Time period of a simple pendulum is T.
Question: Time period of a simple pendulum is $T$. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{\alpha}{\beta} T$. The value of $\alpha$ is Solution: (7) For given $\left(\frac{5}{8}\right)$ oscillation, we can write it as $\rightarrow\left(\frac{1}{2}+\frac{1}{8}\right)$ And we know for half oscillations time $\rightarrow \frac{T}{2}$ For final point $\rightarrow \pi+\frac{\pi}{6} \quad \Rightarrow \frac{7 \pi}{6}$ Time $\rightarrow \frac{7 T}{12} ...
Read More →Time period of a simple pendulum is T.
Question: Time period of a simple pendulum is $T$. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{\alpha}{\beta} T$. The value of $\alpha$ is Solution: (7) For given $\left(\frac{5}{8}\right)$ oscillation, we can write it as $\rightarrow\left(\frac{1}{2}+\frac{1}{8}\right)$ And we know for half oscillations time $\rightarrow \frac{T}{2}$ For final point $\rightarrow \pi+\frac{\pi}{6} \quad \Rightarrow \frac{7 \pi}{6}$ Time $\rightarrow \frac{7 T}{12} ...
Read More →If the mirror image of the point
Question: If the mirror image of the point $(1,3,5)$ with respect to the plane $4 x-5 y+2 z=8$ is $(\alpha, \beta, \gamma)$, then $5(\alpha+\beta+\gamma)$ equals:(1) 47(2) 39(3) 43(4) 41Correct Option: 1 Solution: Image of $(1,3,5)$ in the plane $4 x-5 y+2 z=8$ is $(\alpha, \beta, \gamma)$ $\Rightarrow \frac{\alpha-1}{4}=\frac{\beta-3}{-5}=\frac{\gamma-5}{2}=-2 \frac{(4(1)-5(3)+2(5)-8)}{4^{2}+5^{2}+2^{2}}=\frac{2}{5}$ $\therefore \alpha=1+4\left(\frac{2}{5}\right)=\frac{13}{5}$ $\beta=3-5\left(\...
Read More →A particle excutes S.H.M with amplitude 'a' and time period T.
Question: A particle excutes S.H.M with amplitude 'a' and time period $T$. The displacement of the particle when its speed is half of maximum speed is $\frac{\sqrt{x} a}{2} .$ The value of $x$ is Solution: (3) Fora particle excutes S.H.M $V=\omega \sqrt{a^{2}-x^{2}}$ Given $V=\frac{V_{\max }}{2} \Rightarrow \frac{A \omega}{2}$ $\frac{A^{2} \omega^{2}}{4}=\omega^{2} a^{2}-\omega^{2} x^{2}$ $x=\frac{\sqrt{3}}{2} a$...
Read More →Let L be a line obtained from the intersection
Question: Let $\mathrm{L}$ be a line obtained from the intersection of two planes $x+2 y+z=6$ and $y+2 z=4$. If point $P(\alpha, \beta, \gamma)$ is the foot of perpendicular from $(3,2,1)$ on $L$, then the value of $21(\alpha+\beta+\gamma)$ equals :(1) 142(2) 68(3) 136(4) 102Correct Option: , 4 Solution: Dr's of line $\left|\begin{array}{lll}\hat{i} \hat{j} \hat{k} \\ 1 2 1 \\ 0 1 2\end{array}\right|=3 \hat{i}-2 \hat{j}+\hat{k}$ Dr/s: $-(3,-2,1)$ Points on the line $(-2,4,0)$ Equation of the lin...
Read More →Given below are two statements :
Question: Given below are two statements : Statement (I) :- A second's pendulum has a time period of 1 second. Statement (II) :- It takes precisely one second to move between the two extreme positions. In the light of the above statements, choose the correct answer from the options give below.(1) Both Statement I and Statement II are false(2) Statement I is true but Statement II is false(3) Statement I is false but Statement II is true(4) Both Statement $I$ and Statement II is trueCorrect Option...
Read More →Among the compounds A and B with molecular formula
Question: Among the compounds $A$ and $B$ with molecular formula $\mathrm{C}_{9} \mathrm{H}_{18} \mathrm{O}_{3}, A$ is having higher boiling point the $B$. The possible structures of $A$ and $B$ are: Correct Option: 1 Solution: In (A), $-\mathrm{OH}$ group is present, so inter-molecular H-bonding is possible while in (B), due to methoxy group there is no possibility of Inter-molecular H-bonding. So $\mathrm{A}$ is having higher boiling point than B....
Read More →Prove the following
Question: Let $(\lambda, 2,1)$ be a point on the plane which passes through the point $(4,-2,2) .$ If the plane is perpendicular to the line joining the point $(-2,-21,29)$ and $(-1,-16,23)$, then $\left(\frac{\lambda}{11}\right)^{2}-\frac{4 \lambda}{11}-4$ is equal to Solution: $\overrightarrow{A B} \perp \overrightarrow{P Q}$ $[(4-\lambda) \hat{i}-4 \hat{j}+\hat{k}] \cdot[+\hat{i}+5 \hat{j}-6 \hat{k}]=0$ $4-\lambda-20-6=0$ $N=-22$ Now, $\frac{\lambda}{11}=-2$ $\Rightarrow\left(\frac{\lambda}{1...
Read More →A particle executes S.H.M., the graph of velocity as a function of displacement is:c
Question: A particle executes S.H.M., the graph of velocity as a function of displacement is:(1) a circle(2) a parabola(3) an ellipse(4) a helixCorrect Option: , 3 Solution: (3) For a body performing SHM, relation between velocity and displacement $v=\omega \sqrt{A^{2}-x^{2}}$ now, square both side $v^{2}=w^{2}\left(A^{2}-x^{2}\right)$ $\Rightarrow v^{2}=w^{2} A^{2}-\omega^{2} x^{2}$ $v^{2}+\omega^{2} x^{2}=\omega^{2} A^{2}$ divide whole equation by $\omega^{2} \mathrm{~A}^{2} \frac{\mathrm{v}^{...
Read More →Prove the following
Question: If $(1,5,35),(7,5,5),(1, \lambda, 7)$ and $(2 \lambda, 1,2)$ are coplanar, then the sum of all possible values of $\lambda$ is:(1) $-\frac{44}{5}$(2) $\frac{39}{5}$(3) $-\frac{39}{5}$(4) $\frac{44}{5}$Correct Option: , 4 Solution: Let $\mathrm{P}(1,5,35), \mathrm{Q}(7,5,5), \mathrm{R}(1, \lambda, 7), \mathrm{S}(2 \lambda, 1,2)$ $\begin{array}{ll}\overrightarrow{\mathrm{PQPR}} \overrightarrow{\mathrm{PS}}]=0\end{array}$ $\left|\begin{array}{ccc}6 0 -30 \\ 0 \lambda-5 -28 \\ 2 \lambda-1 ...
Read More →If two similar springs each of spring constant
Question: If two similar springs each of spring constant $\mathrm{K}_{1}$ are joined in series, the new spring constant and time period would be changed by a factor :(1) $\frac{1}{2}, \sqrt{2}$(2) $\frac{1}{4}, 2 \sqrt{2}$(3) $\frac{1}{2}, 2 \sqrt{2}$(4) $\frac{1}{4}, \sqrt{2}$Correct Option: 1 Solution: $T=2 \pi \sqrt{\frac{m}{K_{1}}}$...
Read More →Consider the three planes
Question: Consider the three planes $P_{1}: 3 x+15 y+21 z=9$ $P_{2}: x-3 y-z=5$, and $P_{3}: 2 x+10 y+14 z=5$ Then, which one of the following is true? (1) $\mathrm{P}_{1}$ and $\mathrm{P}_{3}$ are parallel.(2) $\mathrm{P}_{2}$ and $\mathrm{P}_{3}$ are parallel.(3) $\mathrm{P}_{1}$ and $\mathrm{P}_{2}$ are parallel.(4) $\mathrm{P}_{1}, \mathrm{P}_{2}$ and $\mathrm{P}_{3}$ all are parallel.Correct Option: 1 Solution: $P_{1}=x+5 y+7 z=3$ $P_{2}=x-3 y-z=5$ $P_{3}=x+5 y+7 z=5 / 2$ $\Rightarrow P_{1}...
Read More →Assume that a tunnel is dug along a chord of the earth,
Question: Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance $(R / 2)$ from the earth's centre, where ' $R$ ' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period :(1) $2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$(2) $\frac{1}{2 \pi} \sqrt{\frac{g}{R}}$(3) $\frac{2 \pi \mathrm{R}}{\mathrm{g}}$(4) $\frac{g}{2 \pi R}$Correct Option: 1 Solution: (1) $\...
Read More →Prove the following
Question: A line ' $\ell$ passing through origin is perpendicular to the lines $\ell_{1}: \overrightarrow{\mathrm{r}}=(3+\mathrm{t}) \hat{\mathrm{i}}+(-1+2 \mathrm{t}) \hat{\mathrm{j}}+(4+2 \mathrm{t}) \hat{\mathrm{k}}$ $\ell_{2}: \overrightarrow{\mathrm{r}}=(3+2 \mathrm{~s}) \hat{\mathrm{i}}+(3+2 \mathrm{~s}) \hat{\mathrm{j}}+(2+\mathrm{s}) \hat{\mathrm{k}}$ If the co-ordinates of the point in the first octant on ${ }^{\prime} \ell_{2}{ }^{\prime}$ at the distance of $\sqrt{17}$ from the point ...
Read More →Aflask contains a mixture of isohexaneand 3-methylpentane.
Question: Aflask contains a mixture of isohexaneand 3-methylpentane. One of the liquids boils at $63^{\circ} \mathrm{C}$ while the other boils at $60^{\circ} \mathrm{C}$. What is the best way to separate the two liquids and which one will be distilled out first?fractional distillation, isohexanesimple distillation, 3-methylpentanesimple distillation, isohexanefractional distillation, 3-methylpentaneCorrect Option: 1 Solution: Liquid having lower boiling point comes out first in fractional distil...
Read More →A chromatography column, packed with silica gel as stationary phase,
Question: A chromatography column, packed with silica gel as stationary phase, was used to separate a mixture of compounds consisting of (A) benzanilide (B) aniline and (C) acetophenone. When the column is eluted with a mixture of solvents, hexane: ethyl acetate $(20: 80)$, the sequence of obtained compounds is:(B), (C) and (A)(B), (A) and (C)(C), (A) and (B)(A), (B) and (C)Correct Option: , 3 Solution: Compounds which are able to form strong $\mathrm{H}$-bond with the stationary phase (silica g...
Read More →Two identical spring of spring constant '2K' are attached to a block of mass m and to fixed support (see figure).
Question: Two identical spring of spring constant '2K' are attached to a block of mass $m$ and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. Then time period of oscillations of this system is: (1) $\pi \sqrt{\frac{m}{k}}$(2) $\pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$(3) $2 \pi \sqrt{\frac{m}{k}}$(4) $2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$Correct Option: 1 Solution: Dut to parallel combination $\mat...
Read More →A plane passes through the points
Question: A plane passes through the points $\mathrm{A}(1,2,3), \mathrm{B}(2,3,1)$ and $\mathrm{C}(2,4,2)$. If $\mathrm{O}$ is the origin and $\mathrm{P}$ is $(2,-1,1)$, then the projection of $\overrightarrow{O P}$ on this plane is of length:(1) $\sqrt{\frac{2}{5}}$(2) $\sqrt{\frac{2}{3}}$(3) $\sqrt{\frac{2}{11}}$(4) $\sqrt{\frac{2}{7}}$Correct Option: , 3 Solution: $\mathrm{A}(1,2,3), \mathrm{B}(2,3,1), \mathrm{C}(2,4,2), \mathrm{O}(0,0,0)$ Equation of plane passing through $A, B, C$ will be $...
Read More →A solution of m-chloroaniline,
Question: A solution of $m$-chloroaniline, $m$-chlorophenol and $m$-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of $\mathrm{NaHCO}_{3}$ to give fraction $A$. The left over organic phase was extracted with dilute $\mathrm{NaOH}$ solution to give fraction $B$. The final organic layer was labelled as fraction $C$. Fractions $A, B$ and $C$, contain respectively:$m$-chlorobenzoic acid, $m$-chloroaniline and $m$-chlorophenol$m$-chlorobenzoic acid, $m$-chloroph...
Read More →Match the following :
Question: Match the following : (I)-(D), (II)-(C), (III)-(B), (IV)-(A)(I)-(B), (II)-(D), (III)-(E), (IV)-(A)(I)-(D), (II)-(C), (III)-(E), (IV)-(A)(I)-(B), (II)-(A), (III)-(C), (IV)-(D)Correct Option: , 3 Solution: (I) Lucas reagent - Conc. $\mathrm{HCl} / \mathrm{ZnCl}_{2}$ (II) Dumas method $-\mathrm{CuO} / \mathrm{CO}_{2}$ (III) $\mathrm{Kjeldahl's~method~-~} \mathrm{H}_{2} \mathrm{SO}_{4}$ (IV) Hinsberg test $-\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl} / \mathrm{aq} . \mathrm{K...
Read More →if
Question: $Y=A \sin \left(\omega t+\phi_{0}\right)$ is the time - displacement equation of a SHM, At $t=0$ the displacement of the particle is $Y=\frac{A}{2}$ and it is moving along negative $x$-direction. Then the initial phase angle $\phi_{0}$ will be(1) $\frac{\pi}{6}$(2) $\frac{\pi}{3}$(3) $\frac{2 \pi}{3}$(4) $\frac{5 \pi}{6}$Correct Option: , 4 Solution: The initial phase angle $\phi_{0}=\pi-\frac{\pi}{6}$ $=\frac{5-\pi}{6}$...
Read More →In an estimation of bromine by Carius method,
Question: In an estimation of bromine by Carius method, $1.6 \mathrm{~g}$ of an organic compound gave $1.88 \mathrm{~g}$ of $\mathrm{AgBr}$. The mass percentage of bromine in the compound is ______________ .(Atomic mass, $\mathrm{Ag}=108, \mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}$ ) Solution: (50)...
Read More →The equation of the line through the
Question: The equation of the line through the point $(0,1,2)$ and perpendicular to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is :(1) $\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$(2) $\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{3}$(3) $\frac{x}{3}=\frac{y-1}{-4}=\frac{z-2}{3}$(4) $\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{-3}$Correct Option: 1 Solution: $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=\lambda$ Any point on this line $(2 \lambda+1,3 \lambda-1,-2 \lambda+1)$ Direction ratio of given line ...
Read More →If the time period of a two meter long simple pendulum is 2 s,
Question: If the time period of a two meter long simple pendulum is $2 \mathrm{~s}$, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :(1) $2 \pi^{2} \mathrm{~ms}^{-2}$(2) $16 \mathrm{~m} / \mathrm{s}^{2}$(3) $9.8 \mathrm{~ms}^{-2}$(4) $\pi^{2} \mathrm{~ms}^{-2}$Correct Option: 1 Solution: (1) $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{g}}}$ $\mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{I}}{\mathrm{g}}$ $\mathrm{g}=\frac{4 \pi^{2} \mathrm{I}}{\mathrm{T}^{2}}$ $...
Read More →The Kjeldahl method of Nitrogen estimation fails for which of the following
Question: The Kjeldahl method of Nitrogen estimation fails for which of the following reaction products? (c) and (d)(a) and (d)(a), (c) and (d)(b) and (c)Correct Option: 1 Solution:...
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