Prove that
Question: $x^{2}+12 x+35=0$ Solution: Given : $x^{2}+12 x+35=0$ $\Rightarrow x^{2}+7 x+5 x+35=0$ $\Rightarrow x(x+7)+5(x+7)=0$ $\Rightarrow(x+5)(x+7)=0$ $\Rightarrow x+5=0$ or $x+7=0$ $\Rightarrow x=-5$ or $x=-7$ Hence, $-5$ and $-7$ are the roots of the equation $x^{2}+12 x+35=0$....
Read More →The magnetic field of a plane electromagnetic wave is given by:
Question: The magnetic field of a plane electromagnetic wave is given by: $\vec{B}=B_{0} \hat{i}[\cos (k z-\omega t)]+B_{1} \hat{j} \cos (k z+\omega t)$ Where $\mathrm{B}_{0}=3 \times 10^{-5} \mathrm{~T}$ and $\mathrm{B}_{1}=2 \times 10^{-6} \mathrm{~T}$. The rms value of the force experienced by a stationary charge $\mathrm{Q}=10^{-4} \mathrm{C}$ at $\mathrm{z}=0$ is closest to:(1) $0.6 \mathrm{~N}$(2) $0.1 \mathrm{~N}$(3) $0.9 \mathrm{~N}$(4) $3 \times 10^{-2} \mathrm{~N}$Correct Option: 1, So...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $9 x^{2}-3 x-2=0$ Solution: We write, $-3 x=3 x-6 x$ as $9 x^{2} \times(-2)=-18 x^{2}=3 x \times(-6 x)$ $\therefore 9 x^{2}-3 x-2=0$ $\Rightarrow 9 x^{2}+3 x-6 x-2=0$ $\Rightarrow 3 x(3 x+1)-2(3 x+1)=0$ $\Rightarrow(3 x+1)(3 x-2)=0$ $\Rightarrow 3 x+1=0$ or $3 x-2=0$ $\Rightarrow x=-\frac{1}{3}$ or $x=\frac{2}{3}$ Hence, the roots of the given equation are $-\frac{1}{3}$ and $\frac{2}{3}$....
Read More →If the angle of elevation of a cloud from a
Question: If the angle of elevation of a cloud from a point $\mathrm{P}$ which is $25 \mathrm{~m}$ above a lake be $30^{\circ}$ and the angle of depression of reflection of the cloud in the lake from P be $60^{\circ}$, then the height of the cloud (in meters) from the surface of the lake is:(1) 60(2) 50(3) 45(4) 42Correct Option: , 2 Solution: (2) Let height of the cloud from the surface of the lake be $h$ meters. $\therefore \quad$ In $\triangle P R Q:$ $\tan 30^{\circ}=\frac{h-25}{P R}$ $\ther...
Read More →The functions of antihistamine are :
Question: The functions of antihistamine are :Antiallergic and AnalgesicAntacid and antiallergicAnalgesic and antacidAntiallergic and antidepressantCorrect Option: Solution: (2) Antacid and antiallergic...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $x^{2}+6 x+5=0$ Solution: We write, $6 x=x+5 x$ as $x^{2} \times 5=5 x^{2}=x \times 5 x$ $\therefore x^{2}+6 x+5=0$ $\Rightarrow x^{2}+x+5 x+5=0$ $\Rightarrow x(x+1)+5(x+1)=0$ $\Rightarrow(x+1)(x+5)=0$ $\Rightarrow x+1=0$ or $x+5=0$ $\Rightarrow x=-1$ or $x=-5$ Hence, the roots of the given equation are 1 and 5....
Read More →The magnetic field of an electromagnetic wave is given by:
Question: The magnetic field of an electromagnetic wave is given by: $\overrightarrow{\mathrm{B}}=1.6 \times 10^{-6} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(2 \hat{i}+\hat{j}) \frac{\mathrm{Wb}}{\mathrm{m}^{2}}$ The associated electric field will be :(1) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(2 \hat{i}+\hat{j}) \frac{V}{m}$(2) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} ...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $2 x^{2}+x-6=0$ Solution: We write, $x=4 x-3 x$ as $2 x^{2} \times(-6)=-12 x^{2}=4 x \times(-3 x)$ $\therefore 2 x^{2}+x-6=0$ $\Rightarrow 2 x^{2}+4 x-3 x-6=0$ $\Rightarrow 2 x(x+2)-3(x+2)=0$ $\Rightarrow(x+2)(2 x-3)=0$ $\Rightarrow x+2=0$ or $2 x-3=0$ $\Rightarrow x=-2$ or $x=\frac{3}{2}$ Hence, the roots of the given equation are $-2$ and $\frac{3}{2}$....
Read More →Consider a triangular plot
Question: Consider a triangular plot $\mathrm{ABC}$ with sides $\mathrm{AB}=7 \mathrm{~m}, \mathrm{BC}=5 \mathrm{~m}$ and $\mathrm{CA}=6 \mathrm{~m}$. A vertical lamp-post at the mid point $\mathrm{D}$ of $\mathrm{AC}$ subtends an angle $30^{\circ}$ at $\mathrm{B}$. The height (in $\mathrm{m}$ ) of the lamp-post is:(1) $\frac{3}{2} \sqrt{21}$(2) $\frac{2}{3} \sqrt{21}$(3) $2 \sqrt{21}$(4) $7 \sqrt{3}$Correct Option: , 2 Solution: Let the height of the lamp-post is $h$. By Appolonius Theorem, $2\...
Read More →Solve this
Question: $3 x^{2}-243=0$ Solution: Given: $3 x^{2}-243=0$ $\Rightarrow 3\left(x^{2}-81\right)=0$ $\Rightarrow(x)^{2}-(9)^{2}=0$ $\Rightarrow(x+9)(x-9)=0$ $\Rightarrow x+9=0$ or $x-9=0$ $\Rightarrow x=-9$ or $x=9$ Hence, $-9$ and 9 are the roots of the equation $3 x^{2}-243=0 .$...
Read More →The magnetic field of an electromagnetic wave is given by:
Question: The magnetic field of an electromagnetic wave is given by: $\overrightarrow{\mathrm{B}}=1.6 \times 10^{-6} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(2 \hat{i}+\hat{j}) \frac{W b}{m^{2}}$ The associated electric field will be :(1) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(2 \hat{i}+\hat{j}) \frac{V}{m}$(2) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(-2 \hat{...
Read More →The magnetic field of an electromagnetic wave is given by:
Question: The magnetic field of an electromagnetic wave is given by: $\overrightarrow{\mathrm{B}}=1.6 \times 10^{-6} \cos \left(2 \times 10^{7} z+6 \times 10^{15} t\right)(2 \hat{i}+\hat{j}) \frac{W b}{m^{2}}$ The associated electric field will be :(1) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(2 \hat{i}+\hat{j}) \frac{V}{m}$(2) $\overrightarrow{\mathrm{E}}=4.8 \times 10^{2} \cos \left(2 \times 10^{7} z-6 \times 10^{15} t\right)(-2 \hat{...
Read More →The angle of elevation of the top of a vertical
Question: The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be $45^{\circ}$ from a point $\mathrm{A}$ on the plane. Let $\mathrm{B}$ be the point $30 \mathrm{~m}$ vertically above the point $\mathrm{A}$. If the angle of elevation of the top of the tower from $\mathrm{B}$ be $30^{\circ}$, then the distance (in $\mathrm{m}$ ) of the foot of the tower from the point $\mathrm{A}$ is:(1) $15(3+\sqrt{3)}$(2) $15(5-\sqrt{3)}$(3) $15(3-\sqrt{3)}$(4) $15(...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $4 x^{2}+5 x=0$ Solution: $4 x^{2}+5 x=0$ $\Rightarrow x(4 x+5)=0$ $\Rightarrow x=0$ or $4 x+5=0$ $\Rightarrow x=0$ or $x=-\frac{5}{4}$ Hence, the roots of the given equation are 0 and $-\frac{5}{4}$....
Read More →Decomposition of X exhibits a rate constant of
Question: Decomposition of X exhibits a rate constant of $0.05 \mu \mathrm{g} /$ year. How many years are required for the decomposition of $5 \mu \mathrm{g}$ of X into $2.5 \mu \mathrm{g}$ ?50252040Correct Option: 1 Solution: Rate constant of decomposition of $X=0.05 \mathrm{mg} /$ year. Unit of rate constant confirms that the decomposition of $X$ is a zero order reaction. For zero order kinetics, $[\mathrm{X}]=[\mathrm{X}]_{0}-k t$ $k t=[\mathrm{X}]_{0}-[\mathrm{X}]$ $t=\frac{[X]_{0}-[X]}{k}$ ...
Read More →A plane electromagnetic wave travels in free space along the x-direction.
Question: A plane electromagnetic wave travels in free space along the $x$-direction. The electric field component of the wave at a particular point of space and time is $\mathrm{E}=6 \mathrm{Vm}^{-1}$ along $y$-direction. Its corresponding magnetic field component, B would be:(1) $2 \times 10^{-8} \mathrm{~T}$ along $z$-direction(2) $6 \times 10^{-8} \mathrm{~T}$ along $x$-direction(3) $6 \times 10^{-8} \mathrm{~T}$ along $z$-direction(4) $2 \times 10^{-8} \mathrm{~T}$ along $y$-directionCorrec...
Read More →Solve each of the following quadratic equations:
Question: Solve each of the following quadratic equations: $(2 x-3)(3 x+1)=0$ Solution: $(2 x-3)(3 x+1)=0$ $\Rightarrow 2 x-3=0$ or $3 x+1=0$ $\Rightarrow 2 x=3$ or $3 x=-1$ $\Rightarrow x=\frac{3}{2}$ or $x=-\frac{1}{3}$ Hence, the roots of the given equation are $\frac{3}{2}$ and $-\frac{1}{3}$....
Read More →ABC is a triangular park with
Question: $\mathrm{ABC}$ is a triangular park with $\mathrm{AB}=\mathrm{AC}=100$ metres. $\mathrm{A}$ vertical tower is situated at the mid-point of $\mathrm{BC}$. If the angles of elevation of the top of the tower at $A$ and $B$ are $\cot ^{-1}(3 \sqrt{2})$ and $\operatorname{cosec}^{-1}(2 \sqrt{2})$ respectively, then the height of the tower (in metres) is :(1) $\frac{100}{3 \sqrt{3}}$(2) $10 \sqrt{5}$(3) 20(4) 25Correct Option: , 3 Solution: Let the height of the vertical tower situated at th...
Read More →For a reaction, consider the plot of
Question: For a reaction, consider the plot of $\ln \mathrm{k}$ versus $1 / \mathrm{T}$ given in the figure. If the rate constant of this reaction at $400 \mathrm{~K}$ is $10^{-5} \mathrm{~s}^{-1}$, then the rate constant at $500 \mathrm{~K}$ is : For a reaction, consider the plot of$2 \times 10^{-4} \mathrm{~s}^{-1}$$10^{-4} \mathrm{~s}^{-1}$$4 \times 10^{-4} \mathrm{~s}^{-2}$Correct Option: , 3 Solution: From Arrhenius equation,...
Read More →A plane electromagnetic wave is propagating along the direction
Question: A plane electromagnetic wave is propagating along the direction $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$, with its polarization along the direction $\hat{k}$. The correct form of the magnetic field of the wave would be (here $\mathrm{B}_{0}$ is an appropriate constant):(1) $\mathrm{B}_{0} \frac{\hat{i}-\hat{j}}{\sqrt{2}} \cos \left(\omega t-k \frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)$(2) $\mathrm{B}_{0} \frac{\hat{j}-\hat{i}}{\sqrt{2}} \cos \left(\omega t+k \frac{\hat{i}+\hat{j}}{\sqrt{2}}\rig...
Read More →Two poles standing on a horizontal ground
Question: Two poles standing on a horizontal ground are of heights $5 \mathrm{~m}$ and $10 \mathrm{~m}$ respectively. The line joining their tops makes an angle of $15^{\circ}$ with the ground. Then the distance (in $\mathrm{m}$ ) between the poles, is:(1) $5(2+\sqrt{3})$(2) $5(\sqrt{3}+1)$(3) $\frac{5}{2}(2+\sqrt{3})$(4) $10(\sqrt{3}-1)$Correct Option: 1 Solution: $\tan 15^{\circ}=\frac{5}{d} \Rightarrow d=\frac{5}{\tan 15^{\circ}}=\frac{5(\sqrt{3}+1)}{\sqrt{3}-1}$ $=\frac{5(4+2 \sqrt{3})}{2}=5...
Read More →Solve the following
Question: The reaction $2 X \rightarrow B$ is a zeroth order reaction. If the initial concentration of $X$ is $0.2 \mathrm{M}$, the half-life is $6 \mathrm{~h}$. When the initial concentration of $\mathrm{X}$ is $0.5 \mathrm{M}$, the time required to reach its final concentration of $0.2 \mathrm{M}$ will be :$9.0 \mathrm{~h}$$12.0 \mathrm{~h}$$18.0 \mathrm{~h}$$7.2 \mathrm{~h}$Correct Option: , 3 Solution: For the reaction $2 \mathrm{X} \rightarrow \mathrm{B}$, follow zeroth order Rate equation ...
Read More →Show that
Question: Show that $x=-\frac{b c}{a d}$ is a solution of the quadratic equation $a d^{2}\left(\frac{a x}{b}+\frac{2 c}{d}\right) x+b c^{2}=0$. Solution: LHS;Consider the quadratic equation; $a d^{2}\left(\frac{a x}{b}+\frac{2 c}{d}\right) x+b c^{2}=0$ Put $x=-\frac{b c}{a d}$ in the given equation. $a d^{2}\left[\frac{a\left(-\frac{b c}{a d}\right)}{b}+\frac{2 c}{d}\right]\left(-\frac{b c}{a d}\right)+b c^{2}$ $=\frac{a d^{2}}{b d}\left[a d\left(-\frac{b c}{a d}\right)+2 b c\right]\left(-\frac{...
Read More →If a reaction follows the Arrhenius equation,
Question: If a reaction follows the Arrhenius equation, the plot lnk vs $\frac{1}{(\mathrm{RT})}$ gives straight line with a gradient $(-\mathrm{y})$ unit. The energy required to activate the reactant is:$\mathrm{y} / \mathrm{R}$ unity unityR unit- y unitCorrect Option: , 2 Solution: From Arrhenius equation, $\mathrm{k}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}$ $\ln k=\ln A-\frac{E_{a}}{R T}$ slope $=-y$ (given) $-y=-\mathrm{E}_{\mathrm{a}}$ $\Rightarrow \mathrm{E}_{\mathrm{a}}=y$...
Read More →(i) Find the value of k for which x = 1 is a root of the equation
Question: (i) Find the value of $k$ for which $x=1$ is a root of the equation $x^{2}+k x+3=0$. Also, find the other root. (ii) Find the values of $a$ and $b$ for which $x=\frac{3}{4}$ and $x=-2$ are the roots of the equation $a x^{2}+b x-6=0$. Solution: (i) It is given that $(x=1)$ is a root of $\left(x^{2}+k x+3=0\right)$. Therefore, $(x=1)$ must satisfy the equation. $\Rightarrow(1)^{2}+k \times 1+3=0$ $\Rightarrow k+4=0$ $\Rightarrow k=-4$ Hence, the required value of $k$ is $-4$. So, the equ...
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