The integral
Question: The integral $\int \cos \left(\log _{e} x\right) d x$ is equal to : (where $\mathrm{C}$ is a constant of integration)(1) $\frac{x}{2}\left[\sin \left(\log _{e} x\right)-\cos \left(\log _{e} x\right)\right]+\mathrm{C}$(2) $x\left[\cos \left(\log _{e} x\right)+\sin \left(\log _{e} x\right)\right]+\mathrm{C}$(3) $\frac{x}{2}\left[\cos \left(\log _{e} x\right)+\sin \left(\log _{e} x\right)\right]+\mathrm{C}$(4) $x\left[\cos \left(\log _{e} x\right)-\sin \left(\log _{e} x\right)\right]+\mat...
Read More →In comparison to boron, berylium has :
Question: In comparison to boron, berylium has :lesser nuclear charge and lesser first ionisation enthalpy.greater nuclear charge and lesser first ionisation enthalpy.greater nulear charge and greater first ionisation enthalpy.lesser nuclear charge and greater first ionisation enthalpy.Correct Option: , 4 Solution: Nuclear charge : $\mathrm{B}\mathrm{Be}$ $\mathrm{Be}=1 s^{2} 2 s^{2}$ (more stable) $\mathrm{~B}=1 s^{2} 2 s^{2} 2 p^{1}$ $\therefore$ Ionisation energy of Be is greater than B due t...
Read More →The integral
Question: The integral $\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x$ is equal to: (where $\mathrm{C}$ is a constant of integration) (1) $\frac{x^{4}}{6\left(2 x^{4}+3 x^{2}+1\right)^{3}}+\mathrm{C}$(2) $\frac{x^{12}}{6\left(2 x^{4}+3 x^{2}+1\right)^{3}}+\mathrm{C}$(3) $\frac{x^{4}}{\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C$(4) $\frac{x^{12}}{\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C$Correct Option: , 2 Solution: $I=\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4...
Read More →Two isolated conducting spheres
Question: Two isolated conducting spheres $S_{1}$ and $S_{2}$ of radius $\frac{2}{3} R$ and $\frac{1}{3} R$ have $12 \mu \mathrm{C}$ and $-3 \mu \mathrm{C}$ charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on $S_{1}$ and $S_{2}$ are respectively :(1) $4.5 \mu \mathrm{C}$ on both(2) $+4.5 \mu \mathrm{C}$ and $-4.5 \mu \mathrm{C}$(3) $3 \mu \mathrm{C}$ and $6 \mu \mathrm{C}$(4) $6 \mu \mathrm...
Read More →The pair that has similar atomic radii is :
Question: The pair that has similar atomic radii is :$\mathrm{Mn}$ and $\mathrm{Re}$Ti and Hf$\mathrm{Sc}$ and $\mathrm{Ni}$Mo and WCorrect Option: , 4 Solution: Mo and W belong to group-6 and period 5 (4d series) and $6(5 d$ series) respectively. Due to lanthanoid contraction, radius of Mo and W are almost same i.e. $0.140 \mathrm{~nm}$ and $0.141 \mathrm{~nm}$ respectively....
Read More →The group number, number of valence electrons,
Question: The group number, number of valence electrons, and valency of an element with atomic number 15, respectively, are :16,5 and 215,5 and 316,6 and 315,6 and 2Correct Option: , 2 Solution: Phosphorus has atomic number 15 . Its group number is 15 , number of valence electrons is 5 and valency is 3 ....
Read More →The correct order of the first ionization enthalpies is :
Question: The correct order of the first ionization enthalpies is :$\mathrm{Ti}\mathrm{Mn}\mathrm{Zn}\mathrm{Ni}$$\mathrm{Ti}\mathrm{Mn}\mathrm{Ni}\mathrm{Zn}$$\mathrm{Mn}\mathrm{Ti}\mathrm{Zn}\mathrm{Ni}$$\mathrm{Zn}\mathrm{Ni}\mathrm{Mn}\mathrm{Ti}$Correct Option: , 2 Solution: I.E. increases on moving left to right in a period. $\therefore \mathrm{Ti}\mathrm{Mn}\mathrm{Ni}\mathrm{Zn}_{\text {, }}$...
Read More →A charge Q is distributed over two concentric conducting
Question: A charge $Q$ is distributed over two concentric conducting thin spherical shells radii $r$ and $R(Rr)$. If the surface charge densities on the two shells are equal, the electric potential at the common centre is : (1) $\frac{1}{4 \pi \varepsilon_{0}} \frac{(R+r)}{2\left(R^{2}+r^{2}\right)} Q$(2) $\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 R+r)}{\left(R^{2}+r^{2}\right)} Q$(3) $\frac{1}{4 \pi \varepsilon_{0}} \frac{(R+2 r) Q}{2\left(R^{2}+r^{2}\right)}$(4) $\frac{1}{4 \pi \varepsilon_{0}}...
Read More →The integral
Question: The integral $\int_{1}^{e}\left\{\left(\frac{x}{e}\right)^{2 x}-\left(\frac{e}{x}\right)^{x}\right\} \log _{e} x d x$ is equal to :(1) $\frac{1}{2}-e-\frac{1}{e^{2}}$(2) $-\frac{1}{2}+\frac{1}{e}-\frac{1}{2 e^{2}}$(3) $\frac{3}{2}-\frac{1}{e}-\frac{1}{2 e^{2}}$(4) $\frac{3}{2}-e-\frac{1}{2 e^{2}}$Correct Option: , 4 Solution: $I=\int_{1}^{e}\left\{\left(\frac{x}{e}\right)^{2 x}-\left(\frac{e}{x}\right)^{x}\right\} \log _{e} x d x$ Let $\left(\frac{x}{e}\right)^{x}=t$ $\Rightarrow \quad...
Read More →The isoelectronic set of ions is :
Question: The isoelectronic set of ions is :$\mathrm{N}^{3-}, \mathrm{O}^{2-}, \mathrm{F}^{-}$and $\mathrm{Na}^{+}$$\mathrm{N}^{3-}, \mathrm{Li}^{+}, \mathrm{Mg}^{2+}$ and $\mathrm{O}^{2-}$$\mathrm{F}^{-}, \mathrm{Li}^{+}, \mathrm{Na}^{+}$and $\mathrm{Mg}^{2+}$$\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{O}^{2-}$ and $\mathrm{F}^{-}$Correct Option: 1 Solution: Atomic numbers of $\mathrm{N}, \mathrm{O}, \mathrm{F}$ and $\mathrm{Na}$ are $7,8,9$ and 11 respectively. Therefore, total number of electr...
Read More →A small point mass carrying some positive charge on it,
Question: A small point mass carrying some positive charge on it, is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of the following options then correctly describe the trajectory of the mass? (Curves are drawn schematically and are not to scale). Correct Option: , 4 Solution: (4) Net force acting on the particle, $\vec{F}=q E \hat{i}+m g \hat{j}$ Net acceleration of particle is constant, initial velocity is zero therefore p...
Read More →if
Question: If $\int \frac{x+1}{\sqrt{2 x-1}} \mathrm{~d} x=f(x) \sqrt{2 x-1}+\mathrm{C}$, where $\mathrm{C}$ is a constant of integration, then $f(x)$ is equal to:(1) $\frac{1}{3}(x+1)$(2) $\frac{2}{3}(x+2)$(3) $\frac{2}{3}(x-4)$(4) $\frac{1}{3}(x+4)$Correct Option: , 3 Solution: Let $I=\int \frac{x+1}{\sqrt{2 x-1}} d x$ Put $\sqrt{2 x-1}=t$ $\therefore \quad 2 x-1=t^{2} \Rightarrow d x=t d t$ $I=\int \frac{\left(t^{2}+3\right)}{2} d t=\frac{t^{3}}{6}+\frac{3 t}{2}+C$ $=\frac{(2 x-1)^{\frac{3}{2}...
Read More →Consider the hydrated ions of
Question: Consider the hydrated ions of $\mathrm{Ti}^{2+}, \mathrm{V}^{2+}, \mathrm{Ti}^{3+}$, and $\mathrm{Sc}^{3+}$. The correct order of their spin-only magnetic moments is :$\mathrm{V}^{2+}\mathrm{Ti}^{2+}\mathrm{Ti}^{3+}\mathrm{Sc}^{3+}$$\mathrm{Sc}^{3+}\mathrm{Ti}^{3+}\mathrm{Ti}^{2+}\mathrm{V}^{2+}$$\mathrm{Ti}^{3+}\mathrm{Ti}^{2+}\mathrm{Sc}^{3+}\mathrm{V}^{2+}$$\mathrm{Sc}^{3+}\mathrm{Ti}^{3+}\mathrm{V}^{2+}\mathrm{Ti}^{2+}$Correct Option: , 2 Solution: Electronic configuration of the g...
Read More →chosen integer $m$ and a function A(x)
Question: If $\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x=A(\mathrm{x})\left(\sqrt{1-x^{2}}\right)^{m}+C$, for a suitable chosen integer $m$ and a function $A(x)$, where $C$ is a constant of integration, then $(\mathrm{A}(\mathrm{x}))^{\mathrm{m}}$ equals :(1) $\frac{-1}{27 x^{9}}$(2) $\frac{-1}{3 x^{3}}$(3) $\frac{1}{27 x^{6}}$(4) $\frac{1}{9 x^{4}}$Correct Option: 1 Solution: $A(x)\left(\sqrt{1-x^{2}}\right)^{m}+C=\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x$ $=\int \frac{\sqrt{\frac{1}{x^{2}}-1}}{x^{3}} d...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $3\left(\frac{7 x+1}{5 x-3}\right)-4\left(\frac{5 x-3}{7 x+1}\right)=11, \quad x \neq \frac{3}{5}, \quad-\frac{1}{7}$ Solution: $3\left(\frac{7 x+1}{5 x-3}\right)-4\left(\frac{5 x-3}{7 x+1}\right)=11, \quad x \neq \frac{3}{5},-\frac{1}{7}$ $\Rightarrow \frac{3(7 x+1)^{2}-4(5 x-3)^{2}}{(5 x-3)(7 x+1)}=11$ $\Rightarrow \frac{3\left(49 x^{2}+14 x+1\right)-4\left(25 x^{2}-30 x+9\right)}{35 x^{2}-16 x-3}=11$ $\Rightarrow \frac{147 x^{2}+42 x+...
Read More →if
Question: If $\int x^{5} \mathrm{e}^{-4 x^{3}} d x=\frac{1}{48} \mathrm{e}^{-4 x^{3}} f(x)+\mathrm{C}$, where $\mathrm{C}$ is a constant of integration, then $f(x)$ is equal to:(1) $-2 x^{3}-1$(2) $-4 x^{3}-1$(3) $-2 x^{3}+1$(4) $4 x^{3}+1$Correct Option: , 2 Solution: $I=\int x^{5} e^{-4 x^{3}} d x$ Put $-4 x^{3}=\theta$ $\Rightarrow \quad-12 x^{2} d x=d \theta$ $\Rightarrow x^{2} d x=-\frac{d \theta}{12}$ $I=\int \frac{1}{48} \theta e^{\theta} d \theta=\frac{1}{48}\left[\theta e^{\theta}-e^{\t...
Read More →A charged particle (mass m and charge q ) moves along X
Question: A charged particle (mass $m$ and charge $q$ ) moves along $X$ axis with velocity $V_{0}$. When it passes through the origin it enters a region having uniform electric field $\vec{E}=-E \hat{j}$ which extends upto $x=d$. Equation of path of electron in the region $xd$ is : (1) $y=\frac{q E d}{m V_{0}^{2}}(x-d)$(2) $y=\frac{q E d}{m V_{0}^{2}}\left(\frac{d}{2}-x\right)$(3) $y=\frac{q E d}{m V_{0}^{2}} x$(4) $y=\frac{q E d^{2}}{m V_{0}^{2}} x$Correct Option: , 2 Solution: (2) $F_{x}=0, a_...
Read More →let n>2
Question: Let $\mathrm{n} \geq 2$ be a natural number and $0\theta\frac{\pi}{2}$ Then $\int \frac{\left(\sin ^{n} \theta+\sin \theta\right)^{\frac{1}{n}} \cos \theta}{\sin ^{n+1} \theta} d \theta$ is equal to: (where $\mathrm{C}$ is a constant of integration)(1) $\frac{\mathrm{n}}{\mathrm{n}^{2}-1}\left(1-\frac{1}{\sin ^{\mathrm{n}-1} \theta}\right)^{\frac{\mathrm{n}+1}{\mathrm{n}}}+\mathrm{C}$(2) $\frac{\mathrm{n}}{\mathrm{n}^{2}+1}\left(1-\frac{1}{\sin ^{\mathrm{n}-1} \theta}\right)^{\frac{\ma...
Read More →Solve each of the following quadratic equations
Question: Solve each of the following quadratic equations $3\left(\frac{3 x-1}{2 x+3}\right)-2\left(\frac{2 x+3}{3 x-1}\right)=5, \quad x \neq \frac{1}{3},-\frac{3}{2}$ Solution: $3\left(\frac{3 x-1}{2 x+3}\right)-2\left(\frac{2 x+3}{3 x-1}\right)=5, \quad x \neq \frac{1}{3},-\frac{3}{2}$ $\Rightarrow \frac{3(3 x-1)^{2}-2(2 x+3)^{2}}{(2 x+3)(3 x-1)}=5$ $\Rightarrow \frac{3\left(9 x^{2}-6 x+1\right)-2\left(4 x^{2}+12 x+9\right)}{6 x^{2}+7 x-3}=5$ $\Rightarrow \frac{27 x^{2}-18 x+3-8 x^{2}-24 x-18...
Read More →if f(x) =
Question: If $f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x,(x \geq 0)$ and $f(0)=0$, then the value of $f(1)$ is:(1) $-\frac{1}{2}$(2) $-\frac{1}{4}$(3) $\frac{1}{2}$(4) $\frac{1}{4}$Correct Option: , 4 Solution: $f(x)=$ $=\int \frac{5 x^{8}+7 x^{6}}{x^{14}\left(x^{-5}+x^{-7}+2\right)^{2}} d x$ $=\int \frac{5 x^{-6}+7 x^{-8}}{\left(2+x^{-7}+x^{-5}\right)^{2}} d x$ Let $2+x^{-7}+x^{-5}=t$ $\Rightarrow \quad\left(-7 x^{-8}-5 x^{-6}\right) d x=d t$ $\Rightarrow f(x)=\int \...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: (i) $\frac{1}{x+1}+\frac{2}{x+2}=\frac{5}{x+4}, \quad x \neq-1,-2,-4$ (ii) $\frac{1}{x+1}+\frac{3}{5 x+1}=\frac{5}{x+4}, x \neq-1,-\frac{1}{5},-4$ Solution: (i) $\frac{1}{x+1}+\frac{2}{x+2}=\frac{5}{x+4}, \quad x \neq-1,-2,-4$ $\Rightarrow \frac{x+2+2 x+2}{(x+1)(x+2)}=\frac{5}{x+4}$ $\Rightarrow \frac{3 x+4}{x^{2}+3 x+2}=\frac{5}{x+4}$ $\Rightarrow(3 x+4)(x+4)=5\left(x^{2}+3 x+2\right)$ $\Rightarrow 3 x^{2}+16 x+16=5 x^{2}+15 x+10$ $\Rig...
Read More →Given below are two statements :
Question: Given below are two statements : Statement - I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere. Statement - II : If $\mathrm{R}$ is the radius of a solid metallic sphere and $\mathrm{Q}$ be the total charge on it. The electric field at any point on the spherical surface of radius $r(R)$ is zero but theelectric flux passing through this closed spherical surfac...
Read More →for x
Question: For $x^{2} \neq \mathrm{n} \pi+1, \mathrm{n} \in \mathrm{N}$ (the set of natural numbers), the integral $\int x \sqrt{\frac{2 \sin \left(x^{2}-1\right)-\sin 2\left(x^{2}-1\right)}{2 \sin \left(x^{2}-1\right)+\sin 2\left(x^{2}-1\right)}} d x$ is equal to: (where $c$ is a constant of integration)(1) $\log _{e}\left|\frac{1}{2} \sec ^{2}\left(x^{2}-1\right)\right|+c$(2) $\frac{1}{2} \log _{\mathrm{e}}\left|\sec \left(x^{2}-1\right)\right|+\mathrm{c}$(3) $\frac{1}{2} \log _{e}\left|\sec ^{...
Read More →then the functions A(x) and B(x) are respectively:
Question: Let $\alpha \in(0, \pi / 2)$ be fixed. If the integral $\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} d x=\mathrm{A}(x) \cos 2 \alpha+\mathrm{B}(x) \sin 2 \alpha+\mathrm{C}$, where $\mathrm{C}$ is a constant of integration, then the functions $\mathrm{A}(x)$ and $\mathrm{B}(x)$ are respectively:(1) $x+\alpha$ and $\log _{e}|\sin (x+\alpha)|$(2) $x-\alpha$ and $\log _{e}|\sin (x-\alpha)|$(3) $x-\alpha$ and $\log _{e}|\cos (x-\alpha)|$(4) $x+\alpha$ and $\log _{e}|\sin (x-\alpha)|$C...
Read More →The correct statements among I to III regarding group 13 element oxides are,
Question: The correct statements among I to III regarding group 13 element oxides are, (I) Boron trioxide is acidic. (II) Oxides of aluminium and gallium are amphoteric. (III)Oxides of indium and thallium are basic.(I) and (II) only(I), (II) and (III)(I) and (III) only(II) and (III) onlyCorrect Option: Solution: (I) $\mathrm{B}_{2} \mathrm{O}_{3}$-Acidic oxide (II) $\mathrm{Al}_{2} \mathrm{O}_{3} \ \mathrm{Ga}_{2} \mathrm{O}_{3}-$ Amphoteric oxide (III) $\mathrm{In}_{2} \mathrm{O}_{3} \ \mathrm{...
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