Large holes in 'Swiss cheese' are formed by the activity of microbes:
Show Hint
- Streptococcus pneumoniae
- Monascus purpureus
- Trichoderma polysporum
- Propionibacterium shermanii
The Correct Option is D
Solution and Explanation
Top Questions on Microbes in human welfare
- Explain the biological treatment of primary effluent when passed into the large aeration tanks in a sewage treatment plant (STP).
- CBSE CLASS XII - 2025
- Biology
- Microbes in human welfare
- Bioactive molecule Cyclosporin A used for human welfare is derived from:
- CBSE CLASS XII - 2025
- Biology
- Microbes in human welfare
- Assertion (A): Biogas plants are more often built in rural areas.
Reason (R): The excreta or gobar of cattle is rich in Methanobacterium.- CBSE CLASS XII - 2025
- Biology
- Microbes in human welfare
- Large scale industrial production of Butyric acid for human welfare is done using the microbe:
- CBSE CLASS XII - 2025
- Biology
- Microbes in human welfare
- (A) Answer the following questions:
State what do you understand by “MALT”? Where is it located inside our body?
Explain cytokine barriers.
Name the diagnostic test for AIDS. On what principle does it work?
Bone marrow and thymus play an important role in human immune system. Explain how are they able to achieve this. \begin{center} OR \end{center} (B)
Study the following table \& fill ‘H’, ‘T’, ‘J’, ‘K’, ‘L’ and ‘M’ in the following table with suitable words:
Why are baculoviruses used as biological control agents?- CBSE CLASS XII - 2025
- Biology
- Microbes in human welfare
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.