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Question

A broadcasting station transmits waves with a frequency of $71 \times 10^4\ Hz$ and a speed of $3 \times 10^8\ m/s$. The wavelength of the wave is:

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is
422.5 m

Calculating Wave Wavelength

To find the wavelength of the waves transmitted by the broadcasting station, we use the fundamental wave equation that relates speed, frequency, and wavelength.

Wave Equation

The relationship between the speed of a wave ($v$), its frequency ($f$), and its wavelength ($\lambda$) is given by the formula:

$ v = f \lambda $

Solving for Wavelength

We need to rearrange the formula to solve for the wavelength ($\lambda$):

$ \lambda = \frac{v}{f} $

Given Values

  • Frequency ($f$): $71 \times 10^4\ Hz$
  • Speed ($v$): $3 \times 10^8\ m/s$

Calculation Steps

  1. Substitute the given values into the rearranged formula:

    $ \lambda = \frac{3 \times 10^8\ m/s}{71 \times 10^4\ Hz} $

  2. Simplify the expression:

    $ \lambda = \frac{3}{71} \times \frac{10^8}{10^4}\ m $

    $ \lambda = \frac{3}{71} \times 10^{(8-4)}\ m $

    $ \lambda = \frac{3}{71} \times 10^4\ m $

  3. Calculate the numerical value:

    $ \lambda \approx 0.0422535 \times 10^4\ m $

    $ \lambda \approx 422.535\ m $

  4. Round the result to match the options provided (one decimal place):

    $ \lambda \approx 422.5\ m $

Conclusion

The calculated wavelength of the wave is approximately $422.5\ m$. This corresponds to Option C.

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Important Questions from Waves

  1. Which of the following is related to Doppler effect?

  2. The velocity v(x) of a particle moving in one dimension is given by v(x) = v 0 sin \(\rm\left(\frac{\pi x}{x_0}\right) \) , where v 0  and x 0  are positive constants of appropriate dimensions. If the particle is initially at x/x 0  = ϵ, where |ϵ| ≪ 1, then, in the long time, it
  3. The position of a particle in one dimension changes in discrete steps. With each step it moves to the right, however, the length of the step is drawn from a uniform distribution from the interval \(\left[ {{\rm{λ }}\,{\rm{ - }}\,\frac{{\rm{1}}}{{\rm{2}}}{\rm{w,}}\,{\rm{λ }}\,{\rm{ + }}\,\frac{{\rm{1}}}{{\rm{2}}}{\rm{w}}} \right] \) , where λ and w are positive constants. If X denotes the distance from the starting point after N steps, the standard deviation \(\sqrt {\left\langle {{X^2}} \right\rangle \, - {{\left\langle X \right\rangle }^2}} \)  for large values of N is

  4. A particle of mass m in one dimension is in the ground state of a simple harmonic oscillator described by a Hamiltonian \(\frac{{{{\rm{P}}^{\rm{2}}}}}{{{\rm{2m}}}}{\rm{ + }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}{{\rm{\omega }}^{\rm{2}}}{{\rm{x}}^{\rm{2}}} \) in the standard notation. An impulsive force at time t = 0 suddenly imparts a momentum P0 \(\sqrt {{\rm{hm\omega }}} \) to it. The probability that the particle remains in the original ground state is

  5. In an elastic scattering process at an energy E, the phase shifts satisfy δ 0 ≈ 30°, δ 1≈  10°, while the other phase shifts are zero. The polar angle at which the differential cross-section peaks is closest to

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