The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112.
Question: The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112.If its first term is 11, then find the number of terms. Solution: Let the A.P. be $a, a+d, a+2 d, a+3 d, \ldots a+(n-2) d, a+(n-1) d$ Sum of first four terms $=a+(a+d)+(a+2 d)+(a+3 d)=4 a+6 d$ Sum of last four terms $=[a+(n-4) d]+[a+(n-3) d]+[a+(n-2) d]$ $+[a+n-1) d]$ $=4 a+(4 n-10) d$ According to the given condition, $4 a+6 d=56$ $\Rightarrow 4(11)+6 d=56[$ Since $a=11$ (given) $]$ $\Rightarrow 6 ...
Read More →Evaluate
Question: Evaluate $\left|\begin{array}{ccc}1 x y \\ 1 x+y y \\ x x+y\end{array}\right|$ Solution: $\Delta=\left|\begin{array}{ccc}1 x y \\ 1 x+y y \\ 1 x x+y\end{array}\right|$ Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we have: $\Delta=\left|\begin{array}{lll}1 x y \\ 0 y 0 \\ 0 0 x\end{array}\right|$ Expanding along $\mathrm{C}_{1}$, we have: $\Delta=1(x y-0)=x y$...
Read More →A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places,
Question: A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio. Solution: Let the G.P. be $T_{1}, T_{2}, T_{3}, T_{4}, \ldots T_{2 n}$. Number of terms $=2 n$ According to the given condition, $\mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}+\ldots+\mathrm{T}_{2 n}=5\left[\mathrm{~T}_{1}+\mathrm{T}_{3}+\ldots+\mathrm{T}_{2 n-1}\right]$ $\Rightarrow \mathrm{T}_{1}+\mathrm{T}_{2}+\mathrm{T}_{3}+\ldots+\mat...
Read More →The gravitational attraction between electron and proton
Question: The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 1040. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting. Solution: Radius of the first Bohr orbit is given by the relation, $r_{1}=\frac{4 \pi \in_{0}\left(\frac{h}{2 \pi}\right)^{2}}{m_{e}...
Read More →In the alkane H3C–CH2–C(CH3)2–CH2–CH(CH3)2,
Question: In the alkane $\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}-\mathrm{CH}_{2}-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$, identify $1^{\circ}, 2^{\circ}, 3^{\circ}$ carbon atoms and give the number of $\mathrm{H}$ atoms bonded to each one of these. Solution: 1 carbon atoms are those which are bonded to only one carbon atom, i.e., they have only one carbon atom as their neighbour. The given structure has five 1 carbon atoms and fifteen hydrogen at...
Read More →The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order,
Question: The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers. Solution: Let the three numbers in G.P. be $a, a r$, and $a r^{2}$. From the given condition, $a+a r+a r^{2}=56$ $\Rightarrow a\left(1+r+r^{2}\right)=56$ $\Rightarrow a=\frac{56}{1+r+r^{2}}$(1) $a-1, a r-7, a r^{2}-21$ forms an A.P. $\therefore(a r-7)-(a-1)=\left(a r^{2}-21\right)-(a r-7)$ $\Rightarrow a r-a-6=a r^{2}-a r-14$ $\Rightar...
Read More →Evaluate
Question: Evaluate $\left|\begin{array}{ccc}x y x+y \\ y x+y x \\ x+y x y\end{array}\right|$ Solution: $\Delta=\left|\begin{array}{ccc}x y x+y \\ y x+y x \\ x+y x y\end{array}\right|$ Applying $R, \rightarrow R, R, R$, we have: $\Delta=\left|\begin{array}{ccc}2(x+y) 2(x+y) 2(x+y) \\ y x+y x \\ x+y x y\end{array}\right|$ $=2(x+y)\left|\begin{array}{ccc}1 1 1 \\ y x+y x \\ x+y x y\end{array}\right|$ Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}$ and $\mathrm{C}_{3} \rightarrow...
Read More →Answer the following questions,
Question: Answer the following questions, which help you understand the difference between Thomsons model and Rutherfords model better. (a)Is the average angle of deflection of-particles by a thin gold foil predicted by Thomsons model much less, about the same, or much greater than that predicted by Rutherfords model? (b)Is the probability of backward scattering (i.e., scattering of-particles at angles greater than 90) predicted by Thomsons model much less, about the same, or much greater than t...
Read More →Let verify that
Question: Let $A=\left[\begin{array}{rrr}1 -2 1 \\ -2 3 1 \\ 1 1 5\end{array}\right]$ verify that (i) $[\operatorname{adj} A]^{-1}=\operatorname{adj}\left(A^{-1}\right)$ (ii) $\left(A^{-1}\right)^{-1}=A$ Solution: $A=\left[\begin{array}{rrr}1 -2 1 \\ -2 3 1 \\ 1 1 5\end{array}\right]$ $\therefore|A|=1(15-1)+2(-10-1)+1(-2-3)=14-22-5=-13$ Now, $A_{11}=14, A_{12}=11, A_{13}=-5$ $A_{21}=11, A_{22}=4, A_{23}=-3$ $A_{31}=-5, A_{17}=-3, A_{13}=-1$ $\therefore \operatorname{adj} A=\left[\begin{array}{cc...
Read More →In accordance with the Bohr’s model,
Question: In accordance with the Bohrs model, find the quantum number that characterises the earths revolution around the sun in an orbit of radius 1.5 1011m with orbital speed 3 104m/s. (Mass of earth = 6.0 1024kg.) Solution: Radius of the orbit of the Earth around the Sun,r= 1.5 1011m Orbital speed of the Earth,= 3 104m/s Mass of the Earth,m= 6.0 1024kg According to Bohrs model, angular momentum is quantized and given as: $m v r=\frac{n h}{2 \pi}$ Where, h= Plancks constant = 6.62 1034Js n= Qu...
Read More →How will you convert benzene into
Question: How will you convert benzene into (i)p-nitrobromobenzene (ii)m-nitrochlorobenzene (iii)p-nitrotoluene (iv) acetophenone Solution: (i)Benzene can be converted intop-nitrobromobenzene as: (ii)Benzene can be converted intom-nitrochlorobenzene as: (iii)Benzene can be converted intop-nitrotoulene as: (iv)Benzene can be converted into acetophenone as:...
Read More →In accordance with the Bohr’s model,
Question: In accordance with the Bohrs model, find the quantum number that characterises the earths revolution around the sun in an orbit of radius 1.5 1011m with orbital speed 3 104m/s. (Mass of earth = 6.0 1024kg.) Solution: Radius of the orbit of the Earth around the Sun,r= 1.5 1011m Orbital speed of the Earth,= 3 104m/s Mass of the Earth,m= 6.0 1024kg According to Bohrs model, angular momentum is quantized and given as: $m v r=\frac{n h}{2 \pi}$ Where, h= Plancks constant = 6.62 1034Js n= Qu...
Read More →The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Question: The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P. Solution: Letaandrbe the first term and the common ratio of the G.P. respectively. $\therefore a=1$ $a_{3}=a r^{2}=r^{2}$ $a_{5}=a r^{4}=r^{4}$ $\therefore r^{2}+r^{4}=90$ $\Rightarrow r^{4}+r^{2}-90=0$ $\Rightarrow r^{2}=\frac{-1+\sqrt{1+360}}{2}=\frac{-1 \pm \sqrt{361}}{2}=\frac{-1 \pm 19}{2}=-10$ or 9 $\therefore r=\pm 3$ (Taking real roots) Thus, the common ratio of the G.P....
Read More →A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature.
Question: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted? Solution: It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is 13.6 eV. When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes 13.6 + 12.5 eV i.e., 1.1 eV. Orbital energy is ...
Read More →The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2,
Question: The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms. Solution: Let the sum ofnterms of the G.P. be 315. It is known that, $\mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$ It is given that the first term $a$ is 5 and common ratio $r$ is 2 . $\therefore 315=\frac{5\left(2^{n}-1\right)}{2-1}$ $\Rightarrow 2^{n}-1=63$ $\Rightarrow 2^{n}=64=(2)^{6}$ $\Rightarrow n=6$ $\therefore$ Last term of the ...
Read More →Explain why the following systems are not aromatic?
Question: Explain why the following systems arenotaromatic? (i) (ii) (iii) Solution: (i) For the given compound, the number of -electrons is six. But only four -electrons are present within the ring. Also there is no conjugation of -electrons within the ring and the compound is not planar in shape. Hence, the given compound is not aromatic in nature. (ii) For the given compound, the number of -electrons is four. By Huckels rule, $4 n+2=4$ $4 n=2$ $n=\frac{1}{2}$ For a compound to be aromatic, th...
Read More →If f is a function satisfying such that , find the value of n.
Question: If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in \mathrm{N}$ such that $f(1)=3$ and $\sum_{x=1}^{n} f(x)=120$, find the value of $n$. Solution: It is given that, $f(x+y)=f(x) \times f(y)$ for all $x, y \in \mathrm{N}$ $f(1)=3$ Taking $x=y=1$ in (1), we obtain $f(1+1)=f(2)=f(1) f(1)=3 \times 3=9$ Similarly, $f(1+1+1)=f(3)=f(1+2)=f(1) f(2)=3 \times 9=27$ $f(4)=f(1+3)=f(1) f(3)=3 \times 27=81$ $\therefore f(1), f(2), f(3), \ldots$, that is $3,9,27, \ldots$, forms a G.P...
Read More →The radius of the innermost electron orbit of a hydrogen atom is
Question: The radius of the innermost electron orbit of a hydrogen atom is 5.3 1011m. What are the radii of then= 2 andn=3 orbits? Solution: The radius of the innermost orbit of a hydrogen atom,r1= 5.3 1011m. Letr2be the radius of the orbit atn= 2. It is related to the radius of the innermost orbit as: $r_{2}=(n)^{2} r_{1}$ $=4 \times 5.3 \times 10^{-11}=2.12 \times 10^{-10} \mathrm{~m}$ Forn= 3, we can write the corresponding electron radius as: $r_{3}=(n)^{2} r_{1}$ $=9 \times 5.3 \times 10^{-...
Read More →if find find $(A B)^{-1}$
Question: If $A^{-1}=\left[\begin{array}{crr}3 -1 1 \\ -15 6 -5 \\ 5 -2 2\end{array}\right]$ and $B=\left[\begin{array}{rrr}1 2 -2 \\ -1 3 0 \\ 0 -2 1\end{array}\right]$, find $(A B)^{-1}$ Solution: We know that $(A B)^{-1}=B^{-1} A^{-1}$. $B=\left[\begin{array}{rrc}1 2 -2 \\ -1 3 0 \\ 0 -2 1\end{array}\right]$ $\therefore|B|=1 \times 3-2 \times(-1)-2(2)=3+2-4=5-4=1$ Now, $A_{11}=3, A_{12}=1, A_{13}=2$ $A_{21}=2, A_{22}=1, A_{23}=2$ $A_{31}=6, A_{3,}=2, A_{33}=5$ $\therefore \operatorname{adj} B...
Read More →(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2,
Question: (a) Using the Bohrs model calculate the speed of the electron in a hydrogen atom in then= 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels. Solution: (a)Let1be the orbital speed of the electron in a hydrogen atom in the ground state level,n1= 1. For charge (e)of an electron,1is given by the relation, $v_{1}=\frac{e^{2}}{n_{1} 4 \pi \in_{0}(h / 2 \pi)}=\frac{e^{2}}{2 \in_{0} h}$ Where, e= 1.6 1019C 0= Permittivity of free space = 8.85 1012N1C2m2 h= Plancks co...
Read More →What are the necessary conditions for any system to be aromatic?
Question: What are the necessary conditions for any system to be aromatic? Solution: A compound is said to be aromatic if it satisfies the following three conditions: (i)It should have a planar structure. (ii)The electrons of the compound are completely delocalized in the ring. (iii)The total number of electrons present in the ring should be equal to (4n+ 2), wheren= 0, 1, 2 etc. This is known as Huckels rule....
Read More →Why is benzene extra ordinarily stable though it contains three double bonds?
Question: Why is benzene extra ordinarily stable though it contains three doublebonds? Solution: Benzene is a hybrid of resonating structures given as: All six carbon atoms in benzene aresp2hybridized. The twosp2hybrid orbitals of each carbon atom overlap with thesp2hybrid orbitals of adjacent carbon atoms to form six sigma bonds in the hexagonal plane. The remainingsp2hybrid orbital on each carbon atom overlaps with thes-orbital of hydrogen to form six sigma CH bonds. The remaining unhybridized...
Read More →Prove that
Question: Prove that $\left|\begin{array}{ccc}a^{2} b c a c+c^{2} \\ a^{2}+a b b^{2} a c^{2} \\ a b b^{2}+b c c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$ Solution: $\Delta=\left|\begin{array}{ccc}a^{2} b c a c+c^{2} \\ a^{2}+a b b^{2} a c \\ a b b^{2}+b c c^{2}\end{array}\right|$ Taking out common factors $a, b$, and $c$ from $\mathrm{C}_{1}, \mathrm{C}_{2}$, and $\mathrm{C}_{3}$, we have: $\Delta=a b c\left|\begin{array}{ccc}a c a+c \\ a+b b a \\ b b+c c\end{array}\right|$ Applying $R_{2} \rig...
Read More →A hydrogen atom initially in the ground level absorbs a photon,
Question: A hydrogen atom initially in the ground level absorbs a photon, which excites it to then= 4 level. Determine the wavelength and frequency of the photon. Solution: For ground level,n1= 1 LetE1be the energy of this level. It is known thatE1is related withn1as: $E_{1}=\frac{-13.6}{n_{1}{ }^{2}} \mathrm{eV}$ $=\frac{-13.6}{1^{2}}=-13.6 \mathrm{eV}$ The atom is excited to a higher level,n2= 4. LetE2be the energy of this level. $\therefore E_{2}=\frac{-13.6}{n_{2}{ }^{2}} \mathrm{eV}$ $=\fra...
Read More →Draw the cis and trans structures of hex-2-ene.
Question: Draw thecisandtransstructures of hex-2-ene. Which isomer will have higher b.p. and why? Solution: Hex-2-ene is represented as: $\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$ Geometrical isomers of hex-2-ene are: The dipole moment of cis-compound is a sum of the dipole moments of $\mathrm{C}-\mathrm{CH}_{3}$ and $\mathrm{C}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$ bonds acting in the same direction. The dipole moment of trans-compound i...
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