\(\frac{ π§_1 + π§_2 + π§_3 + π§_4}{ π§_1π§_2π§_3π§_4 }= \frac{4}{13}\)
By Vietaβs formulas for a polynomial, the sum and product of the roots of the polynomial are related to the coefficients of the polynomial.
For the given polynomial:
\[ f(z) = z^4 - 8z^3 + 27z^2 - 38z + 26 \]
The sum of the roots, \( z_1 + z_2 + z_3 + z_4 \), is equal to the coefficient of \( z^3 \) (with the opposite sign):
\[ z_1 + z_2 + z_3 + z_4 = 8 \]
The product of the roots, \( z_1 z_2 z_3 z_4 \), is equal to the constant term (with the opposite sign):
\[ z_1 z_2 z_3 z_4 = 26 \]
The required ratio is:
\[ \frac{z_1 + z_2 + z_3 + z_4}{z_1 z_2 z_3 z_4} = \frac{8}{26} = \frac{4}{13} \]
Thus, the correct answer is option (B).
At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/mΒ³, which is dimensionally incorrect for a spectral density. The correct unit J/(mΒ³Β·Hz) or JΒ·s/mΒ³ is used here for the solution.)