area of the triangle = 4 sq units

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

When we send a signal to be transmitted through sky waves must have a frequency range of 1710 kHz to 40 MHz.

But, the frequency of TV signals are 60 MHz which is beyond the required range.

So, sky waves will not be suitable for transmission of TV signals of 60 MHz frequency.

The area of triangle from by the given points is area ( ΔABC) = $\frac{1}{2}$ $\left|\begin{array}{ccc}\hfill a\hfill & \hfill b+c\hfill & \hfill 1\hfill \\ \hfill b\hfill & \hfill c+a\hfill & \hfill 1\hfill \\ \hfill c\hfill & \hfill a+b\hfill & \hfill 1\hfill \end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill a+b+c+1\hfill & \hfill b+c\hfill & \hfill 1\hfill \\ \hfill b+c+a+1\hfill & \hfill c+a\hfill & \hfill 1\hfill \\ \hfill c+a+b+1\hfill & \hfill a+b\hfill & \hfill 1\hfill \end{array}\right|$C_1→ C₁ + C₂ + C₃

$=\frac{1}{2}(a+b+c+1)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill b+c\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill c+a\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill a+b\hfill & \hfill 1\hfill \end{array}\right|$ Taking (a + b + c + 1) common from C₁

$=\frac{(a+b+c+1)}{2}*0\left\{?c=c_3\right\}$

= 0

Hence the points A, B C are collinear.

(i) Area of triangle is given by,

Δ = $\frac{1}{2}$ $\left|\begin{array}{ccc}\hfill x_1\hfill & \hfill y_1\hfill & \hfill 1\hfill \\ \hfill x_2\hfill & \hfill y_2\hfill & \hfill 1\hfill \\ \hfill x_3\hfill & \hfill y_3\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 6\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 4\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right|$

$=\frac{1}{2}|[-3+18]|=\frac{15}{2}$ =7.5sq. units.

(ii) Area of the triangle is given by,

Δ = $\frac{1}{2}$ $\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 10\hfill & \hfill 8\hfill & \hfill 1\hfill \end{array}\right|$

$=\frac{1}{2}|[2(1-8)-7(1-10)+1(8-10)]|$

$=\frac{1}{2}|(2*(-7)-7*(-9)+21*(-2)]|$

$=\frac{1}{2}|[-14+63-2]|=\frac{1}{2}\left|47\right|$

= $\frac{47}{2}$ =23.5 sq. units

(iii) Area of triangle is given by,

Δ = $\frac{1}{2}$ ${\left|\begin{array}{ccc}\hfill -2\hfill & \hfill -3\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill -8\hfill & \hfill 1\hfill \end{array}\right|}_{.}$

$=\frac{1}{2}|[-2*10+3(4)+(-24+2)]|$

$=\frac{1}{2}|[-20+12-22]|$

$=\frac{1}{2}|-30|=\frac{30}{2}$  = 15 sq. units.

Frequency of career wave is 20MHz

Bandwidth require for modulation is 3kHz/2= 1.5kHz

For demodulating we need to reciprocal it

1/f= 1/20MHz= 0.5 $*$ 10^-7s

For modulation = 1/1.5KHz= 0.7 $*$ 10³s

According to first option =RC= 1 $*$ 0.01= 10^-5s

So it will demodulated

According to 2^nd option- RC= 10^-4s

So it can also be demodulated

According to third option – RC= 10^-8s

So it cannot be modulated

Option 'C' is correct as determinant is a number associated to a square matrix.

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

in first case crowding spectrum is visible

If we want more wave to be modulated then more crowding will occur and more mixing up of signal.

But we want to accommodate this we use higher band width and frequency career wave

Then, KA = $k\left[\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill k{a}_{11}\hfill & \hfill k{a}_{12}\hfill & \hfill k{a}_{13}\hfill \\ \hfill k{a}_4\hfill & \hfill k{a}_{22}\hfill & \hfill k{a}_{23}\hfill \\ \hfill k{a}_{31}\hfill & \hfill k{a}_{32}\hfill & \hfill k{a}_{33}\hfill \end{array}\right]$

$\therefore |KA|=\left|\begin{array}{ccc}\hfill {}_{a11}\hfill & \hfill k{a}_{12}\hfill & \hfill {}_{13}\hfill \\ \hfill {}_{a_{21}}\hfill & \hfill {}_{22}\hfill & \hfill k{a}_{23}\hfill \\ \hfill {}_{31}\hfill & \hfill {}_{a_{32}}\hfill & \hfill k_{a_{33}}\hfill \end{array}\right|$

$=k{}^3\left|\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{22}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right|$

$=k^3|A|$

So, option c is correct.

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

Maximum voltage = 100/2 = 50V

And minimum voltage = 20/2 = 10V

Percentage modulation= max voltage -min voltage/ max voltage + min voltage $*100$

= $\frac{50-10}{50+10}*100$ = 66.67%

Peak career voltage= max voltage+ min voltage/2= 50+10/2=30V

Peak value of information voltage= 66.67/100 $*$  30= 20V

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

height of satellite hs= 600km
As we know velocity = distance/time

 2x/4.0410^-3 = 3⁸

So x=606 km after solving 
According to Pythagoras theorem d²=x²-h²= 606²-600²= 7236

So d= 85.06km so total distance will be double from receiver to transmitter = 170.1km
And d = √2Rh
 h=7236/2*6400 = 565m