A cylindrical vessel of radius 0.5 m
Question: A cylindrical vessel of radius $0.5 \mathrm{~m}$ is filled with oil at the rate of $0.25 \pi \mathrm{m}^{3} /$ minute. The rate at which the surface of the oil is rising, is Solution: (a) 1 m/minute Let $r$ be the radius, $h$ be the height and $V$ be the volume of the cylindrical vessel at any time $t$. Then, $V=\pi r^{2} h$ $\Rightarrow \frac{d V}{d t}=\pi r^{2} \frac{d h}{d t}$ $\Rightarrow \frac{d h}{d t}=\frac{1}{\pi r^{2}} \frac{d V}{d t}$ $\Rightarrow \frac{d h}{d t}=\frac{0.25 \...
Read More →A cylindrical vessel of radius 0.5 m
Question: A cylindrical vessel of radius $0.5 \mathrm{~m}$ is filled with oil at the rate of $0.25 \pi \mathrm{m}^{3} /$ minute. The rate at which the surface of the oil is rising, is Solution: (a) 1 m/minute Let $r$ be the radius, $h$ be the height and $V$ be the volume of the cylindrical vessel at any time $t$. Then, $V=\pi r^{2} h$ $\Rightarrow \frac{d V}{d t}=\pi r^{2} \frac{d h}{d t}$ $\Rightarrow \frac{d h}{d t}=\frac{1}{\pi r^{2}} \frac{d V}{d t}$ $\Rightarrow \frac{d h}{d t}=\frac{0.25 \...
Read More →A cone whose height is always equal
Question: A cone whose height is always equal to its diameter is increasing in volume at the rate of $40 \mathrm{~cm}^{3} / \mathrm{sec}$. At what rate is the radius increasing when its circular base area is $1 \mathrm{~m}^{2}$ ? (a) $1 \mathrm{~mm} / \mathrm{sec}$ (b) $0.001 \mathrm{~cm} / \mathrm{sec}$ (c) $2 \mathrm{~mm} / \mathrm{sec}$ (d) $0.002 \mathrm{~cm} / \mathrm{sec}$ Solution: (d) $0.002 \mathrm{~cm} / \mathrm{sec}$ Let $r$ be the radius, $h$ be the height and $V$ be the volume of th...
Read More →The sum of an infinite geometric series is 20, and the sum of the squares
Question: The sum of an infinite geometric series is 20, and the sum of the squares of these terms is 100. Find the series. Solution: Given: $\frac{\mathrm{a}}{1-\mathrm{r}}=20 \ \frac{\mathrm{a}^{2}}{1-\mathrm{r}^{2}}=100$ (Because on squaring both first term a and common ratio r will be squared.) To find: the series $\mathrm{a}=20(1-\mathrm{r}) \ldots$ (i) $\Rightarrow \frac{a^{2}}{1-r^{2}}=100=\frac{(20 \times(1-r))^{2}}{(1-r)(1+r)} \ldots($ from $(i))$ $\Rightarrow 100=400 \times \frac{1-r}{...
Read More →The radius of a sphere is changing at the rate of 0.1 cm/sec.
Question: The radius of a sphere is changing at the rate of 0.1 cm/sec. The rate of change of its surface area when the radius is 200 cm is (a) $8 \pi \mathrm{cm}^{2} / \mathrm{sec}$ (b) $12 \pi \mathrm{cm}^{2} / \mathrm{sec}$ (c) $160 \pi \mathrm{cm}^{2} / \mathrm{sec}$ (d) $200 \mathrm{~cm}^{2} / \mathrm{sec}$ Solution: (c) $160 \pi \mathrm{cm}^{2} / \mathrm{sec}$ Let $r$ be the radius and $S$ be the surface area of the sphere at any time $t .$ Then, $S=4 \pi r^{2}$ $\Rightarrow \frac{d S}{d t...
Read More →The radius of a sphere is changing at the rate of 0.1 cm/sec.
Question: The radius of a sphere is changing at the rate of 0.1 cm/sec. The rate of change of its surface area when the radius is 200 cm is (a) $8 \pi \mathrm{cm}^{2} / \mathrm{sec}$ (b) $12 \pi \mathrm{cm}^{2} / \mathrm{sec}$ (c) $160 \pi \mathrm{cm}^{2} / \mathrm{sec}$ (d) $200 \mathrm{~cm}^{2} / \mathrm{sec}$ Solution: (c) $160 \pi \mathrm{cm}^{2} / \mathrm{sec}$ Let $r$ be the radius and $S$ be the surface area of the sphere at any time $t .$ Then, $S=4 \pi r^{2}$ $\Rightarrow \frac{d S}{d t...
Read More →The sum of an infinite geometric series is 6.
Question: The sum of an infinite geometric series is 6. If its first term is 2, find its common ratio. Solution: Given: $\frac{a}{1-r}=6, a=2$ To find: $r=$ ? $\therefore \frac{2}{1-r}=6$ $\Rightarrow 1-r=\frac{2}{6}=\frac{1}{3}$ $\Rightarrow 3(1-r)=1$ $\Rightarrow 3-3 r=1$ $\Rightarrow 3 r=3-1$ $\Rightarrow r=\frac{2}{3}$ Common ratio $r=\frac{2}{3}$...
Read More →Write the value of
Question: Write the value of $2 . \overline{134}$ in the form of a simple fraction. Solution: Let, x=2.134134134 (i) Multiplying this equation by 1000 on both the sides so that repetitive terms cancel out and we get: 1000x=2134.134134134 (ii) Equation (ii)-(i), ⇒ 1000x-x=2134.134134134-2.134134134=2132 ⇒ 999x=2132 $\Rightarrow \mathrm{X}=\frac{2132}{999}$ $2 . \overline{134}=\frac{2132}{999}$...
Read More →Write the value of
Question: Write the value of $0 . \overline{423}$ in the form of a simple fraction. Solution: Let, x=0.423423423 (i) Multiplying this equation by 1000 on both the sides so that repetitive terms cancel out and we get: 1000x=423.423423423 (ii) Equation (ii)-(i), ⇒ 1000x-x=423.423423423-0.423423423=423 ⇒ 999x=423 $\Rightarrow \mathrm{X}=\frac{423}{999}=\frac{47}{111}$ $0 . \overline{423}=\frac{47}{111}$...
Read More →Express the recurring decimal 0.125125125 ….
Question: Express the recurring decimal 0.125125125 . $=0 . \overline{125}$ as a rational number. Solution: Let, x=0.125125125 (i) Multiplying this equation by 1000 on both the sides so that repetitive terms cancel out and we get: 1000x=125.125125125 (ii) Equation (ii)-(i), ⇒ 1000x-x=125.125125125-0.125125125=125 ⇒ 999x=125 $\Rightarrow \mathrm{X}=\frac{125}{999}$ $0 . \overline{125}=\frac{125}{999}$...
Read More →Find the rational number whose decimal expansion is given below :
Question: Find the rational number whose decimal expansion is given below : (i) $0 . \overline{3}$ (ii) $0 . \overline{231}$ (iii) $3 . \overline{52}$ Solution: (i) Let, x=0.3333 $\Rightarrow x=0.3+0.03+0.003+\ldots$ $\Rightarrow x=3(0.1+0.01+0.001+0.0001+\ldots \infty)$ $\Rightarrow x=3\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\frac{1}{10000}+\cdots \infty\right)$ This is an infinite geometric series. Here, a=1/10 and r=1/10 $\therefore \operatorname{Sum}=\frac{\mathrm{a}}{1-\mathrm{r}}=\...
Read More →Side of an equilateral triangle expands at the rate of 2 cm/sec.
Question: Side of an equilateral triangle expands at the rate of 2 cm/sec. The rate of increase of its area when each side is 10 cm is (a) $10 \sqrt{2} \mathrm{~cm}^{2} / \mathrm{sec}$ (b) $10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}$ (c) $10 \mathrm{~cm}^{2} / \mathrm{sec}$ (d) $5 \mathrm{~cm}^{2} / \mathrm{sec}$ Solution: (b) $10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}$ Let $x$ be the side and $A$ be the area of the equilateral triangle at any time $t$. Then, $A=\frac{\sqrt{3}}{4} x^{2}$ $\Ri...
Read More →Side of an equilateral triangle expands at the rate of 2 cm/sec.
Question: Side of an equilateral triangle expands at the rate of 2 cm/sec. The rate of increase of its area when each side is 10 cm is (a) $10 \sqrt{2} \mathrm{~cm}^{2} / \mathrm{sec}$ (b) $10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}$ (c) $10 \mathrm{~cm}^{2} / \mathrm{sec}$ (d) $5 \mathrm{~cm}^{2} / \mathrm{sec}$ Solution: (b) $10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}$ Let $x$ be the side and $A$ be the area of the equilateral triangle at any time $t$. Then, $A=\frac{\sqrt{3}}{4} x^{2}$ $\Ri...
Read More →If the solve the problem
Question: If $V=\frac{4}{3} \pi r^{3}$, at what rate in cubic units is $V$ increasing when $r=10$ and $\frac{d r}{d t}=0.01 ?$ (a) $\Pi$ (b) $4 \pi$ (c) $40 \pi$ (d) $4 \pi / 3$ Solution: (b) $4 \pi$ Given: $V=\frac{4}{3} \pi r^{3}, r=10$ and $\frac{d r}{d t}=0.01$ $\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}$ $\Rightarrow \frac{d V}{d t}=4 \pi(10)^{2} \times 0.01$ $\Rightarrow \frac{d V}{d t}=4 \pi$...
Read More →A circular disc of radius 3 cm is being heated. Due to expansion,
Question: A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/sec. Find the rate at which its area is increasing when radius is 3.2 cm. Solution: Let $r$ be the radius and $A$ be the area of the circular disc at any time $t .$ Then, $A=\pi r^{2}$ $\Rightarrow \frac{d A}{d t}=2 \pi r \frac{d r}{d t}$ $\Rightarrow \frac{d A}{d t}=2 \pi \times 3.2 \times 0.05$ $\left[\because r=3.2 \mathrm{~cm}\right.$ and $\left.\frac{d r}{d t}=0.05 \mathrm...
Read More →The length x of a rectangle is decreasing at the rate of 5 cm/minute
Question: The lengthxof a rectangle is decreasing at the rate of 5 cm/minute and the widthyis increasing at the rate of 4 cm/minute. Whenx= 8 cm andy= 6 cm, find the rates of change of (i) the perimeter (ii) the area of the rectangle. Solution: (i) Let $P$ be the perimeter of the rectangle at any time $t$. Then, $P=2(x+y)$ $\Rightarrow \frac{d P}{d t}=2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)$ $\Rightarrow \frac{d P}{d t}=2(-5+4)$ $\left[\because \frac{d x}{d t}=-5 \mathrm{~cm} / \mathrm{min...
Read More →Prove that
Question: Prove that $9^{1 / 3} \times 9^{1 / 9} \times 9^{1 / 27} \times \ldots \ldots^{\infty}=3$ Solution: L.H.S $=9^{1 / 3} \times 9^{1 / 9} \times 9^{1 / 27} \times \ldots \ldots \infty$ $=9^{(1 / 3)+(1 / 9)+(1 / 27)+\ldots \infty}$ The series in the exponent is an infinite geometric series Whose, $a=\frac{1}{3}$ $r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1 \times 3}{1 \times 9}=\frac{1}{3}$ Sum of the series in the exponent $=\frac{a}{1-r}=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1 \times 3}{...
Read More →The volume of a spherical balloon is increasing at the rate
Question: The volume of a spherical balloon is increasing at the rate of $25 \mathrm{~cm}^{3} / \mathrm{sec}$. Find the rate of change of its surface area at the instant when radius is $5 \mathrm{~cm}$. Solution: Let $r$ be the radius and $V$ be the volume of the sphere at any time $t$. Then, $V=\frac{4}{3} \pi r^{3}$ $\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}$ $\Rightarrow \frac{d r}{d t}=\frac{1}{4 \pi r^{2}} \frac{d V}{d t}$ $\Rightarrow \frac{d r}{d t}=\frac{25}{4 \pi(5)^{2}}$ ...
Read More →Find the sum of each of the following infinite series :
Question: Find the sum of each of the following infinite series : $\frac{2}{5}+\frac{3}{5^{2}}+\frac{2}{5^{3}}+\frac{3}{5^{4}}+\ldots \ldots$ Solution: This geometric series is the sum of two geometric series: $\frac{2}{5}+\frac{2}{5^{3}}+\frac{2}{5^{5}}+\cdots \infty \ \frac{3}{5^{2}}+\frac{3}{5^{4}}+\frac{4}{5^{6}}+\cdots \infty$ Sum of geometric series: $\frac{2}{5}+\frac{2}{5^{3}}+\frac{2}{5^{5}}+\cdots \infty$ Here, $a=\frac{2}{5}$ $r=\frac{\frac{2}{5^{3}}}{\frac{2}{5}}=\frac{1}{5^{2}}=\fra...
Read More →Find the sum of each of the following infinite series :
Question: Find the sum of each of the following infinite series : $10-9+8.1-\ldots \ldots \infty$ Solution: It is Infinite Geometric Series Here, $a=10$ $r=\frac{-9}{10}=-0.9$ $\therefore S u m=\frac{a}{1-r}=\frac{10}{1-(-0.9)}=\frac{10}{1+0.9}=\frac{10}{1.9}=\frac{100}{19}$ Sum $=\frac{100}{19}$...
Read More →The volume of a cube is increasing at the rate
Question: The volume of a cube is increasing at the rate of $9 \mathrm{~cm}^{3} / \mathrm{sec}$. How fast is the surface area increasing when the length of an edge is $10 \mathrm{~cm}$ ? Solution: Let $x$ be the side and $V$ be the volume of the cube at any time $t$. Then, $V=x^{3}$ $\Rightarrow \frac{d V}{d t}=3 x^{2} \frac{d x}{d t}$ $\Rightarrow 9=3(10)^{2} \frac{d x}{d t}$ $\left[\because x=10 \mathrm{~cm}\right.$ and $\left.\frac{d V}{d t}=\mathrm{cm}^{3} / \mathrm{sec}\right]$ $\Rightarrow...
Read More →Find the sum of each of the following infinite series :
Question: Find the sum of each of the following infinite series : $\sqrt{2}-\frac{1}{\sqrt{2}}+\frac{1}{2 \sqrt{2}}-\frac{1}{4 \sqrt{2}}+\ldots . .$ Solution: It is Infinite Geometric Series Here, $a=\sqrt{2}$ $r=\frac{\frac{-1}{\sqrt{2}}}{\sqrt{2}}=\frac{-1}{2}$ $\therefore S u m=\frac{a}{1-r}=\frac{\sqrt{2}}{1-\frac{-1}{2}}=\frac{\sqrt{2}}{1+\frac{1}{2}}=\frac{2 \sqrt{2}}{3}$ Sum $=\frac{2 \sqrt{2}}{3}$...
Read More →Find the sum of each of the following infinite series :
Question: Find the sum of each of the following infinite series : $6+1.2+0.24+\ldots \ldots \infty$ Solution: It is Infinite Geometric Series. Here, a=6 $r=\frac{1.2}{6}=\frac{2}{10}=0.2$ The formula used: Sum of an infinite Geometric series $=\frac{\mathrm{a}}{1-\mathrm{r}}$ $\therefore$ Sum $=\frac{6}{1-0.2}=\frac{6}{0.8}=\frac{15}{2}$ Sum $=\frac{15}{2}$...
Read More →Find the point on the curve
Question: Find the point on the curve $y^{2}=8 x$ for which the abscissa and ordinate change at the same rate. Solution: Here, $y^{2}=8 x \quad \ldots(1)$ $\Rightarrow 2 y \frac{d y}{d t}=8 \frac{d x}{d t}$ $\Rightarrow 2 y=8$ $\left[\because \frac{d y}{d t}=\frac{d x}{d t}\right]$ $\Rightarrow y=4$ $\Rightarrow x=\frac{y^{2}}{8}$ [From eq. (1)] $\Rightarrow x=\frac{16}{8}=2$ So, the point is $(2,4)$....
Read More →Find the sum of each of the following infinite series :
Question: Find the sum of each of the following infinite series : $8+4 \sqrt{2}+4+2 \sqrt{2}+\ldots . \infty$ Solution: It is Infinite Geometric Series. Here, a=8 $r=\frac{4 \sqrt{2}}{8}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$ The formula used: Sum of an infinite Geometric series $=\frac{\mathrm{a}}{1-\mathrm{r}}$ $\therefore S u m=\frac{8}{1-\frac{1}{\sqrt{2}}}=\frac{8 \sqrt{2}}{\sqrt{2}-1}$ Sum $=\frac{8 \sqrt{2}}{\sqrt{2}-1}$...
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