(a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68.
Question: (a)Figure 9.35 shows a cross-section of a light pipe made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure. (b)What is the answer if there is no outer covering of the pipe? Solution: (a) Refractive index of the glass fibre, $\mu_{1}=1.68$ Refractive index of ...
Read More →Find the 20th term in the following sequence whose nth term is
Question: Find the20thterm in the following sequence whosenthterm is$a_{n}=\frac{n(n-2)}{n+3} ; a_{20}$ Solution: Substituting $n=20$, we obtain $a_{20}=\frac{20(20-2)}{20+3}=\frac{20(18)}{23}=\frac{360}{23}$...
Read More →Find values of x, if
Question: Find values ofx, if (i) $\left|\begin{array}{ll}2 4 \\ 5 1\end{array}\right|=\left|\begin{array}{cc}2 x 4 \\ 6 x\end{array}\right|$ (ii) $\left|\begin{array}{ll}2 3 \\ 4 5\end{array}\right|=\left|\begin{array}{cc}x 3 \\ 2 x 5\end{array}\right|$ Solution: (i) $\left|\begin{array}{ll}2 4 \\ 5 1\end{array}\right|=\left|\begin{array}{cc}2 x 4 \\ 6 x\end{array}\right|$ $\Rightarrow 2 \times 1-5 \times 4=2 x \times x-6 \times 4$ $\Rightarrow 2-20=2 x^{2}-24$ $\Rightarrow 2 x^{2}=6$ $\Rightar...
Read More →What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite.
Question: What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on physical properties of two allotropes? Solution: Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are calledallotropes. Diamond: The rigid 3-D structure of diamond makes it a very hard substance. In fact, diamond is one of the hardest nat...
Read More →Find the 9th term in the following sequence whose nth term is
Question: Find the9thterm in the following sequence whosenthterm is$a_{n}=(-1)^{n-1} n^{3} ; a_{9}$ Solution: Substitutingn= 9, we obtain $a_{9}=(-1)^{9-1}(9)^{3}=(9)^{3}=729$...
Read More →If, find.
Question: If $A=\left[\begin{array}{rrr}1 1 -2 \\ 2 1 -3 \\ 5 4 -9\end{array}\right]$, find $|\mathrm{A}|$. Solution: Let $A=\left[\begin{array}{rrr}1 1 -2 \\ 2 1 -3 \\ 5 4 -9\end{array}\right]$ By expanding along the first row, we have: $\begin{aligned}|A| =1\left|\begin{array}{ll}1 -3 \\ 4 -9\end{array}\right|-1\left|\begin{array}{ll}2 -3 \\ 5 -9\end{array}\right|-2\left|\begin{array}{ll}2 1 \\ 5 4\end{array}\right| \\ =1(-9+12)-1(-18+15)-2(8-5) \\ =1(3)-1(-3)-2(3) \\ =3+3-6 \\ =6-6 \\ =0 \end...
Read More →Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are
Question: Find the indicated terms in each of the sequences in Exercises 7 to 10 whose $n^{\text {th }}$ terms are : $a_{n}=\frac{n^{2}}{2^{n}} ; a_{7}$ Solution: Here, $a_{n}=\frac{n^{2}}{2^{n}}$. Substituting $n=7$, we obtain $a_{7}=\frac{7^{2}}{2^{7}}=\frac{49}{128}$...
Read More →How would you explain the lower atomic radius of Ga as compared to Al?
Question: How would you explain the lower atomic radius of Ga as compared to Al? Solution: Although Ga has one shell more than Al, its size is lesser than Al. This isbecause of the poor shielding effect of the 3d-electrons. The shielding effect ofd-electrons is very poor and the effective nuclear charge experienced by the valence electrons in gallium is much more than it is in the case of Al....
Read More →Evaluate the determinants
Question: Evaluate the determinants (i) $\left|\begin{array}{ccc}3 -1 -2 \\ 0 0 -1 \\ 3 -5 0\end{array}\right|$ (iii) $\left|\begin{array}{ccc}3 -4 5 \\ 1 1 -2 \\ 2 3 1\end{array}\right|$ (ii) $\left|\begin{array}{ccc}0 1 2 \\ -1 0 -3 \\ -2 3 0\end{array}\right|$ (iv) $\left[\begin{array}{ccc}2 -1 -2 \\ 0 2 -1 \\ 3 -5 0\end{array}\right]$ Solution: (i) Let $A=\left|\begin{array}{ccc}3 -1 -2 \\ 0 0 -1 \\ 3 -5 0\end{array}\right|$. It can be observed that in the second row, two entries are zero. T...
Read More →A small pin fixed on a table top is viewed from above from a distance of 50 cm.
Question: A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab? Solution: Actual depth of the pin,d= 15 cm Apparent dept of the pin $=d^{\prime}$ Refractive index of glass, $\mu=1.5$ Ratio of actual depth to the apparent depth is equal to th...
Read More →Find the 17th term in the following sequence whose nth term is
Question: Find the17thterm in the following sequence whosenthterm is$\mathrm{s} \mathrm{a}_{\mathrm{n}}=4 \mathrm{n}-3 ; \mathrm{a}_{17}, \mathrm{a}_{24}$ Solution: Substitutingn= 17, we obtain $a_{17}=4(17)-3=68-3=65$ Substitutingn= 24, we obtain $\mathrm{a}_{24}=4(24)-3=96-3=93$...
Read More →Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon?
Question: Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon? Solution: Ionisation enthalpy of carbon (the first element of group 14) is very high ( $1086 \mathrm{~kJ} / \mathrm{mol}$ ). This is expected owing to its small size. However, on moving down the group to silicon, there is a sharp decrease in the enthalpy (786 kJ). This is because of an appreciable increase in the atomic sizes of elements on moving down the group....
Read More →Write the first five terms of the sequences whose nth term is
Question: Write the first five terms of the sequences whosenthterm is$\mathrm{a}_{\mathrm{n}}=\mathrm{n} \frac{\mathrm{n}^{2}+5}{4}$ Solution: Substitutingn= 1, 2, 3, 4, 5, we obtain $a_{1}=1 \cdot \frac{1^{2}+5}{4}=\frac{6}{4}=\frac{3}{2}$ $a_{2}=2 \cdot \frac{2^{2}+5}{4}=2 \cdot \frac{9}{4}=\frac{9}{2}$ $a_{3}=3 \cdot \frac{3^{2}+5}{4}=3 \cdot \frac{14}{4}=\frac{21}{2}$ $a_{4}=4 \cdot \frac{4^{2}+5}{4}=21$ $a_{5}=5 \cdot \frac{5^{2}+5}{4}=5 \cdot \frac{30}{4}=\frac{75}{2}$ Therefore, the requi...
Read More →Give reasons:
Question: Give reasons: (i) Conc. HNO3can be transported in aluminium container. (ii) A mixture of dilute NaOH and aluminium pieces is used to open drain. (iii) Graphite is used as lubricant. (iv) Diamond is used as an abrasive. (v) Aluminium alloys are used to make aircraft body. (vi) Aluminium utensils should not be kept in water overnight. (vii) Aluminium wire is used to make transmission cables. Solution: (i)ConcentratedHNO3can be stored and transported in aluminium containers as it reacts w...
Read More →Use the mirror equation to deduce that:
Question: Use the mirror equation to deduce that: (a)an object placed betweenfand 2fof a concave mirror produces a real image beyond 2f. (b)a convex mirror always produces a virtual image independent of the location of the object. (c)the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d)an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note:This exercise helps you deduce a...
Read More →Use the mirror equation to deduce that:
Question: Use the mirror equation to deduce that: (a)an object placed betweenfand 2fof a concave mirror produces a real image beyond 2f. (b)a convex mirror always produces a virtual image independent of the location of the object. (c)the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d)an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note:This exercise helps you deduce a...
Read More →If, then show that
Question: If $\mathrm{A}=\left[\begin{array}{lll}1 0 1 \\ 0 1 2 \\ 0 0 0\end{array}\right]$, then show that $|3 A|=27|A|$. Solution: The given matrix is $\mathrm{A}=\left[\begin{array}{lll}1 0 1 \\ 0 1 2 \\ 0 0 4\end{array}\right]$. It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation. $|\mathrm{A}|=1\left|\begin{array}{ll}1 2 \\ 0 4\end{array}\right|-0\left|\begin{array}{ll}0 1 \\ 0 4\end{array}\right|+0\left|\begi...
Read More →Write the first five terms of the sequences whose nth term is
Question: Write the first five terms of the sequences whosenthterm is$a_{n}=(-1)^{n-1} 5^{n+1}$ Solution: Substitutingn= 1, 2, 3, 4, 5, we obtain $\mathrm{a}_{1}=(-1)^{1-1} 5^{1+1}=5^{2}=25$ $\mathrm{a}_{2}=(-1)^{2-1} 5^{2+1}=-5^{3}=-125$ $\mathrm{a}_{3}=(-1)^{3-1} 5^{3+1}=5^{+}=625$ $\mathrm{a}_{4}=(-1)^{4-1} 5^{4+1}=-5^{5}=-3125$ $\mathrm{a}^{5}=(-1)^{5-1} 5^{5+1}=5^{6}=15625$ Therefore, the required terms are 25, 125, 625, 3125, and 15625....
Read More →Explain the following reactions
Question: Explain the following reactions (a) Silicon is heated with methyl chloride at high temperature in the presence of copper; (b) Silicon dioxide is treated with hydrogen fluoride; (c) CO is heated with ZnO; (d) Hydrated alumina is treated with aqueous NaOH solution. Solution: (a) When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about $537 \mathrm{~K}$, a class of organosilicon polymers called methylsubstituted chlorosilanes $\left(\math...
Read More →Write the first five terms of the sequences whose nth term is
Question: Write the first five terms of the sequences whosenthterm is$\mathrm{a}_{\mathrm{n}}=\frac{2 \mathrm{n}-3}{6}$ Solution: Substituting n= 1, 2, 3, 4, 5, we obtain $a_{1}=\frac{2 \times 1-3}{6}=\frac{-1}{6}$ $a_{2}=\frac{2 \times 2-3}{6}=\frac{1}{6}$ $a_{3}=\frac{2 \times 3-3}{6}=\frac{3}{6}=\frac{1}{2}$ $a_{4}=\frac{2 \times 4-3}{6}=\frac{5}{6}$ $a_{5}=\frac{2 \times 5-3}{6}=\frac{7}{6}$ Therefore, the required terms are $\frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}$, and $\frac{7...
Read More →Write the first five terms of the sequences whose nth term is an = 2n
Question: Write the first five terms of the sequences whose $\mathrm{n}^{\text {th }}$ term is $a_{n}=2^{n}$ Solution: $a_{n}=2^{n}$ Substitutingn= 1, 2, 3, 4, 5, we obtain $a_{1}=2^{1}=2$ $a_{2}=2^{2}=4$ $a_{3}=2^{3}=8$ $a_{4}=2^{4}=16$ $a_{5}=2^{5}=32$ Therefore, the required terms are $2,4,8,16$, and 32 ....
Read More →If, then show that
Question: If $A=\left[\begin{array}{ll}1 2 \\ 4 2\end{array}\right]$, then show that $|2 A|=4|A|$ Solution: The given matrix is $A=\left[\begin{array}{ll}1 2 \\ 4 2\end{array}\right]$. $\therefore 2 A=2\left[\begin{array}{ll}1 2 \\ 4 2\end{array}\right]=\left[\begin{array}{ll}2 4 \\ 8 4\end{array}\right]$ $\therefore$ L.H.S. $=|2 A|=\left|\begin{array}{ll}2 4 \\ 8 4\end{array}\right|=2 \times 4-4 \times 8=8-32=-24$ Now, $|A|=\left|\begin{array}{ll}1 2 \\ 4 2\end{array}\right|=1 \times 2-2 \times...
Read More →Write the first five terms of the sequences whose nth term is
Question: Write the first five terms of the sequences whose nthterm is$a_{n}=\frac{n}{n+1}$ Solution: $a_{n}=\frac{n}{n+1}$ Substituting $n=1,2,3,4,5$, we obtain $a_{1}=\frac{1}{1+1}=\frac{1}{2}, a_{2}=\frac{2}{2+1}=\frac{2}{3}, a_{3}=\frac{3}{3+1}=\frac{3}{4}, a_{4}=\frac{4}{4+1}=\frac{4}{5}, a_{5}=\frac{5}{5+1}=\frac{5}{6}$ Therefore, the required terms are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}$, and $\frac{5}{6}$....
Read More →What happens when
Question: What happens when (a) Borax is heated strongly, (b) Boric acid is added to water, (c) Aluminium is treated with dilute NaOH, (d) BF3is reacted with ammonia? Solution: (a)When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead. (b)When boric acid is added to water, it accepts electrons fromOH ion. $\mathrm{B}(\mathrm{OH})_{3}+2 \mathrm{HOH} \longrigh...
Read More →Evaluate the determinants in Exercises 1 and 2.
Question: Evaluate the determinants in Exercises 1 and 2. (i) $\left|\begin{array}{lr}\cos \theta -\sin \theta \\ \sin \theta \cos \theta\end{array}\right|$ (ii) $\left|\begin{array}{cc}x^{2}-x+1 x-1 \\ x+1 x+1\end{array}\right|$ Solution: (i) $\left|\begin{array}{rr}\cos \theta -\sin \theta \\ \sin \theta \cos \theta\end{array}\right|=(\cos \theta)(\cos \theta)-(-\sin \theta)(\sin \theta)=\cos ^{2} \theta+\sin ^{2} \theta=1$ (ii) $\left|\begin{array}{cc}x^{2}-x+1 x-1 \\ x+1 x+1\end{array}\right...
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