Explain the difference in properties of diamond and graphite on the basis of their structures.
Question: Explain the difference in properties of diamond and graphite on the basis of their structures. Solution:...
Read More →A beam of light converges at a point P.
Question: A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm? Solution: In the given situation, the object is virtual and the image formed is real. Object distance,u= +12 cm (a)Focal length of the convex lens,f= 20 cm Image distance =v According to the lens formula, we have the relation: $\frac{1}{v}-\...
Read More →What is the state of hybridisation of carbon in
Question: What is the state of hybridisation of carbon in (a) $\mathrm{CO}_{3}^{2-}$ (b) diamond (c) graphite? Solution: The state of hybridisation of carbon in: (a) $\mathrm{CO}_{3}^{2-}$ $\mathrm{C}$ in $\mathrm{CO}_{3}^{2-}$ is $s p^{2}$ hybridised and is bonded to three oxygen atoms. (b)Diamond Each carbon in diamond issp3hybridised and is bound to four other carbon atoms. (c)Graphite Each carbon atom in graphite issp2hybridised and is bound to three other carbon atoms....
Read More →Double-convex lenses are to be manufactured from a glass of refractive index 1.55,
Question: Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20cm? Solution: Refractive index of glass, $\mu=1.55$ Focal length of the double-convex lens,f= 20 cm Radiusof curvature of one face of the lens =R1 Radius of curvature of the other face of the lens =R2 Radius of curvature of the double-convex lens =R $\therefore R_{1}=R$ and $R_{2}=-R$ ...
Read More →Find an approximation of (0.99)5 using the first three terms of its expansion.
Question: Find an approximation of $(0.99)^{5}$ using the first three terms of its expansion. Solution: $0.99=1-0.01$ $\therefore(0.99)^{5}=(1-0.01)^{5}$ $={ }^{3} \mathrm{C}_{0}(1)^{5}-{ }^{3} \mathrm{C}_{1}(1)^{4}(0.01)+{ }^{5} \mathrm{C}_{2}(1)^{3}(0.01)^{2} \quad$ (Approximately) $=1-5(0.01)+10(0.01)^{2}$ $=1-0.05+0.001$ $=1.001-0.05$ $=0.951$ Thus, the value of $(0.99)^{5}$ is approximately $0.951$....
Read More →Write the resonance structures of
Question: Write the resonance structures of $\mathrm{CO}_{3}^{2-}$ and $\mathrm{HCO}_{3}^{-}$. Solution: There are only two resonating structures for the bicarbonate ion....
Read More →Find the value of.
Question: Find the value of $\left(a^{2}+\sqrt{a^{2}-1}\right)^{4}+\left(a^{2}-\sqrt{a^{2}-1}\right)^{4}$. Solution: Firstly, the expression $(x+y)^{4}+(x-y)^{4}$ is simplified by using Binomial Theorem. This can be done as $(x+y)^{4}={ }^{4} C_{0} x^{4}+{ }^{+} C_{1} x^{3} y+{ }^{4} C_{2} x^{2} y^{2}+{ }^{4} C_{3} x y^{3}+{ }^{4} C_{4} y^{4}$ $=x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4}$ $(x-y)^{4}={ }^{4} C_{0} x^{4}-{ }^{4} C_{1} x^{5} y+{ }^{4} C_{2} x^{2} y^{2}-{ }^{4} C_{3} x y^{3}+{ }^...
Read More →A prism is made of glass of unknown refractive index.
Question: A prism is made of glass of unknown refractive index. A parallelbeam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40. What is the refractive index of the material of the prism? The refracting angle of the prism is 60. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light. Solution: Angle of minimum deviation, $\delta_{\mathrm{m}}=40^{\circ}$ Angle of the prism,A= 6...
Read More →Find the matrix X so that
Question: Find the matrix $X$ so that $X\left[\begin{array}{lll}1 2 3 \\ 4 5 6\end{array}\right]=\left[\begin{array}{rrr}-7 -8 -9 \\ 2 4 6\end{array}\right]$ Solution: It is given that: $X\left[\begin{array}{lll}1 2 3 \\ 4 5 6\end{array}\right]=\left[\begin{array}{rrr}-7 -8 -9 \\ 2 4 6\end{array}\right]$ Now, let $X=\left[\begin{array}{ll}a c \\ b d\end{array}\right]$ Therefore, we have: $\left[\begin{array}{ll}a c \\ b d\end{array}\right]\left[\begin{array}{lll}1 2 3 \\ 4 5 6\end{array}\right]=...
Read More →What are electron deficient compounds?
Question: What are electron deficient compounds? Are BCl3and SiCl4electron deficient species? Explain. Solution: In an electron-deficient compound, theoctet of electrons is not complete, i.e., the central metal atom has an incomplete octet. Therefore, it needs electrons to complete its octet. (i)BCl3 BCl3is an appropriate example of an electron-deficient compound. B has 3 valence electrons. After forming three covalent bonds with chlorine, the number of electrons around it increases to 6. Howeve...
Read More →Write reactions to justify amphoteric nature of aluminium.
Question: Write reactions to justify amphoteric nature of aluminium. Solution: A substance is called amphoteric if it displays characteristics of both acids and bases.Aluminium dissolves in both acids and bases, showing amphoteric behaviour. (i) $2 \mathrm{Al}_{(s)}+6 \mathrm{HCl}_{(a q)} \longrightarrow 2 \mathrm{Al}_{(a q)}^{3+}+6 \mathrm{Cl}_{(a q)}^{-}+3 \mathrm{H}_{2(g)}$ (ii) $2 \mathrm{Al}_{(s)}+2 \mathrm{NaOH}_{(a q)}+6 \mathrm{H}_{2} \mathrm{O}_{(l)} \longrightarrow 2 \mathrm{Na}^{+}\le...
Read More →A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm.
Question: A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.) Solution: Actual depth of the bulbin water,d1= 80 cm = 0.8 m Refractive index of water, $\mu=1.33$ Thegiven situation is shown in the following figure: Where, i= Angle of incidence r= Angle of refraction = 90 Since the bulb is a point sou...
Read More →A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm.
Question: A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.) Solution: Actual depth of the bulbin water,d1= 80 cm = 0.8 m Refractive index of water, Thegiven situation is shown in the following figure: Where, i= Angle of incidence r= Angle of refraction = 90 Since the bulb is a point source, the em...
Read More →Evaluate.
Question: Evaluate $(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$ Solution: Firstly, the expression $(a+b)^{6}-(a-b)^{6}$ is simplified by using Binomial Theorem. This can be done as $(a+b)^{6}={ }^{6} \mathrm{C}_{0} a^{6}+{ }^{6} \mathrm{C}_{1} a^{5} b+{ }^{6} \mathrm{C}_{2} a^{4} b^{2}+{ }^{6} \mathrm{C}_{3} a^{3} b^{3}+{ }^{6} \mathrm{C}_{4} a^{2} b^{4}+{ }^{6} \mathrm{C}_{5} a^{1} b^{5}+{ }^{6} \mathrm{C}_{6} b^{6}$ $=a^{6}+6 a^{5} b+15 a^{4} b^{2}+20 a^{3} b^{3}+15 a^{2} b^{4}+6 a b^{5}+...
Read More →Describe the shapes of BF3 and BH4–.
Question: Describe the shapes of $\mathrm{BF}_{3}$ and $\mathrm{BH}_{4}^{-}$. Assign the hybridisation of boron in these species. Solution: (i)BF3 As a resultof its small size and high electronegativity, boron tends to form monomeric covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by the overlap of threesp2hybridised orbitals of boron with thesporbitals of three halogen atoms. Boron issp2hybridised in BF3. (ii)BH4 Boron-hydride ion (BH4) is form...
Read More →Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively.
Question: Figures 9.34(a) and (b) show refraction of a ray in air incident at 60 with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45 with the normal to a water-glass interface [Fig. 9.34(c)]. Solution: As per the given figure, for the glass air interface: Angle of incidence,i= 60 Angle of refraction,r= 35 The relative refractive index of glass with respect to air is given by Snells law as: $\mu_...
Read More →A manufacturer produces three products x, y, z which he sells in two markets.
Question: A manufacturer produces three productsx,y,zwhich he sells in two markets. Annual sales are indicated below: (a) If unit sale prices ofx,yandzare Rs 2.50, Rs 1.50 and Rs 1.00, respectively, find the total revenue in each market with the help of matrix algebra. (b) If the unit costs of the above three commodities are Rs 2.00, Rs 1.00 and 50 paise respectively. Find the gross profit. Solution: (a)The unit sale prices ofx,y, andzare respectively given as Rs 2.50, Rs 1.50, and Rs 1.00. Cons...
Read More →If a and b are distinct integers, prove that a – b is a factor of an – bn,
Question: If $a$ and $b$ are distinct integers, prove that $a-b$ is a factor of $a^{n}-b^{n}$, whenever $n$ is a positive integer. [Hint: write $a^{n}=(a-b+b)^{n}$ and expand] Solution: In order to prove that $(a-b)$ is a factor of $\left(a^{n}-b^{n}\right)$, it has to be proved that $a^{n}-b^{n}=k(a-b)$, where $k$ is some natural number It can bewritten that,a=ab+b $\therefore a^{n}=(a-b+b)^{n}=[(a-b)+b]^{n}$ $={ }^{n} \mathrm{C}_{0}(a-b)^{n}+{ }^{n} \mathrm{C}_{1}(a-b)^{n-1} b+\ldots+{ }^{n} \...
Read More →A tank is filled with water to a height of 12.5 cm.
Question: A tank is filled with water to a height of 12.5 cm. The apparentdepth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again? Solution: Actual depth ofthe needle in water,h1= 12.5 cm Apparent depth of the needle in water,h2= 9.4 cm Refractive index of wat...
Read More →Explain what happens when boric acid is heated.
Question: Explain what happens when boric acid is heated. Solution: On heating orthoboric acid (H3BO3) at 370 K or above, it changes to metaboric acid (HBO2). On further heating, this yields boric oxide B2O3....
Read More →A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm.
Question: A 4.5 cm needle is placed 12 cm away from a convex mirror of focallength 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror. Solution: Height of theneedle,h1= 4.5 cm Object distance,u= 12 cm Focal length of the convex mirror,f= 15 cm Image distance =v The value ofvcan be obtained using the mirror formula: $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$ $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ $=\frac{1}{15}+\frac{1}{12}=...
Read More →Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.
Question: Find the coefiicient of $x^{5}$ in the product $(1+2 x)^{6}(1-x)^{7}$ using binomial theorem. Solution: Using Binomial Theorem, the expressions, $(1+2 x)^{6}$ and $(1-x)^{7}$, can be expanded as $(1+2 x)^{6}={ }^{6} \mathrm{C}_{0}+{ }^{6} \mathrm{C}_{1}(2 x)+{ }^{6} \mathrm{C}_{2}(2 x)^{2}+{ }^{6} \mathrm{C}_{3}(2 x)^{3}+{ }^{6} \mathrm{C}_{4}(2 x)^{4}$ $+{ }^{6} \mathrm{C}_{5}(2 x)^{5}+{ }^{6} \mathrm{C}_{6}(2 x)^{6}$ $=1+6(2 x)+15(2 x)^{2}+20(2 x)^{3}+15(2 x)^{4}+6(2 x)^{5}+(2 x)^{6}...
Read More →Is boric acid a protic acid? Explain.
Question: Is boric acid a protic acid? Explain. Solution: Boric acid is not a protic acid. It is a weak monobasic acid, behaving as a Lewis acid. $\mathrm{B}(\mathrm{OH})_{3}+2 \mathrm{HOH} \longrightarrow\left[\mathrm{B}(\mathrm{OH})_{4}\right]^{-}+\mathrm{H}_{3} \mathrm{O}^{+}$ It behaves as an acid by accepting a pair of electrons from $-\mathrm{OH}$ ion....
Read More →A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm.
Question: A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved? Solution: Size of the candle,h= 2.5 cm Image size =h Object distance,u= 27 cm Radius of curvature of the concave mirror,R= 36 cm Focal length of the concave mirr...
Read More →Consider the compounds, BCl3 and CCl4.
Question: Consider the compounds, BCl3and CCl4. How will they behave with water? Justify. Solution: Being a Lewis acid, BCl3readily undergoes hydrolysis. Boric acid is formed as a result. $\mathrm{BCl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow 3 \mathrm{HCl}+\mathrm{B}(\mathrm{OH})_{3}$ CCl4completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When CCl4and water are mixed, they form separate layers. $\m...
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