Question:
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:
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:
Show Hint
For small changes in geometrical quantities, differential approximation is extremely useful:
\[
dV \approx \frac{dV}{dr}dr
\]
and
\[
dS \approx \frac{dS}{dr}dr
\]
This avoids lengthy exact calculations and gives very accurate approximations.
Updated On: Jun 17, 2026
- \(0.07\)
- \(0.04\)
- \(0.032\)
- \(0.028\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
For a sphere of radius \(r\),
\[
S = 4\pi r^2
\]
where \(S\) is the surface area.
Also,
\[
V = \frac{4}{3}\pi r^3
\]
where \(V\) is the volume.
When very small changes occur, differentials provide an excellent approximation:
\[
dV = \frac{dV}{dr}dr
\]
and
\[
dS = \frac{dS}{dr}dr
\]
Using these relations, we can connect a small increase in surface area with the corresponding increase in volume.
Step 1: Find the radius of the sphere using the given surface area. Given, \[ S = 49\pi \] Using the formula: \[ 4\pi r^2 = 49\pi \] Dividing both sides by \(\pi\), \[ 4r^2 = 49 \] Therefore, \[ r^2 = \frac{49}{4} \] Hence, \[ r = \frac{7}{2} \]
Step 2: Differentiate the surface area formula. We know that \[ S = 4\pi r^2 \] Differentiating with respect to \(r\), \[ \frac{dS}{dr} = 8\pi r \] Thus, \[ dS = 8\pi r\,dr \] The increase in surface area is: \[ dS = 0.016 \] Substituting \(r=\frac72\), \[ 0.016 = 8\pi \left(\frac72\right)dr \] \[ 0.016 = 28\pi\,dr \] Therefore, \[ dr = \frac{0.016}{28\pi} \]
Step 3: Differentiate the volume formula. The volume of a sphere is: \[ V = \frac43\pi r^3 \] Differentiating, \[ \frac{dV}{dr} = 4\pi r^2 \] Hence, \[ dV = 4\pi r^2\,dr \] Substituting \(r=\frac72\), \[ dV = 4\pi\left(\frac{49}{4}\right)\left(\frac{0.016}{28\pi}\right) \] Simplifying carefully, \[ dV = \frac{49\times0.016}{28} \] \[ dV = 0.028 \] Thus, the approximate increase in volume is: \[ \boxed{0.028} \]
Step 1: Find the radius of the sphere using the given surface area. Given, \[ S = 49\pi \] Using the formula: \[ 4\pi r^2 = 49\pi \] Dividing both sides by \(\pi\), \[ 4r^2 = 49 \] Therefore, \[ r^2 = \frac{49}{4} \] Hence, \[ r = \frac{7}{2} \]
Step 2: Differentiate the surface area formula. We know that \[ S = 4\pi r^2 \] Differentiating with respect to \(r\), \[ \frac{dS}{dr} = 8\pi r \] Thus, \[ dS = 8\pi r\,dr \] The increase in surface area is: \[ dS = 0.016 \] Substituting \(r=\frac72\), \[ 0.016 = 8\pi \left(\frac72\right)dr \] \[ 0.016 = 28\pi\,dr \] Therefore, \[ dr = \frac{0.016}{28\pi} \]
Step 3: Differentiate the volume formula. The volume of a sphere is: \[ V = \frac43\pi r^3 \] Differentiating, \[ \frac{dV}{dr} = 4\pi r^2 \] Hence, \[ dV = 4\pi r^2\,dr \] Substituting \(r=\frac72\), \[ dV = 4\pi\left(\frac{49}{4}\right)\left(\frac{0.016}{28\pi}\right) \] Simplifying carefully, \[ dV = \frac{49\times0.016}{28} \] \[ dV = 0.028 \] Thus, the approximate increase in volume is: \[ \boxed{0.028} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Application of derivatives Questions
- If the normal drawn at a point $ P $ on the curve $ 3y = 6x - 5x^3 $ passes through the origin $(0, 0)$, then the positive integral value of the abscissa of point $ P $ is:
- The slope of the normal to the curve $y = 2x^2 + 3\sin x$ at $x = 0$ is:
- A balloon, which always remains spherical, has a variable radius. The rate at which its volume is increasing with respect to its radius $r$ when $r = 5$ cm is:
- The minimum value of the function $f(x) = x^2 + \frac{250}{x}$ for $x > 0$ is:
- If the length of the tangent at a point on the parabola $y^{2}=4ax$ is $4a\sqrt{5}$, then the length of the sub-normal at that point is
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions