RankLab
Back to Questions
PhysicsMediumNumerical2022 · 25 Jul Shift 1

Q30.The required height of a TV tower which can cover the population of 6 . 03 lakh is ℎ. If the average population density is 100 per square km and the radius of earth is 6400 km, then the value of ℎ will be _____ m.

What This Question Tests

This problem combines the formula for the range of a TV tower with the calculation of area covered and population density to determine the required height of the tower. It tests understanding of communication system range.

Concepts Tested

Line of Sight CommunicationRange of TV TowerEarth CurvaturePopulation Coverage

Formulas Used

d = √(2hR)

Area = πd^2

📚 NCERT Sections This Tests

9.27(A) M = ( Fo/Fe) = 28

Physics Class 12 · Chapter 9

70% match

9.27 (a) m = ( fO/fe) = 28 f O  f O  (b) m = 1 + = 33.6 f e  25  349 Reprint 2025-26 Physics 9.28 (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective h h = = f O 140 Equating the two, h = 4.7 cm. (c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. 9.29 The image formed by the larger (concave) mirror acts as virtual object for the smaller (convex) mirror. Parallel rays coming from the object at infinity will focus at a distance of 110 mm from the larger mirror. The distance of virtual object for the smaller mirror = (110 –20) = 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror formula, image is formed at 315 mm from the smaller mirror. 9.30 The reflected rays get deflected by twice the angle of rotation of the mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm. 9.31 n = 1.33 CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) l = 589 nm, n = 5.09 ´ 1014 Hz, c = 3.00 ´ 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) n = 5.09 ´ 1014Hz v = (c/n) = 2.26 × 108 m s–1, l = (v/n) = 444 nm 10.2 (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar). 10.3 (a) 2.0 × 108 m s–1 (b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. [When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour.] Now we know violet colour deviates more than red in a glass prism, i.e. nv > nr. Therefore, the violet component of white light travels slower than the red component. 1.2 10 – 2  0.28 10 – 3 10.4  m = 600 nm 4 14. 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 350 10.8 tan–1(1.5) ~ 56.3o Reprint 2025-26 Answers

9.5A Small Bulb Is Placed At The Bottom Of A Tank Containing Water To A

Physics Class 12 · Chapter 9

70% match

9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

8.2A Parallel Plate Capacitor (Fig. 8.6) Made Of Circular Plates Each Of Radius

Physics Class 11 · Chapter 8

70% match

8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to 213 a 230 V ac supply with a (angular) frequency of 300 rad s–1. Reprint 2025-26 Physics (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. FIGURE 8.6 8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? 8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? 8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? 8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? 8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B. 8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? 8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s–1.] Reprint 2025-26

📋 Question Details

Chapter
Communication Systems
Topic
Antenna and Range
Year
2022
Shift
25 Jul Shift 1
Q Number
Q30
Type
Numerical
NCERT Ref
Class 12 Physics Ch 15: Communication Systems

More from this Chapter

Q78.Consider telecommunication through optical fibres. Which of the following statements is not true? (1) Optical fibres can be of graded refractive index (2) Optical fibres are subjective to electromagnetic interference from outside (3) Optical fibres have extremely low transmission (4) Optical fibres may have homogeneous core with loss a suitable cladding.

2003
Easy

Q29.This question has Statement −1 and Statement −2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1 : Sky wave signals are used for long distance radio communication. These signals are in general, less stable than ground wave signals. Statement-2 : The state of ionosphere varies from hour to hour, day to day and season to season. (1) Statement-1 is true, Statement-2 is true; (2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-2 is not the correct explanation of Statement-1. Statement-1. (3) Statement- 1 is false, Statement- 2 is true. (4) Statement-1 is true, Statement-2 is false. JEE Main 2011 JEE Main Previous Year Paper

2011
Easy

Q30.Broadcasting antennas are generally (1) vertical type (2) both vertical and horizontal type (3) omni directional type (4) horizontal type

2012
Easy

Q29.Given the electric field of a complete amplitude modulated wave as Em →E = ^iEc + cos cos ωct (1 Ec ωmt) Where the subscript c stands for the carrier wave and m for the modulating signal. The frequencies present in the modulated wave are (1) ωc and √ω2c + ω2m (2) ωc, ωc + ωm and ωc −ωm (3) ωc and ωm (4) ωc and √ωcωm

2012
Medium
More Physics questions