In an electrical circuit, a battery is connected to pass 20 C of charge through it in a certain given time.
Question: In an electrical circuit, a battery is connected to pass $20 \mathrm{C}$ of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at $15 \mathrm{~V}$. The work done by the battery is______$\mathrm{J} .$ Solution: $(300)$ Charge flown $(Q)=20 c$ Potential difference $(\mathrm{V})=15 \mathrm{~V}$ Work done $(w)=Q . V$ $=20 \times 15=300 \mathrm{~J}$ $w=300 \mathrm{~J}$...
Read More →The IUPAC symbol for the element with atomic number 119 would be :
Question: The IUPAC symbol for the element with atomic number 119 would be :uueuneunhuunCorrect Option: 1 Solution: Symbol for 1 is $u$ and for 9 is e. $\therefore$IUPAC symbol for 119 is uue....
Read More →Find the electric field at point $mathrm{P}$ (as shown in figure) on the perpendicular
Question: Find the electric field at point $\mathrm{P}$ (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $\mathrm{L}$ carrying a charge $\mathrm{Q}$. The distance of the point $\mathrm{P}$ from the centre of the rod is $a=\frac{\sqrt{3}}{2} L$ (1) $\frac{Q}{2 \sqrt{3} \pi \varepsilon_{0} L^{2}}$(2) $\frac{\sqrt{3} Q}{4 \pi \varepsilon_{0} L^{2}}$(3) $\frac{Q}{3 \pi \varepsilon_{0} L^{2}}$(4) $\frac{Q}{4 \pi \varepsilon_{0} L^{2}}$Correct Option: 1 Sol...
Read More →The size of the iso-electronic species
Question: The size of the iso-electronic species $\mathrm{Cl}^{-}, \mathrm{Ar}$ and $\mathrm{Ca}^{2+}$ is affected by :azimulthal quantum number of valence shellelectron-electron interaction in the outer orbitalsprincipal quantum number of valence shellnuclear chargeCorrect Option: , 4 Solution: Iso-electronic species differ in size due to different effective nuclear charge....
Read More →The integral
Question: The integral $\int \frac{2 x^{3}-1}{x^{4}+x} d x$ is equal to : (Here $\mathrm{C}$ is a constant of integration)(1) $\frac{1}{2} \log _{e} \frac{\left|x^{3}+1\right|}{x^{2}}+C$(2) $\frac{1}{2} \log _{e} \frac{\left(x^{3}+1\right)^{2}}{\left|x^{3}\right|}+C$(3) $\log _{e}\left|\frac{x^{3}+1}{x}\right|+C$(4) $\log _{e} \frac{\left|x^{3}+1\right|}{x^{2}}+C$Correct Option: 3, Solution: Given integral, $I=\int \frac{\left(2 x^{3}-1\right) d x}{x^{4}+x}=\int \frac{\left(2 x-x^{-2}\right) d x...
Read More →B has a smaller first ionization enthalpy than Be.
Question: B has a smaller first ionization enthalpy than Be. Consider the following statements: (I) it is easier to remove $2 p$ electron than $2 s$ electron (II) $2 p$ electron of B is more shielded from the nucleus by the inner core of electrons than the $2 s$ electrons of $\mathrm{Be}$ (III) $2 s$ electron has more penetration power than $2 p$ electron (IV) atomic radius of $\mathrm{B}$ is more than $\mathrm{Be}$ (atomic number $\mathrm{B}=5, \mathrm{Be}=4$ ) The correct statements are:(I), (...
Read More →if
Question: If $\int x^{5} e^{-x^{2}} d x=g(x) e^{-x^{2}}+c$, where $c$ is a constant of integration, then $g(-1)$ is equal to :(1) $-1$(2) 1(3) $-\frac{5}{2}$(4) $-\frac{1}{2}$Correct Option: , 3 Solution: Let, $1=\int x^{2} \cdot e^{-x^{2}} d x$ Put $-x^{2}=t \Rightarrow-2 x d x=d t$ $1=\int \frac{t^{2} \cdot e^{t} d t}{(-2)}=\frac{-1}{2} e^{t}\left(t^{2}-2 t+2\right) c$ $\therefore g(x)=\frac{-1}{2}\left(x^{4}+2 x^{2}+2\right) \Rightarrow g(-1)=\frac{-5}{2}$...
Read More →Solve this
Question: $\frac{1}{(x-2)}+\frac{2}{(x-1)}=\frac{6}{x},(x \neq 2,1)$ Solution: Given : $\frac{1}{(x-2)}+\frac{2}{(x-1)}=\frac{6}{x}$ $\Rightarrow \frac{(x-1)+2(x-2)}{(x-1)(x-2)}=\frac{6}{x}$ $\Rightarrow \frac{3 x-5}{x^{2}-3 x+2}=\frac{6}{x}$ $\Rightarrow 3 x^{2}-5 x=6 x^{2}-18 x+12 \quad[$ On cross multiplying $]$ $\Rightarrow 3 x^{2}-13 x+12=0$ $\Rightarrow 3 x^{2}-(9+4) x+12=0$ $\Rightarrow 3 x^{2}-9 x-4 x+12=0$ $\Rightarrow 3 x(x-3)-4(x-3)=0$ $\Rightarrow(3 x-4)(x-3)=0$ $\Rightarrow 3 x-4=0$...
Read More →constant of integration, then :
Question: If $\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}$ $=\mathrm{A}\left(\tan ^{-1}\left(\frac{x-1}{3}\right)+\frac{f(x)}{x^{2}-2 x+10}\right)+\mathrm{C}$ where $\mathrm{C}$ is a constant of integration, then :(1) $A=\frac{1}{54}$ and $f(x)=3(x-1)$(2) $A=\frac{1}{81}$ and $f(x)=3(x-1)$(3) $A=\frac{1}{27}$ and $f(x)=9(x-1)$(4) $\mathrm{A}=\frac{1}{54}$ and $\mathrm{f}(\mathrm{x})=9(\mathrm{x}-1)^{2}$Correct Option: 1 Solution: Let $I=\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}=\int \fra...
Read More →The increasing order of the atomic radii of the follo elements is:
Question: The increasing order of the atomic radii of the follo elements is: (a) $\mathrm{C}$ (b) $\mathrm{O}$ (c) $\mathrm{F}$ (d) $\mathrm{Cl}$ (e) $\mathrm{Br}$(b) $(\mathrm{c})(\mathrm{d})(\mathrm{a})(\mathrm{e})$$(\mathrm{d})(\mathrm{c})(\mathrm{b})(\mathrm{a})(\mathrm{e})$$(\mathrm{c})(\mathrm{b})(\mathrm{a})(\mathrm{d})(\mathrm{e})$(a) $(\mathrm{b})(\mathrm{c})(\mathrm{d})(\mathrm{e})$Correct Option: , 3 Solution:...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3 \frac{1}{3}, \quad x \neq 2,4$ Solution: $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3 \frac{1}{3}, \quad x \neq 2,4$ $\Rightarrow \frac{(x-1)(x-4)+(x-2)(x-3)}{(x-2)(x-4)}=\frac{10}{3}$ $\Rightarrow \frac{x^{2}-5 x+4+x^{2}-5 x+6}{x^{2}-6 x+8}=\frac{10}{3}$ $\Rightarrow \frac{2 x^{2}-10 x+10}{x^{2}-6 x+8}=\frac{10}{3}$ $\Rightarrow \frac{x^{2}-5 x+5}{x^{2}-6 x+8}=\frac{5}{3}$ $\Rightarrow 3 x^{2}-15 x+15=5 x^{2}-30...
Read More →if
Question: If $\int e^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x \tan x+\sec ^{2} x\right)\right) d x$ $=e^{\sec x} f(x)+\mathrm{C}$, then a possible choice of $f(x)$ is:(1) $\sec x+\tan x+\frac{1}{2}$(2) $\sec x-\tan x-\frac{1}{2}$(3) $\sec x+x \tan x-\frac{1}{2}$(4) $x \sec x+\tan x+\frac{1}{2}$Correct Option: 1 Solution: Given, $\int e^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x+\tan x+\sec ^{2} x\right)\right) d x$ $=e^{\sec x} f(x)+C \quad \ldots(1)$ $\because \int e^{g(x)}\left(\left(g...
Read More →The first ionization energy
Question: The first ionization energy (in $\mathrm{kJ} / \mathrm{mol}$ ) of $\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}$ and $\mathrm{Si}$ respectively, are:$496,737,577,786$$496,577,737,786$$786,737,577,496$$496,577,786,737$Correct Option: 1 Solution: All given elements belongs to period III and generally their ionisation energy will increase along the period but Mg will show higher ionisation potential compared to $\mathrm{Al}$ due to its stable configuration. Thus correct order of ionisation energ...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $\frac{x-4}{x-5}+\frac{x-6}{x-7}=3 \frac{1}{3}, \quad x \neq 5,7$ Solution: $\frac{x-4}{x-5}+\frac{x-6}{x-7}=3 \frac{1}{3}, \quad x \neq 5,7$ $\Rightarrow \frac{(x-4)(x-7)+(x-5)(x-6)}{(x-5)(x-7)}=\frac{10}{3}$ $\Rightarrow \frac{x^{2}-11 x+28+x^{2}-11 x+30}{x^{2}-12 x+35}=\frac{10}{3}$ $\Rightarrow \frac{2 x^{2}-22 x+58}{x^{2}-12 x+35}=\frac{10}{3}$ $\Rightarrow \frac{x^{2}-11 x+29}{x^{2}-12 x+35}=\frac{5}{3}$ $\Rightarrow 3 x^{2}-33 x+8...
Read More →The integral
Question: The integral $\int \sec ^{2 / 3} x \operatorname{cosec}^{4 / 3} x d x$ is equal to:(1) $-3 \tan ^{-1 / 3} x+C$(2) $-\frac{3}{4} \tan ^{-4 / 3} x+C$(3) $-3 \cot ^{-1 / 3} x+C$(4) $3 \tan ^{-1 / 3} x+C$Correct Option: 1 Solution: $I=\int \sec ^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} d x$ $I=\int \frac{\sec ^{2} x d x}{\tan ^{\frac{4}{3}} x}$ Put $\tan x=z$ $\Rightarrow \sec ^{2} x d x=d z$ $\Rightarrow I=\int z^{-\frac{4}{3}} \cdot d z=\frac{z^{\frac{-1}{3}}}{\left(\frac...
Read More →Within each pair of elements
Question: Within each pair of elements $\mathrm{F} \ \mathrm{Cl}, \mathrm{S} \ \mathrm{Se}$, and $\mathrm{Li} \ \mathrm{Na}$, respectively, the elements that release more energy upon an electron gain are:$\mathrm{Cl}, \mathrm{Se}$ and $\mathrm{Na}$$\mathrm{Cl}, \mathrm{S}$ and $\mathrm{Li}$$\mathrm{F}, \mathrm{S}$ and $\mathrm{Li}$$\mathrm{F}$, Se and $\mathrm{Na}$Correct Option: , 2 Solution: Generally, electron affinity decreases on moving down a group. Chlorine has more electron affinity than...
Read More →A current of 6A enters one corner P of an equilateral triangle
Question: A current of $6 \mathrm{~A}$ enters one corner $\mathrm{P}$ of an equilateral triangle $\mathrm{PQR}$ having 3 wires of resistance $2 \Omega$ each and leaves by the corner $R$. The currents $i_{1}$ in ampere is Solution: The current $\mathrm{i}_{1}=\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}\right)^{\mathrm{i}}$ $=\left(\frac{2}{4+2}\right) \times 6$ $\mathrm{i}_{1}=2 \mathrm{~A}$...
Read More →The atomic radius of
Question: The atomic radius of $\mathrm{Ag}$ is closest to:AuNiCuHgCorrect Option: 1 Solution: Atomic size of elements of $4 d$ and $5 d$ transition series are nearly same due to lanthanide contraction....
Read More →then the function f(x) is equal to :
Question: If $\int \frac{d x}{x^{3}\left(1+x^{6}\right)^{2 / 3}}=x f(x)\left(1+x^{6}\right)^{\frac{1}{3}}+C$, where $\mathrm{C}$ is a constant of integration, then the function $f(x)$ is equal to :(1) $\frac{3}{x^{2}}$(2) $-\frac{1}{6 x^{3}}$(3) $-\frac{1}{2 x^{2}}$(4) $-\frac{1}{2 x^{3}}$Correct Option: , 4 Solution: Let, $\int \frac{d x}{x^{3}\left(1+x^{6}\right)^{\frac{2}{3}}}=\int \frac{d x}{x^{7}\left(1+x^{-6}\right)^{\frac{2}{3}}}$ Put $1+x^{-6}=t^{3} \quad \Rightarrow-6^{-7} d x=3 t^{2} d...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $\frac{x}{x+1}+\frac{x+1}{x}=2 \frac{4}{15}, \quad x \neq 0,-1$ Solution: $\frac{x}{x+1}+\frac{x+1}{x}=2 \frac{4}{15}, \quad x \neq 0, \quad-1$ $\frac{x}{x+1}+\frac{x+1}{x}=2 \frac{4}{15}, \quad x \neq 0,-1$ $\Rightarrow \frac{x^{2}+(x+1)^{2}}{x(x+1)}=\frac{34}{15}$ $\Rightarrow \frac{2 x^{2}+2 x+1}{x^{2}+x}=\frac{34}{15}$ $\Rightarrow 30 x^{2}+30 x+15=34 x^{2}+34 x$ $\Rightarrow 4 x^{2}+4 x-15=0$ $\Rightarrow 4 x^{2}+10 x-6 x-15=0$ $\Ri...
Read More →The atomic number of Unnilunium is
Question: The atomic number of Unnilunium is _______________ . Solution: (101)...
Read More →(where c is a constant of integration.)
Question: $\int \frac{\sin -}{\sin -} d x$ is equal to : (where $c$ is a constant of integration.)(1) $2 x+\sin x+2 \sin 2 x+c$(2) $x+2 \sin x+2 \sin 2 x+c$(3) $x+2 \sin x+\sin 2 x+c$(4) $2 x+\sin x+\sin 2 x+c$Correct Option: , 3 Solution: $\int \frac{\sin \left(\frac{5 x}{2}\right)}{\sin \left(\frac{x}{2}\right)} d x=\int \frac{2 \cos \frac{x}{2} \cdot \sin \frac{5 x}{2}}{2 \cos \frac{x}{2} \cdot \sin \frac{x}{2}} d x$ $=\int \frac{\sin 3 x+\sin 2 x}{\sin x} d x$ $=\int\left(3-4 \sin ^{2} x+2 \...
Read More →The correct order of the ionic radii of
Question: The correct order of the ionic radii of $\mathrm{O}^{2-}, \mathrm{N}^{3-}, \mathrm{F}^{-}, \mathrm{Mg}^{2+}$, $\mathrm{Na}^{+}$and $\mathrm{Al}^{3+}$ is :$\mathrm{N}^{3-}\mathrm{O}^{2-}\mathrm{F}^{-}\mathrm{Na}^{+}\mathrm{Mg}^{2+}\mathrm{Al}^{3+}$$\mathrm{Al}^{3+}\mathrm{Na}^{+}\mathrm{Mg}^{2+}\mathrm{O}^{2-}\mathrm{F}^{-}\mathrm{N}^{3-}$$\mathrm{Al}^{3+}\mathrm{Mg}^{2+}\mathrm{Na}^{+}\mathrm{F}^{-}\mathrm{O}^{2-}\mathrm{N}^{3-}$$\mathrm{N}^{3-}\mathrm{F}^{-}\mathrm{O}^{2-}\mathrm{Mg}^...
Read More →Two identical conducting spheres with negligible volume have 2.1 nC and -0.1 nC charges,
Question: Two identical conducting spheres with negligible volume have $2.1 \mathrm{nC}$ and $-0.1 \mathrm{nC}$ charges, respectively. They are brought into contact and then separated by a distance of $0.5 \mathrm{~m}$. The electrostatic force acting between the spheres is $\times 10^{-9} \mathrm{~N}$ [Given : $4 \pi \varepsilon_{0}=\frac{1}{9 \times 10^{9}}$ SI unit] Solution: When they are brought into contact \ then separated by a distance $=0.5 \mathrm{~m}$ Then charge distribution will be T...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: (i) $\frac{x}{x-1}+\frac{x-1}{x}=4 \frac{1}{4}, \quad x \neq 0,1$ (ii) $\frac{x-1}{2 x+1}+\frac{2 x+1}{x-1}=2, x \neq-\frac{1}{2}, 1$ Solution: (i) $\frac{x}{x-1}+\frac{x-1}{x}=4 \frac{1}{4}, \quad x \neq 0,1$ $\Rightarrow \frac{x^{2}+(x-1)^{2}}{x(x-1)}=\frac{17}{4}$ $\Rightarrow \frac{x^{2}+x^{2}-2 x+1}{x^{2}-x}=\frac{17}{4}$ $\Rightarrow \frac{2 x^{2}-2 x+1}{x^{2}-x}=\frac{17}{4}$ $\Rightarrow 8 x^{2}-8 x+4=17 x^{2}-17 x$ $\Rightarrow ...
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