Solve 3x + 8 > 2, when
Question: Solve $3 x+82$, when (i)xis an integer (ii)xis a real number Solution: The given inequality is $3 x+82$. $3 x+82$ $\Rightarrow 3 x+8-82-8$ $\Rightarrow 3 x-6$ $\Rightarrow \frac{3 x}{3}\frac{-6}{3}$ $\Rightarrow x-2$ (i) The integers greater than $-2$ are $-1,0,1,2, \ldots$ Thus, when $x$ is an integer, the solutions of the given inequality are $-1,0,1,2 \ldots$ Hence, in this case, the solution set is $\{-1,0,1,2, \ldots\}$. (ii) When $x$ is a real number, the solutions of the given i...
Read More →A long straight wire carries a current of 35 A.
Question: A long straight wire carries a current of 35 A. What is the magnitude of the fieldBat a point 20 cm from the wire? Solution: Current in the wire,I= 35 A Distance of a point from the wire,r= 20 cm = 0.2 m Magnitude of the magnetic field at this point is given as: $B=\frac{\mu_{0}}{4 \pi} \frac{2 I}{r}$ Where, $\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1}$ $B=\frac{4 \pi \times 10^{-7} \times 2 \times 35}{4 \pi \times 0.2}$ $=3.5 \t...
Read More →Solve 5x– 3 < 7, when
Question: Solve $5 x-37$, when (i) $x$ is an integer (ii) $x$ is a real number Solution: The given inequality is $5 x-37$. $5 x-37$ $\Rightarrow 5 x-3+37+3$ $\Rightarrow 5 x10$ $\Rightarrow \frac{5 x}{5}\frac{10}{5}$ $\Rightarrow x2$ (i) The integers less than 2 are $\ldots,-4,-3,-2,-1,0,1$. Thus, whenxis an integer, the solutions of the given inequality are , 4, 3, 2, 1, 0, 1. Hence, in this case, the solution set is $\{\ldots,-4,-3,-2,-1,0,1\}$. (ii) When $x$ is a real number, the solutions of...
Read More →Show that one and only one out of $n, n+4, n+8, n+12$ and $n+16$ is divisible by 5, where $n$ is any positive integer.
Question: Show that one and only one out of $n, n+4, n+8, n+12$ and $n+16$ is divisible by 5, where $n$ is any positive integer. Solution: Consider the numbers $n,(n+4),(n+8),(n+12)$ and $(n+16)$, where $n$ is any positive integer. Suppose $n=5 q+r$, where $0 \leq r5$ $n=5 q, 5 q+1,5 q+2,5 q+3,5 q+4$ (By Euclid's division algorithm) Case: 1 Whenn=5q. n=5qisdivisibleby5.n=5qisdivisibleby5. $n=5 q$ is divisible by 5 $n+4=5 q+4$ is not divisible by 5 $n+8=5 q+5+5+3=5(q+1)+3$ is not divisible by 5 ....
Read More →A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A.
Question: A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic fieldBat the centre of the coil? Solution: Number of turns on the circular coil,n= 100 Radius of each turn,r= 8.0 cm = 0.08 m Current flowing in the coil,I= 0.4 A Magnitude of the magnetic field at the centre of the coil is given by the relation, $|\mathbf{B}|=\frac{\mu_{0}}{4 \pi} \frac{2 \pi n I}{r}$ Where, $\mu_{0}=$ Permeability of free space $=4...
Read More →Solve –12x > 30, when
Question: Solve $-12 x30$, when (i)xis a natural number (ii)xis an integer Solution: The given inequality is $-12 x30$. $-12 x30$ $\Rightarrow \frac{-12 x}{-12}\frac{30}{-12} \quad$ [Dividing both sides by same negative number] $\Rightarrow x-\frac{5}{2}$ (i) There is no natural number less than $\left(-\frac{5}{2}\right)$. Thus, whenxis a natural number, there is no solution of the given inequality. (ii) The integers less than $\left(-\frac{5}{2}\right)$ are ..., $-5,-4,-3$. Thus, whenxis an in...
Read More →A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A.
Question: A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic fieldBat the centre of the coil? Solution: Number of turns on the circular coil,n= 100 Radius of each turn,r= 8.0 cm = 0.08 m Current flowing in the coil,I= 0.4 A Magnitude of the magnetic field at the centre of the coil is given by the relation, $|\mathbf{B}|=\frac{\mu_{0}}{4 \pi} \frac{2 \pi n I}{r}$ Where, $\mu_{0}=$ Permeability of free space $=4...
Read More →Solve
Question: Solve $\sin \left(\tan ^{-1} x\right),|x|1$ is equal to (A) $\frac{x}{\sqrt{1-x^{2}}}$ (B) $\frac{1}{\sqrt{1-x^{2}}}$ (C) $\frac{1}{\sqrt{1+x^{2}}}$ (D) $\frac{x}{\sqrt{1+x^{2}}}$ Solution: Let $\tan ^{-1} x=y$. Then, $\tan y=x \Rightarrow \sin y=\frac{x}{\sqrt{1+x^{2}}}$. $\therefore y=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right) \Rightarrow \tan ^{-1} x=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)$ $\therefore \sin \left(\tan ^{-1} x\right)=\sin \left(\sin ^{-1} \frac{x}{\sqrt{...
Read More →Solve 24x < 100,
Question: Solve 24x 100, when (i)xis a natural number (ii) $x$ is an integer Solution: The given inequality is $24 x100$. $24 x100$ $\Rightarrow \frac{24 x}{24}\frac{100}{24}$ [Dividing both sides by same positive number] (i) It is evident that $1,2,3$, and 4 are the only natural numbers less than $\frac{25}{6}$ Thus, whenxis a natural number, the solutions of the given inequality are 1, 2, 3, and 4. Hence, in this case, the solution set is {1, 2, 3, 4}. (ii) The integers less than $\frac{25}{6}...
Read More →Solve
Question: Solve $\tan ^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan ^{-1} x,(x0)$ Solution: $\tan ^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan ^{-1} x$ $\Rightarrow \tan ^{-1} 1-\tan ^{-1} x=\frac{1}{2} \tan ^{-1} x \quad\left[\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}\right]$ $\Rightarrow \frac{\pi}{4}=\frac{3}{2} \tan ^{-1} x$ $\Rightarrow \tan ^{-1} x=\frac{\pi}{6}$ $\Rightarrow x=\tan \frac{\pi}{6}$ $\therefore x=\frac{1}{\sqrt{3}}$...
Read More →Solve
Question: Solve $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$ Solution: $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$ $\Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)=\tan ^{-1}(2 \operatorname{cosec} x)$ $\left[2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right]$ $\Rightarrow \frac{2 \cos x}{1-\cos ^{2} x}=2 \operatorname{cosec} x$ $\Rightarrow \frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x}$ $\Rightarrow \cos x=\sin x$ $\Rightarrow \tan x=1$ $\there...
Read More →Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell.
Question: Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell. Solution: Internal resistance of the cell =r Balance point of the cell in open circuit,l1= 76.3 cm An external resistance (R) is connected...
Read More →Figure 3.34 shows a potentiometer circuit for comparison of two resistances.
Question: Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistorR= 10.0 Ω is found to be 58.3 cm, while that with the unknown resistanceXis 68.5 cm. Determine the value ofX. What might you do if you failed to find a balance point with the given cell of emf? Solution: Resistance of the standard resistor,R= 10.0 Ω Balance point for this resistance,l1= 58.3 cm Current in the potentiometer wire =i Hence, potential drop acrossR,E1=iR Re...
Read More →Balance the following redox reactions by ion-electron method:
Question: Balance the following redox reactions by ion-electron method: (a) $\mathrm{MnO}_{4}^{-}$(aq) $+\mathrm{I}^{-}$(aq) $\rightarrow \mathrm{MnO}_{2}$ (s) $+\mathrm{I}_{2}$ (s) (in basic medium) (b) $\mathrm{MnO}_{4}^{-}$(aq) $+\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{HSO}_{4}^{-}$(aq) (in acidic solution) (c) $\mathrm{H}_{2} \mathrm{O}_{2}$ (aq) $+\mathrm{Fe}^{2+}$ (aq) $\rightarrow \mathrm{Fe}^{3+}$ (aq) $+\mathrm{H}_{2} \mathrm{O}$ (I) (in acidic sol...
Read More →Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB.
Question: Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then repl...
Read More →Prove
Question: Prove $\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$ Solution: L.H.S. $=\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}$ $=\frac{9}{4}\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{3}\right)$ $=\frac{9}{4}\left(\cos ^{-1} \frac{1}{3}\right)$....(1) $\left[\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$ Now, let $\cos ^{-1} \frac{1}{3}=x$. Then, $\cos x=\frac{1}{3} \Rightarrow \sin x=\sqrt{1-\left(\frac{1}{3}\right)^{2}}=\frac{2 \sqrt{2}}{3}...
Read More →If, then find the least positive integral value of m.
Question: If $\left(\frac{1+i}{1-i}\right)^{m}=1$ then find the least positive integral value of $m$. Solution: $\left(\frac{1+i}{1-i}\right)^{m}=1$ $\Rightarrow\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{m}=1$ $\Rightarrow\left(\frac{(1+i)^{2}}{1^{2}+1^{2}}\right)^{m}=1$ $\Rightarrow\left(\frac{1^{2}+i^{2}+2 i}{2}\right)^{m}=1$ $\Rightarrow\left(\frac{1-1+2 i}{2}\right)^{m}=1$ $\Rightarrow\left(\frac{2 i}{2}\right)^{w}=1$ $\Rightarrow i^{m}=1$ $\therefore m=4 k$, where $k$ is some inte...
Read More →Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32.
Question: Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1 Ω resistance. Solution: The resistance of each resistor connected in the given circuit,R= 1 Ω Equivalent resistance of the given circuit =R The network is infinite. Hence, equivalent resistance is given by the relation, $\therefore R^{\prime}=2+\frac{R^{\prime}}{\left(R^{\prime}+1\right)}$ $\left(R^{\prime}\right)^{2}-2 R^{\prime}-2=0$ $R^{\prime...
Read More →Show that the cube of a positive integer is of the form $6 q+r$, where $q$ is ana integer and $r=0,1,2,3,4,5$.
Question: Show that the cube of a positive integer is of the form $6 q+r$, where $q$ is ana integer and $r=0,1,2,3,4,5$. Solution: Supposeabe any arbitrary positive integer, then by Euclid's division algorithm, corresponding to the positive integersaand 6, there exists non-negative integersaandrsuch that $a=6 q+r$, where $0 \leq r6$ $\Rightarrow a^{3}=(6 q+r)^{3}=216 q^{3}+r^{3}+3 \times 6 q \times r(6 q+r)$ $\Rightarrow a^{3}=6\left(216 q^{3}+108 q^{2} r+18 q r^{2}\right)+r^{3} \quad \ldots \ld...
Read More →If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that
Question: If $(a+i b)(c+i d)(e+i l)(g+i h)=A+i B$, then show that $\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)\left(e^{2}+f^{2}\right)\left(g^{2}+h^{2}\right)=A^{2}+B^{2}$ Solution: $(a+i b)(c+i d)(e+i f)(g+i h)=\mathrm{A}+i \mathrm{~B}$ $\therefore|(a+i b)(c+i d)(e+i f)(g+i h)|=|\mathrm{A}+i \mathrm{~B}|$ $\Rightarrow|(a+i b)| \times|(c+i d)| \times|(e+i f)| \times|(g+i h)|=|\mathrm{A}+i \mathrm{~B}| \quad\left[\left|z_{1} z_{2}\right|=\left|z_{1}\right|\left|z_{2}\right|\right]$ $\Rightar...
Read More →Prove
Question: Prove $\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}}\right)=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq 1$ [Hint: put $x=\cos 2 \theta$ ] Solution: Put $x=\cos 2 \theta$ so that $\theta=\frac{1}{2} \cos ^{-1} x$. Then, we have: L.H.S. $=\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$ $=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}\right)$ $=\tan ^{-1}\...
Read More →Find the number of non-zero integral solutions of the equation.
Question: Find the number of non-zero integral solutions of the equation$|1-i|^{x}=2^{x}$ Solution: $|1-i|^{x}=2^{x}$ $\Rightarrow\left(\sqrt{1^{2}+(-1)^{2}}\right)^{x}=2^{x}$ $\Rightarrow(\sqrt{2})^{x}=2^{x}$ $\Rightarrow 2^{\frac{x}{2}}=2^{x}$ $\Rightarrow \frac{x}{2}=x$ $\Rightarrow x=2 x$ $\Rightarrow 2 x-x=0$ $\Rightarrow x=0$ Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-zero integral solutions of the given equation is 0....
Read More →If α and β are different complex numbers with = 1, then find.
Question: If and are different complex numbers with$|\beta|=1$, then find $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$. Solution: Let $\alpha=a+i b$ and $\beta=x+i y$ It is given that, $|\beta|=1$ $\therefore \sqrt{x^{2}+y^{2}}=1$ $\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=1$ $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|=\left|\frac{(x+i y)-(a+i b)}{1-(a-i b)(x+i y)}\right|$ $=\left|\frac{(x-a)+i(y-b)}{1-(a x+a i y-i b x+b y)}\right|$ $=\left|\frac{(x-a)+i(y-b)}{(1-a x-b y)...
Read More →Prove
Question: Prove $\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}, x \in\left(0, \frac{\pi}{4}\right)$ Solution: Consider $\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$ $=\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x})^{2}-(\sqrt{1-\sin x})^{2}}$ (by rationalizing) $=\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1+\sin x)(1-\sin x)}}{1+\sin x-1+\sin x}$ $=\frac{2\left(1+\sqrt{1-\sin ^{2} x}\right)}{2 \sin ...
Read More →If (x + iy)3 = u + iv, then show that.
Question: If $(x+i y)^{3}=u+i v$, then show that $\frac{u}{x}+\frac{y}{y}=4\left(x^{2}-y^{2}\right)$. Solution: $(x+i y)^{3}=u+i v$ $\Rightarrow x^{3}+(i y)^{3}+3 \cdot x \cdot i y(x+i y)=u+i v$ $\Rightarrow x^{3}+i^{3} y^{3}+3 x^{2} y i+3 x y^{2} i^{2}=u+i v$ $\Rightarrow x^{3}-i y^{3}+3 x^{2} y i-3 x y^{2}=u+i v$ $\Rightarrow\left(x^{3}-3 x y^{2}\right)+i\left(3 x^{2} y-y^{3}\right)=u+i v$ On equating real and imaginary parts, we obtain $u=x^{3}-3 x y^{2}, v=3 x^{2} y-y^{3}$ $\therefore \frac{...
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