Solve the following
Question: For the cell $\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(\mathrm{aq}) \| \mathrm{M}^{\mathrm{x}+}(\mathrm{aq})\right| \mathrm{M}(\mathrm{s})$, different half cells and their standard electrode potentials are given below: If $\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\mathrm{O}}=-0.76 \mathrm{~V}$, which cathode will give a maximum value of $E_{\text {cell }}^{\circ}$ per electron transferred?$\mathrm{Ag}^{+} / \mathrm{Ag}$$\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$$\mathrm{Au}^{3+} / \...
Read More →The value of
Question: The value of $\left|\begin{array}{lll}(a+1)(a+2) a+2 1 \\ (a+2)(a+3) a+3 1 \\ (a+3)(a+4) a+4 1\end{array}\right|$ is(1) $-2$(2) $(a+1)(a+2)(a+3)$ (3) 0(4) $(a+2)(a+3)(a+4)$Correct Option: 1 Solution: $\mathrm{C}_{1} \rightarrow \mathrm{c}_{1}-\mathrm{c}_{2}, \mathrm{c}_{2} \rightarrow \mathrm{c}_{2}-\mathrm{c}_{3}$ $=\left|\begin{array}{ccc}(a+2) a a+1 1 \\ (a+3)(a+1) a+2 1 \\ (a+4)(a+2) a+3 1\end{array}\right|$ $R_{2} \rightarrow R_{2}-R_{1} \quad \ R_{3} \rightarrow R_{3}-R_{1}$ $=\l...
Read More →Divide 16 into two parts such that twice the squares of the larger part exceeds the square of the smaller part by 164.
Question: Divide 16 into two parts such that twice the squares of the larger part exceeds the square of the smaller part by 164. Solution: Let the larger and smaller parts be $x$ and $y$, respectively. According to the question: $x+y=16 \ldots$ (i) $2 x^{2}=y^{2}+164 \ldots$ (ii) From (i), we get: $x=16-y \ldots$ (iii) From (ii) and (iii), we get: $2(16-y)^{2}=y^{2}+164$ $\Rightarrow 2\left(256-32 y+y^{2}\right)=y^{2}+164$ $\Rightarrow 512-64 y+2 y^{2}=y^{2}+164$ $\Rightarrow y^{2}-64 y+348=0$ $...
Read More →Prove the following
Question: Let $\mathrm{A}$ be a $3 \times 3$ matrix with $\operatorname{det}(\mathrm{A})=4$. Let $\mathrm{R}_{\mathrm{i}}$ denote the $\mathrm{i}^{\text {th }}$ row of $\mathrm{A}$. If a matrix $\mathrm{B}$ is obtained by performing the operation $\mathrm{R}_{2} \rightarrow 2 \mathrm{R}_{2}+5 \mathrm{R}_{3}$ on $2 \mathrm{~A}$, then $\operatorname{det}(\mathrm{B})$ is equal to:(1) 64(2) 16(3) 80(4) 128Correct Option: 1 Solution: $A=\left[\begin{array}{lll}\mathrm{R}_{11} \mathrm{R}_{12} \mathrm{...
Read More →Solve the following
Question: In the cell $\mathrm{Pt}(\mathrm{s}) \mid \mathrm{H}_{2}(\mathrm{~g}, 1$ bar $) / \mathrm{HCl}(\mathrm{aq}) \| \mathrm{AgCl}(\mathrm{s}) / \mathrm{Ag}(\mathrm{s}) \mid \mathrm{Pt}(\mathrm{s})$, the cell potential is $0.92 \mathrm{~V}$ when a $10^{-6}$ molal $\mathrm{HCl}$ solution is used. The standard electrode potential of $\left(\mathrm{AgCl} / \mathrm{Ag}, \mathrm{Cl}^{-}\right)$ electrode is: $\left\{\right.$ Given $: \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$ at $\lef...
Read More →A satellite is in an elliptical orbit around a planet P.
Question: A satellite is in an elliptical orbit around a planet $P$. It is observed that the velocity of the satellite when it is farthest from the planet is 6 times less than that when it is closest to the planet. The ratio of distances between the satellite and the planet at closest and farthest points is :(1) $1: 6$(2) $1: 3$(3) $1: 2$(4) $3: 4$Correct Option: 1 Solution: (1) By angular momentum conservation $r_{\min } v_{\max }=m r_{\max } v_{\min }$ Given, $v_{\min }=\frac{v_{\max }}{6}$ $\...
Read More →The following system of linear equations
Question: The following system of linear equations $3 x+3 y+2 z=9$ $3 x+2 y+2 z=9$ $x-y+4 z=8$ (1) does not have any solution(2) has a unique solution(3) has a solution $(\alpha, \beta, \gamma)$ satisfying $\alpha+\beta^{2}+\gamma^{3}=12$(4) has infinitely many solutionsCorrect Option: , 2 Solution: $\Delta=\left|\begin{array}{ccc}2 3 2 \\ 3 2 2 \\ 1 -1 4\end{array}\right|=-20 \neq 0 \quad \therefore$ unique solution $\Delta_{x}=\left|\begin{array}{ccc}9 3 2 \\ 9 2 2 \\ 8 -1 4\end{array}\right|=...
Read More →Divide 27 into two parts such that the sum of their reciprocals is
Question: Divide 27 into two parts such that the sum of their reciprocals is $\frac{3}{20}$. Solution: Let the two parts bexand (27 x).According to the given condition, $\frac{1}{x}+\frac{1}{27-x}=\frac{3}{20}$ $\Rightarrow \frac{27-x+x}{x(27-x)}=\frac{3}{20}$ $\Rightarrow \frac{27}{27 x-x^{2}}=\frac{3}{20}$ $\Rightarrow 27 x-x^{2}=180$ $\Rightarrow x^{2}-27 x+180=0$ $\Rightarrow x^{2}-15 x-12 x+180=0$ $\Rightarrow x(x-15)-12(x-15)=0$ $\Rightarrow(x-12)(x-15)=0$ $\Rightarrow x-12=0$ or $x-15=0$ ...
Read More →The acceleration due to gravity on the earth's surface at the poles is g
Question: The acceleration due to gravity on the earth's surface at the poles is $g$ and angular velocity of the earth about the axis passing through the pole is $\omega$. An object is weighed at the equator and at a height $h$ above the poles by using a spring balance. If the weights are found to be same, then $h$ is : $(hR$, where $R$ is the radius of the earth)(1) $\frac{R^{2} \omega^{2}}{2 g}$(2) $\frac{R^{2} \omega^{2}}{g}$(3) $\frac{R^{2} \omega^{2}}{4 g}$(4) $\frac{R^{2} \omega^{2}}{8 g}$...
Read More →If the standard electrode potential for a cell is
Question: If the standard electrode potential for a cell is $2 \mathrm{~V}$ at 300 $\mathrm{K}$, the equilibrium constant $(\mathrm{K})$ for the reaction $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$ at $300 \mathrm{~K}$ is approximately $\left(\mathrm{R}=8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~F}=96000 \mathrm{C} \mathrm{mol}^{-1}\right)$$\mathrm{e}^{-80}$$\mathrm{e}^{-160}$$e^{320}$$e^{160}$Correct Optio...
Read More →If for the matrix,
Question: If for the matrix, $\mathrm{A}=\left[\begin{array}{cc}1 -\alpha \\ \alpha \beta\end{array}\right], \mathrm{AA}^{\top}=\mathrm{I}_{2}$, then the value of $\alpha^{4}+\beta^{4}$ is: (1) 1(2) 3(3) 2(4) 4Correct Option: 1 Solution: $\left[\begin{array}{cc}1 -\alpha \\ \alpha \beta\end{array}\right]\left[\begin{array}{cc}1 \alpha \\ -\alpha \beta\end{array}\right]=\left[\begin{array}{cc}1+\alpha^{2} \alpha-\alpha \beta \\ \alpha-\alpha \beta \alpha^{2}+\beta^{2}\end{array}\right]=\left[\beg...
Read More →Divide 57 into two parts whose product is 680.
Question: Divide 57 into two parts whose product is 680. Solution: Let the two parts bexand (57 x).According to the given condition, $x(57-x)=680$ $\Rightarrow 57 x-x^{2}=680$ $\Rightarrow x^{2}-57 x+680=0$ $\Rightarrow x^{2}-40 x-17 x+680=0$ $\Rightarrow x(x-40)-17(x-40)=0$ $\Rightarrow(x-40)(x-17)=0$ $\Rightarrow x-40=0$ or $x-17=0$ $\Rightarrow x=40$ or $x=17$ Whenx= 40,57 x= 57 40 = 17Whenx= 17,57 x= 57 17 = 40Hence, the required parts are 17 and 40....
Read More →Prove the following
Question: Let $A=\left[\begin{array}{lll}x y z \\ y z x \\ z x y\end{array}\right]$, where $x, y$ and $z$ are real numbers such that $x+y+z^{2}0$ and $x y z=2$. If $A^{2}=I_{3}$, then the value of $x^{3}+y^{3}+z^{3}$ is Solution: $\mathrm{A}=\left[\begin{array}{ccc}\mathrm{x} \mathrm{y} \mathrm{z} \\ \mathrm{y} \mathrm{z} \mathrm{x} \\ \mathrm{z} \mathrm{x} \mathrm{y}\end{array}\right] \quad \therefore|\mathrm{A}|=\left(\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xyz}\right)$ $\mathrm...
Read More →The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday.
Question: The anodic half-cell of lead-acid battery is recharged using electricity of $0.05$ Faraday. The amount of $\mathrm{PbSO}_{4}$ electrolyzed in $g$ during the process is : (Molar mass of $\mathrm{PbSO}_{4}=303 \mathrm{~g} \mathrm{~mol}^{-1}$ )$22.8$$15.2$$7.6$$11.4$Correct Option: , 3 Solution: Half cell reaction: $\mathrm{PbSO}_{4} \rightarrow \mathrm{Pb}^{4+}+2 \mathrm{e}^{-}$ According to the reaction: $\mathrm{PbSO}_{4} \rightarrow \mathrm{Pb}^{4+}+2 \mathrm{e}^{-}$ We require $2 \ma...
Read More →The value of acceleration due to gravity
Question: The value of acceleration due to gravity is $g_{1}$ at a height $h=\frac{R}{2}(R=$ radius of the earth $)$ from the surface of the earth. It is again equal to $g_{1}$ and a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals :(1) $\frac{4}{9}$(2) $\frac{5}{9}$(3) $\frac{1}{3}$(4) $\frac{7}{9}$Correct Option: , 2 Solution: (2) According to question, $g_{h}=g_{d}=g_{1}$ $g_{h}=\frac{G M}{\left(R+\frac{R}{2}\right)^{2}}$ and $g_{d}=\frac{G M(R-d)}{R^{3}}$...
Read More →The sum of a natural number and its reciprocal is
Question: The sum of a natural number and its reciprocal is $\frac{65}{8}$. Find the number. Solution: Let the natural number bex.According to the given condition, $x+\frac{1}{x}=\frac{65}{8}$ $\Rightarrow \frac{x^{2}+1}{x}=\frac{65}{8}$ $\Rightarrow 8 x^{2}+8=65 x$ $\Rightarrow 8 x^{2}-65 x+8=0$ $\Rightarrow 8 x^{2}-64 x-x+8=0$ $\Rightarrow 8 x(x-8)-1(x-8)=0$ $\Rightarrow(x-8)(8 x-1)=0$ $\Rightarrow x-8=0$ or $8 x-1=0$ $\Rightarrow x=8$ or $x=\frac{1}{8}$ x= 8 (xis a natural number)Hence, the r...
Read More →If the system of equations
Question: If the system of equations $k x+y+2 z=1$ $3 x-y-2 z=2$ $-2 x-2 y-4 z=3$ has infinitely many solutions, then $\mathrm{k}$ is equal to Solution: $D=0$ $\Rightarrow\left|\begin{array}{ccc}\mathrm{k} 1 2 \\ 3 -1 -2 \\ -2 -2 -4\end{array}\right|=0$ $\Rightarrow \mathrm{k}(4-4)-1(-12-4)+2(-6-2)$ $\Rightarrow 16-16=0$ Also. $\quad \mathrm{D}_{1}=\mathrm{D}_{2}=\mathrm{D}_{3}=0$ $\Rightarrow \mathrm{D}_{2}=\left|\begin{array}{ccc}\mathrm{k} 1 2 \\ 3 2 -2 \\ -2 3 -4\end{array}\right|=0$ $\Right...
Read More →oxidizing power of the species will increase in the order:
Question: Given: $\mathrm{Co}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Co}^{2+} ; \mathrm{E}^{\mathrm{o}}=+1.81 \mathrm{~V}$ $\mathrm{~Pb}^{4+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}^{2+} ; \mathrm{E}^{\mathrm{o}}=+1.67 \mathrm{~V}$ $\mathrm{Ce}^{4+}+\mathrm{e}^{-} \rightarrow \mathrm{Ce}^{3+} ; \mathrm{E}^{0}=+1.61 \mathrm{~V}$ $\mathrm{Bi}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Bi} ; \mathrm{E}^{\mathrm{o}}=+0.20 \mathrm{~V}$ oxidizing power of the species will increase in the order:$\math...
Read More →The sum of the squares of two consecutive multiples of 7 is 1225
Question: The sum of the squares of two consecutive multiples of 7 is 1225. Find the multiples. Solution: Let the required consecutive multiples of 7 be 7xand 7(x+ 1).According to the given condition, $(7 x)^{2}+[7(x+1)]^{2}=1225$ $\Rightarrow 49 x^{2}+49\left(x^{2}+2 x+1\right)=1225$ $\Rightarrow 49 x^{2}+49 x^{2}+98 x+49=1225$ $\Rightarrow 98 x^{2}+98 x-1176=0$ $\Rightarrow x^{2}+x-12=0$ $\Rightarrow x^{2}+4 x-3 x-12=0$ $\Rightarrow x(x+4)-3(x+4)=0$ $\Rightarrow(x+4)(x-3)=0$ $\Rightarrow x+4=0...
Read More →A body is moving in a low circular orbit about a planet of mass M and radius R.
Question: A body is moving in a low circular orbit about a planet of mass $M$ and radius $R$. The radius of the orbit can be taken to be $R$ itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is :(1) $\frac{1}{\sqrt{2}}$(2) 2(3) 1(4) $\sqrt{2}$Correct Option: 1 Solution: (1) Orbital speed of the body when it revolves very close to the surface of planet $V_{0}=\sqrt{\frac{G M}{R}}$ ....(1) Here, $G=$ gravitational constant Escape speed from the su...
Read More →Prove the following
Question: If $A=\left[\begin{array}{lr}0 -\tan \left(\frac{\theta}{2}\right) \\ \tan \left(\frac{\theta}{2}\right) 0\end{array}\right]$ and $\left(I_{2}+A\right)\left(I_{2}-A\right)^{-1}$ $=\left[\begin{array}{cc}a -b \\ b a\end{array}\right]$, then $13\left(a^{2}+b^{2}\right)$ is equal to Solution: $\mathrm{A}=\left[\begin{array}{cc}0 -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} 0\end{array}\right]$ $\Rightarrow I+A=\left[\begin{array}{cc}1 -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} 1\e...
Read More →The difference of two natural numbers is 5 and the difference of their reciprocals is
Question: The difference of two natural numbers is 5 and the difference of their reciprocals is $\frac{5}{14}$. Find the numbers. Solution: Let the required natural numbers bexand(x+ 5).Now,xx+ 5 $\therefore \frac{1}{x}\frac{1}{x+5}$ According to the given condition, $\frac{1}{x}-\frac{1}{x+5}=\frac{5}{14}$ $\Rightarrow \frac{x+5-x}{x(x+5)}=\frac{5}{14}$ $\Rightarrow \frac{5}{x^{2}+5 x}=\frac{5}{14}$ $\Rightarrow x^{2}+5 x=14$ $\Rightarrow x^{2}+5 x-14=0$ $\Rightarrow x^{2}+7 x-2 x-14=0$ $\Right...
Read More →A solution of
Question: A solution of $\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}$ is electrolysed between platinum electrodes using $0.1$ Faraday electricity. How many mode of Ni will be deposited at the cathode?$0.05$$0.20$$0.15$$0.10$Correct Option: 1 Solution: According to the Faraday's law of electrolysis, $\mathrm{nF}$ of current is required for the deposition of $1 \mathrm{~mol}$ According to the reaction, $\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \longrightarrow \mathrm{Ni}^{2+}+2 \mathrm{NO}_{3}^...
Read More →Let A and B be 3x3 real matrices such
Question: Let $\mathrm{A}$ and $\mathrm{B}$ be $3 \times 3$ real matrices such that $\mathrm{A}$ is symmetric matrix and $\mathrm{B}$ is skew-symmetric matrix. Then the system of linear equations $\left(A^{2} B^{2}-B^{2} A^{2}\right) X=O$, where $X$ is a $3 \times 1$ column matrix of unknown variables and $\mathrm{O}$ is a $3 \times 1$ null matrix, has :(1) a unique solution(2) exactly two solutions(3) infinitely many solutions(4) no solutionCorrect Option: , 3 Solution: $A^{\top}=A, B^{\top}=-B...
Read More →The difference of two natural numbers is 3 and the difference of their reciprocals is
Question: The difference of two natural numbers is 3 and the difference of their reciprocals is $\frac{3}{28}$. Find the numbers. Solution: Let the required natural numbers bexand(x+ 3).Now,xx+ 3 $\therefore \frac{1}{x}\frac{1}{x+3}$ According to the given condition, $\frac{1}{x}-\frac{1}{x+3}=\frac{3}{28}$ $\Rightarrow \frac{x+3-x}{x(x+3)}=\frac{3}{28}$ $\Rightarrow \frac{3}{x^{2}+3 x}=\frac{3}{28}$ $\Rightarrow x^{2}+3 x=28$ $\Rightarrow x^{2}+3 x-28=0$ $\Rightarrow x^{2}+7 x-4 x-28=0$ $\Right...
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