Examine the consistency of the system of equations.
Question: Examine the consistency of the system of equations. 2xy= 5 x+y= 4 Solution: The given system of equations is: 2xy= 5 x+y= 4 The given system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{cc}2 -1 \\ 1 1\end{array}\right], X=\left[\begin{array}{l}x \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}5 \\ 4\end{array}\right]$ Now, $|A|=2(1)-(-1)(1)=2+1=3 \neq 0$ $\therefore A$ is non-singular. Therefore, $A^{-1}$ exists. Hence, the given system of equations...
Read More →What is the de Broglie wavelength of
Question: What is the de Broglie wavelength of (a)a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, (b)a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and (c)a dust particle of mass 1.0 109kg drifting with a speed of 2.2 m/s? Solution: (a)Mass of the bullet,m= 0.040 kg Speed of the bullet,v= 1.0 km/s = 1000 m/s Plancks constant,h= 6.6 1034Js De Broglie wavelength of the bullet is given by the relation: $\lambda=\frac{h}{m v}$ $=\frac{6.6 \times 10^{-34}}{0.040 \times 1000}=1.6...
Read More →Examine the consistency of the system of equations
Question: Examine the consistency of the system of equations $x+2 y=2$ $2 x+3 y=3$ Solution: The given system of equations is: $x+2 y=2$ $2 x+3 y=3$ The given system of equations can be written in the form ofAX=B, where $A=\left[\begin{array}{ll}1 2 \\ 2 3\end{array}\right], X=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $B=\left[\begin{array}{l}2 \\ 3\end{array}\right]$ Now, $|A|=1(3)-2(2)=3-4=-1 \neq 0$ $\therefore A$ is non-singular. Therefore, $A^{-1}$ exists. Hence, the given system ...
Read More →The wavelength of light from the spectral emission line of sodium is 589 nm.
Question: The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a)an electron, and (b)a neutron, would have the same de Broglie wavelength. Solution: Wavelength of light of a sodium line,= 589 nm = 589 109m Mass of an electron,me= 9.1 1031kg Mass of a neutron,mn= 1.66 1027kg Plancks constant,h= 6.6 1034Js (a)For the kinetic energyK, of an electron accelerating with a velocityv, we have the relation: $K=\frac{1}{2} m_{e} v^{2}$ ....(1) We h...
Read More →Differentiate between the principle of estimation of nitrogen in an organic compound by
Question: Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahls method. Solution: In Dumas method, a known quantity of nitrogen containing organic compound is heated strongly with excess of copper oxide in an atmosphere of carbon dioxide to produce free nitrogen in addition to carbon dioxide and water. The chemical equation involved in the process can be represented as $\mathrm{CxH} y \mathrm{Nz}+(2 x+y / 2) \mathrm{CuO} \long...
Read More →Find the value of n so that may be the geometric mean between a and b.
Question: Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$ may be the geometric mean between $a$ and $b$. Solution: G. M. of $a$ and $b$ is $\sqrt{a b}$. By the given condition, $\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=\sqrt{a b}$ Squaring both sides, we obtain $\frac{\left(a^{n+1}+b^{n+1}\right)^{2}}{\left(a^{n}+b^{n}\right)^{2}}=a b$ $\Rightarrow a^{2 n+2}+2 a^{n+1} b^{n+1}+b^{2 n+2}=(a b)\left(a^{2 n}+2 a^{n} b^{n}+b^{2 n}\right)$ $\Rightarrow a^{2 n+2}+2 a^{n+1} b^{n+1}+b^{2 n...
Read More →If A is an invertible matrix of order 2, then det (A−1) is equal to
Question: IfAis an invertible matrix of order 2, then det (A1) is equal to A. $\operatorname{det}(A)$ B. $\frac{1}{\operatorname{det}(A)}$ C. 1 D. 0 Solution: Since $A$ is an invertible matrix, $A^{-1}$ exists and $A^{-1}=\frac{1}{|A|}$ adjA. As matrix $A$ is of order 2, let $A=\left[\begin{array}{ll}a b \\ c d\end{array}\right]$. Then, $|A|=a d-b c$ and adj $A=\left[\begin{array}{cc}d -b \\ -c a\end{array}\right]$. Now, $A^{-1}=\frac{1}{|A|}$ adjA $=\left[\begin{array}{cc}\frac{d}{|A|} \frac{-b...
Read More →What is the
Question: What is the (a)momentum, (b)speed, and (c)de Broglie wavelength of an electron with kinetic energy of 120 eV. Solution: Kinetic energy of the electron,Ek= 120 eV Plancks constant,h= 6.6 1034Js Mass of an electron,m= 9.1 1031kg Charge on an electron,e= 1.6 1019C (a)For the electron, we can write the relation for kinetic energy as: $E_{k}=\frac{1}{2} m v^{2}$ Where, v= Speed of the electron $\therefore v^{2}=\sqrt{\frac{2 e E_{k}}{m}}$ $=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 12...
Read More →Discuss the chemistry of Lassaigne’s test.
Question: Discuss the chemistry of Lassaignes test. Solution: Lassaignes test This test is employed to detect the presence of nitrogen, sulphur, halogens, and phosphorous in an organic compound. These elements are present in the covalent form in an organic compound. These are converted into the ionic form by fusing the compound with sodium metal. $\mathrm{Na}+\mathrm{C}+\mathrm{N} \stackrel{\Delta}{\longrightarrow} \mathrm{NaCN}$ $2 \mathrm{Na}+\mathrm{S} \stackrel{\Delta}{\longrightarrow} \math...
Read More →Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Question: Insert two numbers between 3 and 81 so that the resulting sequence is G.P. Solution: Let $G_{1}$ and $G_{2}$ be two numbers between 3 and 81 such that the series, $3, G_{1}, G_{2}, 81$, forms a G.P. Let $a$ be the first term and $r$ be the common ratio of the G.P. $\therefore 81=(3)(r)^{3}$ $\Rightarrow r^{3}=27$ $\therefore r=3$ (Taking real roots only) For $r=3$, $G_{1}=a r=(3)(3)=9$ $G_{2}=a r^{2}=(3)(3)^{2}=27$ Thus, the required two numbers are 9 and 27 ....
Read More →Let A be a nonsingular square matrix of order 3 × 3. Then is equal to
Question: LetAbe a nonsingular square matrix of order 3 3. Thenis equal to A. $|A|$ B. $|A|^{2}$ C. $|A|^{3}$ D. $3|A|$ Solution: Answer: B We know that, $(\operatorname{adj} A) A=|A| I=\left[\begin{array}{ccc}|A| 0 0 \\ 0 |A| 0 \\ 0 0 |A|\end{array}\right]$ $\Rightarrow|(\operatorname{adj} A) A|=\left|\begin{array}{lll}A \mid 0 0 \\ 0 |A| 0 \\ 0 0 |A|\end{array}\right|$ $\left.\Rightarrow|\operatorname{adj} A| A|=| A\right|^{3}\left|\begin{array}{ccc}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right|=|A...
Read More →If a, b, c and d are in G.P. show that
Question: Ifa, b, canddare in G.P. show that$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2} .$ Solution: a,b,c,dare in G.P. Therefore, bc=ad (1) $b^{2}=a c \ldots(2)$ $c^{2}=b d \ldots(3)$ It has to be proved that, $\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c-c d)^{2}$ R.H.S. $=(a b+b c+c d)^{2}$ $=(a b+a d+c d)^{2}[\cup \sin g(1)]$ $=[a b+d(a+c)]^{2}$ $=a^{2} b^{2}+2 a b d(a+c)+d^{2}(a+c)^{2}$ $=a^{2} b^{2}+2 a^{2} b d+2 a c b d+d^{2}\le...
Read More →Calculate the
Question: Calculate the (a)momentum, and (b)de Broglie wavelength of the electrons accelerated through a potential difference of 56 V. Solution: Potential difference,V= 56 V Planck's constant, $h=6.6 \times 10^{-34} \mathrm{Js}$ Mass of an electron, $m=9.1 \times 10^{-31} \mathrm{~kg}$ Charge on an electron, $e=1.6 \times 10^{-19} \mathrm{C}$ (a)At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each ele...
Read More →If verify that A3 − 6A2 + 9A − 4I = O and hence find A−1
Question: If $A=\left[\begin{array}{ccc}2 -1 1 \\ -1 2 -1 \\ 1 -1 2\end{array}\right]$ verify that $A^{3}-6 A^{2}+9 A-4 I=0$ and hence find $A^{-1}$ Solution: $A=\left[\begin{array}{ccc}2 -1 1 \\ -1 2 -1 \\ 1 -1 2\end{array}\right]$ $A^{2}=\left[\begin{array}{ccr}2 -1 1 \\ -1 2 -1 \\ 1 -1 2\end{array}\right]\left[\begin{array}{ccc}2 -1 1 \\ -1 2 -1 \\ 1 -1 2\end{array}\right]$ $=\left[\begin{array}{lll}4+1+1 -2-2-1 2+1+2 \\ -2-2-1 1+4+1 -1-2-2 \\ 2+1+2 -1-2-2 1+1+4\end{array}\right]$ $=\left[\be...
Read More →Calculate the
Question: Calculate the (a)momentum, and (b)de Broglie wavelength of the electrons accelerated through a potential difference of 56 V. Solution: Potential difference,V= 56 V Planck's constant, $h=6.6 \times 10^{-34} \mathrm{Js}$ Mass of an electron, $m=9.1 \times 10^{-31} \mathrm{~kg}$ Charge on an electron, $e=1.6 \times 10^{-19} \mathrm{C}$ (a)At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each ele...
Read More →Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from
Question: Show that the ratio of the sum of firstnterms of a G.P. to the sum of terms from$(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term is $\frac{1}{r^{n}}$. Solution: Let $a$ be the first term and $r$ be the common ratio of the G.P. Sum of first $\mathrm{n}$ terms $=\frac{\mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right)}{(1-\mathrm{r})}$ Since there are $n$ terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term, Sum of terms from $(n+1)^{\text {th }}$ to $(2 n)^{\text {th }}$ term...
Read More →If the first and the nth term of a G.P. are a ad b, respectively,
Question: If the first and the $n^{\text {th }}$ term of a G.P. are $a$ ad $b$, respectively, and if $P$ is the product of $n$ terms, prove that $P^{2}=(a b)^{n}$. Solution: The first term of the G.P isaand the last term isb. Therefore, the G.P. is $a, a r, a r^{2}, a r^{3}, \ldots a r^{n-1}$, where $r$ is the common ratio. $b=a r^{n-1} \ldots(1)$ $P=$ Product of $n$ terms $=(a)(a r)\left(a r^{2}\right) \ldots\left(a r^{n-1}\right)$ $=(a \times a \times \ldots a)\left(r \times r^{2} \times \ldot...
Read More →Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect.
Question: Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made. Solution: Wavelength of light produced by the argon laser,= 488 nm = 488 109m Stopping potential of the photoelectrons,V0= 0.38 V 1eV = 1.6 1019J $\therefore V_{0}=\frac{0.38}{1.6 \times 10^{-19...
Read More →If the terms of a G.P. are a, b and c, respectively.
Question: If the $p^{b \prime}, q^{t h}$ and $r^{\prime \prime}$ terms of a G.P. are $a, b$ and $c$, respectively. Prove that $a^{a-r} b^{t-p} c^{p-q}=1$ Solution: LetAbe the first term andRbe the common ratio of the G.P. According to the given information, $A R^{p-1}=a$ $A R^{q-1}=b$ $A R^{r-1}=c$ $a^{q-r} b^{r-p} c^{p-q}$ $=A^{q-r} \times R^{(p-1)(q-r)} \times A^{r-p} \times R^{(q-1)(r-p)} \times A^{p-q} \times R^{(r-1)(p-q)}$ $=A q^{-r+r-p+p-q} \times R^{(p r-p r-q+r)+(r q-r+p-p q)+(p r-p-q r...
Read More →Light of frequency
Question: Light of frequency $7.21 \times 10^{14} \mathrm{~Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^{5} \mathrm{~m} / \mathrm{s}$ are ejected from the surface. What is the threshold frequency for photoemission of electrons? Solution: Frequency of the incident photon, $v=488 \mathrm{~nm}=488 \times 10^{-9} \mathrm{~m}$ Maximum speed of the electrons,v= 6.0105m/s Plancks constant,h= 6.6261034Js Mass of an electron,m= 9.11031kg For threshold frequency0, ...
Read More →Find four numbers forming a geometric progression in which third term is greater than the first term by 9,
Question: Find four numbers forming a geometric progression in which third term is greater than the first term by 9,and the second term is greater than the 4th by 18. Solution: Letabe thefirst term andrbe the common ratio of the G.P. $a_{1}=a, a_{2}=a r, a_{3}=a r^{2}, a_{4}=a r^{3}$ By the given condition, $a_{3}=a_{1}+9$ $\Rightarrow a r^{2}=a+9 \ldots(1)$ $a_{2}=a_{4}+18$ $\Rightarrow a r=a r^{3}+18 \ldots(2)$ From $(1)$ and $(2)$, we obtain $a\left(r^{2}-1\right)=9 \ldots(3)$ $\operatorname{...
Read More →For the matrixshow that A3 − 6A2 + 5A + 11 I = O. Hence, find A−1.
Question: For the matrix $A=\left[\begin{array}{ccc}1 1 1 \\ 1 2 -3 \\ 2 -1 3\end{array}\right]$ show that $A^{3}-6 A^{2}+5 A+11 I=0 .$ Hence, find $A^{-1}$ Solution: $A=\left[\begin{array}{ccc}1 1 1 \\ 1 2 -3 \\ 2 -1 3\end{array}\right]$ $A^{2}=\left[\begin{array}{ccc}1 1 1 \\ 1 2 -3 \\ 2 -1 3\end{array}\right]\left[\begin{array}{ccc}1 1 1 \\ 1 2 -3 \\ 2 -1 3\end{array}\right]$ $=\left[\begin{array}{ccc}1+1+2 1+2-1 1-3+3 \\ 1+2-6 1+4+3 1-6-9 \\ 2-1+6 2-2-3 2+3+9\end{array}\right]=\left[\begin{a...
Read More →For the matrixshow that A3 − 6A2 + 5A + 11 I = O. Hence, find A−1.
Question: For the matrix $A=\left[\begin{array}{ccc}1 1 1 \\ 1 2 -3 \\ 2 -1 3\end{array}\right]$ show that $A^{3}-6 A^{2}+5 A+11 I=0 .$ Hence, find $A^{-1}$ Solution: $A=\left[\begin{array}{ccc}1 1 1 \\ 1 2 -3 \\ 2 -1 3\end{array}\right]$ $A^{2}=\left[\begin{array}{ccc}1 1 1 \\ 1 2 -3 \\ 2 -1 3\end{array}\right]\left[\begin{array}{ccc}1 1 1 \\ 1 2 -3 \\ 2 -1 3\end{array}\right]$ $=\left[\begin{array}{ccc}1+1+2 1+2-1 1-3+3 \\ 1+2-6 1+4+3 1-6-9 \\ 2-1+6 2-2-3 2+3+9\end{array}\right]=\left[\begin{a...
Read More →The work function for a certain metal is 4.2 eV.
Question: The work function for a certain metal is 4.2 eV. Will this metal givephotoelectric emission for incident radiation of wavelength 330 nm? Solution: No Work function of the metal, $\phi_{0}=4.2 \mathrm{eV}$ Charge onan electron,e= 1.61019C Plancksconstant,h= 6.6261034Js Wavelength of the incident radiation,= 330 nm = 330 109m Speed of light,c= 3108m/s The energy of the incident photon is given as: $E=\frac{h c}{\lambda}$ $=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{330 \times 10...
Read More →The work function for a certain metal is 4.2 eV.
Question: The work function for a certain metal is 4.2 eV. Will this metal givephotoelectric emission for incident radiation of wavelength 330 nm? Solution: No Work function of the metal, $\phi_{0}=4.2 \mathrm{eV}$ Charge onan electron,e= 1.61019C Plancksconstant,h= 6.6261034Js Wavelength of the incident radiation,= 330 nm = 330 109m Speed of light,c= 3108m/s The energy of the incident photon is given as: $E=\frac{h c}{\lambda}$ $=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{330 \times 10...
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