Q.If the area of a circle increases at a uniform rate, then prove that the perimeter varies inversely as the radius.

Showing the 12 most recent of 15 on this concept.

1. CBSE 2025Set 65/2/11 markMCQ
   Q.A cylindrical tank of radius 10 cm is being filled with sugar at the rate of $100\pi$ cm$^3$/s. The rate at which the height of the sugar inside the tank is increasing is: (A) $0.1$ cm/s (B) $0.5$ cm/s (C) $1$ cm/s (D) $1.1$ cm/s

   ›Reveal solutionSolution

   The volume of a cylinder is $V = \pi r^2 h$. Since the radius is constant, the rate of change of volume with respect to time is $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$. Given $\frac{dV}{dt} = 100\pi$ cm³/s and $r = 10$ cm, solving gives $\frac{dh}{dt} = 1$ cm/s. The correct option is (C).

   This is a classic Related Rates problem. The core idea is that when two quantities are linked by a geometric formula (here, volume and height of a cylinder), their rates of change with respect to time are also linked. You differentiate the relationship with respect to time, plug in what you know, and solve for the unknown rate.

   The key insight: the tank’s radius is fixed at 10 cm. So as sugar pours in, the height increases, but the cross-sectional area stays the same. That means the volume increases at a constant rate per unit height — specifically, each 1 cm rise in height adds $\pi (10)^2 = 100\pi$ cm³ of volume. Since sugar is being added at exactly $100\pi$ cm³/s, the height must be rising at 1 cm/s.

   Let’s work it out formally.

   1. Write the relationship between volume and height. For a cylinder, $V = \pi r^2 h$. Here $r = 10$ cm, so

   $V = \pi (10)^2 h = 100\pi h.$

   2. Differentiate both sides with respect to time $t$.

      Since $r$ is constant, $\frac{dV}{dt} = 100\pi \frac{dh}{dt}$.

      This is the related rates equation — it tells us how fast the volume changes in terms of how fast the height changes.

   3. Substitute the given rate.

      We know $\frac{dV}{dt} = 100\pi$ cm³/s. So:

   $100\pi = 100\pi \frac{dh}{dt}.$

   4. Solve for $\frac{dh}{dt}$. Divide both sides by $100\pi$:

   $\frac{dh}{dt} = 1 \text{ cm/s}.$

   Watch out

   A common mistake is to forget that the radius is constant and try to differentiate $V = \pi r^2 h$ using the product rule, treating $r$ as a variable. Here $r$ is fixed, so it’s just a constant factor. If the radius were also changing (e.g., a conical tank), you’d need a different approach.

   Tip

   You can often avoid calculus entirely for constant-cross-section tanks: the rate of height increase is simply (volume flow rate) ÷ (cross-sectional area). Here, area = $\pi (10)^2 = 100\pi$ cm², so $\frac{dh}{dt} = \frac{100\pi}{100\pi} = 1$ cm/s. This shortcut works because the shape is a right cylinder.

   ✓Final answer

   The height increases at 1 cm/s, so the correct option is (C).

2. CBSE 2026Set ANNUAL1 mark
   Q.The edge of a variable cube is increasing at the rate of 3 cm/s. The volume of the cube is increasing at the rate of __________ while the edge is 10 cm long.

   ›Reveal solutionSolution

   Use $V=e^3$ and the chain rule $dV/dt = 3e^2\,de/dt$.

   Let $e$ be the edge; $V=e^3$, so $\dfrac{dV}{dt}=3e^2\dfrac{de}{dt}$.

   Given $\dfrac{de}{dt}=3$ cm/s and $e=10$ cm:

   $\dfrac{dV}{dt}=3(10)^2(3)=900$ cm³/s.

   ✓Final answer

   The volume is increasing at $900\text{ cm}^3/\text{s}$.

3. CBSE 2026Set ANNUAL1 mark
   Q.The radius of an air bubble is increasing at the rate of $1/2$ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

   ›Reveal solutionSolution

   Use $V=\dfrac43\pi r^3$ and $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.

   Given $\dfrac{dr}{dt}=\dfrac12$ cm/s, at $r=1$ cm:

   $\dfrac{dV}{dt}=4\pi(1)^2\left(\dfrac12\right)=2\pi$ cm³/s.

   ✓Final answer

   The volume increases at $2\pi\text{ cm}^3/\text{s}$.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Radius of a circle is increasing at the rate of 1/π m/s. Rate of change of its circumference is:

   (a) 4π m/s

   (b) 2 m/s

   (c) 2π m/s

   (d) 4 m/s

   ›Reveal solutionSolution

   Since $C = 2\pi r$, differentiating both sides w.r.t. time gives $\frac{dC}{dt} = 2\pi \frac{dr}{dt}$ directly.

   The circumference of a circle of radius $r$ is $C = 2\pi r$.

   Differentiating with respect to time $t$:

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

   Given $\frac{dr}{dt} = \frac{1}{\pi}$ m/s, substitute:

   $\frac{dC}{dt} = 2\pi \times \frac{1}{\pi} = 2 \text{ m/s}$

   ✓Final answer

   The rate of change of the circumference is 2 m/s (option b).

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.Radius of a circle is increasing at the rate of 2 m/s. Rate of change of its circumference is:

   (a) 4π m/s

   (b) 2 m/s

   (c) 2π m/s

   (d) 4 m/s

   ›Reveal solutionSolution

   Differentiate the circumference formula $C=2\pi r$ with respect to time and plug in $\dfrac{dr}{dt}$.

   Given $\dfrac{dr}{dt} = 2$ m/s. Circumference $C = 2\pi r$.

   Differentiating both sides with respect to time $t$:

   $\frac{dC}{dt} = 2\pi \frac{dr}{dt} = 2\pi(2) = 4\pi \text{ m/s}.$

   Note the rate is constant — it does not depend on the actual value of $r$.

   ✓Final answer

   $4\pi$ m/s — option (a).

6. CBSE 2024Set ANNUAL1 mark
   Q.The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

   ›Reveal solutionSolution

   Use related rates: differentiate $A=\pi r^2$ w.r.t. time and substitute the given $\dfrac{dr}{dt}$ and $r$.

   Given $\dfrac{dr}{dt} = 3$ cm/s, find $\dfrac{dA}{dt}$ at $r = 10$ cm.

   $A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$

   $\dfrac{dA}{dt} = 2\pi (10)(3) = 60\pi$ cm² per second

   ✓Final answer

   The area is increasing at the rate of $60\pi$ cm²/s.

7. CBSE 2024Set ANNUAL1 mark
   Q.The radius of a circle is increasing at the rate of $0.7$ cm/s. What is the rate of increase in its circumference?

   ›Reveal solutionSolution

   Differentiate the circumference formula $C=2\pi r$ with respect to time.

   The circumference of a circle of radius $r$ is

   $C = 2\pi r.$

   Given $\dfrac{dr}{dt} = 0.7$ cm/s. Differentiating with respect to $t$:

   $\dfrac{dC}{dt} = 2\pi\,\dfrac{dr}{dt} = 2\pi(0.7) = 1.4\pi\ \text{cm/s}.$

   ✓Final answer

   The circumference increases at $\boldsymbol{1.4\pi\ \text{cm/s}\ (\approx 4.4\ \text{cm/s})}$

8. CBSE 2023Set ANNUAL1 markMCQ
   Q.The radius of a circle is increasing at the rate of $0.3$ cm/sec. The rate of increase of its perimeter is

   (a) $0.4\pi$ cm/sec

   (b) $0.6\pi$ cm/sec

   (c) $0.8\pi$ cm/sec

   (d) none of these

   ›Reveal solutionSolution

   This is a related-rates problem: differentiate the perimeter formula w.r.t. time.

   Perimeter (circumference) $P = 2\pi r$. Differentiating w.r.t. time $t$: $\dfrac{dP}{dt} = 2\pi\dfrac{dr}{dt}$.

   Given $\dfrac{dr}{dt}=0.3$ cm/sec: $\dfrac{dP}{dt} = 2\pi(0.3) = 0.6\pi$ cm/sec.

   ✓Final answer

   (b) $0.6\pi$ cm/sec.

9. CBSE 2023Set ANNUAL1 mark
   Q.Radius of a circle is increasing at the rate of $3$ cm/sec. Find the rate of change of area when radius of circle is $10$ cm.

   ›Reveal solutionSolution

   Differentiate $A=\pi r^2$ with respect to time and substitute $r=10$, $dr/dt=3$.

   Area of circle: $A=\pi r^2$. Differentiating both sides with respect to $t$:

   $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$

   Given $\dfrac{dr}{dt}=3$ cm/sec and $r=10$ cm:

   $\dfrac{dA}{dt}=2\pi(10)(3)=60\pi$ cm²/sec.

   ✓Final answer

   $\dfrac{dA}{dt}=60\pi$ cm²/sec.

10. CBSE 2023Set ANNUAL1 markMCQ
    Q.The radius of a circle is increasing at the rate of 0.7 cm/s. The rate of increase of its circumference is –

    (a) 7.1 cm/s

    (b) 4.0 cm/s

    (c) 3.9 cm/s

    (d) 4.4 cm/s

    ›Reveal solutionSolution

    Differentiate $C = 2\pi r$ with respect to time and substitute the given rate of change of the radius.

    Circumference $C = 2\pi r$. Differentiating with respect to time $t$:

    $\frac{dC}{dt} = 2\pi\frac{dr}{dt}.$

    Given $\dfrac{dr}{dt} = 0.7$ cm/s:

    $\frac{dC}{dt} = 2\pi(0.7) = 1.4\pi \approx 4.4 \text{ cm/s}.$

    ✓Final answer

    Rate of increase of circumference ≈ 4.4 cm/s — option (d).

11. CBSE 2018Set ANNUAL1 markMCQ
    Q.The volume of a spherical balloon is increasing at the rate of 25 $cm^3$/sec. The rate of increase of its surface area when its radius is 5cm is

    (a) 5 $cm^2$/sec.

    (b) 10 $cm^2$/sec.

    (c) 15 $cm^2$/sec.

    (d) 20 $cm^2$/sec.

    ›Reveal solutionSolution

    related rates: relate dV/dt to dr/dt, then dS/dt to dr/dt

    Volume $V=\dfrac43\pi r^3\Rightarrow \dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$.

    At $r=5$: $25=4\pi(25)\dfrac{dr}{dt}\implies \dfrac{dr}{dt}=\dfrac{25}{100\pi}=\dfrac{1}{4\pi}$.

    Surface area $S=4\pi r^2\Rightarrow \dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$.

    At $r=5$: $\dfrac{dS}{dt}=8\pi(5)\left(\dfrac{1}{4\pi}\right)=\dfrac{40\pi}{4\pi}=10\ \text{cm}^2/\text{sec}$

    ✓Final answer

    $\dfrac{dS}{dt}=10\ \text{cm}^2/\text{sec}$, option (b).

12. CBSE 2018Set ANNUAL1 markMCQ
    Q.The angle $x$ which increases twice as fast as its sine is

    (a) $\dfrac{\pi}{3}$

    (b) $\dfrac{\pi}{2}$

    (c) $\pi$

    (d) $\dfrac{3\pi}{2}$

    ›Reveal solutionSolution

    Translate "x increases twice as fast as sin x" into $dx/dt = 2\,d(\sin x)/dt$ and solve for x.

    "x increases twice as fast as its sine" means $\dfrac{dx}{dt} = 2\dfrac{d(\sin x)}{dt} = 2\cos x\dfrac{dx}{dt}$

    Since $\dfrac{dx}{dt}\ne0$, divide both sides by it: $1=2\cos x \Rightarrow \cos x=\dfrac12$

    $x=\dfrac\pi3$ (the standard angle with $\cos x=1/2$).

    ✓Final answer

    $x=\dfrac{\pi}{3}$, option (a).