(a) Write the full form of BOD.
(b) Define BOD. Explain how it is a measure of the organic matter present in the water body.
(b) Define BOD. Explain how it is a measure of the organic matter present in the water body.
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Solution and Explanation
Step 2: (b) BOD is the amount of dissolved oxygen needed by aerobic microorganisms to break down organic matter present in a water sample.
- A higher BOD indicates more organic pollution and microbial activity.
- It is an indicator of the level of biodegradable organic matter in the water body.
Top Questions on Ecology and Environment
- Correctly depict (also indicate the trophic level) and describe the ecological pyramid of biomass in sea with 40 standing crop of phytoplankton supporting 90 standing crop of zooplankton which further supports 120 small fishes.
- CBSE CLASS XII - 2025
- Biology
- Ecology and Environment
Highly conserved proteins such as Haemoglobin and Cytochrome-C provide the best biochemical evidence to trace evolutionary relationships between different groups. Cytochrome-C is formed of 104 amino acids. Cytochrome-C is the respiratory pigment present in all eukaryotic cells. It has evolved at a constant rate during evolution. In chimpanzees and humans, Cytochrome-C genes are identical. The given data shows the evolution of the Cytochrome-C gene in different mammals from kangaroos, cows, rodents to humans:
Groups Nucleotide substitution in the gene of Cytochrome-C Millions of years ago Human/Kangaroo 100 125 mya Human/Cow 75 120 mya Human/Rodent 60 75 mya (a) Select the correct option for the time of separation of two groups and the number of nucleotide substitutions in the gene of Cytochrome-C:
Options Time of separation of two groups during evolution Number of nucleotide substitutions (i) Lesser Greater (ii) Greater Lesser (iii) Greater Greater (b) What do you infer about the type of evolution (convergent or divergent) for the given pair of groups and why?
(i) Human and Kangaroo
(ii) Human and Rodent
(c)
(i) Define convergent evolution.
OR
(ii) Define divergent evolution.- CBSE CLASS XII - 2025
- Biology
- Ecology and Environment
- Explain how the loss of habitat and fragmentation drives plants and animals to extinction with the help of an example of habitat loss in the Tropical Rain Forest. Also write the effect of fragmentation of a habitat on the population decline.
- CBSE CLASS XII - 2025
- Biology
- Ecology and Environment
- Cattle never feed on the leaves of Calotropis because of the presence of these chemicals
- AP EAPCET - 2025
- Zoology
- Ecology and Environment
- Study the following and pick up the correct statements:
[I)] \Daphnia exhibits cyclomorphosis
[II)] In a thermally stratified lake of temperate region, the upper layer in winter is epilimnion
[III)] Niche is the functional role of an organism in an ecosystem
[IV)] Bergmann’s rule states the relationship between temperature and metabolic rate of animals- AP EAPCET - 2025
- Zoology
- Ecology and Environment
Questions Asked in CBSE CLASS XII exam
- An electric dipole of dipole moment \( \vec{p} = (0.8\,\hat{i} + 0.6\,\hat{j}) \times 10^{-29}\,\text{Cm} \) is placed in an electric field \( \vec{E} = 1.0 \times 10^7\,\hat{k}\,\text{V/m} \). Calculate the magnitude of the torque acting on it and the angle it makes with the x-axis, at this instant.
- CBSE CLASS XII - 2025
- Electric Dipole
- Calculate the binding energy per nucleon (in MeV) of a helium nucleus \(\left( ^4_2\text{He} \right)\). Given: \[ \begin{aligned} m\left(^4_2\text{He}\right) &= 4.002603\,\text{u} \\ m_n &= 1.008665\,\text{u} \\ m_H &= 1.007825\,\text{u} \\ 1\,\text{u} &= 931.5\,\text{MeV}/c^2 \end{aligned} \]
- CBSE CLASS XII - 2025
- Nuclear physics
- An electric field \( \vec{E} \) is given by:
\[ \vec{E} = \begin{cases} +100\, \hat{i} \dfrac{\text{N}}{\text{C}} & \text{for } x>0 \\ -100\, \hat{i} \dfrac{\text{N}}{\text{C}} & \text{for } x<0 \end{cases} \] A right circular cylinder of length \( 10\, \text{cm} \) and radius \( 2\, \text{cm} \), is placed such that its axis coincides with the x-axis and its two faces are at \( x = -5\, \text{cm} \) and \( x = 5\, \text{cm} \). Calculate: (a) the net outward flux through the cylinder, and (b) the net charge inside the cylinder.- CBSE CLASS XII - 2025
- Electric Flux
In a metallic conductor, an electron, moving due to thermal motion, suffers collisions with the heavy fixed ions but after collision, it will emerge with the same speed but in random directions. If we consider all the electrons, their average velocity will be zero. When an electric field is applied, electrons move with an average velocity known as drift velocity (vd). The average time between successive collisions is known as relaxation time (τ ). The magnitude of drift velocity per unit electric field is called mobility (μ). An expression for current through the conductor can be obtained in terms of drift velocity, number of electrons per unit volume (n), electronic charge (−e), and the cross-sectional area (A) of the conductor. This expression leads to an expression between current density (-j) and the electric field (E→ ). Hence, an expression for resistivity (ρ) of a metal is obtained. This expression helps us to understand increase in resistivity of a metal with increase in its temperature, in terms of change in the relaxation time (τ ) and change in the number density of electrons (n).
- CBSE CLASS XII - 2025
- Current electricity
When light travels from an optically denser medium to an optically rarer medium, at the interface it is partly reflected back into the same medium and partly refracted to the second medium. The angle of incidence corresponding to an angle of refraction 90° is called the critical angle (ic) for the given pair of media. This angle is related to the refractive index of medium 1 with respect to medium 2. Refraction of light through a prism involves refraction at two plane interfaces. A relation for the refractive index of the material of the prism can be obtained in terms of the refracting angle of the prism and the angle of minimum deviation. For a thin prism, this relation reduces to a simple equation. Laws of refraction are also valid for refraction of light at a spherical interface. When an object is placed in front of a spherical surface separating two media, its image is formed. A relation between object and image distance, in terms of refractive indices of two media and the radius of curvature of the spherical surface can be obtained. Using this relation for two surfaces of lens, ’lensemaker formula’ is obtained.