Physics

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

Explanation- voltage across R_B= 10V

Resistance R_B= 400kohm

V_BE= 0, V_CE= 0, R_c=3kohm

I_B= voltage /R_B= 10/400 $*$  10³= 25 $*$ 10^-6A

Voltage across R_c= 10V

I_c=voltage/R_c= 10/3 $*$ 10³= 3.33 $*$ 10^-3

$\beta =\frac{I_c}{I_B}=\frac{3.33*{10}^{-3}}{25*{10}^{-6}}=133$

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

forward biased resistance = 25ohm

Reverse biased resistance = infinity

As CD branch is reverse biased having infinite resistance.

So I₃= 0

Resistance in branch AB= 25+125= 150 ohm (R₁)

Resistance in branch EF= 25+125= 150 ohm (R₂)

They both are in parallel

 So $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{150}+\frac{1}{150}=\frac{2}{150}$

R= 75ohm

So total resistance = 25+75= 100ohm

Current = V/R= 5/100=0.05A

I₁=I₂+I₃+I₄

I₁=I₄+I₂

Here the resistance R₁ and R₂ is same

I₄=I₂

I₁= 2I₂

I₂= I₁/2=0.05/2=0.025A

I₄= 0.025A

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b, d

Explanation- modulation index m= A_m/A_c

If m>1 then A_m>A_c

maximum modulation frequency m_f= frequency deviation / maximum frequency of modulating wave

here if m>1 then it means overlapping of both sides resulting loss of information.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- production of modulated wave is given by frequency of upper side band – frequency of lower side band so in first three cases it is clearly visible.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b, c, d

Explanation- height of tower= 240m

For line of sight communication maximum distance would be d= $\sqrt[]{2Rh}$

So d= $\sqrt[]{2*6.4*{10}^6*240}$  = 55.4km

So distance under this are communicable

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b, d

Explanation- w_m=3KHz

And w_c= 1.5MHz= 1500KHz

By using these two 1500 $?$ 3= 1503 and 1497

Bandwidth = 2w_m= 2 $*3$ = 6KHz

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, d

Explanation- frequency is given by Vm=15KHz

So wavelength is given by $?=\frac{c}{v}=\frac{3*{10}^8}{15*{10}^3}=\frac{1}{5}*{10}^{5}m$

Also we know that l = $?/4$  = 5km

The audio signals are of low frequency waves. Thus, they cannot be transmitted through sky

Waves as they are absorbed by atmosphere.

If the size of the antenna is less than 5 km, the effective power transmission would be very

Low because of deviation from resonance wavelength of wave and antenna length.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation-equation of modulating wave is, m (t)= A_msinw_mt

Carrier wave equation is C_m (t)= (A_c+A_msinw_mt)sinw_ct

 = A_c [1+ $\frac{A_m}{A_c}sin{w}_{m}t$ ]sinw_ct

Also $\frac{A_m}{A_c}$ = M

C_mt= (A_c+A_{c $*\propto sin{w}_{m}t$} )sinw_ct

Ac $*\propto$ =K

Sinw_mt= V_m

equation having phase $\varnothing$

so equation is A_c+K₂V_mt}+sinw_ct+ $?$ ]

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation- it is represented by a diagram given below