The isotopes of hydrogen are:

Question: The isotopes of hydrogen are:Tritium and protium onlyProtium and deuterium onlyProtium, deuterium and tritiumDeuterium and tritium onlyCorrect Option: Solution: Hydrogen has three isotopes: Protium $\left({ }_{1} \mathrm{H}^{1}\right)$, deuterium $\left({ }_{1} \mathrm{H}^{2}\right)$ and tritium $\left({ }_{1} \mathrm{H}^{3}\right)$....

The temporary hardness of a water sample is due to compound X.

Question: The temporary hardness of a water sample is due to compound $X$. Boiling this sample converts $X$ to compound Y. X and Y, respectively, are :$\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}$ and $\mathrm{Mg}(\mathrm{OH})_{2}$$\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$ and $\mathrm{Ca}(\mathrm{OH})_{2}$$\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}$ and $\mathrm{MgCO}_{3}$$\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$ and $\mathrm{CaO}$Correct Option: 1 Solution: Temporary hardness is ca...

Question: The temporary hardness of a water sample is due to compound $X$. Boiling this sample converts $X$ to compound Y. X and Y, respectively, are :$\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}$ and $\mathrm{Mg}(\mathrm{OH})_{2}$$\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$ and $\mathrm{Ca}(\mathrm{OH})_{2}$$\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}$ and $\mathrm{MgCO}_{3}$$\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$ and $\mathrm{CaO}$Correct Option: 1 Solution: Temporary hardness is ca...

A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure.

Question: A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is $5 \mathrm{~cm}$ and the angular speed of rotation is $\omega \mathrm{rad} \mathrm{s}^{-1}$. The difference in the height, $h$ (in $\mathrm{cm}$ ) of liquid at the centre of vessel and at the side will be : (1) $\frac{2 \omega^{2}}{25 g}$(2) $\frac{5 \omega^{2}}{2 g}$(3) $\frac{25 \omega^{2}}{2 g}$(4) $\frac{2 \omega^{2}}{5 g}$Correct ...

If m arithmetic means (A.Ms) and three geometric

Question: If $m$ arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that $4^{\text {th }}$ A.M. is equal to $2^{\text {nd }}$ G.M., then $m$ is equal to_________. Solution: Let $m$ arithmetic mean be $A_{1}, A_{2} \ldots A_{m}$ and $G_{1}, G_{2}, G_{3}$ be geometric mean. The A.P. formed by arithmetic mean is, $3, A_{1}, A_{2}, A_{3}, \ldots \ldots A_{m}, 243$ $\therefore d=\frac{243-3}{m+1}=\frac{240}{m+1}$ The G.P. formed by geometric mean $3, G_{1}, G...

The number of water molecules(s) not coordinated to copper ion directly in

Question: The number of water molecules(s) not coordinated to copper ion directly in $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$, is:2314Correct Option: Solution: In $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$, four $\mathrm{H}_{2} \mathrm{O}$ molecules are directly coordinated to the central metal ion while one $\mathrm{H}_{2} \mathrm{O}$ molecule is hydrogen bonded....

The strength of 11.2 volume solution of

Question: The strength of $11.2$ volume solution of $\mathrm{H}_{2} \mathrm{O}_{2}$ is : [Given that molar mass of $\mathrm{H}=1 \mathrm{~g} \mathrm{~mol}^{-1}$ and $\mathrm{O}=16 \mathrm{~g} \mathrm{~mol}^{-1}$ ]$13.6 \%$$3.4 \%$$34 \%$$1.7 \%$Correct Option: , 2 Solution: $11.2 \mathrm{~V}$ strength of $\mathrm{H}_{2} \mathrm{O}_{2}$ means, $11.2 \mathrm{~L}$ of $\mathrm{O}_{2}$ is liberated at STP. $\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}...

A large number of water drops,

Question: A large number of water drops, each of radius $\mathrm{r}$, combine to have a drop of radius $\mathrm{R}$. If the surface tension is $T$ and mechanical equivalent of heat is $\mathrm{J}$, the rise in heat energy per unit volume will be :(1) $\frac{2 T}{r J}$(2) $\frac{3 T}{r J}$(3) $\frac{2 T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$(4) $\frac{3 T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$Correct Option: , 4 Solution: (4) $\mathrm{R}$ is the radius of bigger drop. is the radius of $n$ wat...

Solve the following

Question: $100 \mathrm{~mL}$ of a water sample contains $0.81 \mathrm{~g}$ of calcium bicarbonate and $0.73 \mathrm{~g}$ of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of $\mathrm{CaCO}_{3}$ is: (molar mass of calcium bicarbonate is $162 \mathrm{~g} \mathrm{~mol}^{-1}$ and magnesium bicarboante is $146 \mathrm{~g} \mathrm{~mol}^{-1}$ )$5,000 \mathrm{ppm}$$1,000 \mathrm{ppm}$$100 \mathrm{ppm}$$10,000 \mathrm{ppm}$Correct Option: Solution: Moles of $\...

A hydraulic press can lift 100 kg when a mass ' m ' is placed on the smaller piston.

Question: A hydraulic press can lift $100 \mathrm{~kg}$ when a mass ' $\mathrm{m}$ ' is placed on the smaller piston. It can lift______ $\mathrm{kg}$ when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass ' $m$ ' on the smaller piston. Solution: $(25600)$ Atmospheric pressure $P_{0}$ will be acting on both the limbs of hydraulic lift. Applying pascal's law for same liquid level $\Rightarrow \mathrm{P}_{0}+\frac{...

In the sum of the series

Question: In the sum of the series $20+19 \frac{3}{5}+19 \frac{1}{5}+18 \frac{4}{5}+\ldots$ upto $n^{\text {th }}$ term is 488 and then $n^{\text {th }}$ term is negative, then : (1) $n=60$(2) $n^{\text {th }}$ term is $-4$(3) $n=41$(4) $n^{\text {th }}$ term is $-4 \frac{2}{5}$Correct Option: , 2 Solution: $S_{n}=20+19 \frac{3}{5}+19 \frac{1}{5}+18 \frac{4}{5}+\ldots$ $\because S_{n}=488$ $488=\frac{n}{2}\left[2\left(\frac{100}{5}\right)+(n-1)\left(-\frac{2}{5}\right)\right]$ $488=\frac{n}{2}(1...

The value of

Question: The value of $(0.16) \log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots\right.$ to $\left.\infty\right)$ is equal to________. Solution: $(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . \infty\right)}$ $=0.16$ $\log _{2.5}\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)$$\left[\because S_{\infty}=\frac{a}{1-r}\right]$ $=0.16^{\log _{2}}$ $16^{\log _{2.5}\left(\frac{1}{2}\right)}$ $=(2.5)^{-2 \log _{2.5}\left(\frac{1}{2}\right)}=\left(\frac{1}...

The hardness of a water sample containing

Question: The hardness of a water sample containing $10^{-3} \mathrm{M} \mathrm{MgSO}_{4}$ expressed as $\mathrm{CaCO}_{3}$ equivalents (in ppm) is___________. (molar mass of $\mathrm{MgSO}_{4}$ is $120.37 \mathrm{~g} / \mathrm{mol}$ ) Solution: (100.00) $10^{-3}$ molar $\mathrm{MgSO}_{4} \equiv 10^{-3}$ moles of $\mathrm{MgSO}_{4}$ present in $1 \mathrm{~L}$ solutions. $10^{-3} \mathrm{M} \mathrm{MgSO}_{4} \equiv 10^{-3} \mathrm{M} \mathrm{CaCO}_{3}$ $10^{-3} \mathrm{M} \mathrm{CaCO}_{3}=10^{-3...

Consider a water tank as shown in the figure.

Question: Consider a water tank as shown in the figure. It's cross-sectional area is $0.4 \mathrm{~m}^{2}$. The tank has an opening B near the bottom whose cross-section area is $1 \mathrm{~cm}^{2}$. A load of $24 \mathrm{~kg}$ is applied on the water at the top when the height of the water level is $40 \mathrm{~cm}$ above the bottom, the velocity of water coming out the opening $B$ is $v m s^{-1}$. The value of $\mathrm{v}$, to the nearest integer, is [Take value of $\mathrm{g}$ to be $10 \math...

If the first term of an A.P. is 3 and the sum of its first

Question: If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is :(1) $\frac{1}{6}$(2) $\frac{1}{5}$(3) $\frac{1}{4}$(4) $\frac{1}{7}$Correct Option: 1 Solution: Given $a=3$ and $S_{25}=S_{40}-S_{25}$ $\Rightarrow 2 S_{25}=S_{40}$ $\Rightarrow 2 \times \frac{25}{2}[6+24 d]=\frac{40}{2}[6+39 d]$ $\Rightarrow 25[6+24 d]=20[6+39 d]$ $\Rightarrow 5(2+8 d)=4(2+13 d)$ $\Rightarrow 10+40 d=8+52 d$ $\Righta...

Find the sum of each of the following APs:

Question: Find the sum of each of the following APs: (i) $2,7,12,17, \ldots$ to 19 terms (ii) $9,7,5,3, \ldots$ to 14 terms (iii) $-37,-33,-29, \ldots$ to 12 terms (iv) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$ to 11 terms (v) $0.6,1.7,2.8, \ldots$ to 100 terms Solution: (i) The given AP is 2, 7, 12, 17, ... .Here,a= 2 andd= 7 2 = 5 Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we have $S_{19}=\frac{19}{2}[2 \times 2+(19-1) \times 5]$ $=\frac{19}{2} \times(4+90)$ $=\frac{19}{2} \...

Hydrogen has three isotopes (A),(B) and (C).

Question: Hydrogen has three isotopes $(A),(B)$ and $(C)$. If the number of neutron(s) in (A), (B) and (C) respectively, are $(x),(y)$ and $(z)$, the sum of $(x),(y)$ and $(z)$ is:3241Correct Option: Solution:...

Let S be the sum of the first 9 terms of the series

Question: Let $S$ be the sum of the first 9 terms of the series : $\{x+k a\}+\left\{x^{2}+(k+2) a\right\}+\left\{x^{3}+(k+4) a\right\}$ $+\left\{x^{4}+(k+6) a\right\}+\ldots \quad$ where $\quad a \neq 0$ and $x \neq 1$. If $S=\frac{x^{10}-x+45 a(x-1)}{x-1}$, then $k$ is equal to :(1) $-5$(2) 1(3) $-3$(4) 3Correct Option: , 3 Solution: $S=\left(x+x^{2}+x^{3}+\ldots .9\right.$ terms $)$ $+a[k+(k+2)++(k+4)+\ldots 9$ terms $]$ $\Rightarrow S=\frac{x\left(x^{9}-1\right)}{x-1}+\frac{9}{2}[2 a k+8 \tim...

If the sum of first 11 terms of an A.P.

Question: If the sum of first 11 terms of an A.P., $a_{1}, a_{2}, a_{3}, \ldots$ is 0 $\left(a_{1} \neq 0\right)$, then the sum of the A.P., $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ is $k a_{1}$, where $k$ is equal to :(1) $-\frac{121}{10}$(2) $\frac{121}{10}$(3) $\frac{72}{5}$(4) $-\frac{72}{5}$Correct Option: , 4 Solution: Let common difference be $d$. $\because S_{11}=0 \quad \therefore \frac{11}{2}\left\{2 a_{1}+10 \cdot d\right\}=0$ $\Rightarrow a_{1}+5 d=0 \Rightarrow d=-\frac{a_{1}}{5}$ Now,...

Among statements (A)-(D), the correct ones are:

Question: Among statements (A)-(D), the correct ones are: (A) Decomposition of hydrogen peroxide gives dioxygen. (B) Like hydrogen peroxide, compounds, such as $\mathrm{KClO}_{3}$, $\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$ and $\mathrm{NaNO}_{3}$ when heated liberate dioxygen. (C) 2-Ethylanthraquinone is useful for the industrial preparation of hydrogen peroxide. (D) Hydrogen peroxide is used for the manufacture of sodium perborate.(A) (B), (C) and (D)(A), (B) and (C) only(A), (C) and (D) on...

The sum of three numbers in AP is 18.

Question: The sum of three numbers in AP is 18. If the product of first and third number is five times the common difference, find the numbers. Solution: Let the firstthree numbers in an arithmetic progression bea d,a,a + d.The sum of the first three numbers in an arithmetic progression is 18.a d+a+a + d= 18⇒ 3a= 18⇒a= 6The product of the first and third term is 5 times the common difference. $(a-d)(a+d)=5 d$ $\Rightarrow a^{2}-d^{2}=5 d$ $\Rightarrow(6)^{2}-d^{2}=5 d$ $\Rightarrow 36-d^{2}=5 d$...

In comparison to the zeolite process for the removal of permanent hardness,

Question: In comparison to the zeolite process for the removal of permanent hardness, the synthetic resins method is:less efficient as it exchanges only anionsmore efficient as it can exchange both cations as well as anionsless efficient as the resins cannot be regeneratedmore efficient as it can exchange only cationsCorrect Option: , 2 Solution: Synthetic resin method is more efficient than zeolite process as it can exchange both cations as well as anions....

Suppose you have taken a dilute solution of oleic acid in such a way

Question: Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes $0.01 \mathrm{~cm}^{3}$ of oleic acid per $\mathrm{cm}^{3}$ of the solution. Then you make a thin film of this solution (monomolecular thickness) of area $4 \mathrm{~cm}^{2}$ by considering 100 spherical drops of radius $\left(\frac{3}{40 \pi}\right)^{\frac{1}{3}} \times 10^{-3} \mathrm{~cm}$. Then the thickness of oleic acid layer will be $x \times 10^{-14} \mathrm{~m}$. Where $\mathrm{...

Suppose you have taken a dilute solution of oleic acid in such a way

Question: Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes $0.01 \mathrm{~cm}^{3}$ of oleic acid per $\mathrm{cm}^{3}$ of the solution. Then you make a thin film of this solution (monomolecular thickness) of area $4 \mathrm{~cm}^{2}$ by considering 100 spherical drops of radius $\left(\frac{3}{40 \pi}\right)^{\frac{1}{3}} \times 10^{-3} \mathrm{~cm}$. Then the thickness of oleic acid layer will be $x \times 10^{-14} \mathrm{~m}$. Where $\mathrm{...

Prove the following

Question: If $|x|1,|y|1$ and $x \neq y$, then the sum to infinity of the following series $(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots .$ is :(1) $\frac{x+y-x y}{(1+x)(1+y)}$(2) $\frac{x+y+x y}{(1+x)(1+y)}$(3) $\frac{x+y-x y}{(1-x)(1-y)}$(4) $\frac{x+y+x y}{(1-x)(1-y)}$Correct Option: , 3 Solution: $\mathrm{S}=(x+y)+\left(x^{2}+y^{2}+x y\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots \infty$ $=\frac{1}{x-y}\left[\left(x^{2}-y^{2}\right)+\left(x^{3}-...