Question:
The surface area of a spherical ball is increasing at the rate of \(4π \text{cm}^2\)/second. The rate at which the radius is increasing when the surface area is \(16π \text{cm}^2\) is
The surface area of a spherical ball is increasing at the rate of \(4π \text{cm}^2\)/second. The rate at which the radius is increasing when the surface area is \(16π \text{cm}^2\) is
Show Hint
Differentiate S = 4 pi r^2 with respect to time.
Updated On: Oct 1, 2026
- \(0.5\) cm/second
- \(0.25\) cm/second
- \(0.125\) cm/second
- \(1\) cm/second
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Surface area of a sphere is \(S = 4\pi r^2\). Differentiating with respect to time, \(\dfrac{dS}{dt} = 8\pi r\dfrac{dr}{dt}\).
Step 2: Find the radius
When \(S = 16\pi\): \(4\pi r^2 = 16\pi\), so \(r = 2\) cm.
Step 3: Solve for dr/dt
\[ 4\pi = 8\pi (2)\frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{4\pi}{16\pi} = 0.25 \text{ cm/s} \]
The value 0.5 comes from forgetting the factor 2 in 8 pi r, and 1 from using r = 0.5.
Final Answer:
The radius increases at 0.25 cm/s, option (B). \[ \boxed{0.25 \text{ cm/s}} \]
Surface area of a sphere is \(S = 4\pi r^2\). Differentiating with respect to time, \(\dfrac{dS}{dt} = 8\pi r\dfrac{dr}{dt}\).
Step 2: Find the radius
When \(S = 16\pi\): \(4\pi r^2 = 16\pi\), so \(r = 2\) cm.
Step 3: Solve for dr/dt
\[ 4\pi = 8\pi (2)\frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{4\pi}{16\pi} = 0.25 \text{ cm/s} \]
The value 0.5 comes from forgetting the factor 2 in 8 pi r, and 1 from using r = 0.5.
Final Answer:
The radius increases at 0.25 cm/s, option (B). \[ \boxed{0.25 \text{ cm/s}} \]
Was this answer helpful?
0
0
Top MHT CET Mathematics Questions
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- If $\int\frac{f\left(x\right)}{log \left(sin\,x\right)}dx = log\left[log\,sin\,x\right]+c$ then $f\left(x\right)=$
- If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is
- If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
View More Questions
Top MHT CET Rate of Change of Quantities Questions
- Let $B\equiv(0,3)$ and $C\equiv(4,0)$. The point A is moving on the line $y=2x$ at the rate of 2 units/second. The area of $\triangle ABC$ is increasing at the rate of
- The diagonal of a square is changing at the rate of $0.5$ cm/sec. Then the rate of change of area when the area is $400$ cm$^{2}$ is equal to}
- The rate of change of $\sqrt{x^{2}+16}$ with respect to $\frac{x}{x-1}$ at $x=5$ is
- An object is moving in the clockwise direction around the unit circle $x^{2}+y^{2}=1$. As it passes through the point $(\frac{1}{2},\frac{\sqrt{3{2})$, its y-coordinate is decreasing at the rate of 3 units per sec. The rate at which the x-coordinate changes at this point is
- A spherical balloon is filled with (4500\pi) cubic meters of helium gas. If a leak causes gas to escape at (72\pi) m(^3)/min, then the rate at which the radius decreases 49 minutes after leakage began is
View More Questions
Top MHT CET Questions
- During r- DNA technology, which one of the following enzymes is used for cleaving DNA molecule ?
- In non uniform circular motion, the ratio of tangential to radial acceleration is (r = radius of circle, $v =$ speed of the particle, $\alpha =$ angular acceleration)
- The temperature of $32^{\circ}C$ is equivalent to
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- The heat of formation of water is $ 260\, kJ $ . How much $ H_2O $ is decomposed by $ 130\, kJ $ of heat ?
View More Questions