The volume of a cone is given by:
\[ V = \frac{1}{3} \pi R^2 H \] where \( R \) is the radius of the base and \( H \) is the height of the cone.
The relative error in the volume \( V \) is: \[ \frac{\Delta V}{V} = 2 \frac{\Delta R}{R} + \frac{\Delta H}{H} \]
The percentage error in measuring the volume is: \[ \% \text{error in volume} = 2 \cdot \frac{\Delta R}{R} + \frac{\Delta H}{H} \cdot 100 \] Substitute \( \Delta R = 0.2 \, \text{cm} \), \( R = 20 \, \text{cm} \), \( \Delta H = 0.2 \, \text{cm} \), and \( H = 20 \, \text{cm} \): \[ \% \text{error in volume} = 2 \cdot \frac{0.2}{20} + \frac{0.2}{20} \cdot 100 \] Simplify: \[ \% \text{error in volume} = \left[ 2 \cdot 0.01 + 0.01 \right] \cdot 100 = 3 \]
The maximum percentage error in the determination of the volume is:
\[ \boxed{3\%} \]
The ratio of the power of a light source \( S_1 \) to that of the light source \( S_2 \) is 2. \( S_1 \) is emitting \( 2 \times 10^{15} \) photons per second at 600 nm. If the wavelength of the source \( S_2 \) is 300 nm, then the number of photons per second emitted by \( S_2 \) is ________________ \( \times 10^{14} \).
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.