Metallic spheres of radii 6 cm, 8 cm and 10 cm,
Question: Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere. Solution: Radii of the given spheres are : $r_{1}=6 \mathrm{~cm}, r_{2}=8 \mathrm{~cm}, r_{3}=10 \mathrm{~cm}$ $\Rightarrow$ Volume of the given spheres are : $V_{1}=\frac{4}{3} \pi r_{1}{ }^{3}, V_{2}=\frac{4}{3} \pi r_{2}{ }^{3}$ and $V_{3}=\frac{4}{3} \pi r_{3}{ }^{3}$ $\therefore$ Total volume of the given spheres $=\mathrm{V}_{1}+\mathrm{...
Read More →A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm.
Question: A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. Solution: Let the height of the cylinder made be equal to h cm. Its radius = 6 cm Volume of the cylinder made = Volume of the given metallic sphere of radius 4.2 cm $\Rightarrow \pi \times(6)^{2} \times \mathrm{h}=\frac{\mathbf{4}}{\mathbf{3}} \pi \times(4.2)^{3} \mathrm{~cm}^{3}$ $\Rightarrow 36 \times \mathrm{h}=\frac{\mathbf{4}}{\mathbf{3}} \times 4.2...
Read More →Show that the Modulus Function
Question: Show that the Modulus Function $f: \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=|x|$, is neither one-one nor onto, where $|x|$ is $x$, if $x$ is positive or 0 and $|x|$ is $-x$, if $x$ is negative. Solution: $f: \mathbf{R} \rightarrow \mathbf{R}$ is given by, $f(x)=|x|=\left\{\begin{array}{l}x, \text { if } x \geq 0 \\ -x, \text { if } x0\end{array}\right.$ It is seen that $f(-1)=|-1|=1, f(1)=|1|=1$. $\therefore f(-1)=f(1)$, but $-1 \neq 1$. $\therefore f$ is not one-one. Now, con...
Read More →Find the values of other five trigonometric functions if sec x = 13/5
Question: Find the values of other five trigonometric functions if $\sec x=\frac{13}{5}, x$ lies in fourth quadrant Solution: $\sec x=\frac{13}{5}$ $\cos x=\frac{1}{\sec x}=\frac{1}{\left(\frac{13}{5}\right)}=\frac{5}{13}$ $\sin ^{2} x+\cos ^{2} x=1$ $\Rightarrow \sin ^{2} x=1-\cos ^{2} x$ $\Rightarrow \sin ^{2} x=1-\left(\frac{5}{13}\right)^{2}$ $\Rightarrow \sin ^{2} x=1-\frac{25}{169}=\frac{144}{169}$ $\Rightarrow \sin x=\pm \frac{12}{13}$ Since $x$ lies in the $4^{\text {th }}$ quadrant, the...
Read More →A spherical glass vessel has a cylindrical neck 8 cm long,
Question: A spherical glass vessel has a cylindrical neck $8 \mathrm{~cm}$ long, $2 \mathrm{~cm}$ in diameter; the diamter of the spherical part is $8.5 \mathrm{~cm}$. By measuring the amount of water it holds, a child finds its volume to be $345 \mathrm{~cm}^{3}$. Check whether she is correct, taking the above as the inside measurements, and $\pi$ $=3.14$. Solution: The cylinder neck has length = 8 cm and radius = 1 cm Volume of the cylinder part $=\pi(1)^{2} \times 8 \mathrm{~cm}^{3}$ $=8 \pi ...
Read More →At 450 K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium.
Question: At $450 \mathrm{~K}, K_{\mathrm{p}}=2.0 \times 10^{10} /$ bar for the given reaction at equilibrium. $2 \mathrm{SO}_{2(g)}+\mathrm{O}_{2(g)} \longleftrightarrow 2 \mathrm{SO}_{3(g)}$ What isKcat this temperature? Solution: For the given reaction, $\Delta n=2-3=-1$ $T=450 \mathrm{~K}$ $R=0.0831$ bar $\mathrm{L}$ bar $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$ $K_{\mathrm{p}}=2.0 \times 10^{10} \mathrm{bar}-1$ We know that, $K_{\mathrm{P}}=K_{\mathrm{C}}(R T) \Delta n$ $\Rightarrow 2.0 \times 1...
Read More →Prove that the Greatest Integer Function
Question: Prove that the Greatest Integer Function $f: \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=[x]$, is neither one-one nor onto, where $[x]$ denotes the greatest integer less than or equal to $x$. Solution: $f: \mathbf{R} \rightarrow \mathbf{R}$ is given by, $f(x)=[x]$ It is seen that $f(1.2)=[1.2]=1, f(1.9)=[1.9]=1$.\ $\therefore f(1.2)=f(1.9)$, but $1.2 \neq 1.9$. $\therefore f$ is not one-one. Now, consider $0.7 \in \mathbf{R}$. It is known that $f(x)=[x]$ is always an integer. Thu...
Read More →A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom.
Question: A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius is 60 cm and its height is 180 cm. Solution: Height of the conical part = 120 cm Base radius of the conical part = 60 cm $\therefore$ Volume of the conical part $=\frac{\mathbf{1}}{\mathbf{3}} \pi \mathrm{r}^{2} \mat...
Read More →Find the values of other five trigonometric functions if $cot x= rac{3}{4}$
Question: Find the values of other five trigonometric functions if $\cot x=\frac{3}{4}, x$ lies in third quadrant. Solution: $\cot x=\frac{3}{4}$ $\tan x=\frac{1}{\cot x}=\frac{1}{\left(\frac{3}{4}\right)}=\frac{4}{3}$ $1+\tan ^{2} x=\sec ^{2} x$ $\Rightarrow 1+\left(\frac{4}{3}\right)^{2}=\sec ^{2} x$ $\Rightarrow 1+\frac{16}{9}=\sec ^{2} x$ $\Rightarrow \frac{25}{9}=\sec ^{2} x$ $\Rightarrow \sec x=\pm \frac{5}{3}$ Since $x$ lies in the $3^{\text {rd }}$ quadrant, the value of $\sec x$ will be...
Read More →Check the injectivity and surjectivity of the following functions:
Question: (i) $f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^{2}$ (ii) $f: \mathbf{Z} \rightarrow \mathbf{Z}$ given by $f(x)=x^{2}$ (iii) $f: \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=x^{2}$ (iv) $f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^{3}$ (v) $f: \mathbf{Z} \rightarrow \mathbf{Z}$ given by $f(x)=x^{3}$ Solution: (i) $f: \mathbf{N} \rightarrow \mathbf{N}$ is given by, $f(x)=x^{2}$ It is seen that for $x, y \in \mathbf{N}, f(x)=f(y) \Rightarrow x^{2}=y^{2} \Right...
Read More →Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:
Question: Nitric oxide reacts with Br2and gives nitrosyl bromide as per reaction given below: $2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g}) \longleftrightarrow 2 \mathrm{NOBr}(\mathrm{g})$ When $0.087 \mathrm{~mol}$ of $\mathrm{NO}$ and $0.0437 \mathrm{~mol}$ of $\mathrm{Br}_{2}$ are mixed in a closed container at constant temperature, $0.0518 \mathrm{~mol}$ of $\mathrm{NOBr}$ is obtained at equilibrium. Calculate equilibrium amount of $\mathrm{NO}$ and $\mathrm{Br}_{2}$. Solution: The...
Read More →A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm,
Question: A solid iron pole consists of a cylinder of height $220 \mathrm{~cm}$ and base diameter $24 \mathrm{~cm}$, which is surmounted by another cylinder of height $60 \mathrm{~cm}$ and radius $8 \mathrm{~cm}$. Find the mass of the pole, given that $1 \mathrm{~cm}^{3}$ of iron has approximately $8 \mathrm{~g}$ mass. (Use $\pi=3.14$ ) Solution: First cylindrical part has height 220 cm and radius 12 cm. Its volume $=\pi \times(12)^{2} \times 220 \mathrm{~cm}^{3}$. Second cylindrical part has he...
Read More →A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open,
Question: A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water upto the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one fourth of the water flows out. Find the number of lead shots dropped in the vessel. Solution: Height of the conical vessel (h) = 8 cm Base radius (r) = 5 cm Volume of water in conical vessel $=\frac{\mathbf{1}}{\mathbf{3}} \pi \mathrm{r}^{2} \...
Read More →Find the values of other five trigonometric functions if
Question: Find the values of other five trigonometric functions if $\sin x=\frac{3}{5}$,x lies in second quadrant. Solution: $\sin x=\frac{3}{5}$ $\operatorname{cosec} x=\frac{1}{\sin x}=\frac{1}{\left(\frac{3}{5}\right)}=\frac{5}{3}$ $\sin ^{2} x+\cos ^{2} x=1$ $\Rightarrow \cos ^{2} x=1-\sin ^{2} x$ $\Rightarrow \cos ^{2} x=1-\left(\frac{3}{5}\right)^{2}$ $\Rightarrow \cos ^{2} x=1-\frac{9}{25}$ $\Rightarrow \cos ^{2} x=\frac{16}{25}$ $\Rightarrow \cos x=\pm \frac{4}{5}$ Since $x$ lies in the ...
Read More →A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens.
Question: A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see fig.). Solution: Radius of conical cavity = 0.5 cm and depth (i.e., vertical height) = 1.4 cm Volume of wood taken out to make one cavity $=\frac{\mathbf{1}}{\mathbf{3}} \pi \mathbf{r}^{2} \times \mathbf{h}=\frac...
Read More →Find the values of other five trigonometric functions if $cos x=- rac{1}{2}$ ,
Question: Find the values of other five trigonometric functions if $\cos x=-\frac{1}{2}, x$ lies in third quadrant. Solution: $\cos x=-\frac{1}{2}$ $\therefore \sec x=\frac{1}{\cos x}=\frac{1}{\left(-\frac{1}{2}\right)}=-2$ $\sin ^{2} x+\cos ^{2} x=1$ $\Rightarrow \sin ^{2} x=1-\cos ^{2} x$ $\Rightarrow \sin ^{2} x=1-\left(-\frac{1}{2}\right)^{2}$ $\Rightarrow \sin ^{2} x=1-\frac{1}{4}=\frac{3}{4}$ $\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2}$ Since $x$ lies in the $3^{\text {rd }}$ quadrant, the ...
Read More →A gulab jamun, contains sugar syrup up to about 30% of its volume.
Question: A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. (see fig.) Solution: Since, a gulab jamun is like a cylinder with hemispherical ends. Total height of the gulab jamun = 5 cm. Diameter = 2.8 cm $\Rightarrow$ Radius $=1.4 \mathrm{~cm}$ $\therefore$ Length (height) of the cylindrical part = 5cm (1.4 + 1.4) cm =...
Read More →Show that the function f: R* → R* defined by
Question: Show that the function $t: \mathbf{R}_{\star} \rightarrow \mathbf{R}_{\star}$ defined by $f(x)=\frac{1}{x}$ is one-one and onto, where $\mathbf{R}_{\star}$ is the set of all non-zero real numbers. Is the result true, if the domain $\mathbf{R}_{\star}$ is replaced by $\mathbf{N}$ with co-domain being same as $\mathbf{R}_{*} ?$ Solution: It is given that $f: \mathbf{R}_{\star} \rightarrow \mathbf{R}_{*}$ is defined by $f(x)=\frac{1}{x}$. One-one: $f(x)=f(y)$ $\Rightarrow \frac{1}{x}=\fra...
Read More →Reaction between N2 and O2 takes place as follows:
Question: Reaction between N2and O2takes place as follows: $2 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longleftrightarrow 2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g})$ If a mixture of $0.482 \mathrm{~mol}$ of $\mathrm{N}_{2}$ and $0.933 \mathrm{~mol}$ of $\mathrm{O}_{2}$ is placed in a $10 \mathrm{~L}$ reaction vessel and allowed to form $\mathrm{N}_{2} \mathrm{O}$ at a temperature for which $K_{c}=2.0 \times 10^{-37}$, determine the composition of equilibrium mixture. Solution: Let...
Read More →Find the angle in radian though which a pendulum swings if its length
Question: Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm Solution: We know that in a circle of radius $r$ unit, if an $\operatorname{arc}$ of length $/$ unit subtends an angle $\theta$ radian at the centre, then $\theta=\frac{l}{r}$. It is given that $r=75 \mathrm{~cm}$ (i) Here, $I=10 \mathrm{~cm}$ $\theta=\frac{10}{75}$ radian $=\frac{2}{15}$ radian (ii) Here, $I=15 \mathrm{~cm}$ $\theta=\fr...
Read More →Let R be the relation in the set N given by R = {(a, b): a = b − 2, b > 6}.
Question: Let $R$ be the relation in the set $N$ given by $R=\{(a, b): a=b-2, b6\}$. Choose the correct answer. (A) $(2,4) \in R(B)(3,8) \in R(C)(6,8) \in R(D)(8,7) \in R$ Solution: $\mathrm{R}=\{(a, b): a=b-2, b6\}$ Now, since $b6,(2,4) \notin \mathrm{R}$ Also, as $3 \neq 8-2,(3,8) \notin \mathrm{R}$ And, as $8 \neq 7-2$ $\therefore(8,7) \notin \mathrm{R}$ Now, consider $(6,8)$. We have $86$ and also, $6=8-2$. $\therefore(6,8) \in \mathrm{R}$ The correct answer is $C$....
Read More →If in two circles, arcs of the same length subtend angles 60° and 75° at the centre,
Question: If in two circles, arcs of the same length subtend angles $60^{\circ}$ and $75^{\circ}$ at the centre, find the ratio of their radii. Solution: Let the radii of the two circles be $r_{1}$ and $r_{2}$. Let an arc of length / subtend an angle of $60^{\circ}$ at the centre of the circle of radius $r_{1}$, while let an arc of length / subtend an angle of $75^{\circ}$ at the centre of the circle of radius $r_{2}$. Now, $60^{\circ}=\frac{\pi}{3}$ radian and $75^{\circ}=\frac{5 \pi}{12}$ radi...
Read More →Rachel, an engineering student,
Question: Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same). Solution: Volume of the cylindrical part $=\pi \times(1.5)^{2} \times 8 \mathrm{~cm}^{3}=18 \pi \mathr...
Read More →Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Question: Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression? Solution: For a pure substance (both solids and liquids), $[$ Pure substance $]=\frac{\text { Number of moles }}{\text { Volume }}$ $=\frac{\text { Mass } / \text { molecular mass }}{\text { Volume }}$ $=\frac{\text { Mass }}{\text { Volume } \times \text { Molecular mass }}$ $=\frac{\text { Density }}{\text { Molecular mass }}$ Now, the molecular mass and density (at a particular temp...
Read More →In a circle of diameter 40 cm, the length of a chord is 20 cm.
Question: In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord. Solution: Diameter of the circle = 40 cm $\therefore$ Radius $(r)$ of the circle $=\frac{40}{2} \mathrm{~cm}=20 \mathrm{~cm}$ Let $A B$ be a chord (length $=20 \mathrm{~cm}$ ) of the circle. In $\triangle \mathrm{OAB}, \mathrm{OA}=\mathrm{OB}=$ Radius of circle $=20 \mathrm{~cm}$ Also, $A B=20 \mathrm{~cm}$ Thus, $\triangle \mathrm{OAB}$ is an equilateral triangle. $\therefore \the...
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