Given:
Density of wire, \( \rho = 6 \times 10^4 \, \mathrm{kg/m^3} \)
Breaking stress, \( \sigma = 1.2 \times 10^8 \, \mathrm{N/m^2} \)
Acceleration due to gravity on the planet, \( g' = \frac{g}{3} = \frac{10}{3} \, \mathrm{m/s^2} \)
The breaking stress (\( \sigma \)) is given by:
\[ \sigma = \frac{T}{A} = \frac{mg}{A} \]
Where \( T \) is the tension, \( m \) is the mass, and \( A \) is the cross-sectional area.
Since \( m = \rho A \ell \) (where \( \ell \) is the length of the wire), we have:
\[ \sigma = \frac{(\rho A \ell) g'}{A} = \rho \ell g' \]
Rearranging for \( \ell \):
\[ \ell = \frac{\sigma}{\rho g'} \]
Substituting the given values:
\[ \ell = \frac{1.2 \times 10^8}{6 \times 10^4 \times \frac{10}{3}} = 600 \, \mathrm{m} \]
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32