Verify A (adj A) = (adj A) A = I .
Question: Verify $A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I .$ $\left[\begin{array}{rrr}1 -1 2 \\ 3 0 -2 \\ 1 0 3\end{array}\right]$ Solution: $A=\left[\begin{array}{ccc}1 -1 2 \\ 3 0 -2 \\ 1 0 3\end{array}\right]$ $|A|=1(0-0)+1(9+2)+2(0-0)=11$ $\therefore|A| I=11\left[\begin{array}{ccc}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right]=\left[\begin{array}{ccc}11 0 0 \\ 0 11 0 \\ 0 0 11\end{array}\right]$ Now, $A_{11}=0, A_{12}=-(9+2)=-11, A_{13}=0$ $A_{21}=-(-3-0)=3, A_{22}=3-2=1, A_{23}=-(...
Read More →What are electrophiles and nucleophiles?
Question: What are electrophiles and nucleophiles? Explain with examples. Solution: An electrophile isa reagent that takes away an electron pair. In other words, an electron-seeking reagent is called an electrophile (E+). Electrophiles are electron-deficient and can receive an electron pair. Carbocations $\left(\mathrm{CH}_{3} \mathrm{CH}_{2}^{+}\right)$and neutral molecules having functional groups such as carbonyl groupare examples of electrophiles. A nulceophile is a reagent that brings an el...
Read More →Find the sum to n terms in the geometric progression
Question: Find the sum tonterms in the geometric progression$1,-a, a^{2},-a^{3} \ldots($ if $a \neq-1)$ Solution: The given G.P. is $1,-a, a^{2},-a^{3}, \ldots \ldots \ldots \ldots . .$ Here, first term $=a_{1}=1$ Common ratio $=r=-a$ $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{a}_{1}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{1-\mathrm{r}}$ $\therefore S_{n}=\frac{1\left[1-(-a)^{n}\right]}{1-(-a)}=\frac{\left[1-(-a)^{n}\right]}{1+a}$...
Read More →The 6563 Å
Question: The 6563 Å$\mathrm{H}_{a}$line emitted by hydrogen in a star is found to be red shifted by 15 Å. Estimate the speed with which the star is receding from the Earth. Solution: Wavelength of $\mathrm{H}_{a}$ line emitted by hydrogen, = 6563 Å = 6563 1010m. Stars red-shift, Speed of light, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Let the velocity of the star receding away from the Earth bev. The red shift is related with velocity as: $\lambda^{\prime}-\lambda=\frac{v}{c} \lambda$ $v=\f...
Read More →The 6563 Å
Question: The 6563 Å$\mathrm{H}_{a}$line emitted by hydrogen in a star is found to be red shifted by 15 Å. Estimate the speed with which the star is receding from the Earth. Solution: Wavelength of $\mathrm{H}_{a}$ line emitted by hydrogen, = 6563 Å = 6563 1010m. Star's red-shift, $\left(\lambda^{\prime}-\lambda\right)=15Å =15 \times 10^{-10} \mathrm{~m}$ Speed of light, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Let the velocity of the star receding away from the Earth bev. The red shift is r...
Read More →Find the sum to n terms in the geometric progression
Question: Find the sum tonterms in the geometric progression$\sqrt{7}, \sqrt{21}, 3 \sqrt{7} \ldots$ Solution: The given G.P. is $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$ Here, $a=\sqrt{7}$ $r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}$ $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{1-\mathrm{r}}$ $\therefore S_{a}=\frac{\sqrt{7}\left[1-(\sqrt{3})^{n}\right]}{1-\sqrt{3}}$ $=\frac{\sqrt{7}\left[1-(\sqrt{3})^{n}\right]}{1-\sqrt{3}} \times \frac{1+\sqrt{3}}{1+\sqrt{3}}$ (...
Read More →The 6563 Å
Question: The 6563 Å$\mathrm{H}_{a}$line emitted by hydrogen in a star is found to be red shifted by 15 Å. Estimate the speed with which the star is receding from the Earth. Solution: Wavelength of $\mathrm{H}_{a}$ line emitted by hydrogen, = 6563 Å = 6563 1010m. Star's red-shift, $\left(\lambda^{\prime}-\lambda\right)=15 \AA=15 \times 10^{-10} \mathrm{~m}$ Speed of light, $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Let the velocity of the star receding away from the Earth bev. The red shift is...
Read More →Verify A (adj A) = (adj A) A = I
Question: Verify $A(\operatorname{adj} A)=(\operatorname{adj} A) A=A \mid$. $\left[\begin{array}{rr}2 3 \\ -4 -6\end{array}\right]$ Solution: $A=\left[\begin{array}{rr}2 3 \\ -4 -6\end{array}\right]$ we have, $|A|=-12-(-12)=-12+12=0$ $\therefore|A| I=0\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]=\left[\begin{array}{ll}0 0 \\ 0 0\end{array}\right]$ Now, $A_{11}=-6, A_{12}=4, A_{21}=-3, A_{22}=2$ $\therefore \operatorname{adj} A=\left[\begin{array}{rr}-6 -3 \\ 4 2\end{array}\right]$ Now, $\...
Read More →Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
Question: Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm. Solution: Fresnels distance (ZF) is the distance for which the ray optics is a good approximation. It is given by the relation, $Z_{F}=\frac{a^{2}}{\lambda}$ Where, Aperture width,a= 4 mm = 4 103m Wavelength of light,= 400 nm = 400 109m $Z_{F}=\frac{\left(4 \times 10^{-3}\right)^{2}}{400 \times 10^{-9}}=40 \mathrm{~m}$ Therefore, the distance for which the ray optics is a goo...
Read More →Find adjoint of each of the matrices.
Question: Find adjoint of each of the matrices. $\left[\begin{array}{lrr}1 -1 2 \\ 2 3 5 \\ -2 0 1\end{array}\right]$ Solution: Let $A=\left[\begin{array}{lrr}1 -1 2 \\ 2 3 5 \\ -2 0 1\end{array}\right]$. We have, $A_{11}=\left|\begin{array}{ll}3 5 \\ 0 1\end{array}\right|=3-0=3$ $A_{12}=-\left|\begin{array}{ll}2 5 \\ -2 1\end{array}\right|=-(2+10)=-12$ $A_{13}=\left|\begin{array}{ll}2 3 \\ -2 0\end{array}\right|=0+6=6$ $A_{21}=-\left|\begin{array}{ll}-1 2 \\ 0 1\end{array}\right|=-(-1-0)=1$ $A_...
Read More →Light of wavelength 5000 Å falls on a plane reflecting surface.
Question: Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray? Solution: Wavelength of incident light, = 5000 Å = 5000 1010m Speed of light,c= 3 108m Frequency of incident light is given by the relation, $v=\frac{c}{\lambda}$ $=\frac{3 \times 10^{8}}{5000 \times 10^{-10}}=6 \times 10^{14} \mathrm{~Hz}$ The wavelength and frequency of incident light...
Read More →Draw the resonance structures for the following compounds.
Question: Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. (a) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$ (b) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}$ (c) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$ (d) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}$ Solution: (a) The structure of $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$ is: The resonating structures of phenol are represented as: (b) The structure of $\mathrm{C}_{6...
Read More →Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …
Question: Find the sum to 20 terms in the geometric progression $0.15,0.015,0.0015$ Solution: The given G.P. is $0.15,0.015,0.00015, \ldots$ Here, $a=0.15$ and $r=\frac{0.015}{0.15}=0.1$ $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{1-\mathrm{r}}$ $\therefore S_{20}=\frac{0.15\left[1-(0.1)^{20}\right]}{1-0.1}$ $=\frac{0.15}{0.9}\left[1-(0.1)^{20}\right]$ $=\frac{15}{90}\left[1-(0.1)^{20}\right]$ $=\frac{1}{6}\left[1-(0.1)^{20}\right]$...
Read More →Find adjoint of each of the matrices.
Question: Find adjoint of each of the matrices. $\left[\begin{array}{ll}1 2 \\ 3 4\end{array}\right]$ Solution: Let $A=\left[\begin{array}{ll}1 2 \\ 3 4\end{array}\right]$ We have, $A_{11}=4, A_{12}=-3, A_{21}=-2, A_{22}=1$ $\therefore \operatorname{adj} A=\left[\begin{array}{ll}A_{11} A_{21} \\ A_{12} A_{22}\end{array}\right]=\left[\begin{array}{lr}4 -2 \\ -3 1\end{array}\right]$...
Read More →What is the Brewster angle for air to glass transition?
Question: What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.) Solution: Refractive index of glass, $\mu=1.5$ Brewster angle = Brewster angle is related to refractive index as: $\tan \theta=\mu$ $\theta=\tan ^{-1}(1.5)=56.31^{\circ}$ Therefore, the Brewster angle for air to glass transition is 56.31....
Read More →For what values of x, the numbers are in G.P?
Question: For what values of $x$, the numbers $\frac{2}{7}+x,-\frac{7}{2}$ are in G.P? Solution: The given numbers are $\frac{-2}{7}, x, \frac{-7}{2}$. Common ratio $=\frac{x}{\frac{-2}{7}}=\frac{-7 x}{2}$ Also, common ratio $=\frac{\frac{-7}{2}}{x}=\frac{-7}{2 x}$ $\therefore \frac{-7 x}{2}=\frac{-7}{2 x}$ $\Rightarrow x^{2}=\frac{-2 \times 7}{-2 \times 7}=1$ $\Rightarrow x=\sqrt{1}$ $\Rightarrow x=\pm 1$ Thus, for $x=\pm 1$, the given numbers will be in G.P....
Read More →In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away.
Question: In a double-slit experiment the angular width of a fringe is found to be 0.2 on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3. Solution: Distance of the screen from the slits,D= 1 m Wavelength of light used, $\lambda_{1}=600 \mathrm{~nm}$ Angular width of the fringe in air, $\theta_{1}=0.2^{\circ}$ Angular width of the frin...
Read More →If and Aij is Cofactors of aij, then value of Δ is given by
Question: If $\Delta=\left|\begin{array}{lll}a_{11} a_{12} a_{13} \\ a_{21} a_{22} a_{23} \\ a_{21} a_{23} a_{23}\end{array}\right|$ and $\mathrm{A}_{i j}$ is Cofactors of $a_{i j}$, then value of $\Delta$ is given by (A) $\quad a_{11} \mathrm{~A}_{31}+a_{12} \mathrm{~A}_{32}+a_{13} \mathrm{~A}_{33}$ (B) $a_{11} \mathrm{~A}_{11}+a_{12} \mathrm{~A}_{21}+a_{13} \mathrm{~A}_{31}$ (C) $a_{21} \mathrm{~A}_{11}+a_{22} \mathrm{~A}_{12}+a_{23} \mathrm{~A}_{13}$ (D) $a_{11} \mathrm{~A}_{11}+a_{21} \mathr...
Read More →Which term of the following sequences:
Question: Which term of the following sequences: (a) $2,2 \sqrt{2}, 4, \ldots$ is 128 ? (b) $\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is 729 ? (c) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$ ? Solution: (a) The given sequence is $2,2 \sqrt{2}, 4, \ldots$ Here, $a=2$ and $r=\frac{2 \sqrt{2}}{2}=\sqrt{2}$ Let the $n^{\text {th }}$ term of the given sequence be 128 . $a_{n}=a r^{n-1}$ $\Rightarrow(2)(\sqrt{2})^{n-1}=128$ $\Rightarrow(2)(2)^{\frac{n-1}{2}}=(2)^{7}$ $\Rightarrow(2)^{...
Read More →Using Cofactors of elements of third column, evaluate
Question: Using Cofactors of elements of third column, evaluate $\Delta=\left|\begin{array}{lll}1 x y z \\ 1 y z x \\ 1 z x y\end{array}\right|$ Solution: The given determinant is $\left|\begin{array}{lll}1 x y z \\ 1 y z x \\ 1 z x y\end{array}\right|$. We have: $\mathrm{M}_{13}=\left|\begin{array}{ll}1 y \\ 1 z\end{array}\right|=z-y$ $\mathrm{M}_{23}=\left|\begin{array}{ll}1 x \\ 1 z\end{array}\right|=z-x$ $\mathrm{M}_{33}=\left|\begin{array}{ll}1 x \\ 1 y\end{array}\right|=y-x$ $\therefore \m...
Read More →A beam of light consisting of two wavelengths,
Question: A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Youngs double-slit experiment. (a)Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b)What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? Solution: Wavelength of the light beam, $\lambda_{1}=650 \mathrm{~nm}$ Wavelength of another light beam, $\lambda_{2}=520...
Read More →Explain why alkyl groups act as electron donors
Question: Explain why alkyl groups act as electron donors when attached to a $\pi$ system. Solution: When an alkyl group is attached to a $\pi$ system, it acts as an electron-donor group by the process of hyperconjugation. To understand this concept better, let us take the example of propene. In hyperconjugation, the sigma electrons of the CH bond of an alkyl group are delocalised. This group is directly attached to an atom of an unsaturated system. The delocalisation occurs because of a partial...
Read More →In Young’s double-slit experiment using monochromatic light of wavelengthλ,
Question: In Youngs double-slit experiment using monochromatic light of wavelength, the intensity of light at a point on the screen where path difference is , isKunits. What is the intensity of light at a point where path difference is /3? Solution: LetI1andI2be the intensity of the two light waves. Their resultant intensities can be obtained as: $I^{\prime}=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \cos \phi$ Where, $\phi=$ Phase difference between the two waves For monochromatic light waves, $I_{1}=I_{...
Read More →The 4th term of a G.P. is square of its second term,
Question: The $4^{\text {th }}$ term of a G.P. is square of its second term, and the first term is $-3$. Determine its $7^{\text {th }}$ term. Solution: Letabe the first term andrbe the common ratio of the G.P. $\therefore a=-3$ It is known that, $a_{n}=a r^{n-1}$ $\therefore a_{4}=a r^{3}=(-3) r^{3}$ $a_{2}=a r^{1}=(-3) r$ According to the given condition, $(-3) r^{3}=[(-3) r]^{2}$ $\Rightarrow-3 r^{3}=9 r^{2}$ $\Rightarrow r=-3$ $a_{7}=a r^{7-1}=a r^{6}=(-3)(-3)^{6}=-(3)^{7}=-2187$ Thus, the s...
Read More →Which of the two: O2NCH2CH2O– or CH3CH2O– is expected
Question: Which of the two: $\mathrm{O}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{O}^{-}$or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}$is expected to be more stable and why? Solution: $\mathrm{NO}_{2}$ group is an electron-withdrawing group. Hence, it shows -l effect. By withdrawing the electrons toward it, the $\mathrm{NO}_{2}$ group decreases the negative charge on the compound, thereby stabilising it. On the other hand, ethyl group is an electron-releasing group. Hence, the ethyl grou...
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