📘 Physics Quiz: Orbital Motion, Waves & Communication

Sharpen your problem-solving skills with these NTA-style conceptual physics questions!

Q1.

A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P', where R is the radius of the planet. The time period of another satellite at a height of 2R from the surface of 'P' is ____ hours, given that the time period of the geostationary satellite is 24 hours.

Options:

1. 6√2
2. 6
3. 3
4. 5

Show Solution

Official Answer: (3)

Explanation:

The time period of a satellite is proportional to \( r^{3/2} \). For the geostationary satellite: \[ T_1 \propto (12R)^{3/2} \] For the second satellite: \[ T_2 \propto (3R)^{3/2} \] \[ \frac{T_2}{T_1} = \left(\frac{3}{12}\right)^{3/2} = \frac{1}{8} \] So, \( T_2 = \frac{24}{8} = 3 \, \text{hours.} \)

Q2.

A sound wave of frequency 245 Hz travels with the speed of 300 m/s along the positive x-axis. Each point of the wave moves to and fro through a total distance of 6 cm. What will be the mathematical expression of this travelling wave?

Options:

1. \( y(x,t) = 0.03 \sin(5.1x - 0.2 \times 10^3 t) \)
2. \( y(x,t) = 0.06 \sin(5.1x - 1.5 \times 10^3 t) \)
3. \( y(x,t) = 0.06 \sin(0.8x - 0.5 \times 10^3 t) \)
4. \( y(x,t) = 0.03 \sin(5.1x - 1.5 \times 10^3 t) \)

Show Solution

Official Answer: (4)

Explanation:

\( f = 245 \, \text{Hz}, \, v = 300 \, \text{m/s}, \, A = 3 \, \text{cm} = 0.03 \, \text{m} \)

Wave number \( k = \frac{2\pi f}{v} = \frac{2\pi \times 245}{300} \approx 5.1 \)
Angular frequency \( \omega = 2\pi f = 1.5 \times 10^3 \)

Equation of wave: \( y(x,t) = 0.03 \sin(5.1x - 1.5 \times 10^3 t) \)

Q3.

The velocity of a particle is given by \( v = v_0 + gt + Ft^2 \). Its position is \( x = 0 \) at \( t = 0 \). What will be its displacement after time \( t = 1 \)?

Options:

1. \( v_0 + g + F \)
2. \( v_0 + \frac{g}{2} + \frac{F}{3} \)
3. \( v_0 + \frac{g}{3} + \frac{F}{2} \)
4. \( v_0 + 2g + 3F \)

Show Solution

Official Answer: (2)

Explanation:

\( v = v_0 + gt + Ft^2 \Rightarrow dx = (v_0 + gt + Ft^2) dt \)

Integrating from \( 0 \) to \( 1 \): \[ x = \int_0^1 (v_0 + gt + Ft^2) dt = v_0 t + \frac{gt^2}{2} + \frac{Ft^3}{3} \] Substituting \( t = 1 \): \[ x = v_0 + \frac{g}{2} + \frac{F}{3} \]

Q4.

A carrier signal \( C(t) = 25 \sin(2.512 \times 10^{10} t) \) is amplitude modulated by a message signal \( m(t) = 5 \sin(1.57 \times 10^8 t) \) and transmitted through an antenna. What will be the bandwidth of the modulated signal?

Options:

1. 8 GHz
2. 2.01 GHz
3. 1987.5 MHz
4. 50 MHz

Show Solution

Official Answer: (4)

Explanation:

Given: \( m(t) = 5 \sin(1.57 \times 10^8 t) \) \[ \omega_m = 1.57 \times 10^8 = 2\pi f_m \Rightarrow f_m = \frac{1.57 \times 10^8}{2\pi} \approx 25 \times 10^6 \, \text{Hz} = 25 \, \text{MHz} \] Bandwidth of AM signal = \( 2f_m = 50 \, \text{MHz} \)

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