Q.In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (loge2=0.6931).
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Exponential Growth Rate
Exponential Growth Rate
A quantity grows exponentially when its rate of change is proportional to its current size: the more there is, the faster it grows. This differs sharply from linear growth, where a fixed amount is added each step. In exponential growth the quantity multiplies by the same factor over equal time intervals.
The differential equation
Let y(t) be the quantity and k>0 the proportionality constant. The rate law "rate of change proportional to the current amount" becomes
dtdy=ky.
This is a separable equation. Integrating,
∫ydy=∫kdt⟹log∣y∣=kt+C⟹y=y0ekt,
where y0=y(0) is the starting value. The constant k is the growth rate: a larger k means faster growth. (If k<0, the very same equation describes exponential decay.)
Reading the growth rate
Over each unit of time, y is multiplied by ek. So if the quantity doubles every unit of time, then ek=2, giving k=log2. This is how a stated doubling time is converted into the constant k.
Linear growth adds the same amount each step; exponential growth multiplies by the same factor. That is why an exponential quantity looks slow at first and then climbs steeply — the increase itself keeps getting bigger.
Where it appears …
The key idea is Exponential Growth Rate: when a quantity increases continuously at a fixed percentage rate, it follows the law dtdP=100rP.
Step 1 – Set up the differential equation
Let P be the principal. Continuous growth at r% per year gives:
dtdP=100rP
Step 2 – Solve
Separate variables and integrate:
∫PdP=∫100rdt⇒logP=100rt+C
So P=P0ert/100, where P0 is the initial principal.
Step 3 – Apply given data …
The problem involves continuous exponential growth of money in a bank. Using the formula for continuous compounding, A=Pert, and the given doubling time of 10 years, we find the rate r=10ln2≈6.931% per year.
The key idea here is that "principal increases continuously at the rate of r% per year" means we are dealing with exponential growth — the same mathematics that governs population growth, radioactive decay, and compound interest compounded every instant. Unlike simple interest or annual compounding, continuous growth means the money is growing at every moment, and the growth itself is proportional to the current amount.
This is a classic differential equation situation: if P(t) is the principal at time t (in years), then the statement "increases continuously at the rate of r% per year" translates to:
dtdP=100rP
The factor 100r converts the percentage rate into a decimal. The solution to this is P(t)=P0ert/100, where P0 is the initial principal.
Now let's work through the problem step by step.
- Set up the continuous growth equation. Let P0=100 (Rs). After t=10 years, the amount doubles to P=200. The continuous growth formula is:
P(t)=P0e100rt
Substituting the known values:
200=100e100r⋅10
- Simplify the equation. Divide both sides by 100:
2=e10010r=er/10
- Solve for r using natural logarithms. Take the natural log of both sides:
ln2=10r
Therefore:
r=10ln2
- Plug in the given value of ln2. …
Method: Continuous growth ("rate proportional to amount")
Use this for interest/population problems where a quantity grows continuously at a rate proportional to itself, and you must find the rate constant.
Steps
Step 1: Write the proportional-rate equation
"Increases continuously at rate r%" means
dtdP=100rP.
Step 2: Separate and integrate to an exponential
PdP=100rdt gives P=P0ert/100.
Step 3: Use the given condition and take logs …
Common Mistakes
Mistake 1: Writing the rate as r instead of 100r
Why it's wrong: "r% per year" means dtdP=100rP; using r directly gives an answer off by a factor of 100. Correct approach: convert the percentage to 100r.
Mistake 2: Not taking logarithms to release the exponent …
- AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQQ.In logistic growth equation dTdN=rN[KK−N] where 'r' represents (A) Population density (B) Intrinsic rate of natural increase (C) Carrying capacity (D) Age distribution
›Reveal solutionSolution
In the standard logistic growth equation, r denotes the intrinsic rate of natural increase, the species-specific maximum per-capita growth rate, while K is separately the carrying capacity — matching option (B).
Concept and Intuition
The logistic growth model describes how a population grows rapidly when resources are abundant (population size N much smaller than carrying capacity K) but growth slows and eventually plateaus as N approaches K, due to increasing resource limitation, captured by the term [KK−N]. The constant r in this equation, called the intrinsic rate of natural increase, represents the population's inherent capacity to grow per individual per unit time in the absence of any limiting factor — it is determined by the species' biology (birth rate minus death rate under ideal conditions) and differs from K, which represents the maximum population size the environment can sustain, not a rate at all.
Step-by-Step Solution
- Identify the variables in the given equation: N = population size, K = carrying capacity, r = the proportionality constant governing growth rate.
- Recall that K is explicitly the environment's carrying capacity — this rules out option (C), since K, not r, represents carrying capacity. …
- AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQQ.The integral form of the exponential growth equation is (A) dtdN=(b−d)N (B) dtdN=rNK(K−N) (C) dtdN=rn (D) Nt=Noert
›Reveal solutionSolution
The differential form is dN/dt=rN; integrating it gives the integral form Nt=N0ert, which is option (D).
Concept and Intuition
Population ecology distinguishes the rate of change (differential form) from the closed-form solution describing population size at any time (integral form). Exponential growth assumes unlimited resources, so the rate of increase is proportional to the current population size.
Step-by-Step Solution
- Start from the differential (instantaneous rate) form: dtdN=rN, where r=b−d (birth rate minus death rate).
- Separate variables: NdN=rdt.
- Integrate both sides: logN=rt+C.
- Apply the condition N=N0 at t=0: C=logN0.
- Exponentiate: Nt=N0ert — this is the integral (solved) form. …
- AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQQ.If a>0,b>0 then n→∞lim(aa+b1/n−1)n= (A) ab (B) ba (C) b1/a (D) a1/b
›Reveal solutionSolution
A (1+x/n)n→ex type limit after expanding b1/n for large n; the answer is b1/a.
Concept and Intuition
Whenever you see (1+nf(n))n with f(n)→L, the limit is eL. Here we must first simplify b1/n−1 using the exponential expansion, since b1/n=e(logb)/n≈1+nlogb for large n.
Step-by-Step Solution
- Write b1/n=enlogb=1+nlogb+O(1/n2).
- Then a+b1/n−1=a+nlogb+O(1/n2).
- So aa+b1/n−1=1+anlogb+O(1/n2). …
- AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQQ.In plants Growth rate is expressed in Arithmetic and Geometric growth equations (A) L0=W1ert W0=L0+rt (B) Lt=L0+rt W1=W0ert (C) W0=W1ert L0=Lt+rt (D) Lt=W0+rt W0=L0+ert
›Reveal solutionSolution
Arithmetic growth increases by a constant amount per unit time (Lt=L0+rt); geometric growth compounds exponentially (W1=W0ert).
Concept and Intuition
Plant growth can follow two mathematical patterns:
- Arithmetic growth: after mitotic division, only one daughter cell continues to divide, the other differentiates — the growth increment per unit time stays constant, giving a linear increase, plotted as a straight line: Lt=L0+rt (length at time t = initial length + rate × time). A classic example is a root elongating at a constant rate.
- Geometric growth: initially both daughter cells retain the capacity to divide, so growth accelerates — first a lag phase, then exponential (log) growth, described by W1=W0ert (final size = initial size × e^(rate × time)). This is the pattern of "unlimited" exponential growth seen early in most growing organisms/populations before resources become limiting.
Step-by-Step Solution
- Identify arithmetic growth's linear form: Lt=L0+rt.
- Identify geometric growth's exponential form: W1=W0ert. …
- AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQQ.Choose the correct statement related to growth rate, conditions and growth substances A. Root elongation is the arithmetic growth B. Geometric growth shows sigmoid growth curve C. Turgidity helps the cells in extension growth D. Natural cytokinins are synthesised in older parts of the plant (A) A, B, C (B) B, C, D (C) A, C, D (D) A, B, D
›Reveal solutionSolution
Three of the four statements — arithmetic growth in roots, the sigmoid curve arising from geometric growth, and turgidity driving extension growth — are standard correct facts; the claim that cytokinins are made in older parts of the plant is the deliberate error (they're made in actively dividing, young tissue).
Concept and Intuition
- Arithmetic growth: after mitosis, if only one daughter cell keeps dividing while the other differentiates, growth proceeds at a constant rate — root elongation at a constant rate is the classic example.
- Geometric growth: initial growth is slow (lag phase), then rapid/exponential; but as resources become limiting, the rate declines and growth plateaus (stationary phase) — plotted over time, this lag → exponential → plateau progression traces out the characteristic sigmoid (S-shaped) curve typical of organisms growing in a natural, resource-limited environment.
- Turgidity and extension growth: plant cells elongate mainly by taking up water, which raises internal turgor pressure against the cell wall, stretching it — turgor-driven extension is the dominant mode of plant cell growth.
- Cytokinins: these are synthesised predominantly in regions of active cell division — root apices, developing shoot buds, young leaves, and developing seeds/fruits — not in old, mature/senescing tissue.
Step-by-Step Solution
- A: root elongation as arithmetic growth — correct. …
- AP EAPCET 2021Set ap-2021-09-06-AN1 markMCQQ.Study the following statements of plant growth. Choose the correct option?(i) One single maize root apical meristem can give rise to mor than 17,500 new cells per hour(ii) A cell in watermelon can increase in its size up to 3,50,000 times(iii) Growth of pollen tube is measured in the terms of its length(iv) Growth in dorsiventral leaf is measured in terms of an increase in its surface area (A)(i) &(ii) only (B)(ii) &(iii) only (C)(iii) &(iv) only (D) (i), (ii),(iii) & (iv)
›Reveal solutionSolution
Tests recall of standard growth-rate/growth-measurement facts from the Plant Growth and Development chapter; all four listed statements are correct.
Concept and Intuition
Growth in plants is not measured by a single universal metric — the appropriate parameter depends on the organ's geometry and growth pattern. A structure that elongates in one direction (like a pollen tube, which grows only at its tip) is best measured by length. A structure that expands in a flat plane (like a dorsiventral/bifacial leaf, which has distinct upper and lower surfaces) is better measured by surface area, since length alone wouldn't capture its lateral expansion. Cell division rates (as in the maize root meristem) and cell enlargement magnitudes (as in watermelon fruit cells) are used to illustrate just how extreme and rapid plant growth can be at the cellular level.
Step-by-Step Solution
- (i) Root apical meristems are highly active in cell division; a commonly cited figure is that a single maize root apical meristem can generate more than 17,500 new cells every hour — a standard NCERT-cited statistic. TRUE.
- (ii) Cell enlargement can be dramatic; watermelon cells are cited as increasing in size up to 3,50,000-fold, illustrating that growth by cell enlargement can far exceed growth by cell division in magnitude. TRUE.
- (iii) A pollen tube grows only from its tip in one direction toward the ovule, so its growth is naturally quantified by the increase in its length. TRUE. …
- AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQQ.The maximum growth rate is seen in _______________ phase (A) Lag phase (B) Exponential phase (C) Stationary phase (D) Senescence phase
›Reveal solutionSolution
The exponential (log) phase is defined by the fastest, unrestricted rate of growth in a population/growth curve.
Concept and Intuition
Growth curves (bacterial growth curves, and analogously the "grand period of growth" in plants) typically show four phases: lag (adjustment, minimal growth), exponential/log (rapid, geometric increase as resources are abundant and unrestricted), stationary (growth rate equals death/limiting-factor rate, net growth plateaus), and senescence/decline (resources exhausted, growth rate falls, population/organism ages and declines). By definition, the steepest slope on the growth curve — i.e., the maximum rate of increase — occurs during the exponential phase, before nutrient depletion, waste accumulation, or space constraints begin to slow things down.
Step-by-Step Solution
- Recall the four standard phases of a growth curve: lag, exponential, stationary, senescence.
- Lag phase: growth is minimal as cells/organisms adapt — not the maximum rate.
- Exponential phase: growth proceeds at its fastest, unimpeded rate — this IS the maximum. …
- AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQQ.A sophisticated instrument disigned by J.C. Bose was so sensitive that is could record even the minute growth of a plant upto a millinth part of millimeter is (A) Crescograph (B) Micrograph (C) Thermograph (D) Monograph
›Reveal solutionSolution
The instrument devised by J.C. Bose to record extremely minute plant growth is the Crescograph — option (A).
Concept and Intuition
Jagadish Chandra Bose pioneered instruments to demonstrate that plants respond to stimuli much like animals do, and needed a device sensitive enough to magnify and record the very slow, minute elongation growth of plant tissues.
Step-by-Step Solution
- The Crescograph was designed by J.C. Bose specifically to magnify and record plant growth movements to an extraordinary degree of precision — down to about a millionth of a millimetre.
- This instrument was central to Bose's demonstrations of plant responsiveness (growth response to stimuli) and remains the standard textbook answer to this fact. …
- AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQQ.Select the incorrect statements of Eichhornia ________ (A) Invasive weed in standing water (B) It drains oxygen from the water (C) It is indigenous to India (D) It can propagate vegetatively
›Reveal solutionSolution
The false statement is (C) — Eichhornia is an introduced (non-native) invasive weed in India, not indigenous.
Concept and Intuition
Eichhornia crassipes, the "terror of Bengal," is a textbook example of an invasive alien species — introduced into India from its native South America for its attractive flowers, it spread uncontrollably, forms dense mats on standing water, drains dissolved oxygen (harming aquatic life), and propagates rapidly via vegetative fragmentation.
Step-by-Step Solution
- (A) invasive weed of standing water bodies — TRUE.
- (B) drains oxygen from water — TRUE, a well-documented ecological effect.
- (C) "indigenous to India" — FALSE; it is an introduced species from South America. …
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