Q.The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ?

Showing the 12 most recent of 30 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If the base of an isosceles triangle is $3\sqrt{2}$ feet and the two equal sides of it are increasing at the rate of $1$ ft/s, then the rate of increase of its area (in sq.ft/sec) when the angle between the equal sides is a right angle is (A) $3\sqrt{3}$ (B) $\sqrt{3}$ (C) $9$ (D) $3$

   ›Reveal solutionSolution

   The area of an isosceles triangle is expressed in terms of the equal side length and the included angle. Using the given rate of change of the side and the fact that the angle is fixed at the instant of interest, the rate of increase of area is found to be $3$ sq.ft/sec.

   The problem gives an isosceles triangle with base $3\sqrt{2}$ feet and equal sides that are increasing at $1$ ft/s. We need the rate of increase of its area at the moment when the angle between the equal sides is a right angle.

   The key is to choose a formula for area that directly involves the changing quantity (the equal side length) and the angle. For any triangle, area is $\frac12 ab \sin C$. Here, the two equal sides are the ones forming the included angle, so that formula is perfect.

   1. Set up the variables. Let the equal sides each have length $s$ feet, and let $\theta$ be the angle between them. The area $A$ of the triangle is

   $A = \frac12 \cdot s \cdot s \cdot \sin\theta = \frac12 s^2 \sin\theta.$

   2. What is given and what is wanted? We know $\frac{ds}{dt} = 1$ ft/s. We want $\frac{dA}{dt}$ at the instant when $\theta = 90^\circ = \frac{\pi}{2}$ radians. But note: the base is fixed at $3\sqrt{2}$ feet. Does that give a relation between $s$ and $\theta$? Yes — by the law of cosines, the base $b$ satisfies

   $b^2 = s^2 + s^2 - 2s^2 \cos\theta = 2s^2(1 - \cos\theta).$

   So $b = 3\sqrt{2}$ is constant, meaning $s$ and $\theta$ are not independent — as $s$ increases, $\theta$ must change to keep the base fixed. However, we only need the rate at a specific instant, not a full functional relation.

   3. Differentiate the area with respect to time. Since both $s$ and $\theta$ can change with time,

   $\frac{dA}{dt} = \frac12 \left( 2s \frac{ds}{dt} \sin\theta + s^2 \cos\theta \cdot \frac{d\theta}{dt} \right) = s \frac{ds}{dt} \sin\theta + \frac12 s^2 \cos\theta \frac{d\theta}{dt}.$

   4. Find $s$ and $\frac{d\theta}{dt}$ at the required instant. At $\theta = \frac{\pi}{2}$, $\sin\theta = 1$, $\cos\theta = 0$. The law of cosines gives

   $(3\sqrt{2})^2 = 2s^2 (1 - \cos\frac{\pi}{2}) = 2s^2 (1 - 0) = 2s^2.$

   So $18 = 2s^2$, hence $s^2 = 9$ and $s = 3$ feet (positive length).

   Now we need $\frac{d\theta}{dt}$ at that instant. Differentiate the law of cosines relation with respect to time. From $b^2 = 2s^2(1 - \cos\theta)$, since $b$ is constant,

   $0 = \frac{d}{dt}\left[2s^2(1 - \cos\theta)\right] = 4s \frac{ds}{dt}(1 - \cos\theta) + 2s^2 \sin\theta \frac{d\theta}{dt}.$

   At $\theta = \frac{\pi}{2}$, $\cos\theta = 0$, $\sin\theta = 1$, $s = 3$, $\frac{ds}{dt} = 1$:

   $0 = 4(3)(1)(1 - 0) + 2(9)(1) \frac{d\theta}{dt} = 12 + 18 \frac{d\theta}{dt}.$

   Thus $18 \frac{d\theta}{dt} = -12$, so $\frac{d\theta}{dt} = -\frac{2}{3}$ rad/s. (The angle is decreasing, which makes sense: as the sides lengthen, the angle must narrow to keep the base fixed.)

   5. Plug into the area rate formula. At the instant: $s = 3$, $\frac{ds}{dt} = 1$, $\sin\theta = 1$, $\cos\theta = 0$, $\frac{d\theta}{dt} = -\frac{2}{3}$.

   $\frac{dA}{dt} = (3)(1)(1) + \frac12 (9)(0)\left(-\frac{2}{3}\right) = 3 + 0 = 3.$

   Watch out

   A common mistake is to forget that $\theta$ changes with time and treat it as constant. That would give $\frac{dA}{dt} = s \frac{ds}{dt} \sin\theta = 3 \cdot 1 \cdot 1 = 3$, which accidentally matches the correct answer here — but only because $\cos\theta = 0$ eliminates the $\frac{d\theta}{dt}$ term. In general, you must include it.

   Tip

   When the included angle is $90^\circ$, the $\cos\theta$ term vanishes, so the rate depends only on the side length and its rate of change. That’s why the answer simplifies so neatly.

   ✓Final answer

   The rate of increase of the area is $\boxed{3}$ sq.ft/sec, which corresponds to option (D).

2. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The height of a cone with semi vertical angle $\dfrac{\pi}{3}$ is increasing at the rate of 2 units/min. The rate at which the radius of the cone is to be decreased so as to have a fixed volume always is (A) $\dfrac{1}{\sqrt{3}}$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $\sqrt{3}$ (D) $\sqrt{2}$

   ›Reveal solutionSolution

   The problem uses related rates with the cone’s volume fixed. Differentiating $V = \frac13 \pi r^2 h$ and using $\frac{dh}{dt}=2$ gives $\frac{dr}{dt} = -\frac{r}{2h} \cdot 2$. With semi-vertical angle $\pi/3$, $\frac{r}{h} = \tan(\pi/3)=\sqrt{3}$, so $\frac{dr}{dt} = -\sqrt{3}$ units/min. The rate of decrease is $\sqrt{3}$.

   Concept & Intuition

   We have a cone whose height is increasing, but we want its volume to stay constant. That means the radius must shrink to compensate. The key is to relate the radius and height through the fixed semi-vertical angle — this gives a constant ratio $r/h = \tan(\pi/3) = \sqrt{3}$. Then we use calculus (related rates) to find how fast the radius must change when the height changes at 2 units/min.

   Step-by-step solution

   1. Volume of a cone

      The volume is $V = \frac13 \pi r^2 h$. Since the volume is fixed, $V$ is constant, so $\frac{dV}{dt} = 0$.

   2. Differentiate implicitly with respect to time

      Using the product rule:

   $\frac{dV}{dt} = \frac13 \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right) = 0.$

   Multiply through by $3/\pi$ (nonzero):

   $2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} = 0.$

   3. Solve for $\frac{dr}{dt}$

   $2rh \frac{dr}{dt} = - r^2 \frac{dh}{dt} \quad\Rightarrow\quad \frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}.$

   4. Use the given rate and geometry We are told $\frac{dh}{dt} = 2$ units/min. The semi-vertical angle is $\pi/3$, so in a right triangle formed by the height, radius, and slant height:

   $\tan\left(\frac{\pi}{3}\right) = \frac{r}{h} = \sqrt{3}.$

   Hence $r = \sqrt{3}\,h$.

   5. Substitute into the rate equation

   $\frac{dr}{dt} = -\frac{\sqrt{3}\,h}{2h} \cdot 2 = -\sqrt{3}.$

   The negative sign means the radius is decreasing. The rate of decrease is $\sqrt{3}$ units/min.

   Tip

   The ratio $r/h$ is constant because the angle is fixed — this lets us avoid needing actual values of $r$ and $h$ at any instant.

   Watch out

   A common mistake is forgetting the factor $2$ from differentiating $r^2$, or mixing up which variable is increasing/decreasing. Always check the sign: if height increases and volume is fixed, radius must decrease.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The height of a cone with semi vertical angle $\pi/3$ is increasing at the rate of 2 units/min. The rate at which the radius of the cone is to be decreased so as to have a fixed volume always is (A) $\sqrt{3}$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $\dfrac{1}{\sqrt{3}}$ (D) $\sqrt{2}$

   ›Reveal solutionSolution

   For a cone of fixed volume, the radius must shrink at a rate that exactly compensates the growth in height. Using the relation $V = \frac13 \pi r^2 h$ and differentiating with respect to time gives $\frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}$. With semi-vertical angle $\pi/3$, we have $r/h = \tan(\pi/3) = \sqrt{3}$, so $\frac{dr}{dt} = -\frac{\sqrt{3}}{2} \cdot 2 = -\sqrt{3}$ units/min. The required rate of decrease is $\sqrt{3}$ units/min, so the correct option is (A).

   The key idea is that "fixed volume" ties the radius and height together through a constraint. When one changes, the other must change in a specific way to keep the product $r^2 h$ constant. The semi-vertical angle gives the instantaneous ratio of radius to height at the moment we are considering — that ratio is not constant over time (since the cone's shape changes), but at the instant we care about, it is fixed by the given angle.

   Let’s work through it step by step.

   1. Write the volume constraint. For a cone, $V = \frac13 \pi r^2 h$. Since the volume is fixed, $V$ is constant. Differentiating both sides with respect to time $t$:

   $\frac{dV}{dt} = \frac13 \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right) = 0.$

   Multiply through by $3/\pi$ (non-zero):

   $2rh \frac{dr}{dt} + r^2 \frac{dh}{dt} = 0.$

   2. Solve for $\frac{dr}{dt}$. Rearranging:

   $2rh \frac{dr}{dt} = - r^2 \frac{dh}{dt}.$

   Assuming $r \neq 0$, divide both sides by $r$:

   $2h \frac{dr}{dt} = - r \frac{dh}{dt}.$

   Hence:

   $\frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}.$

   The negative sign tells us that if height increases, radius must decrease — exactly what we expect.

   3. Use the semi-vertical angle to find $r/h$. The semi-vertical angle is the angle between the axis and the slant height. In a right circular cone, $\tan(\text{semi-vertical angle}) = \frac{\text{radius}}{\text{height}}$. Given the angle is $\pi/3$:

   $\frac{r}{h} = \tan\frac{\pi}{3} = \sqrt{3}.$

   So $r = \sqrt{3}\,h$ at the instant under consideration.

   4. Plug in the given rate. We are told $\frac{dh}{dt} = 2$ units/min. Substituting $r/h = \sqrt{3}$ and $\frac{dh}{dt} = 2$:

   $\frac{dr}{dt} = -\frac{1}{2} \cdot \frac{r}{h} \cdot \frac{dh}{dt} = -\frac{1}{2} \cdot \sqrt{3} \cdot 2 = -\sqrt{3}.$

   The negative sign means the radius is decreasing. The question asks for the rate at which the radius is to be decreased — that is, the magnitude of the decrease. So the required rate is $\sqrt{3}$ units/min.

   Watch out

   A common mistake is to treat $r/h$ as constant over time. It is not — the cone's shape changes as $r$ and $h$ change. But at the instant we are given the semi-vertical angle, the ratio is fixed. The derivative relation $\frac{dr}{dt} = -\frac{r}{2h} \frac{dh}{dt}$ is valid at that instant because $r$ and $h$ are the instantaneous values.

   ✓Final answer

   The rate at which the radius must be decreased is $\sqrt{3}$ units/min, so the correct option is (A).

4. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.The volume of a spherical balloon is increasing at the rate of $2\ \text{cm}^3/\text{sec}$. When its radius is $4\ \text{cm}$, the rate of change of its surface area (in $\text{cm}^2/\text{sec}$) is (A) 1 (B) 2 (C) 3 (D) 4

   ›Reveal solutionSolution

   We use related rates to connect the given rate of change of volume to the rate of change of surface area via the radius. The rate of change of the surface area is $\boxed{1\ \text{cm}^2/\text{sec}}$.

   This problem asks us to find the rate of change of the surface area of a spherical balloon, given the rate of change of its volume at a specific instant. This is a classic application of "related rates" in differential calculus. The core idea is that if two or more quantities are related by an equation, and they are all changing with respect to a common variable (usually time), then their rates of change are also related. We use the chain rule to establish these relationships.

   For a sphere, both its volume ($V$) and surface area ($S$) depend on its radius ($r$). If the radius changes over time, then both the volume and surface area will also change over time.

   • The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.
   • The surface area of a sphere is given by $S = 4\pi r^2$.

   We are given $\frac{dV}{dt}$ and need to find $\frac{dS}{dt}$. Both these rates depend on $\frac{dr}{dt}$, the rate at which the radius is changing. So, our strategy will be:

   1. Use the given rate of change of volume ($\frac{dV}{dt}$) and the volume formula to calculate $\frac{dr}{dt}$ at the specified radius.
   2. Use this calculated $\frac{dr}{dt}$ and the surface area formula to find $\frac{dS}{dt}$ at that same radius.

   Here's the step-by-step solution:

   1. Identify the given information and what needs to be found.

      We are given:

      • The rate at which the volume of the spherical balloon is increasing: $\frac{dV}{dt} = 2\ \text{cm}^3/\text{sec}$.
      • The radius of the balloon at the specific instant we are interested in: $r = 4\ \text{cm}$. We need to find:
      • The rate of change of its surface area, $\frac{dS}{dt}$, at that instant.

   2. Write down the formulas for the volume and surface area of a sphere.

      The volume of a sphere with radius $r$ is $V = \frac{4}{3}\pi r^3$.

      The surface area of a sphere with radius $r$ is $S = 4\pi r^2$.

   3. Differentiate the volume formula with respect to time ($t$) to find $\frac{dr}{dt}$.

      Since $V$ is a function of $r$, and $r$ is a function of $t$, we apply the chain rule to differentiate $V$ with respect to $t$:

   $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$

   $\frac{dV}{dt} = \frac{4}{3}\pi \cdot (3r^2) \cdot \frac{dr}{dt}$

   $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

   Now, substitute the given values: $\frac{dV}{dt} = 2\ \text{cm}^3/\text{sec}$ and $r = 4\ \text{cm}$.
   

   $2 = 4\pi (4)^2 \frac{dr}{dt}$

   $2 = 4\pi (16) \frac{dr}{dt}$

   $2 = 64\pi \frac{dr}{dt}$

   Solving for $\frac{dr}{dt}$:
   

   $\frac{dr}{dt} = \frac{2}{64\pi} = \frac{1}{32\pi}\ \text{cm/sec}$

   This is the rate at which the radius is increasing at the instant when $r=4\ \text{cm}$.
   

   4. Differentiate the surface area formula with respect to time ($t$) to find $\frac{dS}{dt}$.

   Similarly, $S$ is a function of $r$, and $r$ is a function of $t$. We use the chain rule to differentiate $S$ with respect to $t$:

   $\frac{dS}{dt} = \frac{d}{dt}(4\pi r^2)$

   $\frac{dS}{dt} = 4\pi \cdot (2r) \cdot \frac{dr}{dt}$

   $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$

   Now, substitute the value of $r = 4\ \text{cm}$ and the value of $\frac{dr}{dt} = \frac{1}{32\pi}\ \text{cm/sec}$ that we found in the previous step:
   

   $\frac{dS}{dt} = 8\pi (4) \left(\frac{1}{32\pi}\right)$

   $\frac{dS}{dt} = 32\pi \left(\frac{1}{32\pi}\right)$

   $\frac{dS}{dt} = 1\ \text{cm}^2/\text{sec}$

   > [!TIP]
   > A useful observation in this problem is that $\frac{dV}{dr} = 4\pi r^2$, which is the surface area $S$. Also, $\frac{dS}{dr} = 8\pi r$.
   > We have $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = S \cdot \frac{dr}{dt}$.
   > And $\frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt}$.
   > From the first relation, $\frac{dr}{dt} = \frac{1}{S} \frac{dV}{dt}$.
   > Substituting this into the second relation:
   > $\frac{dS}{dt} = \frac{dS}{dr} \cdot \left(\frac{1}{S} \frac{dV}{dt}\right) = (8\pi r) \cdot \left(\frac{1}{4\pi r^2} \frac{dV}{dt}\right) = \frac{2}{r} \frac{dV}{dt}$.
   > Using this shortcut with $r=4$ and $\frac{dV}{dt}=2$:
   > $\frac{dS}{dt} = \frac{2}{4} \cdot 2 = \frac{1}{2} \cdot 2 = 1\ \text{cm}^2/\text{sec}$. This confirms our result efficiently.
   

   The rate of change of the surface area when the radius is $4\ \text{cm}$ is $1\ \text{cm}^2/\text{sec}$.

   ✓Final answer

   The rate of change of the surface area is $\boxed{1\ \text{cm}^2/\text{sec}}$.

5. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the rate of change of volume of a cube and that of its surface area are numerically equal, then the length of its diagonal is (A) $2\sqrt{3}$ (B) $\sqrt{3}$ (C) $4\sqrt{3}$ (D) $6\sqrt{3}$

   ›Reveal solutionSolution

   Equating $\dfrac{dV}{dt}$ and $\dfrac{dS}{dt}$ numerically gives edge $x=4$, so the diagonal is $4\sqrt{3}$.

   Let the edge length be $x$. Then

   $V = x^3 \implies \frac{dV}{dt} = 3x^2\,\frac{dx}{dt},$

   $S = 6x^2 \implies \frac{dS}{dt} = 12x\,\frac{dx}{dt}.$

   Numerically equal:

   $3x^2 = 12x \implies x = 4.$

   The space diagonal of a cube of edge $x$ is $x\sqrt{3}$, so

   $\text{diagonal} = 4\sqrt{3}.$

   ✓Final answer

   Length of the diagonal $= 4\sqrt{3}$ — option (C).

6. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.A ladder of length 13 mts has one end resting against a vertical wall and the other on the ground. If the lower end moves away from the wall at a speed of 2 mts/minute, then the speed (in mts/min) at which upper end falls when the bottom is 5 mts away from the wall is (A) $\dfrac{6}{5}$ (B) $\dfrac{12}{5}$ (C) $\dfrac{5}{6}$ (D) $\dfrac{5}{12}$

   ›Reveal solutionSolution

   This is a classic related-rates problem: use the Pythagorean theorem to relate the ladder’s height and base distance, then differentiate with respect to time. The upper end falls at $\frac{5}{6}$ m/min when the bottom is 5 m from the wall.

   We have a ladder of fixed length 13 m leaning against a vertical wall. The bottom slides away from the wall at a constant speed of 2 m/min. We need the speed at which the top slides down the wall at the instant the bottom is 5 m from the wall.

   Concept & Intuition

   The ladder, wall, and ground form a right triangle: the ladder is the hypotenuse (always 13 m), the distance from the wall to the bottom is one leg, and the height of the top along the wall is the other leg. As the bottom moves, both legs change, but the hypotenuse stays fixed. This gives a relationship between the rates of change of the two legs — a classic related rates problem. Differentiating the Pythagorean relation with respect to time lets us connect the known speed (bottom moving away) to the unknown speed (top moving down).

   ---

   1. Set up variables and the fixed relation Let $x$ = distance from the wall to the bottom of the ladder (in m). Let $y$ = height of the top of the ladder on the wall (in m). The ladder length is constant:

   $x^2 + y^2 = 13^2 = 169.$

   2. Differentiate with respect to time Both $x$ and $y$ change with time $t$. Differentiate implicitly:

   $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0.$

   Divide by 2:

   $x \frac{dx}{dt} + y \frac{dy}{dt} = 0.$

   3. Identify known and unknown rates

      We are given $\frac{dx}{dt} = 2$ m/min (positive because $x$ increases).

      We want $\frac{dy}{dt}$ when $x = 5$ m.

      Note: $\frac{dy}{dt}$ will be negative because $y$ decreases (top falls). The problem asks for the speed (magnitude), so we will take the absolute value at the end.

   4. Find $y$ when $x = 5$

      From $x^2 + y^2 = 169$:

   $5^2 + y^2 = 169 \quad\Rightarrow\quad 25 + y^2 = 169 \quad\Rightarrow\quad y^2 = 144 \quad\Rightarrow\quad y = 12 \text{ (positive height)}.$

   5. Plug into the differentiated equation

   $(5)(2) + (12) \frac{dy}{dt} = 0 \quad\Rightarrow\quad 10 + 12 \frac{dy}{dt} = 0.$

   Solve:

   $12 \frac{dy}{dt} = -10 \quad\Rightarrow\quad \frac{dy}{dt} = -\frac{10}{12} = -\frac{5}{6}.$

   6. Interpret the result The negative sign means the top is moving downward. The speed (magnitude) is $\frac{5}{6}$ m/min.

   Tip

   A common mistake is forgetting the negative sign or mixing up which rate is given. Always check: if the bottom moves away, the top must move down, so $\frac{dy}{dt}$ should be negative.

   Watch out

   Another pitfall: using the given $x=5$ before differentiating. You must differentiate the general relation first, then substitute the specific values — otherwise you lose the relationship between the rates.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

7. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If Water is poured into a cylindrical tank of radius $3.5$ ft at the rate of $1$ cu ft/min, then the rate at which the level of the water in the tank increases (in ft/min) is (A) $\dfrac{1}{154}$ (B) $\dfrac{8}{77}$ (C) $\dfrac{2}{77}$ (D) $\dfrac{1}{11}$

   ›Reveal solutionSolution

   The water level rises at a constant rate because the tank’s cross‑sectional area is constant; the rate is the inflow divided by the area. The answer is $\frac{2}{77}$ ft/min, option (C).

   Concept & Intuition

   When you pour water into a cylinder, the volume added is directly proportional to the increase in height, because the cross‑sectional area doesn’t change with depth. So the rate of change of height is simply the volumetric flow rate divided by the area of the base. No calculus chain‑rule gymnastics needed — just a straightforward division.

   ---

   1. Identify the relationship The volume of water in a cylinder of radius $r$ and height $h$ is

   $V = \pi r^2 h.$

   Here $r = 3.5$ ft, so the base area is

   $A = \pi (3.5)^2 = \pi \times 12.25 = \frac{49\pi}{4} \ \text{ft}^2.$

   2. Differentiate with respect to time Since $r$ is constant,

   $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt} = A \frac{dh}{dt}.$

   We are given $\frac{dV}{dt} = 1$ cu ft/min.

   3. Solve for $\frac{dh}{dt}$

   $\frac{dh}{dt} = \frac{1}{A} = \frac{1}{\frac{49\pi}{4}} = \frac{4}{49\pi}.$

   4. Simplify numerically Use $\pi \approx \frac{22}{7}$ (common in such problems):

   $\frac{dh}{dt} = \frac{4}{49 \cdot \frac{22}{7}} = \frac{4}{49 \times \frac{22}{7}} = \frac{4}{7 \times 22} = \frac{4}{154} = \frac{2}{77}.$

   Tip

   If you use $\pi = 22/7$, the arithmetic simplifies neatly. The exact value $\frac{4}{49\pi}$ is fine, but the multiple‑choice options are given as rational numbers, so the approximation is intended.

   Watch out

   A common mistake is to forget that the radius is $3.5$, not $7$, and accidentally use $r=7$ — that would give $\frac{1}{154}$, which is option (A). Always square the radius correctly.

   ---

   ✓Final answer

   The correct option is (C).

   ANSWER: C

8. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.The side of an equilateral triangle is $5$ units. In measuring the side, an error of $0.05$ units is made. Then the percentage error in measuring the area of the triangle is (A) $2$ (B) $3$ (C) $4$ (D) $5$

   ›Reveal solutionSolution

   We use the concept of differentials to approximate the error in the area of an equilateral triangle. For a small error in the side, the percentage error in the area is twice the percentage error in the side. The percentage error in the area is $\boxed{2}$.

   When a quantity is calculated using a measured value, and there's an error in the measurement, this error propagates through the calculation, leading to an error in the final calculated quantity. For small errors, we can use the concept of differentials to approximate how these errors propagate.

   Consider a function $y = f(x)$. If there is a small error $\Delta x$ in the measurement of $x$, it causes a corresponding small error $\Delta y$ in the calculated value of $y$. For sufficiently small $\Delta x$, the change $\Delta y$ can be approximated by the differential $dy$:

   $\Delta y \approx dy = \frac{dy}{dx} \Delta x$

   The fractional error in $y$ is $\frac{\Delta y}{y}$, and the percentage error is $\frac{\Delta y}{y} \times 100\%$.

   In this problem, we are dealing with the area of an equilateral triangle, which depends on its side length.

   1. Identify the given information:

      The side of the equilateral triangle is $s = 5$ units.

      The error in measuring the side is $\Delta s = 0.05$ units.

   2. State the formula for the area of an equilateral triangle:

      The area $A$ of an equilateral triangle with side $s$ is given by:

   $A = \frac{\sqrt{3}}{4} s^2$

   3. Find the differential of the area with respect to the side: To understand how a small change in $s$ affects $A$, we differentiate $A$ with respect to $s$:

   $\frac{dA}{ds} = \frac{d}{ds} \left( \frac{\sqrt{3}}{4} s^2 \right) = \frac{\sqrt{3}}{4} (2s) = \frac{\sqrt{3}}{2} s$

   Now, we can express the approximate error in the area, $\Delta A$, using the differential $dA$:
   

   $\Delta A \approx dA = \frac{dA}{ds} \Delta s = \frac{\sqrt{3}}{2} s \cdot \Delta s$

   4. Calculate the fractional error in the area: The fractional error in the area is $\frac{\Delta A}{A}$. We substitute the expressions for $\Delta A$ and $A$:

   $\frac{\Delta A}{A} = \frac{\frac{\sqrt{3}}{2} s \cdot \Delta s}{\frac{\sqrt{3}}{4} s^2}$

   We can simplify this expression:
   

   $\frac{\Delta A}{A} = \frac{\frac{1}{2} s \cdot \Delta s}{\frac{1}{4} s^2} = \frac{1}{2} \cdot \frac{4}{1} \cdot \frac{s \cdot \Delta s}{s^2} = 2 \frac{\Delta s}{s}$

   > [!IMPORTANT]
   > For a quantity $y$ that depends on another quantity $x$ as $y = kx^n$ (where $k$ is a constant), the fractional error in $y$ is approximately $n$ times the fractional error in $x$:
   > $$ \frac{\Delta y}{y} \approx n \frac{\Delta x}{x} $$
   > In our case, $A = \frac{\sqrt{3}}{4} s^2$, so $n=2$. This confirms our derived relationship $\frac{\Delta A}{A} = 2 \frac{\Delta s}{s}$.
   

   5. Substitute the given values and calculate the percentage error:

   We have $s = 5$ units and $\Delta s = 0.05$ units.

   First, calculate the fractional error in the side:

   $\frac{\Delta s}{s} = \frac{0.05}{5} = \frac{5/100}{5} = \frac{1}{100} = 0.01$

   Now, use the relationship for the fractional error in the area:
   

   $\frac{\Delta A}{A} = 2 \frac{\Delta s}{s} = 2 \times 0.01 = 0.02$

   Finally, convert this fractional error to a percentage error:
   

   $\text{Percentage error in area} = \frac{\Delta A}{A} \times 100\% = 0.02 \times 100\% = 2\%$

   ✓Final answer

   The percentage error in measuring the area of the triangle is $\boxed{2}$.

9. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If $\sin y = \sin 3t$ and $x = \sin t$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{3}{\sqrt{4-x^2}}$ (B) $\dfrac{3}{\sqrt{1-x^2}}$ (C) $\dfrac{1}{\sqrt{4-x^2}}$ (D) $\dfrac{-1}{\sqrt{4-x^2}}$

   ›Reveal solutionSolution

   Treat both $y$ and $x$ as functions of $t$ (parametric differentiation): $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$. With $y=3t$ and $x=\sin t$, this gives $\dfrac{3}{\sqrt{1-x^2}}$ — option (B).

   Concept. We are not given $y$ as a function of $x$ directly; instead both are tied to the parameter $t$. Parametric differentiation says $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ whenever $dx/dt\neq 0$.

   Step 1 — read off $y$ in terms of $t$.

   The relation $\sin y=\sin 3t$ has the principal solution $y=3t$, so

   $\frac{dy}{dt}=3.$

   Step 2 — differentiate $x=\sin t$.

   $\frac{dx}{dt}=\cos t.$

   Step 3 — form the ratio.

   $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3}{\cos t}.$

   Step 4 — express in terms of $x$.

   Since $x=\sin t$, $\cos t=\sqrt{1-\sin^2 t}=\sqrt{1-x^2}$ (principal branch). Hence

   $\frac{dy}{dx}=\frac{3}{\sqrt{1-x^2}}.$

   ✓Final answer

   $\dfrac{dy}{dx}=\dfrac{3}{\sqrt{1-x^2}}$ — option (B).

10. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
    Q.The area of a triangle is obtained with lengths of two sides and included angle between them. If the angle is measured as $60^\circ 20'$ instead of $60^\circ$, then the percentage error in its area is (A) $\dfrac{5\pi}{54}$ (B) $\dfrac{5\pi}{27\sqrt{3}}$ (C) $\dfrac{5\sqrt{3}\pi}{27}$ (D) $\dfrac{5\pi}{27}$

    ›Reveal solutionSolution

    The percentage error in area due to a small angular error is found by differentiating the area formula $A = \frac12 ab\sin C$; the relative error is $\cot C \cdot dC$ (in radians). With $C=60^\circ$, $dC=20'=\frac{\pi}{540}$ rad, the percentage error becomes $\frac{5\pi}{27\sqrt{3}}$, matching option (B).

    Concept & Intuition

    When a quantity is computed from measured values, a small error in one measurement propagates into the result. Here, area $A = \frac12 ab \sin C$ depends on the included angle $C$. If the sides $a$ and $b$ are exact, the only source of error is the angle. For small errors, we use differentials: the change in area $dA \approx \frac{dA}{dC} \cdot dC$, and the relative error is $\frac{dA}{A} = \cot C \cdot dC$ (with $dC$ in radians). This turns a messy trigonometric problem into a simple calculus step.

    Step-by-step solution

    1. Write the exact area formula The area of a triangle with two sides $a$, $b$ and included angle $C$ is

    $A = \frac12 ab \sin C.$

    Here $a$ and $b$ are assumed error‑free; only $C$ is measured incorrectly.

    2. Find the differential of $A$ with respect to $C$ Differentiate:

    $\frac{dA}{dC} = \frac12 ab \cos C.$

    Hence a small error $dC$ in the angle causes an error in area:

    $dA \approx \frac12 ab \cos C \cdot dC.$

    3. Compute the relative error The relative error is

    $\frac{dA}{A} = \frac{\frac12 ab \cos C \cdot dC}{\frac12 ab \sin C} = \cot C \cdot dC.$

    This is the key formula: the relative error in area equals $\cot C$ times the angular error (in radians).

    4. Convert the angular error to radians The measured angle is $60^\circ 20'$ instead of $60^\circ$, so the error is

    $dC = 20' = \frac{20}{60}^\circ = \frac13^\circ.$

    Convert degrees to radians:

    $1^\circ = \frac{\pi}{180} \text{ rad} \quad\Rightarrow\quad dC = \frac13 \cdot \frac{\pi}{180} = \frac{\pi}{540} \text{ rad}.$

    5. Evaluate $\cot C$ at $C = 60^\circ$

    $\cot 60^\circ = \frac{\cos 60^\circ}{\sin 60^\circ} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}.$

    6. Compute the relative error

    $\frac{dA}{A} = \frac{1}{\sqrt{3}} \cdot \frac{\pi}{540} = \frac{\pi}{540\sqrt{3}}.$

    7. Convert to percentage error Multiply by 100:

    $\text{Percentage error} = \frac{\pi}{540\sqrt{3}} \times 100 = \frac{100\pi}{540\sqrt{3}} = \frac{5\pi}{27\sqrt{3}}.$

    Tip

    Notice that the $100$ cancels with $540$ to give $5/27$ — a clean fraction. Always remember to convert minutes to degrees first, then to radians.

    Watch out

    A common mistake is to forget to convert $20'$ to radians, or to use degrees directly in $\cot$ — that would give a wrong numerical value. The derivative formula $\frac{d}{dC}\sin C = \cos C$ only works when $C$ is in radians.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

11. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.A right circular cone is inscribed in a sphere of radius 3 units. If the volume of the cone is maximum, then semi vertical angle of the cone is (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{6}$ (C) $\tan^{-1}(\sqrt{2})$ (D) $\tan^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$

    ›Reveal solutionSolution

    Maximum cone volume in a sphere of radius $3$ occurs at semi-vertical angle $\tan^{-1}\!\big(\tfrac{1}{\sqrt2}\big)$ — option (D).

    Let the sphere have radius $R = 3$ and centre $O$. Let the cone have height $h$ (apex to base) and base radius $r$. The base circle lies at distance $|h-R|$ from the centre, so

    $r^{2} = R^{2} - (h - R)^{2} = 2Rh - h^{2}.$

    Volume:

    $V = \frac{1}{3}\pi r^{2} h = \frac{\pi}{3}\left(2Rh^{2} - h^{3}\right).$

    Maximise:

    $\frac{dV}{dh} = \frac{\pi}{3}\left(4Rh - 3h^{2}\right) = 0 \;\Rightarrow\; h = \frac{4R}{3}.$

    With $R = 3$: $h = 4$. Then

    $r^{2} = 2Rh - h^{2} = 2(3)(4) - 16 = 8 \;\Rightarrow\; r = 2\sqrt{2}.$

    The semi-vertical angle $\alpha$ satisfies

    $\tan\alpha = \frac{r}{h} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}} \;\Rightarrow\; \alpha = \tan^{-1}\!\left(\frac{1}{\sqrt{2}}\right).$

    ✓Final answer

    Semi-vertical angle $= \tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right)$ — option (D).

12. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.If an error of $0.02$ sq.cm is found in the surface area of a sphere when its radius is measured as $10$ cm, then the approximate error that occurs in the volume of the sphere, in cubic centimetres, is (A) $0.2$ (B) $0.01$ (C) $0.3$ (D) $0.1$

    ›Reveal solutionSolution

    The error in volume is found by relating differentials: $dV = \frac{r}{2} \, dS$. With $r = 10$ cm and $dS = 0.02$ sq.cm, the approximate error in volume is $0.1$ cubic cm.

    The key idea here is that when a small error is made in measuring a quantity (here, the radius), that error propagates into any other quantity calculated from it. We are not asked for the exact error — only an approximate error, which is exactly what differentials give us. The surface area and volume of a sphere are both functions of the radius, so a small change $\Delta r$ in radius produces small changes $\Delta S$ and $\Delta V$ that are well approximated by the differentials $dS$ and $dV$.

    We are told the error in surface area ($dS = 0.02$) and the measured radius ($r = 10$). We need the corresponding error in volume ($dV$). The direct link is through the radius: find $dr$ from $dS$, then use that $dr$ to find $dV$.

    1. Relate surface area error to radius error. Surface area of a sphere: $S = 4\pi r^2$. Differentiate: $dS = 8\pi r \, dr$. With $r = 10$ and $dS = 0.02$:

    $0.02 = 8\pi (10) \, dr = 80\pi \, dr$

    So

    $dr = \frac{0.02}{80\pi} = \frac{0.00025}{\pi}$

    This is the approximate error in the radius measurement.

    2. Relate volume error to the same radius error. Volume of a sphere: $V = \frac{4}{3}\pi r^3$. Differentiate: $dV = 4\pi r^2 \, dr$. Substitute $r = 10$ and the $dr$ we found:

    $dV = 4\pi (100) \cdot \frac{0.00025}{\pi} = 400\pi \cdot \frac{0.00025}{\pi}$

    The $\pi$ cancels:

    $dV = 400 \times 0.00025 = 0.1$

    Tip

    You can skip finding $dr$ explicitly by combining the two differentials. From $dS = 8\pi r \, dr$ and $dV = 4\pi r^2 \, dr$, divide: $\frac{dV}{dS} = \frac{4\pi r^2}{8\pi r} = \frac{r}{2}$. So $dV = \frac{r}{2} \, dS$. With $r=10$ and $dS=0.02$, $dV = 5 \times 0.02 = 0.1$ — much faster.

    Watch out

    A common mistake is to compute the error in volume by directly plugging $r = 10 \pm dr$ into the volume formula and subtracting. That gives the exact change, not the approximate change via differentials. For small errors, the differential method is the intended approach in such problems, and it matches the given options.

    ✓Final answer

    The approximate error in the volume is $\boxed{0.1}$ cubic cm, which corresponds to option (D).