Q.A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The volume of a spherical ball is increasing at a rate of $4\pi \text{ cm}^3.\text{s}^{-1}$. The rate at which its radius increases, when its volume is $288\pi \text{ cm}^3$, is ______ $\text{cm}.\text{s}^{-1}$. (A) $\dfrac{1}{6}$ (B) $\dfrac{1}{36}$ (C) $\dfrac{1}{9}$ (D) $\dfrac{1}{24}$

   ›Reveal solutionSolution

   Related rates on $V=\tfrac43\pi r^3$ give $\dfrac{dr}{dt}=\dfrac{1}{r^2}$, and at the given volume $r=6$, so $\dfrac{dr}{dt}=\dfrac1{36}$.

   Concept and Intuition

   This is a classic related-rates problem: differentiate the volume formula with respect to time using the chain rule, then substitute the known rate and the radius at the instant of interest (found from the given volume).

   Step-by-Step Solution

   1. $V=\dfrac43\pi r^3$.
   2. Differentiate w.r.t. time: $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.
   3. Given $\dfrac{dV}{dt}=4\pi$: $4\pi=4\pi r^2\dfrac{dr}{dt} \Rightarrow \dfrac{dr}{dt}=\dfrac{1}{r^2}$.
   4. Find $r$ when $V=288\pi$: $\dfrac43\pi r^3=288\pi \Rightarrow r^3=216 \Rightarrow r=6$.
   5. $\dfrac{dr}{dt}=\dfrac{1}{6^2}=\dfrac{1}{36}$ cm/s.

   Common Mistakes

   • Forgetting to first solve for $r$ from the given volume before substituting into $dr/dt$.
   • Cubing/uncubing errors when solving $r^3=216$ (note $6^3=216$).

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{36}$.

   ANSWER: B

2. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.The volume of a spherical balloon is increasing at the rate of $30\ cc$ per minute. Find the rate of change of surface area of the balloon, when its radius is $6\ cm$. (A) $5\ cm^2.min^{-1}$ (B) $30\ cm^2.min^{-1}$ (C) $10\ cm^2.min^{-1}$ (D) $20\ cm^2.min^{-1}$

   ›Reveal solutionSolution

   A standard related-rates chain: volume rate → radius rate → surface-area rate. The answer is $10\ cm^2/\text{min}$.

   Concept and Intuition

   Both $V$ and $S$ of a sphere depend only on $r$, so their rates of change are linked through $dr/dt$ via the chain rule. First use the given $dV/dt$ to find $dr/dt$ at the specified radius, then plug that into $dS/dt$.

   Step-by-Step Solution

   1. $V=\dfrac43\pi r^3\Rightarrow\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.
   2. Given $\dfrac{dV}{dt}=30\ cc/min$ and $r=6$ cm: $30=4\pi(6)^2\dfrac{dr}{dt}=144\pi\dfrac{dr}{dt}$.
   3. $\dfrac{dr}{dt}=\dfrac{30}{144\pi}=\dfrac{5}{24\pi}$ cm/min.
   4. $S=4\pi r^2\Rightarrow\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$.
   5. At $r=6$: $\dfrac{dS}{dt}=8\pi(6)\left(\dfrac{5}{24\pi}\right)=\dfrac{48\pi\cdot5}{24\pi}=\dfrac{240\pi}{24\pi}=10$ cm²/min.

   Common Mistakes

   • Trying to relate $S$ and $V$ directly without going through $r$ as the common variable.
   • Arithmetic slip in simplifying $\dfrac{240\pi}{24\pi}$.

   ✓Final answer

   The correct option is (C) — $10\ cm^2.min^{-1}$.

   ANSWER: C

3. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If the volume of a sphere is increasing at the rate of 12 c.c./sec, then the rate (in sq. cm/sec) at which its surface area is increasing, when the diameter of the sphere is 12 cm is (A) 2 (B) 3 (C) 4 (D) 6

   ›Reveal solutionSolution

   This is a related-rates problem: given how fast volume grows, find how fast surface area grows at a specific radius; the answer is $4\text{ cm}^2/\text{s}$.

   Concept and Intuition

   Both $V$ and $S$ of a sphere depend only on $r$, so differentiating each with respect to time and eliminating $\frac{dr}{dt}$ (found from the volume-rate condition) gives the surface-area rate.

   Step-by-Step Solution

   1. $V=\dfrac43\pi r^3$, so $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.
   2. Given $\dfrac{dV}{dt}=12$ and diameter $=12\Rightarrow r=6$: $12=4\pi(36)\dfrac{dr}{dt}\Rightarrow \dfrac{dr}{dt}=\dfrac{12}{144\pi}=\dfrac{1}{12\pi}$.
   3. $S=4\pi r^2\Rightarrow \dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$.
   4. Substituting $r=6$, $\dfrac{dr}{dt}=\dfrac{1}{12\pi}$: $\dfrac{dS}{dt}=8\pi(6)\cdot\dfrac{1}{12\pi}=\dfrac{48\pi}{12\pi}=4$.

   Common Mistakes

   • Using the diameter (12) directly as the radius in the formulas instead of halving it first.
   • Forgetting to substitute the specific radius before simplifying, leading to an expression instead of a number.

   ✓Final answer

   The correct option is (C) — 4.

   ANSWER: C

4. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.A spherical iron ball $10\ cm$ in radius is coated with a layer of ice of uniform thickness, which melts at a rate of $50\ cm^3/min$. When the thickness of the ice is $15\ cm$, the rate at which the thickness of ice decreases is ______ $cm/min$ (A) $\dfrac{5}{6\pi}$ (B) $\dfrac{1}{54\pi}$ (C) $\dfrac{1}{18\pi}$ (D) $\dfrac{1}{36\pi}$

   ›Reveal solutionSolution

   This is a related-rates problem: only the outer radius of the ice matters for relating the rate of volume loss to the rate of thickness loss. The answer is $\dfrac{1}{18\pi}$ cm/min.

   Concept and Intuition

   The iron ball's own radius (10 cm) is fixed and drops out of the derivative — only the total outer radius $R=10+x$, where $x$ is the ice thickness, matters, since $V_{ice}=\tfrac43\pi R^3-\tfrac43\pi(10)^3$ and the constant term vanishes on differentiating. This reduces the problem to the standard "rate of change of a sphere's volume vs. its radius" relation, $dV/dt=4\pi R^2\,dR/dt$.

   Step-by-Step Solution

   1. Let $x(t)$ be the ice thickness at time $t$; outer radius $R=10+x$.
   2. $V_{ice}=\dfrac43\pi(10+x)^3-\dfrac43\pi(10)^3$.
   3. $\dfrac{dV_{ice}}{dt}=4\pi(10+x)^2\dfrac{dx}{dt}$ (the constant term differentiates to zero).
   4. The ice melts (volume decreases) at $50\text{ cm}^3/\text{min}$, so $\dfrac{dV_{ice}}{dt}=-50$.
   5. At the outer radius $R=15$ cm: $4\pi(15)^2\dfrac{dx}{dt}=-50\Rightarrow4\pi(225)\dfrac{dx}{dt}=-50\Rightarrow900\pi\dfrac{dx}{dt}=-50$.
   6. $\dfrac{dx}{dt}=-\dfrac{50}{900\pi}=-\dfrac{1}{18\pi}$; the thickness decreases at rate $\dfrac{1}{18\pi}$ cm/min.

   Common Mistakes

   • Using the iron ball's fixed radius (10 cm) instead of the full outer radius in $4\pi R^2$.
   • Sign confusion between the volume decreasing and the thickness decreasing (both are negative rates, and the magnitude is what's asked).

   ✓Final answer

   The correct option is (C) — $\dfrac{1}{18\pi}$.

   ANSWER: C

5. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If the surface area of a spherical bubble is increasing at the rate of 4 sq.cm/sec, then the rate of change in its volume (in cubic cm/sec) when its radius is 8 cms is (A) $8$ (B) $12$ (C) $15$ (D) $16$

   ›Reveal solutionSolution

   A related-rates problem: given $\dfrac{dS}{dt}$, find $\dfrac{dV}{dt}$ by eliminating $\dfrac{dr}{dt}$ through the common variable $r$. Answer: $16$ cubic cm/sec.

   Concept and Intuition

   Both surface area $S$ and volume $V$ of a sphere depend on the single variable $r$ (radius), which itself changes with time. The strategy in any related-rates problem is: express both quantities in terms of the shared variable, differentiate each with respect to time using the chain rule, and then eliminate the unknown rate ($\dfrac{dr}{dt}$ here) between the two equations.

   Step-by-Step Solution

   1. Surface area: $S=4\pi r^2$. Differentiating w.r.t. $t$: $\dfrac{dS}{dt}=8\pi r\,\dfrac{dr}{dt}$.
   2. We're given $\dfrac{dS}{dt}=4$, so $8\pi r\,\dfrac{dr}{dt}=4 \Rightarrow \dfrac{dr}{dt}=\dfrac{4}{8\pi r}=\dfrac{1}{2\pi r}$.
   3. Volume: $V=\dfrac43\pi r^3$. Differentiating w.r.t. $t$: $\dfrac{dV}{dt}=4\pi r^2\,\dfrac{dr}{dt}$.
   4. Substitute $\dfrac{dr}{dt}$ from step 2:

   $\frac{dV}{dt}=4\pi r^2\cdot\frac{1}{2\pi r}=2r.$

   5. At $r=8$: $\dfrac{dV}{dt}=2(8)=16$ cubic cm/sec.

   Common Mistakes

   • Forgetting the chain rule factor $\dfrac{dr}{dt}$ when differentiating $S$ and $V$ with respect to time (treating $r$ as if it were the independent variable directly).
   • Arithmetic slip in simplifying $4\pi r^2\cdot\dfrac{1}{2\pi r}$ (the $\pi$ and one power of $r$ cancel, leaving $2r$).

   ✓Final answer

   The correct option is (D) — $16$.

   ANSWER: D

6. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.The diameter and altitude of a right circular cone, at a certain instant, were found to be $10\ cm$ and $20\ cm$ respectively. If its diameter is increasing at a rate of $2\ cm/s$, then at what rate must its altitude change, in order to keep its volume constant? (A) $4\ cm/s$ (B) $6\ cm/s$ (C) $-4\ cm/s$ (D) $-8\ cm/s$

   ›Reveal solutionSolution

   This tests related rates: differentiating the volume of a cone with respect to time and setting the total rate of change to zero to keep volume constant. Answer: $-8$ cm/s.

   Concept and Intuition

   When a quantity built from several changing variables must stay constant, differentiate the formula with respect to time and set the total derivative to zero — this links the individual rates of change together.

   Step-by-Step Solution

   1. Diameter $=10$ cm $\Rightarrow$ radius $r=5$ cm; height $h=20$ cm.
   2. Diameter increasing at $2$ cm/s $\Rightarrow \dfrac{d(2r)}{dt}=2 \Rightarrow \dfrac{dr}{dt}=1$ cm/s.
   3. Volume: $V = \dfrac13\pi r^2 h$. Differentiate with respect to $t$: $\dfrac{dV}{dt} = \dfrac13\pi\left(2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt}\right)$.
   4. For constant volume, $\dfrac{dV}{dt}=0 \Rightarrow 2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt} = 0$.
   5. Solve for $\dfrac{dh}{dt}$: $\dfrac{dh}{dt} = -\dfrac{2h}{r}\dfrac{dr}{dt} = -\dfrac{2(20)}{5}(1) = -8$ cm/s.
   6. The negative sign means the altitude must decrease at $8$ cm/s to compensate for the increasing radius.

   Common Mistakes

   • Using the rate of change of the diameter directly as $\tfrac{dr}{dt}$ instead of halving it first.

   ✓Final answer

   The correct option is (D) — $-8\ cm/s$.

   ANSWER: D

7. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If the rate of increase in the surface area of a cube is $6\ sq.cm./sec$, then the rate of increase in its volume (in $c.c./sec$), when the length of its edge is $12\ cm$, is (A) $6$ (B) $12$ (C) $18$ (D) $9$

   ›Reveal solutionSolution

   A related-rates problem: convert the given rate of change of surface area into the rate of change of the edge length, then into the rate of change of volume.

   Concept and Intuition

   Both surface area and volume of a cube are functions of the single variable $a$ (the edge length), so differentiating each with respect to time and using the chain rule links their rates through $da/dt$ — the one quantity actually changing independently.

   Step-by-Step Solution

   1. Surface area: $S=6a^2\Rightarrow\dfrac{dS}{dt}=12a\dfrac{da}{dt}$.
   2. Given $\dfrac{dS}{dt}=6$: $12a\dfrac{da}{dt}=6\Rightarrow\dfrac{da}{dt}=\dfrac{1}{2a}$.
   3. Volume: $V=a^3\Rightarrow\dfrac{dV}{dt}=3a^2\dfrac{da}{dt}$.
   4. Substitute: $\dfrac{dV}{dt}=3a^2\cdot\dfrac{1}{2a}=\dfrac{3a}{2}$.
   5. At $a=12$: $\dfrac{dV}{dt}=\dfrac{3\times12}{2}=18$.

   Common Mistakes

   • Forgetting the factor of $6$ faces when writing $S=6a^2$ (or using $S=a^2$).
   • Plugging $a=12$ into $da/dt$ before simplifying $dV/dt=3a\,/\,2$ algebraically, risking an arithmetic slip.

   ✓Final answer

   The correct option is (C) — $18$.

   ANSWER: C

8. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If a cylindrical tank of radius 3 m is filled with water at the rate of $\dfrac{3}{2}$ $m^3$/sec, then the rate of change of its water level in (m/sec) is (A) $\dfrac{1}{3\pi}$ (B) $\dfrac{1}{2\pi}$ (C) $\dfrac{1}{\pi}$ (D) $\dfrac{1}{6\pi}$

   ›Reveal solutionSolution

   This tests related rates for a cylinder of fixed radius; the water level rises at $\dfrac{1}{6\pi}$ m/s.

   Concept and Intuition

   Since the radius doesn't change as the tank fills, the volume $V=\pi r^2h$ is a function of $h$ alone (with $r$ a constant), so differentiating with respect to time directly links $dV/dt$ to $dh/dt$ through the constant cross-sectional area $\pi r^2$.

   Step-by-Step Solution

   1. $V = \pi r^2 h$, with $r=3$ constant, so $V=9\pi h$.
   2. Differentiate w.r.t. time: $\dfrac{dV}{dt} = 9\pi\dfrac{dh}{dt}$.
   3. Given $\dfrac{dV}{dt}=\dfrac32\ \text{m}^3/\text{s}$: $\dfrac32 = 9\pi\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt} = \dfrac{3}{18\pi} = \dfrac{1}{6\pi}$.

   Common Mistakes

   • Forgetting that $r$ is constant and mistakenly differentiating it too (product rule on $r^2$ as if it varied).
   • Arithmetic slip simplifying $\dfrac{3/2}{9\pi}$.

   ✓Final answer

   The correct option is (D) — $\dfrac{1}{6\pi}$.

   ANSWER: D

9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If the velocity of a particle moving on a straight line is proportional to the cube root of its displacement, then its acceleration is (A) constant (B) inversely proportional to its velocity (C) proportional to its velocity (D) proportional to its displacement

   ›Reveal solutionSolution

   Express acceleration as $v\,dv/dx$, substitute the given proportionality, and eliminate $x$ in favour of $v$. Answer: acceleration is inversely proportional to velocity.

   Concept and Intuition

   For motion along a straight line, acceleration can be written as $a = \dfrac{dv}{dt} = v\dfrac{dv}{dx}$ (chain rule through position). Given a relation between $v$ and $x$, this lets us compute $a$ purely as a function of $x$, and then re-express it in terms of $v$ using the same original relation.

   Step-by-Step Solution

   1. Given $v \propto x^{1/3}$, write $v = kx^{1/3}$ for some constant $k$.
   2. $\dfrac{dv}{dx} = \dfrac{k}{3}x^{-2/3}$.
   3. $a = v\dfrac{dv}{dx} = kx^{1/3}\cdot\dfrac{k}{3}x^{-2/3} = \dfrac{k^2}{3}x^{-1/3}$.
   4. From $v=kx^{1/3}$, cube both sides: $v^3 = k^3x \Rightarrow x = \dfrac{v^3}{k^3}$, so $x^{-1/3} = \dfrac{k}{v}$.
   5. Substitute: $a = \dfrac{k^2}{3}\cdot\dfrac{k}{v} = \dfrac{k^3}{3v}$.
   6. Hence $a \propto \dfrac{1}{v}$ — acceleration is inversely proportional to velocity.

   Common Mistakes

   • Trying to differentiate $v$ with respect to $t$ directly without the $v\,dv/dx$ trick, which gets stuck because $x(t)$ isn't known explicitly.
   • Forgetting to eliminate $x$ in favour of $v$ at the end — the question asks for the dependence on velocity, not on displacement.

   ✓Final answer

   The correct option is (B) — inversely proportional to its velocity.

   ANSWER: B

10. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.If the vertical angle of a cone is $60^\circ$ and the rate of change of its total surface area is $2\sqrt{3}\ cm^2/sec$, then the rate of change of its volume (in $cm^3/sec$) when its radius is $5\ cm$, is (A) $15$ (B) $10$ (C) $5$ (D) $9$

    ›Reveal solutionSolution

    A related-rates problem: using the $60°$ vertical angle to fix $h$ and slant height $l$ in terms of $r$, then chaining $\dfrac{dS}{dt}\to\dfrac{dr}{dt}\to\dfrac{dV}{dt}$ gives $5\ \text{cm}^3/\text{sec}$.

    Concept and Intuition

    When a cone's vertical (apex) angle is fixed, its shape stays similar as it grows — radius and height stay in a fixed ratio determined by the semi-vertical angle. This lets us express both surface area and volume purely in terms of $r$, so a single related-rates chain (through $dr/dt$) connects the given rate of surface-area change to the unknown rate of volume change.

    Step-by-Step Solution

    1. Vertical angle $=60°$, so semi-vertical angle $\alpha=30°$. In the cone's cross-section, $\tan\alpha = r/h$, so $r = h\tan30° = h/\sqrt3$, i.e. $h = r\sqrt3$.
    2. Slant height: $l = \sqrt{r^2+h^2} = \sqrt{r^2+3r^2} = \sqrt{4r^2} = 2r$.
    3. Total surface area: $S = \pi r^2 + \pi r l = \pi r^2 + \pi r(2r) = 3\pi r^2$.
    4. Differentiate: $\dfrac{dS}{dt} = 6\pi r\dfrac{dr}{dt}$.
    5. Given $\dfrac{dS}{dt}=2\sqrt3$ and $r=5$: $2\sqrt3 = 6\pi(5)\dfrac{dr}{dt} = 30\pi\dfrac{dr}{dt}$, so $\dfrac{dr}{dt} = \dfrac{2\sqrt3}{30\pi} = \dfrac{\sqrt3}{15\pi}$.
    6. Volume: $V = \dfrac13\pi r^2 h = \dfrac13\pi r^2(r\sqrt3) = \dfrac{\pi r^3}{\sqrt3}$.
    7. Differentiate: $\dfrac{dV}{dt} = \dfrac{3\pi r^2}{\sqrt3}\dfrac{dr}{dt} = \sqrt3\pi r^2\dfrac{dr}{dt}$.
    8. Substitute $r=5$, $\dfrac{dr}{dt}=\dfrac{\sqrt3}{15\pi}$: $\dfrac{dV}{dt} = \sqrt3\pi(25)\cdot\dfrac{\sqrt3}{15\pi} = \dfrac{3\times25}{15} = \dfrac{75}{15} = 5$.

    Common Mistakes

    • Using the wrong trig ratio for the semi-vertical angle (mixing up $r/h$ with $h/r$).
    • Forgetting that total surface area includes both the base ($\pi r^2$) and the lateral surface ($\pi r l$), not just the lateral part.

    ✓Final answer

    The correct option is (C) — $5$.

    ANSWER: C

11. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.The surface area of a sphere is $49\pi\ sq.cm$. If it is increased by $0.016\ sq.cm$, then the approximate increase in its volume (in $c.c.$) is (A) $0.07$ (B) $0.04$ (C) $0.032$ (D) $0.028$

    ›Reveal solutionSolution

    Using differentials to connect a small change in surface area to the corresponding small change in volume, via the shared variable $r$, gives an approximate volume increase of $0.028$ c.c.

    Concept and Intuition

    When a small change in one geometric quantity (surface area) causes a small change in another (volume), and both depend on a common variable ($r$), we can relate their differentials directly: $dS = 8\pi r\,dr$ and $dV = 4\pi r^2\,dr$ share the same $dr$, so we solve for $dr$ from the given $dS$ and substitute into the $dV$ formula — this is the standard "approximate change" technique using derivatives.

    Step-by-Step Solution

    1. Surface area of sphere: $S=4\pi r^2 = 49\pi \Rightarrow r^2 = \dfrac{49}{4} \Rightarrow r = \dfrac72 = 3.5\ \text{cm}$.
    2. Differentiate $S=4\pi r^2$: $dS = 8\pi r\, dr$.
    3. Given $dS = 0.016$, and $r=3.5$: $dr = \dfrac{0.016}{8\pi(3.5)} = \dfrac{0.016}{28\pi}$.
    4. Volume: $V = \dfrac43\pi r^3$. Differentiate: $dV = 4\pi r^2\, dr$.
    5. $r^2 = 12.25$, so $dV = 4\pi(12.25)\,dr = 49\pi\, dr$.
    6. Substitute $dr$: $dV = 49\pi\times\dfrac{0.016}{28\pi} = \dfrac{49\times0.016}{28}$.
    7. $\dfrac{49}{28} = 1.75$, so $dV = 1.75\times0.016 = 0.028$.

    Common Mistakes

    • Computing $r$ incorrectly from $4\pi r^2=49\pi$ (forgetting to divide by $4\pi$ correctly, or taking the wrong square root).
    • Forgetting that both $dS$ and $dV$ formulas share the same $dr$ — trying to compute them independently instead of chaining through $dr$.

    ✓Final answer

    The correct option is (D) — $0.028$.

    ANSWER: D

12. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.If a man of height 1.8 mt. is walking away from the foot of a light pole of height 6 mt. with a speed of 7 km per hour on a straight horizontal road opposite to the pole, then the rate of change of the length of his shadow is (in kmph) (A) $7$ (B) $5$ (C) $3$ (D) $2$

    ›Reveal solutionSolution

    Similar triangles link shadow length to distance walked; the shadow grows at $3$ kmph.

    Concept and Intuition

    The tip of the shadow, the top of the pole, and the top of the man's head are collinear (that's what casts the shadow). This gives a similar-triangles relationship between the man's distance from the pole and his shadow's length, which can be differentiated with respect to time (related rates).

    Step-by-Step Solution

    1. Let $x$ = distance of the man from the pole, $s$ = length of his shadow. The tip of the shadow is at distance $x+s$ from the pole.
    2. Similar triangles (pole-to-shadow-tip vs man-to-shadow-tip): $\dfrac{\text{pole height}}{x+s}=\dfrac{\text{man height}}{s}\Rightarrow \dfrac{6}{x+s}=\dfrac{1.8}{s}$.
    3. Cross-multiply: $6s=1.8(x+s)=1.8x+1.8s\Rightarrow 4.2s=1.8x\Rightarrow s=\dfrac{1.8}{4.2}x=\dfrac37x$.
    4. Differentiate w.r.t. time: $\dfrac{ds}{dt}=\dfrac37\dfrac{dx}{dt}$.
    5. Given $\dfrac{dx}{dt}=7$ kmph: $\dfrac{ds}{dt}=\dfrac37(7)=3$ kmph.

    Common Mistakes

    • Confusing the shadow's length $s$ with the distance from the pole to the tip of the shadow ($x+s$) and differentiating the wrong quantity.
    • Setting up the similar triangles with man-height and pole-height swapped.

    ✓Final answer

    The correct option is (C) — $3$.

    ANSWER: C