To differentiate between classical and photochemical smog:
• Classical smog occurs in cool, humid conditions and is reducing in nature.
• Photochemical smog occurs in sunny conditions and contains strong oxidizing agents like ozone.
NO2 is present in classical smog
Both NO2 and SO2 are present in classical smog
1. Classical Smog:
Also known as "reducing smog."
It is caused by high concentrations of \(\text{SO}_2\) and particulate matter under cool, humid conditions.
Classical smog does not contain oxidizing agents like \(\text{NO}_2\).
2. Photochemical Smog:
Formed in warm, sunny conditions.
It is caused by the reaction of \(\text{NO}_2\) and volatile organic compounds (VOCs) under sunlight, forming ozone and other oxidizing agents.
This type of smog has a high concentration of oxidizing agents like \(\text{O}_3\) and peroxyacetyl nitrate (PAN).
Final Answer: (3) Photochemical smog has a high concentration of oxidizing agents.
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to: