The density \( \rho \) of a cubical block is given by: \[ \rho = \frac{{mass}}{{volume}} = \frac{m}{a^3} \] where \( m \) is the mass and \( a \) is the side length of the cube. The relative error in density is given by: \[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \cdot \frac{\Delta a}{a} \] Given:
Error in mass \( \frac{\Delta m}{m} = 2% \)
Error in side length \( \frac{\Delta a}{a} = 1% \)
Substituting the values: \[ \frac{\Delta \rho}{\rho} = 2% + 3 \cdot 1% = 2% + 3% = 5% \] Thus, the error in the determination of the density is 5%.
Final Answer: 5%
Mass Defect and Energy Released in the Fission of \( ^{235}_{92}\text{U} \)
When a neutron collides with \( ^{235}_{92}\text{U} \), the nucleus gives \( ^{140}_{54}\text{Xe} \) and \( ^{94}_{38}\text{Sr} \) as fission products, and two neutrons are ejected. Calculate the mass defect and the energy released (in MeV) in the process.
Given: