Prove the following by using the principle of mathematical induction for all n ∈ N:
Question: Prove the following by using the principle of mathematical induction for all $n \in N:\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{n}\right)=(n+1)$ Solution: Let the given statement be P(n), i.e., $\mathrm{P}(n):\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{n}\right)=(n+1)$ Forn= 1, we have $P(1):\left(1+\frac{1}{1}\right)=2=(1+1)$, which is true. Let $\mathrm{P}(k)$ be true...
Read More →Predict if the solutions of the following salts are neutral, acidic or basic:
Question: Predict if the solutions of the following salts are neutral, acidic or basic: $\mathrm{NaCl}, \mathrm{KBr}, \mathrm{NaCN}, \mathrm{NH}_{4} \mathrm{NO}_{3}, \mathrm{NaNO}_{2}$ and $\mathrm{KF}$ Solution: (i) NaCl: Therefore, it is a neutral solution. (ii) KBr: Therefore, it is a neutral solution. (iii) NaCN: Therefore, it is a basic solution. (iv) $\mathrm{NH}_{4} \mathrm{NO}_{3}$ Therefore, it is an acidic solution. (v) NaNO2 Therefore, it is a basic solution. (vi) KF Therefore, it is ...
Read More →Find the value of if $sin ^{-1} x=y$, then
Question: Find the value of if $\sin ^{-1} x=y$, then (A) $0 \leq y \leq \pi$ (B) $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$ (C) $0y\pi$ (D) $-\frac{\pi}{2}y\frac{\pi}{2}$ Solution: It is given that $\sin ^{-1} x=y$. We know that the range of the principal value branch of $\sin ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$....
Read More →Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart.
Question: Two tiny spheres carrying charges 1.5 C and 2.5 C are located 30 cm apart. Find the potential and electric field: (a)at the mid-point of the line joining the two charges, and (b)at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point. Solution: Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges. Magnitude of charge located at A,q1= 1.5 C Magnitude of charge located ...
Read More →Prove the following by using the principle of mathematical induction for all n ∈ N:
Question: Prove the following by using the principle of mathematical induction for allnN:$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{(2 n+1)}{n^{2}}\right)=(n+1)^{2}$ Solution: Let the given statement be P(n), i.e., $\mathrm{P}(n):\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{(2 n+1)}{n^{2}}\right)=(n+1)^{2}$ Forn= 1, we have $P(1):\left(1+\frac{3}{1}\right)=4=(1+1)^{2}=2^{2}=4$, which is...
Read More →Find the value of $\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$
Question: Find the value of $\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$ Solution: Let $\cos ^{-1}\left(\frac{1}{2}\right)=x$. Then, $\cos x=\frac{1}{2}=\cos \left(\frac{\pi}{3}\right)$. $\therefore \cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$ Let $\sin ^{-1}\left(\frac{1}{2}\right)=y$. Then, $\sin y=\frac{1}{2}=\sin \left(\frac{\pi}{6}\right)$. $\therefore \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$ $\therefore \cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{...
Read More →Prove the following by using the principle of mathematical induction for all n ∈ N:
Question: Prove the following by using the principle of mathematical induction for all $n \in N: a+a r+a r^{2}+\ldots+a r^{n-1}=\frac{a\left(r^{n}-1\right)}{r-1}$ Solution: Let the given statement be P(n), i.e., $\mathrm{P}(n): a+a r+a r^{2}+\ldots+a r^{n-1}=\frac{a\left(r^{n}-1\right)}{r-1}$ For $n=1$, we have $\mathrm{P}(1): a=\frac{a\left(r^{1}-1\right)}{(r-1)}=a$, which is true. Let P(k) be true for some positive integerk, i.e., $a+a r+a r^{2}+\ldots \ldots+a r^{k-1}=\frac{a\left(r^{k}-1\rig...
Read More →A 0.02 M solution of pyridinium hydrochloride has pH = 3.44.
Question: A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine Solution: $\mathrm{pH}=3.44$ We know that, $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$ $\therefore\left[\mathrm{H}^{+}\right]=3.63 \times 10^{-4}$ Then, $K_{h}=\frac{\left(3.63 \times 10^{-4}\right)^{2}}{0.02}$ $(\because$ concentration $=0.02 \mathrm{M})$ $\Rightarrow K_{h}=6.6 \times 10^{-6}$ Now, $K_{b}=\frac{K_{w}}{K_{a}}$ $\Rightarrow K_{a}=\frac{K_{w}}{K_{h}}=\frac{10^{-...
Read More →A cube of side b has a charge q at each of its vertices.
Question: A cube of sidebhas a chargeqat each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube. Solution: Length of the side of a cube =b Charge at each of its vertices =q A cube of sidebis shown in the following figure. d= Diagonal of one of the six faces of the cube $l^{2}=\sqrt{d^{2}+b^{2}}$ $d=b \sqrt{2}$ l = Length of the diagonal of the cube $l^{2}=\sqrt{d^{2}+b^{2}}$ $=\sqrt{(\sqrt{2} b)^{2}}+b^{2}=\sqrt{2 b^{2}+b^{2}}=\sqrt{3 ...
Read More →A charge of 8 mC is located at the origin.
Question: A charge of $8 \mathrm{mC}$ is located at the origin. Calculate the work done in taking a small charge of $-2 \times 10^{-9} \mathrm{C}$ from a point $\mathrm{P}(0,0,3 \mathrm{~cm})$ to a point $Q(0,4 \mathrm{~cm}, 0)$, via a point $R(0,6 \mathrm{~cm}, 9 \mathrm{~cm})$. Solution: Charge located at the origin, $q=8 \mathrm{mC}=8 \times 10^{-3} \mathrm{C}$ Magnitude of a small charge, which is taken from a point P to point R to point Q,q1= 2 109C All the points are represented in the giv...
Read More →The ionization constant of nitrous acid is 4.5 × 10–4.
Question: The ionization constant of nitrous acid is $4.5 \times 10^{-4}$ Calculate the $\mathrm{pH}$ of $0.04 \mathrm{M}$ sodium nitrite solution and also its degree of hydrolysis. Solution: NaNO2is the salt of a strong base (NaOH) and a weak acid (HNO2). Now, Ifxmoles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be: $\left[\mathrm{NO}_{2}^{-}\right]=.04-x ; 0.04$ $\left[\mathrm{HNO}_{2}\right]=x$ $\left[\mathrm{OH}^{-}\right]=x$ $K_{h}=...
Read More →Prove the following by using the principle of mathematical induction for all n ∈ N:
Question: Prove the following by using the principle of mathematical induction for allnN:$\frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ Solution: Let the given statement be P(n), i.e., $\mathrm{P}(n): \frac{1}{1.2 .3}+\frac{1}{2.3 .4}+\frac{1}{3.4 .5}+\ldots+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ Forn= 1, we have $P(1): \frac{1}{1 \cdot 2 \cdot 3}=\frac{1 \cdot(1+3)}{4(1+1)(1+2)}=\frac{1 \cdot 4}{4 \cdot 2 \cdot 3}=\frac{1}...
Read More →Find the value of Find the value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$
Question: Find the value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$ Solution: Let $\tan ^{-1}(1)=x$. Then, $\tan x=1=\tan \frac{\pi}{4}$. $\therefore \tan ^{-1}(1)=\frac{\pi}{4}$ Let $\cos ^{-1}\left(-\frac{1}{2}\right)=y$. Then, $\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)$. $\therefore \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}$ Let $\sin ^{-1}\left(-\frac{1}{...
Read More →A 600 pF capacitor is charged by a 200 V supply.
Question: A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process? Solution: Capacitance of the capacitor,C= 600 pF Potential difference,V= 200 V Electrostatic energy stored in the capacitor is given by, $E=\frac{1}{2} C V^{2}$ $=\frac{1}{2} \times\left(600 \times 10^{-12}\right) \times(200)^{2}$ $=1.2 \times 10^{-5} \mathrm{~J}$ If supply is disconnected ...
Read More →The pH of 0.1M solution of cyanic acid (HCNO) is 2.34.
Question: The $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of cyanic acid (HCNO) is $2.34$. Calculate the ionization constant of the acid and its degree of ionization in the solution. Solution: $c=0.1 \mathrm{M}$ $\mathrm{pH}=2.34$ $-\log \left[\mathrm{H}^{+}\right]=\mathrm{pH}$ $-\log \left[\mathrm{H}^{+}\right]=2.34$ $\left[\mathrm{H}^{+}\right]=4.5 \times 10^{-3}$ Also, $\left[\mathrm{H}^{+}\right]=c \alpha$ $4.5 \times 10^{-3}=0.1 \times \alpha$ $\frac{4.5 \times 10^{-3}}{0.1}=\alpha$ $\alpha=...
Read More →Find the principal value of
Question: Find the principal value of $\operatorname{cosec}^{-1}(-\sqrt{2})$ Solution: Let $\operatorname{cosec}^{-1}(-\sqrt{2})=y$. Then, $\operatorname{cosec} y=-\sqrt{2}=-\operatorname{cosec}\left(\frac{\pi}{4}\right)=\operatorname{cosec}\left(-\frac{\pi}{4}\right)$. We know that the range of the principal value branch of $\operatorname{cosec}^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$ and $\operatorname{cosec}\left(-\frac{\pi}{4}\right)=-\sqrt{2}$. Therefore, the principal v...
Read More →A 12 pF capacitor is connected to a 50V battery.
Question: A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor? Solution: Capacitor of the capacitance, $C=12 \mathrm{pF}=12 \times 10^{-12} \mathrm{~F}$ Potential difference,V= 50 V Electrostatic energy stored in the capacitor is given by the relation, $E=\frac{1}{2} C V^{2}$ $=\frac{1}{2} \times 12 \times 10^{-12} \times(50)^{2}$ $=1.5 \times 10^{-8} \mathrm{~J}$ Therefore, the electrostatic energy stored in the capacitor is $1.5 \times 10^{...
Read More →Find the principal value of
Question: Find the principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ Solution: Let $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=y$. Then, $\cos y=-\frac{1}{\sqrt{2}}=-\cos \left(\frac{\pi}{4}\right)=\cos \left(\pi-\frac{\pi}{4}\right)=\cos \left(\frac{3 \pi}{4}\right)$. We know that the range of the principal value branch of $\cos ^{-1}$ is $[0, \pi]$ and $\cos \left(\frac{3 \pi}{4}\right)=-\frac{1}{\sqrt{2}}$ Therefore, the principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\ri...
Read More →Find the principal value of
Question: Find the principal value of $\cot ^{-1}(\sqrt{3})$ Solution: Let $\cot ^{-1}(\sqrt{3})=y$. Then, $\cot y=\sqrt{3}=\cot \left(\frac{\pi}{6}\right)$. We know that the range of the principal value branch of $\cot ^{-1}$ is $(0, \pi)$ and $\cot \left(\frac{\pi}{6}\right)=\sqrt{3}$ Therefore, the principal value of $\cot ^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$....
Read More →The ionization constant of propanoic acid is 1.32 × 10–5.
Question: The ionization constant of propanoic acid is $1.32 \times 10^{-5}$. Calculate the degree of ionization of the acid in its $0.05 \mathrm{M}$ solution and also its pH. What will be its degree of ionization if the solution is $0.01 \mathrm{M}$ in $\mathrm{HCl}$ also? Solution: Let the degree of ionization of propanoic acid be . Then, representing propionic acid as HA, we have: In the presence of 0.1M of HCl, let be the degree of ionization. Then, $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right...
Read More →Explain what would happen if in the capacitor given in Exercise 2.8,
Question: Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a)While the voltage supply remained connected. (b)After the supply was disconnected Solution: (a)Dielectric constant of the mica sheet,k= 6 Initial capacitance, $C=1.771 \times 10^{-11} \mathrm{~F}$ New capacitance, $C^{\prime}=k C=6 \times 1.771 \times 10^{-11}=106 \mathrm{pF}$ Supply voltage, $V=100 \mathrm{~V}$ New charge, $q^{\...
Read More →Find the principal value of
Question: Find the principal value of $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ Solution: Let $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=y$. Then, $\sec y=\frac{2}{\sqrt{3}}=\sec \left(\frac{\pi}{6}\right)$. We know that the range of the principal value branch of $\mathrm{sec}^{-1}$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$ and $\sec \left(\frac{\pi}{6}\right)=\frac{2}{\sqrt{3}}$ Therefore, the principal value of $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi}{6}$....
Read More →In a parallel plate capacitor with air between the plates,
Question: In a parallel plate capacitor with air between the plates, each plate has an area of $6 \times 10^{-3} \mathrm{~m}^{2}$ and the distance between the plates is $3 \mathrm{~mm}$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100 \mathrm{~V}$ supply, what is the charge on each plate of the capacitor? Solution: Area of each plate of the parallel plate capacitor,A= 6 103m2 Distance between the plates,d= 3 mm = 3 103m Supply voltage,V= 100 V CapacitanceCof ...
Read More →The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution.
Question: The solubility of $\mathrm{Sr}(\mathrm{OH})_{2}$ at $298 \mathrm{~K}$ is $19.23 \mathrm{~g} / \mathrm{L}$ of solution. Calculate the concentrations of strontium and hydroxyl ions and the $\mathrm{pH}$ of the solution. Solution: Solubility of $\mathrm{Sr}(\mathrm{OH})_{2}=19.23 \mathrm{~g} / \mathrm{L}$ Then, concentration of $\mathrm{Sr}(\mathrm{OH})_{2}$ $=\frac{19.23}{121.63} \mathrm{M}$ $=0.1581 \mathrm{M}$ $\mathrm{Sr}(\mathrm{OH})_{2(a q)} \longrightarrow \mathrm{Sr}^{2+}{ }_{(a q...
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