The formula for resistivity is given by:
\(\rho = R \frac{\pi r^2}{l}\)
The relative error in resistivity is given by:
\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta r}{r} + \frac{\Delta l}{l}\)
Substituting the given values:
\(\frac{\Delta \rho}{\rho} = \frac{10}{100} + 2 \times \frac{0.05}{0.35} + \frac{0.2}{15}\)
Calculating each term:
\(\frac{\Delta \rho}{\rho} = 0.1 + 2 \times 0.1429 + 0.0133\)
\(\frac{\Delta \rho}{\rho} \approx 0.1 + 0.2858 + 0.0133 = 0.3991 \approx 39.9\%\)
If \( T = 2\pi \sqrt{\frac{L}{g}} \), \( g \) is a constant and the relative error in \( T \) is \( k \) times to the percentage error in \( L \), then \( \frac{1}{k} = \) ?
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