If A and B are matrices of the same order,

Question: If $A$ and $B$ are matrices of the same order, then $(3 A-2 B)^{\top}$ is equal to________ Solution: $(3 A-2 B)^{\top}$ $=(3 A)^{\top}-(2 B)^{\top}$ [For any two matrices $X$ and $Y,(X+Y)^{\top}=X^{\top}+Y^{\top}$ ] $=3 A^{\top}-2 B^{\top}$ $\left[(k X)^{\top}=k X^{\top}\right.$, where $k$ is any constant $]$ If $A$ and $B$ are matrices of the same order, then $(3 A-2 B)^{\top}$ is equal to $3 A^{T}-2 B^{T}$...

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If A and B are two skew-symmetric matrices of same order,

Question: If $A$ and $B$ are two skew-symmetric matrices of same order, then $A B$ is symmetric if______ Solution: It is given that, $A$ and $B$ are two skew-symmetric matrices of same order. $\therefore A^{T}=-A$ and $B^{T}=-B$ ....(1) Now, the matrix $A B$ is symmetric if $(A B)^{T}=A B$ (A matrix $X$ is symmetric if $X^{T}=X$ ) $\Rightarrow B^{T} A^{T}=A B$ $\Rightarrow(-B)(-A)=A B \quad[$ Using (1) $]$ $\Rightarrow B A=A B$ Thus, if $A$ and $B$ are two skew-symmetric matrices of same order, ...

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Solve this

Question: If $\tan (A-B)=\frac{1}{\sqrt{3}}$ and $\tan (A+B)=\sqrt{3}, 0^{\circ}(A+B)90^{\circ}$ and $AB$, then find $A$ and $B$. Solution: Here, $\tan (A-B)=\frac{1}{\sqrt{3}}$ $\Rightarrow \tan (A-B)=\tan 30^{\circ} \quad\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]$ ⇒AB=30o ...(i) Also, $\tan (A+B)=\sqrt{3}$ $\Rightarrow \tan (A+B)=\tan 60^{\circ} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right]$ ⇒A+B= 60o...(ii) Solving (i) and (ii), we get:A =45oand B =15o...

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Prove that

Question: If $\sin (A+B)=\frac{\sqrt{3}}{2}$ and $\cos (A-B)=\frac{\sqrt{3}}{2}, 0^{\circ}(A+B) \leq 90^{\circ}$ and $AB$ then find the values of $A$ and $B$. Solution: As we know that, $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ Thus, if $\sin (A+B)=\frac{\sqrt{3}}{2}$ $\Rightarrow A+B=60^{\circ} \quad \ldots(1)$ and $\cos 30^{\circ}=\frac{\sqrt{3}}{2}$ Thus, if $\cos (A-B)=\frac{\sqrt{3}}{2}$ $\Rightarrow A-B=30^{\circ} \quad \ldots(2)$ Solving (1) and (2), we get $A=45^{\circ}$ and $B=15^{\circ}$ He...

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Solve this

Question: If $A=\left[\begin{array}{cc}4 x+2 \\ 2 x-3 x+1\end{array}\right]$ is a symmetric matrix, then $x=$_______ Solution: The given matrix $A=\left[\begin{array}{cc}4 x+2 \\ 2 x-3 x+1\end{array}\right]$ is symmetric. $\therefore A^{T}=A$ $\Rightarrow\left[\begin{array}{cc}4 x+2 \\ 2 x-3 x+1\end{array}\right]^{T}=\left[\begin{array}{cc}4 x+2 \\ 2 x-3 x+1\end{array}\right]$ $\Rightarrow\left[\begin{array}{cc}4 2 x-3 \\ x+2 x+1\end{array}\right]=\left[\begin{array}{cc}4 x+2 \\ 2 x-3 x+1\end{ar...

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In photosynthesis, 6 molecules of carbon dioxide

Question: In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex sereies of reactions to give a molecule of glucose having a molecular formula $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$ and a molecule of oxygen with molecular formula $\mathrm{O}_{2}$. How many grams of water would be required to produce $18 \mathrm{~g}$ of glucose? Compute the volume of water so consumed assuming the density of water to be $1 \mathrm{~g} \mathrm{~c...

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In photosynthesis, 6 molecules of carbon dioxide

Question: In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex sereies of reactions to give a molecule of glucose having a molecular formula $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$ and a molecule of oxygen with molecular formula $\mathrm{O}_{2}$. How many grams of water would be required to produce $18 \mathrm{~g}$ of glucose? Compute the volume of water so consumed assuming the density of water to be $1 \mathrm{~g} \mathrm{~c...

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If sin (A + B) = 1 and tan (A−B)

Question: If $\sin (A+B)=1$ and $\tan (A-B)=\frac{1}{\sqrt{3}}, 0^{\circ}(A+B) \leq 90^{\circ}$ and $AB$ then find the values of $A$ and $B$. Solution: As we know that, $\sin 90^{\circ}=1$ Thus, if $\sin (A+B)=1$ $\Rightarrow A+B=90^{\circ} \quad \ldots(1)$ and $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$ Thus, if $\tan (A-B)=\frac{1}{\sqrt{3}}$ $\Rightarrow A-B=30^{\circ} \quad \ldots(2)$ Solving (1) and (2), we get $A=60^{\circ}$ and $B=30^{\circ}$ Hence, the values of $A$ and $B$ are $60^{\circ}$ and...

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Write the formulae for the following and calculate the molecular mass for each one of them.

Question: Write the formulae for the following and calculate the molecular mass for each one of them. (a) Custic potash (b) Baking powder (c) Lime stone (d) Caustic soda (e) Ethanol (f) Common salt. Solution: (a) $\mathrm{KOH}=39+16+1=56 \mathrm{u}$ (b) $\mathrm{NaHCO}_{3}=23+1+12+3 \times 16=84 \mathrm{u}$ (c) $\mathrm{CaCO}_{3}=12+12 \times 3 \times 16=100 \mathrm{u}$ (d) $\mathrm{NaOH}=23+16+1=40 \mathrm{u}$ (e) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}=2 \times 12+6 \times 1+16=46 \mathrm{u...

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If the matrices

Question: If the matrices $A=\left[\begin{array}{ll}a b \\ c d\end{array}\right]$ and $B=\left[\begin{array}{ll}1 1 \\ 0 1\end{array}\right]$ commute with each other, then $C=$_______ Solution: It is given that, the matrices $A=\left[\begin{array}{ll}a b \\ c d\end{array}\right]$ and $B=\left[\begin{array}{ll}1 1 \\ 0 1\end{array}\right]$ commute with each other. $\therefore A B=B A$ $\Rightarrow\left[\begin{array}{ll}a b \\ c d\end{array}\right]\left[\begin{array}{ll}1 1 \\ 0 1\end{array}\right...

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Complete the following crossword puzzle

Question: Complete the following crossword puzzle by using the name of the chemical elements. Use the data given in table. Solution:...

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Fill in the blanks

Question: Fill in the blanks (a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called .. . (b) A group of atoms carrying a fixed charge on them is called .. . (c) The formula unit mass of Ca3(PO4)2is .. . (d) Formula of sodium carbonate is and that of ammonium sulphate is . Solution: (a) Law of conservation of mass (b) Polyatomic ions (c) $3 \times$ Atomic mass of $\mathrm{Ca}+2 \times$ Atomic mass of $\mathrm{P}+8 \times$ Atomic mass of $...

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Solve the following equations

Question: If $A=\left[\begin{array}{lll}2 0 0 \\ 0 2 0 \\ 0 0 2\end{array}\right]$ such that $A^{5}=\lambda A$, then $\lambda=$ Solution: The given matrix is $A=\left[\begin{array}{lll}2 0 0 \\ 0 2 0 \\ 0 0 2\end{array}\right]$. $\therefore A^{2}=A \cdot A=\left[\begin{array}{lll}2 0 0 \\ 0 2 0 \\ 0 0 2\end{array}\right]\left[\begin{array}{lll}2 0 0 \\ 0 2 0 \\ 0 0 2\end{array}\right]=\left[\begin{array}{lll}4+0+0 0+0+0 0+0+0 \\ 0+0+0 0+4+0 0+0+0 \\ 0+0+0 0+0+0 0+0+4\end{array}\right]=\left[\beg...

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A sample of ethane

Question: A sample of ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$ gas has the same mass as $1.5 \times 10^{20}$ molecules of methane $\left(\mathrm{CH}_{4}\right)$. How many $\mathrm{C}_{2} \mathrm{H}_{6}$ molecules does the sample of gas contain? Solution: No. of molecules of methane $\left(\mathrm{CH}_{4}\right)=1.5 \times 10^{20}$ Mass of one molecule of methane $=\frac{\text { Gram moleculer mass }}{\text { Avogadro's number }}=\frac{16 \mathrm{~g}}{\mathrm{~N}_{\mathrm{A}}}$ Mass of...

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Prove that

Question: If $\frac{\cot \theta-1}{\cot \theta+1}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$ then find the acute angle $\theta$. Solution: Given : $\frac{\cot \theta-1}{\cot \theta+1}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$ $\frac{\cot \theta-1}{\cot \theta+1}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$ $\Rightarrow \frac{\frac{1}{\tan \theta}-1}{\frac{1}{\tan \theta}+1}=\frac{1-\sqrt{3}}{1+\sqrt{3}} \quad\left(\because \cot \theta=\frac{1}{\tan \theta}\right)$ $\Rightarrow \frac{\frac{1-\tan \theta}{\tan \theta}}{\frac{1+\tan \...

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If A, B and C are m × n, n × p and p × q matrices respectively

Question: If $A, B$ and $C$ are $m \times n, n \times p$ and $p \times q$ matrices respectively such that $(B C) A$ is defined, then $m=$_________ Solution: Let $X=\left[x_{i j}\right]_{m \times n}$ and $Y=\left[y_{i j}\right]_{p \times q}$ be two matrices of order $m \times n$ and $p \times q$. The multiplication of matrices $X$ and $Y$ is defined if number of columns of $X$ is same as the number of rows of $Y$ i.e. $n=p$. Also, $X Y$ is a matrix of order $m \times q$. The order of matrices $A,...

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If tan (x + 30°) = 1 then find the value of x.

Question: If tan (x+ 30) = 1 then find the value ofx. Solution: As we know that, $\tan 45^{\circ}=1$ Thus, if $\tan \left(x+30^{\circ}\right)=1$ $\Rightarrow x+30^{\circ}=45^{\circ}$ $\Rightarrow x=45^{\circ}-30^{\circ}$ $\Rightarrow x=15^{\circ}$ Hence, the value of $x$ is $15^{\circ}$....

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A silver ornament of mass 'm' gram

Question: A silver ornament of mass ' $m$ ' gram is polished with gold equivalent to $1 \%$ of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament. Solution: Mass of silver $=m \mathrm{~g}$ Mass of gold deposited $=\frac{m \times 1}{100}=\frac{m}{100} \mathrm{~g}$ No. of atoms of silver $=\frac{\text { Mass }}{\text { Atomic mass }} \times \mathrm{N}_{\mathrm{A}}=\frac{m}{108} \times \mathrm{N}_{\mathrm{A}}$ No. of atoms of gold $=\frac{\text { Mass }}...

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Compute the difference in masses of one

Question: Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass of an electron is $9.1 \times 10^{-28} \mathrm{~g}$ ). Which one is heavier? Solution: $\mathrm{Al}$ atom and $\mathrm{Al}^{3+}$ ion differ by electrons $=3$ 1 mole of $\mathrm{Al}$ atoms and 1 mole of $\mathrm{Al}^{3+}$ ions differ (by) electrons $=3 \times \mathrm{N}_{\mathrm{A}}$ Mass of one electrons $=9.1 \times 10^{-28} \mathrm{~g}$ (given) Mass of $3 \times \mathrm{N}_{\mathrm{A}...

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Solve the following equations

Question: If $f(x)=\left[\begin{array}{cc}\cos x \sin x \\ -\sin x \cos x\end{array}\right]$ and $f(x) f(y)=f(z)$, then $z=$______ Solution: $f(x)=\left[\begin{array}{cc}\cos x \sin x \\ -\sin x \cos x\end{array}\right]$ $\therefore f(y)=\left[\begin{array}{cc}\cos y \sin y \\ -\sin y \cos y\end{array}\right]$ Now, $f(x) f(y)$ $=\left[\begin{array}{cc}\cos x \sin x \\ -\sin x \cos x\end{array}\right]\left[\begin{array}{cc}\cos y \sin y \\ -\sin y \cos y\end{array}\right]$ $=\left[\begin{array}{c...

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If sin(A+B)=sinA cosB+cosA sinB and cos(A−B)

Question: If $\sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}$ and $\cos (\mathrm{A}-\mathrm{B})=\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B}$, find the values of (i) $\sin 75^{\circ}$ and (ii) $\cos 15^{\circ}$. Solution: Let $\mathrm{A}=45^{\circ}$ and $\mathrm{B}=30^{\circ}$ (i) As, $\sin (A+B)=\sin A \cos B+\cos A \sin B$ $\Rightarrow \sin \left(45^{\circ}+30^{\circ}\right)=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30...

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What are ionic and molecular compounds ?

Question: What are ionic and molecular compounds ? Give examples. Solution: Simple ionic compounds are of binary nature. It means that both the positive and negative ions have one atom only. The symbols of these ions are written side by side with their valencies at their bottom. A common factor if any, is removed to get a simple ratio of the valencies of the combining atoms. The crisscross method is then applied to arrive at the final chemical formula of the compound. Let us write the formulae o...

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A gold sample contains 90% of gold and the rest copper.

Question: A gold sample contains $90 \%$ of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold? Solution: Mass of pure gold in one gram of the sample $=\frac{1 \times 90}{100}=0.9 \mathrm{~g}$ No. of moles of gold present $=\frac{\text { Mass of gold }}{\text { Molar atomic mass }}=\frac{(0.9 \mathrm{~g})}{\left(0197 \mathrm{~g} \mathrm{~mol}^{-1}\right)}=0.0046 \mathrm{~mol}$ One mole of gold contains atoms $=\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{...

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Using the formula, sin A

Question: Using the formula, $\sin A=\sqrt{\frac{1-\cos 2 A}{2}}$, find the value of $\sin 30^{\circ}$, it being given that $\cos 60^{\circ}=\frac{1}{2}$. Solution: A = 30o⇒2A= 230o=60oBy substituting the value of the given T-ratio, we get: $\sin A=\sqrt{\frac{1-\cos 2 A}{2}}$ $\Rightarrow \sin 30^{\circ}=\sqrt{\frac{1-\cos 60^{\circ}}{2}}=\sqrt{\frac{1-\frac{1}{2}}{2}}=\sqrt{\frac{\frac{1}{2}}{2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$ $\therefore \sin 30^{\circ}=\frac{1}{2}$...

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Compare the number of ions present in

Question: Compare the number of ions present in $5.85 \mathrm{~g}$ of sodium chloride. Solution: Gram formula mass of $\mathrm{NaCl}=23+35.5=58.5 \mathrm{~g}$ $58.5 \mathrm{~g}$ of $\mathrm{NaCl}$ have ions $=2 \times \mathrm{N}_{\mathrm{A}}$ $5.85 \mathrm{~g}$ of $\mathrm{NaCl}$ have ions $=\frac{2 \times \mathrm{N}_{\mathrm{A}} \times(5.85 \mathrm{~g})}{(58.5 \mathrm{~g})}$ $=\frac{2 \times 6.022 \times 10^{23}}{10}=1.2042 \times 10^{23}$ ions....

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