The reaction of
Question: The reaction of $\mathrm{H}_{3} \mathrm{~N}_{3} \mathrm{~B}_{3} \mathrm{Cl}_{3}(\mathrm{~A})$ with $\mathrm{LiBH}_{4}$ in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of $(\mathrm{A})$ with $(\mathrm{C})$ leads to $\mathrm{H}_{3} \mathrm{~N}_{3} \mathrm{~B}_{3}(\mathrm{Me})_{3}$. Compounds (B) and (C) respectively, are:Borazine and $\mathrm{MeBr}$Diborane and $\mathrm{MeMgBr}$Boron nitride and MeBrBorazine and $\mathrm{MeMgBr}$Correct Option: , 4 Solution:...
Read More →The equation of the plane which contains
Question: The equation of the plane which contains the $\mathrm{y}$-axis and passes through the point $(1,2,3)$ is :(1) $x+3 z=10$(2) $x+3 z=0$(3) $3 x+z=6$(4) $3 x-z=0$Correct Option: , 4 Solution: $\overrightarrow{\mathrm{n}}=\hat{\mathrm{j}} \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ $=-3 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ So, $(-3)(x-1)+0(y-2)+(1)(z-3)=0$ $\Rightarrow-3 x+z=0$ Option 4 Alternate : Required plane is $\left|\begin{array}{lll}x y z \\ 0 1...
Read More →A particle performs simple harmonic motion with a period of 2 second.
Question: A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a} \mathrm{~s}$. The value of 'a' to the nearest integer is______ Solution: $t=\frac{2}{12}=\frac{1}{6}$ $\therefore$ Correct answer $=6.00$...
Read More →The compound that cannot act both as oxidising and reducing agent is:
Question: The compound that cannot act both as oxidising and reducing agent is:$\mathrm{H}_{3} \mathrm{PO}_{4}$$\mathrm{HNO}_{2}$$\mathrm{H}_{2} \mathrm{SO}_{3}$$\mathrm{H}_{2} \mathrm{O}_{2}$Correct Option: 1 Solution: In $\mathrm{H}_{3} \mathrm{PO}_{4}$ oxidation state of $\mathrm{P}$ is $+5$, which cannot be oxidised further to a higher oxidation state. Hence, it cannot act as reducing agent....
Read More →Two particles A and B of equal masses are suspended from two massless springs of spring constants
Question: Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ respectively.If the maximum velocities during oscillations are equal, the ratio of the amplitude of $\mathrm{A}$ and $\mathrm{B}$ is(1) $\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$(2) $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}$(3) $\sqrt{\frac{K_{1}}{K_{2}}}$(4) $\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$Correct Option: , 4 Solution: (4) $\mathrm{A}_{...
Read More →Two particles A and B of equal masses are suspended from two massless springs of spring constants
Question: Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $\mathrm{K}_{1}$ and $\mathrm{K}_{2}$ respectively.If the maximum velocities during oscillations are equal, the ratio of the amplitude of $\mathrm{A}$ and $\mathrm{B}$ is(1) $\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$(2) $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}$(3) $\sqrt{\frac{K_{1}}{K_{2}}}$(4) $\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$Correct Option: , 4 Solution: (4) $\mathrm{A}_{...
Read More →The acidic, basic and amphoteric oxides,
Question: The acidic, basic and amphoteric oxides, respectively, are:$\mathrm{Na}_{2} \mathrm{O}, \mathrm{SO}_{3}, \mathrm{Al}_{2} \mathrm{O}_{3}$$\mathrm{Cl}_{2} \mathrm{O}, \mathrm{CaO}, \mathrm{P}_{4} \mathrm{O}_{10}$$\mathrm{N}_{2} \mathrm{O}_{3}, \mathrm{Li}_{2} \mathrm{O}, \mathrm{Al}_{2} \mathrm{O}_{3}$$\mathrm{MgO}, \mathrm{Cl}_{2} \mathrm{O}, \mathrm{Al}_{2} \mathrm{O}_{3}$Correct Option: , 3 Solution: Generally, non-metal oxides are acidic in nature and metal oxides are basic in nature...
Read More →If the distance of the point
Question: If the distance of the point $(1,-2,3)$ from the plane $x+2 y-3 z+10=0$ measured parallel to the line, $\frac{x-1}{3}=\frac{2-y}{m}=\frac{z+3}{1}$ is $\sqrt{\frac{7}{2}}$, then the value of Iml is equal to Solution: DC of line $\equiv\left(\frac{3}{\sqrt{m^{2}+10}}, \frac{-m}{\sqrt{m^{2}+10}}, \frac{1}{\sqrt{m^{2}+10}}\right)$ Q lies on $x+2 y-3 z+10=0$ $1+\frac{3 \mathrm{r}}{\sqrt{\mathrm{m}^{2}+10}}-4-\frac{2 \mathrm{mr}}{\sqrt{\mathrm{m}^{2}+10}}-9-\frac{3 \mathrm{r}}{\sqrt{\mathrm{...
Read More →White phosphorus on reaction with concentrated
Question: White phosphorus on reaction with concentrated $\mathrm{NaOH}$ solution in an inert atmosphere of $\mathrm{CO}_{2}$ gives phosphine and compound $(X) .(X)$ on acidification with $\mathrm{HCl}$ gives compound $(Y)$. The basicity of compound $(Y)$ is:2143Correct Option: , 2 Solution:...
Read More →A block of mass 1 kg attached to a spring is made to oscillate
Question: A block of mass $1 \mathrm{~kg}$ attached to a spring is made to oscillate with an initial amplitude of $12 \mathrm{~cm}$. After 2 minutes the amplitude decreases to $6 \mathrm{~cm}$. Determine the value of the damping constant for this motion. (take In $2=0.693$ )(1) $0.69 \times 10^{2} \mathrm{~kg} \mathrm{~s}^{-1}$(2) $3.3 \times 10^{2} \mathrm{~kg} \mathrm{~s}^{-1}$(3) $1.16 \times 10^{-2} \mathrm{~kg} \mathrm{~s}^{-1}$(4) $5.7 \times 10^{-3} \mathrm{~kg} \mathrm{~s}^{-1}$Correct O...
Read More →Arrange the following bonds according to their average bond energies in descending order:
Question: Arrange the following bonds according to their average bond energies in descending order: $\mathrm{C}-\mathrm{Cl}, \mathrm{C}-\mathrm{Br}, \mathrm{C}-\mathrm{F}, \mathrm{C}-\mathrm{I}$$\mathrm{C}-\mathrm{F}\mathrm{C}-\mathrm{Cl}\mathrm{C}-\mathrm{Br}\mathrm{C}-\mathrm{I}$$\mathrm{C}-\mathrm{Br}\mathrm{C}-\mathrm{I}\mathrm{C}-\mathrm{C} 1\mathrm{C}-\mathrm{F}$$\mathrm{C}-\mathrm{I}\mathrm{C}-\mathrm{Br}\mathrm{C}-\mathrm{Cl}\mathrm{C}-\mathrm{F}$$\mathrm{C}-\mathrm{Cl}\mathrm{C}-\mathrm...
Read More →If the foot of the perpendicular from point $(4,3,8)$ on the line
Question: If the foot of the perpendicular from point $(4,3,8)$ on the line $L_{1}: \frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4}, l \neq 0$ is $(3,5,7)$, then the shortest distance between the line $\mathrm{L}_{1}$ and line $\mathrm{L}_{2}: \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-4}{4}=\frac{\mathrm{z}-5}{5}$ is equal to : (1) $\frac{1}{2}$(2) $\frac{1}{\sqrt{6}}$(3) $\sqrt{\frac{2}{3}}$(4) $\frac{1}{\sqrt{3}}$Correct Option: , 2 Solution: $(3,5,7)$ satisfy the line $\mathrm{L}_{1}$ $\frac{3-\mathr...
Read More →Consider two identical springs each of spring constant k
Question: Consider two identical springs each of spring constant $\mathrm{k}$ and negligible mass compared to the mass $\mathrm{M}$ as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is $\frac{T_{b}}{T_{a}}=\sqrt{x}$,_____ where value of $x$ is (Round off to the Nearest Integer) Solution: (2) $\mathrm{T}_{\mathrm{a}}=2 \pi \sqrt{\frac{\mathrm{M}}{\mathrm{K}}}$ $\mathrm{T}_{\mathrm{b}}=2 \pi \sqrt{\frac{\mathrm{M}}...
Read More →The number of bonds between sulphur and oxygen atoms in
Question: The number of bonds between sulphur and oxygen atoms in $\mathrm{S}_{2} \mathrm{O}_{8}^{2-}$ and the number of bonds between sulphur and sulphur atoms in rhombic sulphur, respectively, are:4 and 68 and 88 and 64 and 8Correct Option: , 2 Solution:...
Read More →For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?
Question: For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?(1) $x=0$(2) $x=\pm A$(3) $x=\pm \frac{A}{\sqrt{2}}$(4) $x=\frac{A}{2}$Correct Option: , 3 Solution: (3) $\mathrm{KE}=\mathrm{PE}$ $\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\frac{1}{2} m \omega^{2} x^{2}$ $A^{2}-x^{2}=x^{2}$ $2 x^{2}=A^{2}$ $\mathrm{x}=\pm \frac{\mathrm{A}}{\sqrt{2}}$...
Read More →When gypsum is heated to 393 K,
Question: When gypsum is heated to $393 \mathrm{~K}$, it forms:Anhydrous $\mathrm{CaSO}_{4}$$\mathrm{CaSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$$\mathrm{CaSO}_{4} \cdot 0.5 \mathrm{H}_{2} \mathrm{O}$Dead burnt plasterCorrect Option: , 3 Solution: Gypsum on heating to $393 \mathrm{~K}$ forms plaster of Paris....
Read More →When gypsum is heated to 393 K},
Question: When gypsum is heated to $393 \mathrm{~K}$, it forms:Anhydrous $\mathrm{CaSO}_{4}$$\mathrm{CaSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$$\mathrm{CaSO}_{4} \cdot 0.5 \mathrm{H}_{2} \mathrm{O}$Dead burnt plasterCorrect Option: , 3 Solution: Gypsum on heating to $393 \mathrm{~K}$ forms plaster of Paris....
Read More →Prove the following
Question: If $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ be an arbitrary point lying on a plane $\mathrm{P}$ which passes through the point $(42,0,0)(0,42,0)$ and $(0,0,42)$, then the value of expression $3+\frac{x-11}{(y-19)^{2}(z-12)^{2}}+\frac{y-19}{(x-11)^{2}(z-12)^{2}}$ $+\frac{z-12}{(x-11)^{2}(y-19)^{2}}-\frac{x+y+z}{14(x-11)(y-19)(z-12)}$ (1) 0(2) 3(3) 39(4) $-45$Correct Option: , 2 Solution: Plane passing through $(42,0,0),(0,42,0)$, $(0,0,42)$ From intercept from, equation of plane is $x+y+z...
Read More →Amplitude of a mass-spring system, which is
Question: Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass $=500 \mathrm{~g}$, Decay constant $=20 \mathrm{~g} / \mathrm{s}$ then how much time is required for the amplitude of the system to drop to half of its initial value ? $(\ln 2=0.693)$ (1) $34.65 \mathrm{~s}$(2) $17.32 \mathrm{~s}$(3) $0.034 \mathrm{~s}$(4) $15.01 \mathrm{~s}$Correct Option: 1 Solution: (1) $\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{-\gamma \mathrm{t}}=\mathrm{A}_{0} \m...
Read More →Prove the following
Question: Let $\mathrm{P}$ be a plane $l_{\mathrm{x}}+\mathrm{my}+\mathrm{nz}=0$ containing the line, $\frac{1-\mathrm{x}}{1}=\frac{\mathrm{y}+4}{2}=\frac{\mathrm{z}+2}{3}$. If plane $\mathrm{P}$ divides the line segment $\mathrm{AB}$ joining points $\mathrm{A}(-3,-6,1)$ and $\mathrm{B}(2,4,-3)$ in ratio $\mathrm{k}: 1$ then the value of $\mathrm{k}$ is equal to :(1) $1.5$(2) 3(3) 2(4) 4Correct Option: , 3 Solution: Point $\mathrm{C}$ is $\left(\frac{2 k-3}{k+1}, \frac{4 k-6}{k+1}, \frac{-3 k+1}...
Read More →Time period of a simple pendulum is $T$ inside a lift when the lift is stationary.
Question: Time period of a simple pendulum is $T$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $\mathrm{g} / 2$, the time period of pendulum will be :(1) $\sqrt{3} \mathrm{~T}$(2) $\frac{\mathrm{T}}{\sqrt{3}}$(3) $\sqrt{\frac{3}{2}} \mathrm{~T}$(4) $\sqrt{\frac{2}{3}} \mathrm{~T}$Correct Option: , 4 Solution: (4) When lift is stationary $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$ When lift is moving upwards $\Rightarrow$ Pseudo force acts ...
Read More →In the following reactions, products (A) and (B), respectively, are:
Question: In the following reactions, products (A) and (B), respectively, are: $\mathrm{NaOH}+\mathrm{Cl}_{2} \rightarrow(\mathrm{A})+$ side products (hot and conc.) $\mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{Cl}_{2} \rightarrow(\mathrm{B})+$ side products (dry)$\mathrm{NaClO}_{3}$, and $\mathrm{Ca}(\mathrm{OCl})_{2}$$\mathrm{NaClO}_{3}$ and $\mathrm{Ca}\left(\mathrm{ClO}_{3}\right)_{2}$$\mathrm{NaOCl}$ and $\mathrm{Ca}(\mathrm{OCl})_{2}$$\mathrm{NaOCl}$ and $\mathrm{Ca}\left(\mathrm{ClO}_{3}\right)_...
Read More →Chlorine reacts with hot and concentrated
Question: Chlorine reacts with hot and concentrated $\mathrm{NaOH}$ and produces compounds $(X)$ and $(Y)$. Compound $(X)$ gives white precipitate with silver nitrate solution. The average bond order between $\mathrm{Cl}$ and $\mathrm{O}$ atoms in $(Y)$ is __________________ Solution: (1.67)...
Read More →Let the position vectors of two points
Question: Let the position vectors of two points $\mathrm{P}$ and $\mathrm{Q}$ be $3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$, respectively. LetR and $\mathrm{S}$ be two points such that the direction ratios of lines $\mathrm{PR}$ and $\mathrm{QS}$ are $(4,-1,2)$ and $(-2,1,-2)$, respectively. Let lines $\mathrm{PR}$ andQS intersect at $\mathrm{T}$. If the vector $\overrightarrow{\mathrm{TA}}$ is perpendicular to both $\ov...
Read More →Two radioactive materials A and B have decay constants
Question: Two radioactive materials $\mathrm{A}$ and $\mathrm{B}$ have decay constants $10 \lambda$ and $\lambda$, respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of A to that of B will be $1 / \mathrm{e}$ after a time :(1) $\frac{1}{9 \lambda}$(2) $\frac{1}{11 \lambda}$(3) $\frac{11}{10 \lambda}$(4) $\frac{1}{10 \lambda}$Correct Option: 1 Solution: (1) $\mathrm{As}, \mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ so, $\frac{N_{A...
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