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APPLICATION OF DERIVATIVES — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSAPPLICATION OF DERIVATIVES4 Marks
Question
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 propotional to the surface. Prove that the radius is decreasing at a constant rate.
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Answer

We have, rate of decrease of the volume of spherical ball of salt at any instant is surface. Let the radius of the spherical ball of the salt be r.
$\therefore$ Volume of the ball $(\text{V})=\frac{4}{3}\pi\text{r}^3$
and surface area $(\text{S})=4\pi\text{r}^2$
$\because\ \frac{\text{dV}}{\text{dT}}\propto\text{S}$
$\Rightarrow\ \frac{\text{d}}{\text{dt}}\Big(\frac{4}{3}\pi\text{r}^3\Big)\propto4\pi\text{r}^2$
$\Rightarrow\ \frac{4}{3}\pi3\text{r}^2\frac{\text{dr}}{\text{dt}}\propto4\pi\text{r}^2$
$\Rightarrow\ \frac{\text{dr}}{\text{dt}}\propto\frac{4\pi\text{r}^2}{4\pi\text{r}^2}$
$\Rightarrow\ \frac{\text{dr}}{\text{dt}}=\text{k.1}$ [where, k is the proportionality constant]
$\Rightarrow\ \frac{\text{dr}}{\text{dt}}=\text{k}$
Hence, the radius of ball is decreasing at a constant rate.

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  2. $0.04$
  3. $0.47$
  4. $0.06$
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  2. $0.006$
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  2. $0.047$
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  4. $1$
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  3. $R - {0}$
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  1. $\frac{\text{x}+2}{\text{x}}$
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  2. $0$
  3. $\frac{1}{2}$
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Image

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OR

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  1. The function gof(x) is defined as:
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  2. $\text{gof}(\text{x})=\begin{cases}\text{e}^{\sin\text{x}}&,\text{x}\leq0\\\text{e}^{1-\cos\text{x}}&,\text{x}\geq0\end{cases}$
  3. $\text{gof}(\text{x})=\begin{cases}\text{e}^{\sin\text{x}}&,\text{x}\leq0\\1-\text{e}^{\cos\text{x}}&,\text{x}\geq0\end{cases}$
  4. $\text{gof}(\text{x})=\begin{cases}\text{e}^{\sin\text{x}}&,\text{x}\geq0\\\text{e}^{1-\cos\text{x}}&,\text{x}\leq0\end{cases}$
  1. $\frac{\text{d}}{\text{dx}}\{\text{gof}(\text{x})\}=$
  1. $[\text{gof}(\text{x})]'=\begin{cases}\cos\text{x}\cdot\text{e}^{\sin\text{x}}&,\text{x}\geq0\\\text{e}^{1-\cos\text{x}}\cdot\sin\text{x}&,\text{x}\leq0\end{cases}$
  2. $[\text{gof}(\text{x})]'=\begin{cases}\cos\text{x}\cdot\text{e}^{\sin\text{x}}&,\text{x}\geq0\\-\sin\text{x}\cdot\text{e}^{1-\cos\text{x}}&,\text{x}\leq0\end{cases}$
  3. $[\text{gof}(\text{x})]'=\begin{cases}\cos\text{x}\cdot\text{e}^{\sin\text{x}}&,\text{x}\geq0\\\sin\text{x}\cdot({1-\cos\text{x}})&,\text{x}\leq0\end{cases}$
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  2. $1$
  3. $-1$
  4. $2$
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  2. $\frac{1}{\sqrt2}$
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Based on the above information, answer the following questions.
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  3. $\text{x}=\frac{1}{3}$
  4. $\text{x}=\frac{1}{4}$
  1. Graph of given two curves can be drawn as.
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  2. $\frac{\pi}{6}+\frac{\sqrt{3}}{8}$
  3. $\frac{\pi}{2}+\frac{\sqrt{3}}{4}$
  4. $\frac{\pi}{2}-\frac{\sqrt{3}}{4}$
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  1. $\frac{\pi}{6}+\frac{\sqrt{3}}{4}$
  2. $\frac{\pi}{6}+\frac{\sqrt{3}}{8}$
  3. $\frac{\pi}{6}-\frac{\sqrt{3}}{8}$
  4. $\frac{\pi}{2}-\frac{\sqrt{3}}{4}$
  1. Area of hidden portion of lower circle is.
  1. $\bigg(\frac{2\pi}{3}+\frac{\sqrt{3}}{2}\bigg)\text{ sq.units}$
  2. $\bigg(\frac{\pi}{3}-\frac{\sqrt{3}}{8}\bigg)\text{ sq.units}$
  3. $\bigg(\frac{\pi}{3}+\frac{\sqrt{3}}{8}\bigg)\text{ sq.units}$
  4. $\bigg(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\bigg)\text{ sq.units}$
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