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6 public questions tagged with this topic.

A material has \( B = 0.3 \, \text{T} \) and \( M = 1.5 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \mu

**Potential energy of magnetic dipole** U = -m B cosθ explains stability. Given m = 0.9 A·m², B = 0.5 T, θ = 90°, sin90° = 1, τ = 0.45 N·m. For 60°, sin60° = √3/2 ≈0.866, reducing torque proportionally. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.3 T , M = 1.5 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.3/4π × 10⁻⁷) ≈ 2.387 × 10⁵ A m⁻¹ . H = 2.387 × 10⁵ - 1.5 × 10⁵ = 8.87 × 10⁴ A m⁻¹ ≈ 8.9 × 10⁴

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

The formula for calculating generation time (g) in bacterial growth is:

Quantitative estimation of bacterial growth uses relationship between elapsed time and number of doublings. Definition generation time g equals total incubation time t divided by number of generations n realized in that interval, g equals t over n. Generation number derived from cell counts using formula n equals log2 Nt over N0 equals log10 Nt minus log10 N0 over 0.301, where N0 initial density, Nt final. This equation derives from geometric progression Nt equals N0 times 2 to n. Therefore knowing t and n yields g, alternatively g predicts expected increase. Reciprocal relationship gives specific growth rate mu equals ln2 over g equals 0.693 over g. Correct dimensionality requires time units. Formulas multiplying N0 and Nt or adding lack theoretical basis. Laboratory application: plot log viable count versus time slope gives mu, intercept N0, g calculated. Accurate derivation essential for designing chemostat dilution rates equal to mu to maintain steady state, evaluating bacteriostatic agents reducing mu, and modeling infection dynamics where generation time dictates time to reach pathogenic threshold load.

Ref: Brock Biology of Microorganisms, 16th ed., Chapter 6: Formula for generation time g=t/n.

The specific growth rate (μ) is calculated using the formula:

Specific growth rate mu quantifies fractional increase in biomass per unit time during balanced exponential phase and is central to microbial physiology and bioprocess engineering. Derived from exponential model Nt equals N0 times e to power mu t where Nt is cell number at time t and N0 initial value. Taking natural logs gives ln Nt equals ln N0 plus mu t rearranged to mu equals ln Nt minus ln N0 over t equals ln ratio Nt over N0 divided by t with units reciprocal hour. If common logs are used factor 2.303 conversion is required because ln x equals 2.303 log10 x. Parameter links directly to generation time g via mu equals ln 2 over g about 0.693 over g. Accurate determination requires cells in balanced exponential growth with excess nutrients, constant temperature and log-phase inoculum to eliminate lag. Alternative expressions like t over Nt are dimensionally incorrect giving time over cells, additive forms violate mass balance and fail to describe exponential kinetics. Therefore natural logarithmic ratio form remains standard textbook definition routinely used for modeling, scale-up and predictive microbiology.

Ref: Madigan et al., Brock Biology of Microorganisms, 16th ed., Chapter 4: Specific Growth Rate Formula μ = ln(Nt/N0)/t.

In Beer-Lambert’s law, the absorbance A equals:

Beer-Lambert law integrates Lambert observation that absorbance proportional to path length and Beer observation proportional to concentration. Resulting expression equates absorbance to product of molar absorptivity ε reflecting transition probability, molar concentration c and path length l in centimeters. Mathematically A = log10(I0/I) = ε c l, valid under dilute, non-scattering, monochromatic light conditions. Deviation occurs at high concentration, polychromatic radiation or scattering. This fundamental equation underlies spectrophotometric determination of proteins, nucleic acids, NADH kinetics, enzyme assays, equilibrium constant measurement, essential for NEET, CBSE and CSIR-NET quantitative problem solving.

Ref: NCERT Biology Class XII Principles on Klenow fill-in labeling, Lehninger Chapter 9 DNA cloning techniques, and Molecular Cloning by Sambrook Chapter 10 documenting end-labeling of cohesive termini.

Which is the correct formula for NPP?

Net primary productivity is the rate at which producers accumulate new organic matter after meeting their own respiratory costs. Gross primary productivity represents all carbon fixed or energy captured by autotrophs. Autotrophic respiration consumes part of that material to support cellular maintenance, active transport, tissue construction, and repair. The balance is therefore NPP = GPP − R, where R in this formula means respiration by primary producers, often written Ra. NPP supplies plant growth and reproduction and is the energy potentially available to herbivores and decomposers. Adding respiration would count respired carbon as retained production, while reversing the subtraction could yield a biologically meaningless negative value. Division by respiration is a ratio, not a production rate. This equation must also be distinguished from net ecosystem production, which subtracts both autotrophic and heterotrophic respiration from GPP. The carbon-balance mechanism explains the sign: photosynthesis adds organic carbon to producer biomass, whereas respiration oxidizes some organic carbon and returns carbon dioxide, leaving the difference as net production.

Ref: Ecology: From Individuals to Ecosystems, Begon et al., 5th Ed., Ch. 17