Write the first five terms of the sequences whose nth term is
Question: Write the first five terms of the sequences whose$\mathrm{n}^{\text {th }}$term is $a_{n}=n(n+2)$ Solution: $a_{n}=n(n+2)$ Substituting $n=1,2,3,4$, and 5 , we obtain $a_{1}=1(1+2)=3$ $a_{2}=2(2+2)=8$ $a_{3}=3(3+2)=15$ $a_{4}=4(4+2)=24$ $a_{5}=5(5+2)=35$ Therefore, the required terms are 3, 8, 15, 24, and 35....
Read More →(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m.
Question: (a)A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? (b)If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 106m, and the radius of lunar orbit is 3.8 108m. Solution: Focal length of the objective lens,fo= 15 m= 15102cm Focal length of the eyepiece,...
Read More →Evaluate the determinants in Exercises 1 and 2.
Question: Evaluate the determinants in Exercises 1 and 2. $\left|\begin{array}{cc}2 4 \\ -5 -1\end{array}\right|$ Solution: $\left|\begin{array}{cc}2 4 \\ -5 -1\end{array}\right|=2(-1)-4(-5)=-2+20=18$...
Read More →Explain structures of diborane and boric acid.
Question: Explain structures of diborane and boric acid. Solution: (a)Diborane $\mathrm{B}_{2} \mathrm{H}_{6}$ is an electron-deficient compound. $\mathrm{B}_{2} \mathrm{H}_{6}$ has only 12 electrons $-6 \mathrm{e}^{-}$from $6 \mathrm{H}$ atoms and $3 \mathrm{e}^{-}$each from $2 \mathrm{~B}$ atoms. Thus, after combining with $3 \mathrm{H}$ atoms, none of the boron atoms has any electrons left. X-ray diffraction studies have shown the structure of diborane as: 2 boron and 4 terminal hydrogen atom...
Read More →A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm.
Question: A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece? Solution: Focal length of the objective lens,fo= 144 cm Focal length of the eyepiece,fe= 6.0 cm The magnifying power of the telescope is given as: $m=\frac{f_{\mathrm{o}}}{f_{\mathrm{e}}}$ $=\frac{144}{6}=24$ The separation between the objective lens and the eyepiece is calcul...
Read More →Find the expansion of using binomial theorem.
Question: Find the expansion of $\left(3 \mathrm{x}^{2}-2 \mathrm{ax}+3 \mathrm{a}^{2}\right)^{3}$ using binomial theorem. Solution: Using Binomial Theorem, the given expression $\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$ can be expanded as $\left[\left(3 x^{2}-2 a x\right)+3 a^{2}\right]^{3}$ $={ }^{3} C_{0}\left(3 x^{2}-2 a x\right)^{3}+{ }^{3} C_{1}\left(3 x^{2}-2 a x\right)^{2}\left(3 a^{2}\right)+{ }^{3} C_{2}\left(3 x^{2}-2 a x\right)\left(3 a^{2}\right)^{2}+{ }^{3} C_{3}\left(3 a^{2}\right)^...
Read More →A person with a normal near point (25 cm)
Question: A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope, Solution: Focal length of the objective lens,fo= 8 mm = 0.8 cm Focal length of the eyepiece,fe= 2.5 cm Object distance for the objective lens,uo= 9.0 mm = 0.9 cm Least distan...
Read More →How is excessive content of CO2 responsible for global warming?
Question: How is excessive content of $\mathrm{CO}_{2}$ responsible for global warming? Solution: Carbon dioxide is a very essential gas for our survival. However, an increased content of $\mathrm{CO}_{2}$ in the atmosphere poses a serious threat. An increment in the combustion of fossil fuels, decomposition of limestone, and a decrease in the number of trees has led to greater levels of carbon dioxide. Carbon dioxide has the property of trapping the heat provided by sunrays. Higher the level of...
Read More →A person with a normal near point (25 cm)
Question: A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope, Solution: Focal length of the objective lens,fo= 8 mm = 0.8 cm Focal length of the eyepiece,fe= 2.5 cm Object distance for the objective lens,uo= 9.0 mm = 0.9 cm Least distan...
Read More →A person with a normal near point (25 cm)
Question: A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope, Solution: Focal length of the objective lens,fo= 8 mm = 0.8 cm Focal length of the eyepiece,fe= 2.5 cm Object distance for the objective lens,uo= 9.0 mm = 0.9 cm Least distan...
Read More →Suggest a reason as to why CO is poisonous.
Question: Suggest a reason as to why CO is poisonous. Solution: Carbon monoxide is highly-poisonous because of its ability to form a complex with haemoglobin. The $\mathrm{CO}-\mathrm{Hb}$ complex is more stable than the $\mathrm{O}_{2}-\mathrm{Hb}$ complex. The former prevents $\mathrm{Hb}$ from binding with oxygen. Thus, a person dies because of suffocation on not receiving oxygen. It is found that the CO-Hb complex is about 300 times more stable than the $\mathrm{O}_{2}-\mathrm{Hb}$ complex....
Read More →If A is square matrix such that then is equal to
Question: If $A$ is square matrix such that $A^{2}=A$, then $(I+A)^{3}-7 A$ is equal to A.A B.IA C.I D.3A Solution: Answer: C $(I+A)^{3}-7 A=I^{3}+A^{3}+3 I^{2} A+3 A^{2} I-7 A$ $=I+A^{3}+3 A+3 A^{2}-7 A$ $=I+A^{2} \cdot A+3 A+3 A-7 A \quad\left[A^{2}=A\right]$ $=I+A \cdot A-A$ $=I+A^{2}-A$ $=I+A-A$ $=I$ $\therefore(I+A)^{3}-7 A=I$...
Read More →Expand using Binomial Theorem.
Question: Expand using Binomial Theorem$\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0$ Solution: Using Binomial Theorem, the given expression $\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}$ can be expanded as $\left[\left(1+\frac{x}{2}\right)-\frac{2}{x}\right]^{4}$ $={ }^{4} \mathrm{C}_{0}\left(1+\frac{\mathrm{x}}{2}\right)^{4}-{ }^{4} \mathrm{C}_{1}\left(1+\frac{\mathrm{x}}{2}\right)^{3}\left(\frac{2}{\mathrm{x}}\right)+{ }^{4} \mathrm{C}_{2}\left(1+\frac{\mathrm{x}}{2}\right)^{2}\left(\fr...
Read More →Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF.
Question: Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3is bubbled through. Give reasons. Solution: Hydrogen fluoride(HF) is a covalent compound and has a very strong intermolecular hydrogen-bonding. Thus, it does not provide ions and aluminium fluoride (AlF) does not dissolve in it. Sodium fluoride (NaF) is an ionic compound and when it is added to the mixture, AlF dissolves. ...
Read More →If B–Cl bond has a dipole moment, explain
Question: If $\mathrm{B}-\mathrm{Cl}$ bond has a dipole moment, explain why $\mathrm{BCl}_{3}$ molecule has zero dipole moment. Solution: As a resultof the difference in the electronegativities of B and Cl, the BCl bond is polar in nature. However, the BCl3molecule is non-polar. This is because BCl3is trigonal planar in shape. It is a symmetrical molecule. Hence, the respective dipole-moments of the BCl bond cancel each other, thereby causing a zero-dipole moment....
Read More →A compound microscope consists of an objective lens of focal length 2.0 cm
Question: A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case? Solution: Focal length of the objective lens,f1= 2.0 cm Focal length of the eyepiece,f2= 6.25 cm Distance between th...
Read More →A compound microscope consists of an objective lens of focal length 2.0 cm
Question: A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case? Solution: Focal length of the objective lens,f1= 2.0 cm Focal length of the eyepiece,f2= 6.25 cm Distance between th...
Read More →Suggest reasons why the B–F bond lengths in
Question: Suggest reasons why the B-F bond lengths in $\mathrm{BF}_{3}(130 \mathrm{pm})$ and $\mathrm{BF}_{4}^{-}(143 \mathrm{pm})$ differ. Solution: The $\mathrm{B}-\mathrm{F}$ bond length in $\mathrm{BF}_{3}$ is shorter than the $\mathrm{B}-\mathrm{F}$ bond length in $\mathrm{BF}_{4}^{-} . \mathrm{BF}_{3}$ is an electron-deficient species. With a vacant $p$-orbital on boron, the fluorine and boron atoms undergo $p \pi-p \pi$ back-bonding to remove this deficiency. This imparts a double-bond ch...
Read More →If the matrix A is both symmetric and skew symmetric, then
Question: If the matrixAis both symmetric and skew symmetric, then A.Ais a diagonal matrix B.Ais a zero matrix C.Ais a square matrix D.None of these Solution: Answer: B IfAis both symmetric and skew-symmetric matrix, then we should have $A^{\prime}=A$ and $A^{\prime}=-A$ $\Rightarrow A=-A$ $\Rightarrow A+A=O$ $\Rightarrow 2 A=O$ $\Rightarrow A=O$ Therefore,Ais a zero matrix....
Read More →Choose the correct answer in the following questions:
Question: Choose the correct answer in the following questions: If $A=\left[\begin{array}{cc}\alpha \beta \\ \gamma -\alpha\end{array}\right]$ is such that $A^{2}=I$ then A. $1+\alpha^{2}+\beta \gamma=0$ B. $1-\alpha^{2}+\beta \gamma=0$ C. $1-\alpha^{2}-\beta \gamma=0$ D. $1+\alpha^{2}-\beta \gamma=0$ Solution: Answer: C $A=\left[\begin{array}{cc}\alpha \beta \\ \gamma -\alpha\end{array}\right]$ $\therefore A^{2}=A \cdot A=\left[\begin{array}{cc}\alpha \beta \\ \gamma -\alpha\end{array}\right]\l...
Read More →Rationalise the given statements and give chemical reactions:
Question: Rationalise the given statements and give chemical reactions: - Lead(II) chloride reacts with $\mathrm{Cl}_{2}$ to give $\mathrm{PbCl}_{4}$. - Lead(IV) chloride is highly unstable towards heat. - Lead is known not to form an iodide, $\mathrm{Pbl}_{4}$. Solution: (a) Lead belongs to group 14 of the periodic table. The two oxidation states displayed by this group is $+2$ and $+4$. On moving down the group, the $+2$ oxidation state becomes more stable and the $+4$ oxidation state becomes ...
Read More →What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm?
Question: What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses. Solution: Focal length of the convex lens,f1= 30 cm Focal length of the concave lens,f2= 20 cm Focal length of the system of lenses =f The equivalent focal length of a system of two lenses in contact is given as: $\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$ $\frac{1}{f}=\frac{1}{30}-\frac{1...
Read More →An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm.
Question: An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens? Solution: Size of the object,h1= 3 cm Object distance,u= 14 cm Focal length of the concave lens,f= 21 cm Image distance =v $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ $\frac{1}{v}=-\frac{1}{21}-\frac{1}{14}=\frac{-2-3}{42}=\frac{-5}{42}$ $\therefore v=-\frac{42}{5}=-8.4 \mathrm{~cm}$ Hence, the imag...
Read More →Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of
Question: Findn, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}$ is $\sqrt{6}: 1$ Solution: In the expansion, $(a+b)^{n}={ }^{n} C_{v} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$, Fifth term from the beginning $={ }^{n} \mathrm{C}_{4} \mathrm{a}^{n-4} \mathrm{~b}^{4}$ Fifth term from the end $={ }^{n} \mathrm{C}_{n-4} \mathr...
Read More →If A and B are square matrices of the same order such that AB = BA, then prove by induction that.
Question: If $A$ and $B$ are square matrices of the same order such that $A B=B A$, then prove by induction that $A B^{n}=B^{n} A$. Further, prove that $(A B)^{n}=A^{n} B^{n}$ for all $n \in \mathbf{N}$ Solution: AandBare square matrices of the same order such thatAB=BA. To prove: $\quad \mathrm{P}(n): A B^{n}=B^{n} A, n \in \mathbf{N}$ Forn= 1, we have: $\begin{aligned} \mathrm{P}(1): A B=B A \\ \Rightarrow A B^{1}=B^{1} A \end{aligned}$ Therefore, the result is true forn= 1. Let the result be ...
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