Dihydrogen of high purity (>99.95%) is obtained through:
Question: Dihydrogen of high purity $(99.95 \%)$ is obtained through:the reaction of $\mathrm{Zn}$ with dilute $\mathrm{HCl}$.the electrolysis of acidified water using Pt electrodes.the electrolysis of bringe solution.the electrolysis of warm $\mathrm{Ba}(\mathrm{OH})_{2}$ solution using $\mathrm{Ni}$ electrodes.Correct Option: , 4 Solution: Dihydrogen of high purity $(99.95 \%)$ is obtained by the electrolysis of $\mathrm{Ba}(\mathrm{OH})_{2}$ using Ni electrodes....
Read More →The sum of first three terms of an AP is 48.
Question: The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12, find the AP. Solution: Let the first three terms of the AP be (ad),aand (a+d). Then, $(a-d)+a+(a+d)=48$ $\Rightarrow 3 a=48$ $\Rightarrow a=16$ Now, $(a-d) \times a=4(a+d)+12$ (Given) $\Rightarrow(16-d) \times 16=4(16+d)+12$ $\Rightarrow 256-16 d=64+4 d+12$ $\Rightarrow 16 d+4 d=256-76$ $\Rightarrow 20 d=180$ $\Rightarrow d=9$ Whena= 16 andd= 9, $a-d=16-9=7$ $a+d=...
Read More →The one that is NOT suitable for the removal of
Question: The one that is NOT suitable for the removal of permanent hardness of water is :Clark's methodIon-exchange methodCalgon's methodTreatment with sodium carbonateCorrect Option: Solution: Clark's method is used to remove temporary hardness, using lime water (or) $\mathrm{Ca}(\mathrm{OH})_{2}$ from water....
Read More →The sum of the first three terms of a G.P.
Question: The sum of the first three terms of a G.P. is $S$ and their product is 27 . Then all such $S$ lie in :(1) $(-\infty,-9] \cup[3, \infty)$(2) $[-3, \infty)$(3) $(-\infty,-3] \cup[9, \infty)$(4) $(-\infty, 9]$Correct Option: , 3 Solution: Let terms of G.P. be $\frac{a}{r}, a, a r$ $\therefore a\left(\frac{1}{r}+1+r\right)=S$$\ldots$ (i) and $a^{3}=27$ $\Rightarrow a=3$.....(ii) Put $a=3$ in eqn. (1), we get $S=3+3\left(r+\frac{1}{r}\right)$ If $f(x)=x+\frac{1}{x}$, then $f(x) \in(-\infty,...
Read More →An object is located at 2 km beneath the surface of the water.
Question: An object is located at $2 \mathrm{~km}$ beneath the surface of the water. If the fractional compression $\frac{\Delta \mathrm{V}}{\mathrm{V}}$ is $1.36 \%$, the ratio of hydraulic stress to the corresponding hydraulic strain will be [Given : density of water is $1000 \mathrm{~kg} \mathrm{~m}^{-3}$ and $\mathrm{g}=9.8 \mathrm{~ms}^{-2} .$ ] (1) $1.96 \backslash$ times $10^{\wedge}\{7\} \backslash$ mathrm $\{\mathrm{Nm}\}^{\wedge}\{-2\} \$$(1) $1.96 \backslash$ times $10^{\wedge}\{7\} \...
Read More →Hydrogen peroxide, in the pure state, is :
Question: Hydrogen peroxide, in the pure state, is :non-planar and almost colorlesslinear and blue in colorlinear and almost colorlessplanar and blue in colorCorrect Option: 1 Solution: $\mathrm{H}_{2} \mathrm{O}_{2}$ has open book like structure, which is nonplanar. It is a colourless viscous liquid but in large quantity appears blue in colour....
Read More →Divide 32 into four parts which are the four terms of an AP such that the product of
Question: Divide 32 into four parts which are the four terms of an AP such that the product of the first and the fourth term is to the product of the second and the third term is 7 : 15. Solution: Let the four parts in AP be (a 3d), (ad), (a+d) and (a+ 3d). Then, $(a-3 d)+(a-d)+(a+d)+(a+3 d)=32$ $\Rightarrow 4 a=32$ $\Rightarrow a=8 \quad \ldots \ldots(1)$ Also, $(a-3 d)(a+3 d):(a-d)(a+d)=7: 15$ $\Rightarrow \frac{(8-3 d)(8+3 d)}{(8-d)(8+d)}=\frac{7}{15}$ [From (1)] $\Rightarrow \frac{64-9 d^{2}...
Read More →The strengths of 5.6 volume hydrogen peroxide
Question: The strengths of $5.6$ volume hydrogen peroxide (of density $1 \mathrm{~g} / \mathrm{mL})$ in terms of mass percentage and molarity $(\mathrm{M})$, respectively, are: (Take molar mass of hydrogen peroxide as $34 \mathrm{~g} / \mathrm{mol}$ )$1.7$ and $0.5$$0.85$ and $0.25$$1.7$ and $0.25$$0.85$ and $0.5$Correct Option: 1 Solution:...
Read More →When two soap bubbles of radii a and b(b>a) coalesce,
Question: When two soap bubbles of radii a and $\mathrm{b}(\mathrm{b}\mathrm{a})$ coalesce, the radius of curvature of common surface is :(1) $\frac{a b}{b-a}$(2) $\frac{a+b}{a b}$(3) $\frac{\mathrm{b}-\mathrm{a}}{\mathrm{ab}}$(4) $\frac{a b}{a+b}$Correct Option: 1 Solution: (1) Excess pressure at common surface is given by $P_{e x}=4 T\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{4 T}{r}$ $\therefore \frac{1}{\mathrm{r}}=\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}$ $r=\frac{a b}{b-a}$...
Read More →If the arithmetic mean and geometric
Question: If the arithmetic mean and geometric mean of the $p^{\text {th }}$ and $q^{\text {th }}$ terms of the sequence $-16,8,-4,2, \ldots \ldots \ldots \ldots$ satisfy the equation $4 x^{2}-9 x+5=0$, then $\mathrm{p}+\mathrm{q}$ is equal to Solution: Given, $4 x^{2}-9 x+5=0$ $\Rightarrow(x-1)(4 x-5)=0$ $\Rightarrow$ A. $M=\frac{5}{4}$, G. $M=1 \quad$ (QA. $M$ G. $M$ ) Again, for the series $-16,8,-4,2 \ldots \ldots$ $p^{\text {th }}$ term $t_{p}=-16\left(\frac{-1}{2}\right)^{p-1}$ $q^{\text {...
Read More →Find four numbers in AP whose sum is 28 and the sum of whose squares is 216.
Question: Find four numbers in AP whose sum is 28 and the sum of whose squares is 216. Solution: Let the four numbers be $a-3 d, a-d, a+d, a+3 d$ According to question: $(a-3 d)+(a-d)+(a+d)+(a+3 d)=28$ $\Rightarrow 4 a-4 d+4 d=28$ $\Rightarrow a=7$ And $(a-3 d)^{2}+(a-d)^{2}+(a+d)^{2}+(a+3 d)^{2}=216$ $a^{2}+9 d^{2}-6 a d+a^{2}+d^{2}-2 a d+a^{2}+d^{2}+2 a d+a^{2}+9 d^{2}+6 a d=216$ $\Rightarrow 4 a^{2}+20 d^{2}=216$ $\Rightarrow 4(7)^{2}+20 d^{2}=216$ $\Rightarrow d=\pm 1$ Four numbers are : $d=...
Read More →What will be the nature of flow of water from a circular tap,
Question: What will be the nature of flow of water from a circular tap, when its flow rate increased from $0.18 \mathrm{~L} / \mathrm{min}$ to $0.48 \mathrm{~L} / \mathrm{min}$ ? The radius of the tap and viscosity of water are $0.5 \mathrm{~cm}$ and $10^{-3} \mathrm{~Pa}$ s, respectively. (Density of water: $10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ )(1) Unsteady to steady flow(2) Remains steady flow(3) Remains turbulent flow(4) Steady flow to unsteady flowCorrect Option: , 4 Solution: (4) The natu...
Read More →The volume strength of
Question: The volume strength of $8.9 \mathrm{M} \mathrm{H}_{2} \mathrm{O}_{2}$ solution calculated at $273 \mathrm{~K}$ and $1 \mathrm{~atm}$ is_____________. $\left(\mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$ (rounded off to the nearest integer) Solution: (100) Molarity of $\mathrm{H}_{2} \mathrm{O}_{2}$ solution $=\left\{\frac{\text { Volume strength }}{11.2}\right\}$ Volume strength $=8.9 \times 11.2=99.68 \mathrm{~V} \simeq 100 \mathrm{~V}$...
Read More →The sum of the series
Question: The sum of the series $\sum_{n=1}^{\infty} \frac{n^{2}+6 n+10}{(2 n+1) !}$ is equal to:(1) $\frac{41}{8} e+\frac{19}{8} e^{-1}-10$(2) $-\frac{41}{8} e+\frac{19}{8} e^{-1}-10$(3) $\frac{41}{8} e-\frac{19}{8} e^{-1}-10$(4) $\frac{41}{8} e+\frac{19}{8} e^{-1}+10$Correct Option: , 3 Solution: $\sum_{n=1}^{\infty} \frac{n^{2}+6 n+10}{(2 n+1) !}$ Put $2 n+1=r$, where $r=3,5,7, \ldots$ $\Rightarrow n=\frac{r-1}{2}$ $\frac{n^{2}-6 n+10}{(2 n+1) !}=\frac{\left(\frac{r-1}{2}\right)^{2}+3 r-3+10}...
Read More →Which of the following forms of
Question: Which of the following forms of hydrogen emits low energy $\beta^{-}$particles?Proton $\mathrm{H}^{+}$Deuterium ${ }_{1}^{2} \mathrm{H}$Protium ${ }_{1}^{1} \mathrm{H}$Tritium ${ }_{1}^{3} \mathrm{H}$Correct Option: , 4 Solution: Tritium isotope of hydrogen is radioactive and emits low energy $\beta^{-}$particles. It is because of high $\mathrm{n} / \mathrm{p}$ ratio of tritium which makes nucleus unstable....
Read More →The angle of a quadrilateral are in AP whose common difference is 10°.
Question: The angle of a quadrilateral are in AP whose common difference is 10. Find the angles. Solution: Let the required angles be (a- 15)o, (a- 5)o,(a+ 5)oand (a+ 15)o, as the common difference is 10 (given).Then (a- 15)o+ (a- 5)o+ (a+ 5)o+ (a+ 15)o= 360o⇒ 4a=360⇒a =90 Hence, the required angles of a quadrilateral are $(90-15)^{\circ},(90-5)^{\circ},(90+5)^{\circ}$ and $(90+15)^{\circ} ;$ or $75^{\circ}, 85^{\circ}, 95^{\circ}$ and $105^{\circ}$....
Read More →The sum of the infinite series
Question: The sum of the infinite series $1+\frac{2}{3}+\frac{7}{3^{2}}+\frac{12}{3^{3}}+\frac{17}{3^{4}}+\frac{22}{3^{5}}+\ldots$ is equal to(1) $\frac{9}{4}$(2) $\frac{15}{4}$(3) $\frac{13}{4}$(4) $\frac{11}{4}$Correct Option: , 3 Solution: $s=1+\frac{2}{3}+\frac{7}{3^{2}}+\frac{12}{3^{3}}+\frac{17}{3^{4}}+\frac{22}{3^{5}}+\ldots$ $\frac{s}{3}=\frac{1}{3}+\frac{2}{3^{2}}+\frac{7}{3^{3}}+\ldots \ldots \infty$ $\frac{2 s}{3}=1+\frac{1}{3}+\frac{5}{3^{2}}+\frac{5}{3^{3}}+\ldots \ldots \infty$ $\f...
Read More →The pressure acting on a submarine
Question: The pressure acting on a submarine is $3 \times 10^{5} \mathrm{~Pa}$ at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be : (Assume that atmospheric pressure is $1 \times 10^{5} \mathrm{~Pa}$ density of water is $10^{3} \mathrm{~kg} \mathrm{~m}^{-3}, \mathrm{~g}=10 \mathrm{~ms}^{-2}$ )(1) $\frac{200}{3} \%$(2) $\frac{200}{5} \%$(3) $\frac{5}{200} \%$(4) $\frac{3}{200} \%$Correct Option: 1 Solution: (1) $\mathrm{P}_{1}=\rh...
Read More →Calgon is used for water treatment.
Question: Calgon is used for water treatment. Which of the following statement is NOT true about calgon?Calgon contains the $2^{\text {nd }}$ most abundant element by weight in the earth's crust.It is also known as Graham's salt.It is polymeric compound and is water soluble.It doesnot remove Cat ion by precipitation.Correct Option: 1 Solution: $\mathrm{Na}_{6}\left(\mathrm{PO}_{3}\right)_{6}$ or $\mathrm{Na}_{6} \mathrm{P}_{6} \mathrm{O}_{18}$ Order of abundance of element in earth crust is $\ma...
Read More →The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165.
Question: The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms. Solution: Let the required terms be (a-d),aand (a+d).Then (a-d) +a+ (a+d) = 21⇒3a=21⇒a =7Also, (a-d)2+a2+ (a+d)2 = 165⇒ 3a2+ 2d2=165 $\Rightarrow\left(3 \times 49+2 d^{2}\right)=165$ $\Rightarrow 2 d^{2}=165-147=18$ $\Rightarrow d^{2}=9$ $\Rightarrow d=\pm 3$ Thus, $a=7$ and $d=\pm 3$ Hence, the required terms are ( 4, 7,10) or ( 10, 7, 4)....
Read More →In an increasing, geometric series,
Question: In an increasing, geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is 25 . Then, the sum of $4^{\text {th }}, 6^{\text {th }}$ and $8^{\text {th }}$ terms is equal to :(1) 35(2) 30(3) 26(4) 32Correct Option: 1 Solution: $a r+a r^{5}=\frac{25}{2}$ $a r^{2} \times a r^{4}=25$ $a^{2} r^{6}=25$ $a r^{3}=5$ $a=\frac{5}{r^{3}}$ $\frac{5 r}{r^{3}}+\frac{5 r^{5}}{r^{3}}=\frac{25}{2}$ $\frac{1}{r^{2}}+r^{2}=\frac{5}{2}$ Put...
Read More →Statement about heavy water are given below
Question: Statement about heavy water are given below A. Heavy water is used in exchange reactions for the study of reaction mechanisms B. Heavy water is prepared by exhaustive electrolysis of water C. Heavy water has higher boiling point than ordinary water D. Viscosity of $\mathrm{H}_{2} \mathrm{O}$ is greater than $\mathrm{D}_{2} \mathrm{O}$$\mathrm{A}$ and $\mathrm{B}$ onlyA and D only$A, B$ and $C$ only$\mathrm{A}$ and $\mathrm{C}$ onlyCorrect Option: , 3 Solution: (3) Fact...
Read More →Divide 24 in three parts such that they are in AP and their product is 440.
Question: Divide 24 in three parts such that they are in AP and their product is 440. Solution: Let the required parts of 24 be (a-d),aand (a+d) such that they are in AP.Then (a-d) +a+ (a+d) = 24⇒3a=24⇒a =8Also, (a-d).a.(a+d) = 440 $\Rightarrow a\left(a^{2}-d^{2}\right)=440$ $\Rightarrow 8\left(64-d^{2}\right)=440$ $\Rightarrow d^{2}=64-55=9$ $\Rightarrow d=\pm 3$ Thus, $a=8$ and $d=\pm 3$ Hence, the required parts of 24 are (5, 8,11) or (11, 8, 5)....
Read More →Which of the following equation depicts the oxidizing nature of
Question: Which of the following equation depicts the oxidizing nature of $\mathrm{H}_{2} \mathrm{O}_{2}$ ?$\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{HCl}+\mathrm{O}_{2}$$\mathrm{KlO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{KlO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}$$2 \mathrm{I}^{-}+\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+} \rightarrow \mathrm{I}_{2}+2 \mathrm{H}_{2} \mathrm{O}$$\mathrm{I}_{2}+\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{OH}^{-} \r...
Read More →The sum of three numbers in AP is 3 their product is −35.
Question: The sum of three numbers in AP is 3 their product is 35. Find the numbers. Solution: Let the required numbers be (a-d),aand (a+d).Then (a-d) +a+ (a+d) = 3⇒3a=3⇒a =1Also, (a-d).a.(a+d) = -35 $\Rightarrow a\left(a^{2}-d^{2}\right)=-35$ $\Rightarrow 1 \cdot\left(1-d^{2}\right)=-35$ $\Rightarrow d^{2}=36$ $\Rightarrow d=\pm 6$ Thus, $a=1$ and $d=\pm 6$ Hence, the required numbers are ( -5, 1 and 7) or ( 7, 1 and -5)....
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