In general, the property (magnitudes only) that shows an opposite trend in comparison
Question: In general, the property (magnitudes only) that shows an opposite trend in comparison to other properties across a period is :Ionization enthalpyElectronegativityElectron gain enthalpyAtomic radiusCorrect Option: , 4 Solution: On moving left to right along a period in the periodic table atomic radius decreases while electronegativity, electron gain enthalpy and ionisation enthalpy increases, along a period....
Read More →The correct order of electron gain enthalpy is:
Question: The correct order of electron gain enthalpy is:$\mathrm{S}\mathrm{Se}\mathrm{Te}\mathrm{O}$$\mathrm{O}\mathrm{S}\mathrm{Se}\mathrm{Te}$$\mathrm{S}\mathrm{O}\mathrm{Se}\mathrm{Te}$$\mathrm{Te}\mathrm{Se}\mathrm{S}\mathrm{O}$Correct Option: 1 Solution: Electron gain enthalpy of $\mathrm{O}$ is very low due to small size....
Read More →let f(x)
Question: Let $f(x)=\int \frac{\sqrt{x}}{(1+x)^{2}} d x(x \geq 0)$. Then $f(3)-f(1)$ is equal to :(1) $-\frac{\pi}{12}+\frac{1}{2}+\frac{\sqrt{3}}{4}$(2) $\frac{\pi}{6}+\frac{1}{2}-\frac{\sqrt{3}}{4}$(3) $-\frac{\pi}{6}+\frac{1}{2}+\frac{\sqrt{3}}{4}$(4) $\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$Correct Option: , 4 Solution: $\int \frac{\sqrt{x}}{(1+x)^{2}} d x(x0)$ Put $x=\tan ^{2} \theta \Rightarrow 2 x d x=2 \tan \theta \sec ^{2} \theta d \theta$ $I=\int \frac{2 \tan ^{2} \theta \cdot \s...
Read More →Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Question: Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $\mathrm{R}$. Assertion $\mathrm{A}: \mathrm{In} \mathrm{TII}_{3}$, isomorphous to $\mathrm{CsI}_{3}$, the metal is present in $+1$ oxidation state. Reason R: TI metals has fourteen $f$ electrons in its electronic configuration. In the light of the above statements, choose the most appropriate answer from the options given below:Both $A$ and $R$ are correct and $R$ is the correct explan...
Read More →A point charge
Question: A point charge of $+12 \mu \mathrm{C}$ is at a distance $6 \mathrm{~cm}$ vertically above the centre of a square of side $12 \mathrm{~cm}$ as shown in figure. The magnitude of the electric flux through the square will be_____ $\times 10^{3} \mathrm{Nm}^{2} / \mathrm{C}$ Solution: (226) Using Gauss law, it is a part of cube of side $12 \mathrm{~cm}$ and charge at centre so; $\phi=\frac{Q}{6 \varepsilon_{0}}=\frac{12 \mu \mathrm{c}}{6 \varepsilon_{0}}=2 \times 4 \pi \times 9 \times 10^{9...
Read More →The integral
Question: The integral $\int\left(\frac{x}{x \sin x+\cos x}\right)^{2} d x$ is equal to (where $C$ is a constant of integration):(1) $\tan x-\frac{x \sec x}{x \sin x+\cos x}+C$(2) $\sec x+\frac{x \tan x}{x \sin x+\cos x}+C$(3) $\sec x-\frac{x \tan x}{x \sin x+\cos x}+C$(4) $\tan x+\frac{x \sec x}{x \sin x+\cos x}+C$Correct Option: 1 Solution: $\int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$ $\because \frac{d}{d x}(x \sin x+\cos x)=x \cos x$ $=\int \frac{x \cos x}{(x \sin x+\cos x)^{2}}\left(\frac{...
Read More →Match List-I with List-II
Question: Match List-I with List-II $(\mathrm{a})-(\mathrm{ii}),(\mathrm{b})-(\mathrm{iii}),(\mathrm{c})-(\mathrm{iv}),(\mathrm{d})-(\mathrm{i})$(a) $-$ (iv), (b) $-$ (i), (c) $-$ (ii), (d) $-$ (iii)(a) $-$ (i), (b) $-$ (iv), (c) $-$ (iii), (d) $-$ (ii)(a) $-($ i $),($ b $)-($ iii $),($ c $)-($ iv $),($ d $)-($ ii $)$Correct Option: 1 Solution: Order of I.E. in second period $\mathrm{Li}\mathrm{B}\mathrm{Be}\mathrm{C}\mathrm{O}\mathrm{N}\mathrm{F}\mathrm{Ne}$...
Read More →if
Question: If $\int \sin ^{-1}\left(\sqrt{\frac{x}{1+x}}\right) d x=A(x) \tan ^{-1}(\sqrt{x})+B(x)+C$, where $C$ is a constant of integration, then the ordered pair $(A(x)$, $B(x))$ can be: (1) $(x+1,-\sqrt{x})$(2) $(x+1, \sqrt{x})$(3) $(x-1,-\sqrt{x})$(4) $(x-1, \sqrt{x})$Correct Option: 1 Solution: $I=\int \sin ^{-1}\left(\frac{\sqrt{x}}{\sqrt{1+x}}\right) d x=\int \tan ^{-1} \sqrt{x} \cdot 1 d x$ $=x \tan ^{-1} \sqrt{x}-\int \frac{1}{1+x} \cdot \frac{1}{2 \sqrt{x}} \cdot x d x+C$ $=x \tan ^{-1...
Read More →Two electrons each are fixed at a distance '2d'.
Question: Two electrons each are fixed at a distance '2d'. A third charge proton placed at the midpoint is displaced slightly by a distance $x(xd)$ perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency: $(\mathrm{m}=$ mass of charged particle $)$(1) $\left(\frac{q^{2}}{2 \pi \varepsilon_{0} m d^{3}}\right)^{\frac{1}{2}}$(2) $\left(\frac{\pi \varepsilon_{0} \mathrm{md}^{3}}{2 \mathrm{q}^{2}}\right)^{\frac{1}{2}}$(3) $\left(\fr...
Read More →Let [t]
Question: Let $[t]$ denote the greatest integer less than or equal to $t$. Then the value of $\int_{1}^{2}|2 x-[3 x]| d x$ is__________. Solution: $\int_{1}^{2}|2 x-[3 x]| d x=\int_{1}^{2}|3 x-[3 x]-x| d x$ $=\int_{1}^{2}|\{3 x\}-x| d x=\int_{1}^{2}(x-\{3 x\}) d x$ $=\int_{1}^{2} x d x-\int_{1}^{2}\{3 x\} d x=\left[\frac{x^{2}}{2}\right]_{1}^{2}-3 \int_{0}^{1 / 3} 3 x d x$ $=\frac{(4-1)}{2}-9\left[\frac{x^{2}}{2}\right]_{0}^{1 / 3}=\frac{3}{2}-\frac{1}{2}=1$...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $\frac{4}{x}-3=\frac{5}{2 x+3}, \quad x \neq 0,-\frac{3}{2}$ Solution: $\frac{4}{x}-3=\frac{5}{2 x+3}, \quad x \neq 0,-\frac{3}{2}$ $\Rightarrow \frac{4}{x}-\frac{5}{2 x+3}=3$ $\Rightarrow \frac{8 x+12-5 x}{x(2 x+3)}=3$ $\Rightarrow \frac{3 x+12}{2 x^{2}+3 x}=3$ $\Rightarrow \frac{x+4}{2 x^{2}+3 x}=1$ $\Rightarrow 2 x^{2}+3 x=x+4 \quad$ (Cross multiplication) $\Rightarrow 2 x^{2}+2 x-4=0$ $\Rightarrow x^{2}+x-2=0$ $\Rightarrow x^{2}+2 x-...
Read More →The correct order of bond dissociation enthalpy of halogen is :
Question: The correct order of bond dissociation enthalpy of halogen is :$\mathrm{F}_{2}\mathrm{Cl}_{2}\mathrm{Br}_{2}\mathrm{I}_{2}$$\mathrm{Cl}_{2}\mathrm{F}_{2}\mathrm{Br}_{2}\mathrm{I}_{2}$$\mathrm{Cl}_{2}\mathrm{Br}_{2}\mathrm{F}_{2}\mathrm{I}_{2}$$\mathrm{I}_{2}\mathrm{Br}_{2}\mathrm{Cl}_{2}\mathrm{F}_{2}$Correct Option: , 3 Solution: Fact based $\mathrm{F}_{2}$ has $\mathrm{F}-\mathrm{F}, \mathrm{F}_{2}$ involves repulsion of non-bonding electrons $\backslash$ more over its size is small ...
Read More →The integral
Question: The integral $\int \frac{e^{3 \log _{e} 2 x}+5 e^{2 \log _{e} 2 x}}{e^{4 \log _{e} x}+5 e^{3 \log _{e} x}-7 e^{2 \log _{e} x}} d x, x0$, is equal to: (where $c$ is a constant of integration)(1) $\log _{e}\left|x^{2}+5 x-7\right|+c$(2) $\frac{1}{4} \log _{e}\left|x^{2}+5 x-7\right|+c$(3) $4 \log _{\mathrm{e}}\left|\mathrm{x}^{2}+5 \mathrm{x}-7\right|+\mathrm{c}$(4) $\log _{e} \sqrt{x^{2}+5 x-7}+c$Correct Option: , 3 Solution: $\int \frac{e^{3 \log _{e} 2 x}+5 e^{2 \log _{e} 2 x}}{e^{4 \...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $\frac{16}{x}-1=\frac{15}{x+1}, x \neq 0,-1$ Solution: $\frac{16}{x}-1=\frac{15}{x+1}, x \neq 0,-1$ $\Rightarrow \frac{16}{x}-\frac{15}{x+1}=1$ $\Rightarrow \frac{16 x+16-15 x}{x(x+1)}=1$ $\Rightarrow \frac{x+16}{x^{2}+x}=1$ $\Rightarrow x^{2}+x=x+16 \quad$ (Cross multiplication) $\Rightarrow x^{2}-16=0$ $\Rightarrow(x+4)(x-4)=0$ $\Rightarrow x+4=0$ or $x-4=0$ $\Rightarrow x=-4$ or $x=4$ Hence, 4 and 4 are the roots of the given equation...
Read More →Consider the elements Mg, AI, S, P, and Si,
Question: Consider the elements $\mathrm{Mg}, \mathrm{Al}, \mathrm{S}, \mathrm{P}$ and $\mathrm{Si}$, the correct increasing order of their first ionization enthalpy is:$\mathrm{Al}\mathrm{Mg}\mathrm{Si}\mathrm{S}\mathrm{P}$$\mathrm{Al}\mathrm{Mg}\mathrm{S}\mathrm{Si}\mathrm{P}$$\mathrm{Mg}\mathrm{Al}\mathrm{Si}\mathrm{S}\mathrm{P}$$\mathrm{Mg}\mathrm{Al}\mathrm{Si}\mathrm{P}\mathrm{S}$Correct Option: 1 Solution: Order of IE, in $3^{\text {rd }}$ period is $\mathrm{Na}\mathrm{Mg}\mathrm{Al}\math...
Read More →The value of the integral
Question: The value of the integral $\int \frac{\sin \theta \cdot \sin 2 \theta\left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta$ is (where $c$ is a constant of integration)(1) $\frac{1}{18}\left[9-2 \sin ^{6} \theta-3 \sin ^{4} \theta-6 \sin ^{2} \theta\right]^{\frac{3}{2}}+c$(2) $\frac{1}{18}\left[11-18 \sin ^{2} \theta+9 \sin ^{4} \theta-2 \sin ^{6} \theta\right]^{\frac{3}{2}}+c$(3) $\frac{1}{18}\left[11-1...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $9 x^{2}-9(a+b) x+\left(2 a^{2}+5 a b+2 b^{2}\right)=0$ Solution: We write, $-9(a+b) x=-3(2 a+b) x-3(a+2 b) x$ as $9 x^{2} \times\left(2 a^{2}+5 a b+2 b^{2}\right)=9\left(2 a^{2}+5 a b+2 b^{2}\right) x^{2}=[-3(2 a+b) x] \times[-3(a+2 b) x]$ $\therefore 9 x^{2}-9(a+b) x+\left(2 a^{2}+5 a b+2 b^{2}\right)=0$ $\Rightarrow 9 x^{2}-3(2 a+b) x-3(a+2 b) x+(2 a+b)(a+2 b)=0$ $\Rightarrow 3 x[3 x-(2 a+b)]-(a+2 b)[3 x-(2 a+b)]=0$ $\Rightarrow[3 x-(...
Read More →A cube of side ' a '
Question: A cube of side ' $\mathrm{a}$ ' has point charges $+\mathrm{Q}$ located at each of its vertices except at the origin where the charge is -Q. The electric field at the centre of cube is: (1) $\frac{2 Q}{3 \sqrt{3} \pi \varepsilon_{0} a^{2}}(\hat{x}+\hat{y}+\hat{z})$(2) $\frac{\mathrm{Q}}{3 \sqrt{3} \pi \varepsilon_{0} \mathrm{a}^{2}}(\hat{\mathrm{x}}+\hat{\mathrm{y}}+\hat{\mathrm{z}})$(3) $\frac{-2 \mathrm{Q}}{3 \sqrt{3} \pi \varepsilon_{0} \mathrm{a}^{2}}(\hat{\mathrm{x}}+\hat{\mathrm{...
Read More →The first ionization energy of magnesium is smaller as compared to that of
Question: The first ionization energy of magnesium is smaller as compared to that of elements $\mathrm{X}$ and $\mathrm{Y}$, but higher than that of $Z$. the elements $X, Y$ and $Z$, respectively, are :chlorine, lithium and sodiumargon, lithium and sodiumargon, chlorine and sodiumneon, sodium and chlorineCorrect Option: , 3 Solution: The $1^{\text {st }}$ IE order of $3^{\text {rd }}$ period is $\mathrm{Na}\mathrm{Al}\mathrm{Mg}\mathrm{Si}\mathrm{S}\mathrm{P}\mathrm{Cl}\mathrm{Ar}$ $X \ Y$ are $...
Read More →if
Question: If $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x=a \sin ^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c$, where $c$ is a constant of integration, then the ordered pair $(a, b)$ is equal to :(1) $(1,-3)$(2) $(1,3)$(3) $(-1,3)$(4) $(3,1)$Correct Option: , 2 Solution: Put $\sin x+\cos x=t$ $\Rightarrow 1+\sin 2 x=t^{2}$ $\Rightarrow(\cos x-\sin x) d x=d t$ $\therefore I=\int \frac{d t}{\sqrt{8-\left(t^{2}-1\right)}}=\int \frac{d t}{\sqrt{9-t^{2}}}$ $=\sin ^{-1}\left(\frac{t}{3}\right)+C=\...
Read More →Solve this
Question: $a^{2} b^{2} x^{2}+b^{2} x-a^{2} x-1=0$ Solution: Given : $a^{2} b^{2} x^{2}+b^{2} x-a^{2} x-1=0$ $\Rightarrow b^{2} x\left(a^{2} x+1\right)-1\left(a^{2} x+1\right)=0$ $\Rightarrow\left(b^{2} x-1\right)\left(a^{2} x+1\right)=0$ $\Rightarrow\left(b^{2} x-1\right)=0$ or $\left(a^{2} x+1\right)=0$ $\Rightarrow x=\frac{1}{b^{2}}$ or $x=\frac{-1}{a^{2}}$ Hence, $\frac{1}{b^{2}}$ and $\frac{-1}{a^{2}}$ are the roots of the given equation....
Read More →if f(x)
Question: If $f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x,(x \geq 0), f(0)=0 \mathrm{v}$ and $f(1)=\frac{1}{\mathrm{~K}}$, then the value of $\mathrm{K}$ is Solution: $f(x)=\int \frac{\left(5 x^{8}+7 x^{6}\right) d x}{x^{14}\left(x^{-5}+x^{-7}+2\right)^{2}}$ Let $x^{-5}+x^{-7}+2=t$ $\left(-5 x^{-6}-7 x^{-8}\right) d x=d t$ $\Rightarrow f(x)=\int-\frac{d t}{t^{2}}=\frac{1}{t}+c$ $f(x)=\frac{x^{7}}{x^{2}+1+2 x^{7}}$ $f(1)=\frac{1}{4}$...
Read More →Solve this
Question: $12 a b x^{2}-\left(9 a^{2}-8 b^{2}\right) x-6 a b=0$ Solution: Given : $12 a b x^{2}-\left(9 a^{2}-8 b^{2}\right) x-6 a b=0$ $\Rightarrow 12 a b x^{2}-9 a^{2} x+8 b^{2} x-6 a b=0$ $\Rightarrow 3 a x(4 b x-3 a)+2 b(4 b x-3 a)=0$ $\Rightarrow(3 a x+2 b)(4 b x-3 a)=0$ $\Rightarrow 3 a x+2 b=0$ or $4 b x-3 a=0$ $\Rightarrow x=\frac{-2 b}{3 a}$ or $x=\frac{3 a}{4 b}$ Hence, the roots of the equation are $\frac{-2 b}{3 a}$ and $\frac{3 a}{4 b}$....
Read More →The ionic radius of
Question: The ionic radius of $\mathrm{Na}^{+}$ions is $1.02 \backslash \mathrm{AA}$. The ionic radii (in $\backslash \mathrm{AA}$ ) of $\mathrm{Mg}^{2+}$ and $\mathrm{Al}^{3+}$, respectively, are$1.05$ and $0.99$$0.72$ and $0.54$$0.85$ and $0.99$$0.68$ and $0.72$Correct Option: , 2 Solution: The ionic radii order is $\mathrm{Na}^{+}\mathrm{Mg}^{2+}\mathrm{Al}^{3+}$...
Read More →Solve this
Question: $4 x^{2}-2\left(a^{2}+b^{2}\right) x+a^{2} b^{2}=0$ Solution: Given : $4 x^{2}-2\left(a^{2}+b^{2}\right) x+a^{2} b^{2}=0$ $\Rightarrow 4 x^{2}-2 a^{2} x-2 b^{2} x+a^{2} b^{2}=0$ $\Rightarrow 2 x\left(2 x-a^{2}\right)-b^{2}\left(2 x-a^{2}\right)=0$ $\Rightarrow\left(2 x-b^{2}\right)\left(2 x-a^{2}\right)=0$ $\Rightarrow 2 x-b^{2}=0$ or $2 x-a^{2}=0$ $\Rightarrow x=\frac{b^{2}}{2}$ or $x=\frac{a^{2}}{2}$ Hence, the roots of the equation are $\frac{b^{2}}{2}$ and $\frac{a^{2}}{2}$....
Read More →