Q.A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is $10 \text{ cm}$.

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.By the application of derivatives, the approximate value of $\sqrt{242}$ is (A) 2.9085 (B) 2.9975 (C) 2.9527 (D) 2.8529

   ›Reveal solutionSolution

   The answer choices (all near 3) show this is the fifth root $\sqrt[5]{242}$, not the square root. Linearising $f(x)=x^{1/5}$ about $x=243$ (where $3^5=243$) gives $\sqrt[5]{242}\approx 2.9975$, option (B).

   This is a linear (tangent-line) approximation: near a point $a$ where the value is known exactly,

   $f(a+\Delta x)\approx f(a)+f'(a)\,\Delta x.$

   Because the options cluster around 3 (whereas $\sqrt{242}\approx 15.6$), the quantity being approximated is $\sqrt[5]{242}$, and the natural reference point is $243=3^5$.

   1. Choose the point. Let $f(x)=x^{1/5}$. Then $f(243)=243^{1/5}=3$ exactly. Take $a=243$ and $\Delta x = 242-243 = -1$.

   2. Derivative. $f'(x)=\tfrac15 x^{-4/5}$. Since $243^{1/5}=3$, we have $243^{4/5}=3^4=81$, so

      $f'(243)=\frac15\cdot\frac{1}{81}=\frac{1}{405}.$ …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $y = (e^{2x} - 4)(6e^{2x} - 5e^x + 1)$, then $\left(\frac{dy}{dx}\right)_{x=0} - \left(\frac{d^2y}{dx^2}\right)_{x=0} =$ (A) $0$ (B) $-5$ (C) $4$ (D) $6$

   ›Reveal solutionSolution

   Differentiate the product, evaluate the first and second derivatives at $x=0$, and subtract: the result is $4$.

   Let $u = e^{2x}-4$ and $v = 6e^{2x}-5e^{x}+1$, so $y = uv$.

   Derivatives of the factors:

   $u' = 2e^{2x},\quad u'' = 4e^{2x}$

   $v' = 12e^{2x}-5e^{x},\quad v'' = 24e^{2x}-5e^{x}$

   Values at $x=0$ (where $e^{2x}=1,\ e^{x}=1$):

   $u = -3,\ v = 2,\quad u' = 2,\ v' = 7,\quad u'' = 4,\ v'' = 19$

   First derivative:

   $\left(\frac{dy}{dx}\right)_{0} = u'v + uv' = (2)(2) + (-3)(7) = 4 - 21 = -17$

   Second derivative: …

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3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.

   [!FORMULA] $\left[ \frac{d}{dx} \left( (\sin x)^{\cos x} \right) \right]_{x = \frac{\pi}{4}} =$

   (A) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{\sqrt{2}+1}{\sqrt{2}}} (1 + \log \sqrt{2})$ (B) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{1}{\sqrt{2}}} (1 + \log \sqrt{2})$ (C) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{1}{\sqrt{2}}} (1 - \log \sqrt{2})$ (D) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{\sqrt{2}+1}{\sqrt{2}}} (1 - \log \sqrt{2})$

   ›Reveal solutionSolution

   Logarithmic differentiation gives $\dfrac{dy}{dx} = (\sin x)^{\cos x}\left(\dfrac{\cos^2 x}{\sin x} - \sin x\,\log(\sin x)\right)$. Evaluating at $x = \pi/4$ yields $\left(\dfrac{1}{\sqrt 2}\right)^{\frac{\sqrt 2 + 1}{\sqrt 2}}(1 + \log\sqrt 2)$, option (A).

   ---

   Concept & intuition

   Because the exponent $\cos x$ is itself a function of $x$, the ordinary power rule does not apply. Use logarithmic differentiation: take $\log$ of both sides, differentiate implicitly, then multiply back by $y$.

   ---

   1. Take logarithms Let $y = (\sin x)^{\cos x}$. Then

   $\log y = \cos x \,\log(\sin x).$

   2. Differentiate implicitly (product rule on the right):

   $\frac{1}{y}\frac{dy}{dx} = -\sin x\,\log(\sin x) + \cos x\cdot\frac{\cos x}{\sin x} = -\sin x\,\log(\sin x) + \frac{\cos^2 x}{\sin x}.$

   3. Solve for the derivative

   $\frac{dy}{dx} = (\sin x)^{\cos x}\left(\frac{\cos^2 x}{\sin x} - \sin x\,\log(\sin x)\right).$

   4. Evaluate at $x = \pi/4$ where $\sin x = \cos x = \dfrac{1}{\sqrt 2}$:

      • Base: $(\sin x)^{\cos x} = \left(\dfrac{1}{\sqrt 2}\right)^{1/\sqrt 2}$.
      • $\dfrac{\cos^2 x}{\sin x} = \dfrac{1/2}{1/\sqrt 2} = \dfrac{1}{\sqrt 2}$.
      • $-\sin x\,\log(\sin x) = -\dfrac{1}{\sqrt 2}\log\!\left(\dfrac{1}{\sqrt 2}\right) = -\dfrac{1}{\sqrt 2}(-\log\sqrt 2) = \dfrac{1}{\sqrt 2}\log\sqrt 2.$

      The bracket becomes

   $\frac{1}{\sqrt 2} + \frac{1}{\sqrt 2}\log\sqrt 2 = \frac{1}{\sqrt 2}\,(1 + \log\sqrt 2).$

   5. Combine $$\left.\frac{dy}{dx}\right|_{\pi/4} = \left(\frac{1}{\sqrt 2}\right)^{1/\sqrt 2}\cdot\frac{1}{\sqrt 2}\,(1 + \log\sqrt 2). …$$

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $y = \tan^{-1} \left( \frac{\sqrt{1+9x^2}-1}{3x} \right)$ then $\left( \frac{dy}{dx} \right)_{x = \frac{1}{3}} =$ (A) $\frac{2}{3}$ (B) $\frac{1}{3}$ (C) $\frac{3}{4}$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   The key is to simplify the inverse tangent expression using a trigonometric substitution, turning it into a linear function of $x$, which makes differentiation trivial. The derivative at $x = \frac13$ is $\frac{3}{4}$, so option (C) is correct.

   ---

   The expression inside the arctan looks messy, but the presence of $\sqrt{1+9x^2}$ suggests a substitution like $3x = \tan\theta$ or $3x = \sinh t$. Since we have a square root of a sum of squares, the hyperbolic substitution is cleaner: let $3x = \sinh t$. Then $\sqrt{1+9x^2} = \sqrt{1+\sinh^2 t} = \cosh t$. This turns the fraction into something like $\frac{\cosh t - 1}{\sinh t}$, which simplifies using hyperbolic half-angle identities. The result is that the arctan becomes just $t/2$, i.e., a linear function of $t$, and hence a simple function of $x$.

   ---

   1. Substitute to simplify Let $3x = \sinh t$, so $x = \frac13 \sinh t$. Then

   $\sqrt{1+9x^2} = \sqrt{1+\sinh^2 t} = \cosh t.$

   The argument of the arctan becomes

   $\frac{\sqrt{1+9x^2} - 1}{3x} = \frac{\cosh t - 1}{\sinh t}.$

   2. Use the hyperbolic half-angle identity Recall: $\cosh t - 1 = 2\sinh^2(t/2)$ and $\sinh t = 2\sinh(t/2)\cosh(t/2)$. Hence

   $\frac{\cosh t - 1}{\sinh t} = \frac{2\sinh^2(t/2)}{2\sinh(t/2)\cosh(t/2)} = \frac{\sinh(t/2)}{\cosh(t/2)} = \tanh(t/2).$

   3. Simplify the inverse tangent So

   $y = \tan^{-1}\bigl(\tanh(t/2)\bigr).$

   But $\tanh(t/2)$ is always between $-1$ and $1$, and $\tan^{-1}$ of a hyperbolic tangent is not simply $t/2$ — wait, we need a different identity. Actually, there is a known relation:

   $\tan^{-1}\left(\frac{\cosh t - 1}{\sinh t}\right) = \frac{t}{2}.$

   Let’s verify: $\tan(t/2) = \frac{\sin t}{1+\cos t}$, but here we have hyperbolic functions. The correct identity is:

   $\tanh^{-1} u = \frac12 \log\frac{1+u}{1-u},$

   but we have $\tan^{-1}$, not $\tanh^{-1}$. So let’s check numerically: if $t=1$, $\frac{\cosh 1 - 1}{\sinh 1} \approx 0.462$, $\tan^{-1}(0.462) \approx 0.433$, and $t/2 = 0.5$. Not equal. So the direct half-angle idea needs adjustment.

   Better approach: Use the identity

   $\tan^{-1}\left(\frac{\sqrt{1+a^2}-1}{a}\right) = \frac12 \tan^{-1} a,$

   for $a>0$. Let’s prove it: set $a = \tan\theta$, then $\sqrt{1+a^2} = \sec\theta$, so

   $$$$

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5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.$\lim_{x \to 0} \frac{\log{(4+x)^x} - \log{4^x}}{\sin^2{x}} =$ (A) $4$ (B) $\frac{1}{4}$ (C) $2$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   The numerator reduces to $x\log\!\left(1+\tfrac{x}{4}\right)\sim\tfrac{x^2}{4}$, and $\sin^2 x\sim x^2$, so the limit is $\tfrac14$.

   Simplify the numerator. Using $\log(a^b)=b\log a$:

   $\log\!\big((4+x)^x\big)-\log\!\big(4^x\big)=x\log(4+x)-x\log 4=x\log\!\frac{4+x}{4}=x\log\!\left(1+\frac{x}{4}\right).$

   Take the limit. As $x\to0$, $\log\!\left(1+\tfrac{x}{4}\right)\sim\tfrac{x}{4}$ and $\sin^2 x\sim x^2$: …

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the rate of change of the slope of the tangent drawn to the curve $y = x^3 - 2x^2 + 3x - 2$ at the point $(2, 4)$ is $k$ times the rate of change of its abscissa, then $k=$ (A) $2$ (B) $4$ (C) $6$ (D) $8$

   ›Reveal solutionSolution

   The problem asks for the constant $k$ such that the rate of change of the slope of the tangent equals $k$ times the rate of change of the abscissa. This means we need the second derivative of $y$ with respect to $x$, evaluated at $x=2$, which gives $k=8$.

   We are told that the rate of change of the slope of the tangent is $k$ times the rate of change of the abscissa.

   The “slope of the tangent” is just the first derivative $\frac{dy}{dx}$.

   The “rate of change” of that slope with respect to time (or any parameter) is $\frac{d}{dt}\left(\frac{dy}{dx}\right)$.

   The “rate of change of its abscissa” is $\frac{dx}{dt}$.

   The condition is:

   $\frac{d}{dt}\left(\frac{dy}{dx}\right) = k \cdot \frac{dx}{dt}$

   By the chain rule, $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2} \cdot \frac{dx}{dt}$.

   So the equation becomes:

   $\frac{d^2y}{dx^2} \cdot \frac{dx}{dt} = k \cdot \frac{dx}{dt}$

   Assuming $\frac{dx}{dt} \neq 0$ (the abscissa is changing), we cancel it and get:

   $\frac{d^2y}{dx^2} = k$

   Thus, $k$ is simply the second derivative of $y$ with respect to $x$ at the given point.

   Now we compute:

   1. First derivative:

      $y = x^3 - 2x^2 + 3x - 2$

      $\frac{dy}{dx} = 3x^2 - 4x + 3$

   2. Second derivative:

      $\frac{d^2y}{dx^2} = 6x - 4$

   3. Evaluate at $x=2$: …

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7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The number of all common roots of the equation $x^4 - 10x^3 + 37x^2 - 60x + 36 = 0$ and the transformed equation of it obtained by increasing any two distinct roots of it by 1, keeping the other two roots fixed, is (A) 1 (B) 3 (C) 4 (D) 2

   ›Reveal solutionSolution

   We first find the roots of the given quartic equation. Then, we identify the roots that remain unchanged when two distinct roots are increased by 1. The number of such common roots is $\boxed{3}$.

   Concept and Intuition

   The problem asks for the number of common roots between an original polynomial equation and a new equation formed by a specific transformation of its roots.

   1. Understanding "Common Roots": When we talk about common roots of two polynomial equations, we are looking for the values of $x$ that satisfy both equations. If a root has a certain multiplicity in both equations, its contribution to the "number of common roots" is the minimum of its multiplicities in the two equations. For example, if $x=a$ is a root of multiplicity 3 in the first equation and multiplicity 2 in the second, then $x=a$ contributes 2 to the count of common roots.

   2. The Transformation: The original equation has four roots, say $r_1, r_2, r_3, r_4$. The transformation states that we "increase any two distinct roots of it by 1, keeping the other two roots fixed".

      • "Distinct roots" here refers to distinct values of the roots. For example, if the roots are $2, 2, 3, 3$, the distinct values are 2 and 3. So, we must choose one root with value 2 and one root with value 3 to be increased.
      • Let's say we choose $r_a$ and $r_b$ (where $r_a \neq r_b$) to be increased by 1. The new roots will be $r_a+1$ and $r_b+1$.
      • The "other two roots fixed" means the remaining two roots, say $r_c$ and $r_d$, are unchanged.
      • So, if the original set of roots is $R = \{r_a, r_b, r_c, r_d\}$, the transformed set of roots is $R' = \{r_a+1, r_b+1, r_c, r_d\}$.
      • The common roots are the elements in the intersection of these two multisets, $R \cap R'$.

   The strategy is to first find all roots of the original equation, then apply the transformation to form the new set of roots, and finally count the common roots.

   Step-by-Step Solution

   1. Find the roots of the original equation:

      The given equation is $P(x) = x^4 - 10x^3 + 37x^2 - 60x + 36 = 0$.

      We can try to find rational roots using the Rational Root Theorem (divisors of 36).

      • Let's test $x=1$: $1 - 10 + 37 - 60 + 36 = 4 \neq 0$.
      • Let's test $x=2$: $2^4 - 10(2^3) + 37(2^2) - 60(2) + 36$ $= 16 - 10(8) + 37(4) - 120 + 36$ $= 16 - 80 + 148 - 120 + 36$ $= (16 + 148 + 36) - (80 + 120)$ $= 200 - 200 = 0$. So, $x=2$ is a root. This means $(x-2)$ is a factor.
      • We can perform polynomial division or synthetic division:

        2 | 1  -10   37  -60   36
          |    2   -16   42  -36
          --------------------
            1  -8    21  -18    0
        

        The quotient is $x^3 - 8x^2 + 21x - 18$. Let's call this $Q(x)$.
      • Let's test $x=2$ again for $Q(x)$: $2^3 - 8(2^2) + 21(2) - 18$ $= 8 - 8(4) + 42 - 18$ $= 8 - 32 + 42 - 18$ $= (8 + 42) - (32 + 18)$ $= 50 - 50 = 0$. So, $x=2$ is a root of $Q(x)$ as well, meaning it's a root of multiplicity at least 2 for $P(x)$.
      • Divide $Q(x)$ by $(x-2)$:

        2 | 1  -8   21  -18
          |    2  -12   18
          -----------------
            1  -6    9    0
        

        The quotient is $x^2 - 6x + 9$. This is a perfect square: $(x-3)^2$.
      • Therefore, the original equation can be factored as: $P(x) = (x-2)(x-2)(x-3)^2 = (x-2)^2 (x-3)^2 = 0$.
      • The roots of the original equation are $2, 2, 3, 3$.

   2. Apply the transformation to the roots:

      The original roots are $R = \{2, 2, 3, 3\}$.

      The distinct values of the roots are 2 and 3. …

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8. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The radius of a cone of height 9 units is changed from 2 units to 2.12 units. The exact change and approximate change in the volume of the cone are respectively (A) $(1.4437)\pi, (1.44)\pi$ (B) $(1.4832)\pi, (1.479)\pi$ (C) $(1.4842)\pi, (1.48)\pi$ (D) $(1.4832)\pi, (1.44)\pi$

   ›Reveal solutionSolution

   The exact change in volume is found by subtracting the original volume from the new volume; the approximate change uses the derivative (differential) of the volume with respect to the radius. The exact change is $(1.4832)\pi$ and the approximate change is $(1.44)\pi$, so the correct option is (D).

   Concept & Intuition

   The volume of a cone is $V = \frac{1}{3}\pi r^2 h$. Here height $h$ is fixed at 9, so $V = 3\pi r^2$. When the radius changes from $r=2$ to $r=2.12$, the exact change is simply $V(2.12) - V(2)$. The approximate change uses the differential $dV = V'(r)\, dr$, which gives a linear approximation valid for small changes. This is a classic application of calculus: the derivative tells us the instantaneous rate of change, and multiplying by the small change in radius gives a quick estimate.

   1. Write the volume formula

      $V = \frac{1}{3}\pi r^2 h$. With $h=9$, this becomes $V = 3\pi r^2$.

   2. Compute the exact change

      Original volume: $V(2) = 3\pi (2)^2 = 12\pi$.

      New volume: $V(2.12) = 3\pi (2.12)^2 = 3\pi \times 4.4944 = 13.4832\pi$.

      Exact change: $\Delta V = 13.4832\pi - 12\pi = (1.4832)\pi$.

   3. Compute the approximate change using differentials

      Derivative: $\frac{dV}{dr} = 6\pi r$. At $r=2$, $\frac{dV}{dr} = 12\pi$.

      Change in radius: $dr = 2.12 - 2 = 0.12$.

      Approximate change: $dV = 12\pi \times 0.12 = 1.44\pi$. …

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9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $f : [a, b] \to [c, d]$ is a continuous and strictly increasing function, then $\dfrac{d-c}{b-a}$ is (A) Value of the function at a point $t \in (a, b)$ (B) Value of the function at $t \in (a, b)$ such that $f'(t) = 0$ (C) Slope of the tangent drawn to the curve $y = f(t)$ at a point $t \in (c, d)$ (D) Slope of the tangent drawn to the curve $y = f(t)$ at a point $t \in (a, b)$

   ›Reveal solutionSolution

   The expression $\frac{d-c}{b-a}$ is the average rate of change of $f$ over $[a,b]$, and by the Mean Value Theorem it equals $f'(t)$ for some $t \in (a,b)$, which is the slope of the tangent at that point. So the correct option is (D).

   The key idea is to recognize that $\frac{d-c}{b-a}$ is not just any number — it’s the slope of the secant line connecting the endpoints $(a, c)$ and $(b, d)$ of the graph of $f$. Since $f$ is continuous on $[a,b]$ and differentiable (strictly increasing implies differentiable almost everywhere, but we need the Mean Value Theorem which requires differentiability on $(a,b)$ and continuity on $[a,b]$; here $f$ is continuous and strictly increasing, but not necessarily differentiable everywhere — however, the problem likely assumes $f$ is differentiable, or we interpret the slope of the tangent as the derivative where it exists). The Mean Value Theorem guarantees that for a differentiable function, there is some $t$ in $(a,b)$ where the instantaneous slope $f'(t)$ equals that average slope. That instantaneous slope is precisely the slope of the tangent line to $y=f(t)$ at $t$.

   Now let’s walk through the reasoning step by step.

   1. Interpret the expression geometrically. The points $(a, f(a)) = (a, c)$ and $(b, f(b)) = (b, d)$ lie on the curve $y=f(x)$. The slope of the line joining them is

   $\frac{f(b)-f(a)}{b-a} = \frac{d-c}{b-a}.$

   This is the average rate of change of $f$ over $[a,b]$.

   2. Apply the Mean Value Theorem (MVT). Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ (strictly increasing functions on an interval are differentiable almost everywhere, but we assume differentiability here as typical for such problems), the MVT says there exists some $t \in (a,b)$ such that

   $f'(t) = \frac{f(b)-f(a)}{b-a} = \frac{d-c}{b-a}.$

   3. Interpret $f'(t)$ geometrically. The derivative $f'(t)$ is the slope of the tangent line to the curve $y=f(x)$ at the point $(t, f(t))$. So $\frac{d-c}{b-a}$ equals the slope of that tangent line at some $t$ in $(a,b)$. …

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10. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.If $y(\cos x)^{\sin x} = (\sin x)^{\sin x}$ then the value of $\dfrac{dy}{dx}$ at $x = \dfrac{\pi}{4}$ is (A) $0$ (B) $1$ (C) $\sqrt{2}$ (D) $\dfrac{\sqrt{3}}{2}$

    ›Reveal solutionSolution

    The relation simplifies to $y=(\tan x)^{\sin x}$; logarithmic differentiation gives $\dfrac{dy}{dx}=\sqrt{2}$ at $x=\dfrac{\pi}{4}$.

    Simplify. From $y(\cos x)^{\sin x}=(\sin x)^{\sin x}$,

    $y=\frac{(\sin x)^{\sin x}}{(\cos x)^{\sin x}}=(\tan x)^{\sin x}.$

    Logarithmic differentiation.

    $\log y=\sin x\,\log(\tan x),$

    $\frac{1}{y}\frac{dy}{dx}=\cos x\,\log(\tan x)+\sin x\cdot\frac{\sec^2x}{\tan x}.$

    Since $\dfrac{\sec^2x}{\tan x}=\dfrac{1}{\sin x\cos x}$, the second term is $\dfrac{1}{\cos x}=\sec x$:

    $\frac{dy}{dx}=y\big(\cos x\,\log(\tan x)+\sec x\big).$ …

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11. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.The approximate value of $\sqrt[3]{730}$ obtained by the application of derivatives is (A) 9.0041 (B) 9.01 (C) 9.006 (D) 9.05

    ›Reveal solutionSolution

    Using the linear approximation formula $f(x+\Delta x) \approx f(x) + f'(x)\,\Delta x$ with $f(x)=\sqrt[3]{x}$, we set $x=729$ and $\Delta x=1$ to get $\sqrt[3]{730} \approx 9 + \frac{1}{3\cdot 9^2} = 9.004115...$, which matches option (A) 9.0041.

    We want an approximate value of $\sqrt[3]{730}$ without a calculator, using derivatives. The key idea is linear approximation (also called the tangent‑line approximation): for a small change $\Delta x$ near a point $x$, the function’s value changes at roughly the rate given by its derivative. So we pick a perfect cube close to 730 — namely $729 = 9^3$ — and use the derivative to estimate the small step from 729 to 730.

    1. Choose the function and the known point.

       Let $f(x) = \sqrt[3]{x} = x^{1/3}$. We know exactly that $f(729) = \sqrt[3]{729} = 9$. We want $f(730) = f(729 + 1)$. Here $\Delta x = 1$, which is small relative to 729, so the linear approximation should be accurate to several decimal places.

    2. Recall the linear approximation formula.

       For a differentiable function $f$,

    $f(x + \Delta x) \approx f(x) + f'(x)\,\Delta x.$

    This is just the first‑order Taylor expansion; geometrically, we follow the tangent line at $x$ instead of the curve.

    3. Compute the derivative at $x=729$.

    $f'(x) = \frac{1}{3} x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}.$

    At $x = 729$, $\sqrt[3]{729^2} = (\sqrt[3]{729})^2 = 9^2 = 81$, so

    $f'(729) = \frac{1}{3 \cdot 81} = \frac{1}{243}.$

    4. Apply the approximation.

    $\sqrt[3]{730} = f(729+1) \approx f(729) + f'(729) \cdot 1 = 9 + \frac{1}{243}.$

    Now $\frac{1}{243}$ is about $0.004115226...$ (since $243 \times 4 = 972$, so $1/243 \approx 0.004115$). Adding to 9 gives

    $\sqrt[3]{730} \approx 9.004115...$

    5. Compare with the options. …

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.The vertical angle of a right circular cone is $60^\circ$. If water is being poured into the cone at the rate of $\dfrac{1}{\sqrt{3}}$ m$^3$/min, then the rate (m/min) at which the radius of the water level is increasing when the height of the water level is 3 m is (A) $\dfrac{1}{3\sqrt{3}\pi}$ (B) $\dfrac{1}{9\sqrt{3}\pi}$ (C) $\dfrac{1}{9\pi}$ (D) $\dfrac{1}{3\pi}$

    ›Reveal solutionSolution

    The problem is a classic related‑rates cone problem. Using the vertical angle to relate radius and height, we differentiate the volume formula and substitute the given rate to find $\frac{dr}{dt} = \frac{1}{9\pi}$ when $h=3$. The correct option is (C).

    ---

    Concept & Intuition

    When water is poured into a cone, the volume, radius, and height all change with time. The vertical angle gives a fixed ratio between radius and height: for a right circular cone with vertical angle $60^\circ$, the radius $r$ and height $h$ satisfy $\tan(30^\circ) = r/h$, so $r = h/\sqrt{3}$. This lets us express volume solely in terms of $h$ (or $r$). Then we differentiate with respect to time and plug in the known rate $dV/dt$ to find $dr/dt$ at the instant $h=3$.

    ---

    Step‑by‑step solution

    1. Relate radius and height using the vertical angle The vertical angle is $60^\circ$, meaning the angle at the apex between two opposite generatrices is $60^\circ$. The half‑angle at the apex is $30^\circ$. In the cross‑section (a triangle), $\tan 30^\circ = \frac{r}{h}$. Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we have

    $\frac{r}{h} = \frac{1}{\sqrt{3}} \quad\Rightarrow\quad r = \frac{h}{\sqrt{3}}.$

    2. Write the volume of water in terms of $h$ only Volume of a cone: $V = \frac{1}{3}\pi r^2 h$. Substitute $r = h/\sqrt{3}$:

    $V = \frac{1}{3}\pi \left(\frac{h}{\sqrt{3}}\right)^2 h = \frac{1}{3}\pi \cdot \frac{h^2}{3} \cdot h = \frac{\pi}{9} h^3.$

    3. Differentiate with respect to time $t$

    $\frac{dV}{dt} = \frac{\pi}{9} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{3} h^2 \frac{dh}{dt}.$

    We are given $\frac{dV}{dt} = \frac{1}{\sqrt{3}}$ m³/min.

    4. Find $\frac{dh}{dt}$ when $h = 3$ $$\frac{1}{\sqrt{3}} = \frac{\pi}{3} (3)^2 \frac{dh}{dt} = \frac{\pi}{3} \cdot 9 \frac{dh}{dt} = 3\pi \frac{dh}{dt}. …$$

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