The correct sequence of reagents used in the preparation of 4 -bromo-2-nitroethyl benzene from benezene is :
Question: The correct sequence of reagents used in the preparation of 4 -bromo-2-nitroethyl benzene from benezene is : $\mathrm{CH}_{3} \mathrm{COCl} / \mathrm{AlCl}_{3}, \mathrm{Br}_{2} / \mathrm{AlBr}_{3}, \mathrm{HNO}_{3} / \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{Zn} / \mathrm{HCl}$$\mathrm{CH}_{3} \mathrm{COCl} / \mathrm{AlCl}_{3}, \mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}, \mathrm{Br}_{2} / \mathrm{AlBr}_{3}, \mathrm{HNO}_{3} / \mathrm{H}_{2} \mathrm{SO}_{4}$$\mathrm{Br}_{2} / \mathrm{AlBr}_{3...
Read More →The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs.
Question: The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state?(1) $24.2 \mathrm{~nm}$(2) $11.4 \mathrm{~nm}$(3) $35.8 \mathrm{~nm}$(4) $8.6 \mathrm{~nm}$Correct Option: , 2 Solution: (2) According to Bohr's Theory the wavelength of the radiation emitted from hydrogen atom is given by $\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{...
Read More →The major product of the following reaction is :
Question: The major product of the following reaction is : Correct Option: , 4 Solution:...
Read More →An electron of mass m and magnitude of charge $|e|$ initially at rest gets accelerated by a constant electric field E.
Question: An electron of mass $m$ and magnitude of charge $|e|$ initially at rest gets accelerated by a constant electric field $\mathrm{E}$. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is:(1) $-\frac{h}{|e| \mathrm{E} \sqrt{t}}$(2) $\frac{|e| \mathrm{Et}}{h}$(3) $-\frac{h}{|e| \mathrm{Et}}$(4) $\frac{-h}{|e| \mathrm{Et}^{2}}$Correct Option: , 4 Solution: (4) Acceleration of electron in electric field, $a=\frac{e E}{m}$ Using equation $v...
Read More →Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution
Question: Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are: $\mathrm{B}$ and $\mathrm{C}$ onlyB only$A$ and $B$ only$C$ onlyCorrect Option: 1 Solution: Compounds which are more acidic then $\mathrm{H}_{2} \mathrm{CO}_{3}$, gives $\mathrm{CO}_{2}$ gas on reaction with $\mathrm{NaHCO}_{3} .$ Compound $\mathrm{B}$ i.e. $\mathrm{Benzoic}$ acid and compound $\mathrm{C}$ i.e. picric acid both are more acidic than $\mathrm{H}_{2} \mathrm{CO}_{3}$....
Read More →The correct order of the following compounds showing
Question: The correct order of the following compounds showing increasing tendency towards nucleophilic substitution reaction is : $(i v)(i)(i i i)(i i)$$(\mathrm{i} v)(\mathrm{i})(\mathrm{ii})(\mathrm{iii})$$(i)(i i)(i i i)(i v)$$($ iv $)($ iii $)($ ii $)($ i $)$Correct Option: , 3 Solution:...
Read More →The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3.
Question: The curved surface area of a cylindrical pillar is 264 m2and its volume is 924 m3. The height of the pillar is(a) 4 m(b) 5 m(c) 6 m(d) 7 m Solution: (c) 6 m Curved surface area $=264 \mathrm{~m}^{2}$ Volume = 924 m3Letrm be the radius andhm be the height of the cylinder.Then we have: $2 \pi r h=264$ and $\pi r^{2} h=924$ $\Rightarrow r h=\frac{264}{2 \pi}$ $\Rightarrow h=\frac{264}{2 r \times \pi}$ Now, $\pi r^{2} h=\pi \times r^{2} \times \frac{264}{2 r \times \pi}=924$ $\Rightarrow r...
Read More →The height of a cylinder is 14 cm and its curved surface area is 264 cm2.
Question: The height of a cylinder is 14 cm and its curved surface area is 264 cm2. The volume of the cylinder is(a) 308 cm3(b) 396 cm3(c) 1232 cm3(d) 1848 cm3 Solution: (b) $396 \mathrm{~cm}^{3}$ Curved surface area $=264 \mathrm{~cm}^{2}$. Letrcm be the radius of the cylinder.Then we have: $2 \pi r h=264$ $\Rightarrow 2 \times \frac{22}{7} \times r \times 14=264$ $\Rightarrow r=\frac{264 \times 7}{44 \times 14}=3 \mathrm{~cm}$ $\therefore$ Volume of the cylinder $=\frac{22}{7} \times 3^{2} \ti...
Read More →A particle moving with kinetic energy {E} has de Broglie wavelength
Question: A particle moving with kinetic energy $\mathrm{E}$ has de Broglie wavelength $\lambda$. If energy $\Delta \mathrm{E}$ is added to its energy, the wavelength become $\frac{\lambda}{2}$. Value of $\Delta \mathrm{E}$, is:(1) $\mathrm{E}$(2) $4 \mathrm{E}$(3) $3 \mathrm{E}$(4) $2 \mathrm{E}$Correct Option: , 3 Solution: (3) As per question, when $\mathrm{KE}$ of particle $E$, wavelength $\lambda$ and when $K E$ becomes $E+\Delta E$ wavelength becomes $\lambda / 2$ Using, $\lambda=\frac{h}{...
Read More →In the following reaction the reason why meta-nitro product also formed is:
Question: In the following reaction the reason why meta-nitro product also formed is: Formation of anilinium ion$-\mathrm{NO}_{2}$ substitution always takes place at meta-positionlow temperature$-\mathrm{NH}_{2}$ group is highly meta-directiveCorrect Option: 1 Solution: In acidic medium the $-\mathrm{NH}_{2}$ group in aniline converts into anilinium ion which is meta directing....
Read More →If the curved surface area of a cylinder is 1760 cm2 and its base radius is 14 cm,
Question: If the curved surface area of a cylinder is 1760 cm2and its base radius is 14 cm, then its height is(a) 10 cm(b) 15 cm(c) 20 cm(d) 40 cm Solution: (c) 20 cm Curved surface area $=1760 \mathrm{~cm}^{2}$ Suppose thathcm is the height of the cylinder.Then we have: $2 \pi r h=1760$ $\Rightarrow 2 \times \frac{22}{7} \times 14 \times h=1760$ $\Rightarrow h=\frac{1760 \times 7}{44 \times 14}=20 \mathrm{~cm}$...
Read More →If the diameter of a cylinder is 28 cm and its height is 20 cm
Question: If the diameter of a cylinder is 28 cm and its height is 20 cm, then its curved surface area is(a) 880 cm2(b) 1760 cm2(c) 3520 cm2(d) 2640 cm2 Solution: (b) $1760 \mathrm{~cm}^{2}$ Curved surface area of the cylinder $=2 \pi r h$ $=2 \times \frac{22}{7} \times 14 \times 20$ $=44 \times 40$ $=1760 \mathrm{~cm}^{2}$...
Read More →Radiation, with wavelength $6561 A falls on a metal surface to produce photoelectrons.
Question: Radiation, with wavelength $6561 A falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} \mathrm{~T}$. If the radius of the largest circular path followed by the electrons is $10 \mathrm{~mm}$, the work function of the metal is close to:(1) $1.1 \mathrm{ev}$(2) $0.8 \mathrm{ev}$(3) $1.6 \mathrm{ev}$(4) $1.8 \mathrm{ev}$Correct Option: 1 Solution: (1) Using Einstein's photoelectric equation, $E=\omega_{0}+K E_{\...
Read More →The area (in sq. units) of the region bounded by the parabola,
Question: The area (in sq. units) of the region bounded by the parabola, $y=x^{2}+2$ and the lines, $y=x+1, x=0$ and $x=3$, is :(1) $\frac{15}{4}$(2) $\frac{21}{2}$(3) $\frac{17}{4}$(4) $\frac{15}{2}$Correct Option: , 4 Solution: Area of the bounded region $\int_{0}^{3}\left[\left(x^{2}+2\right)-(x+1)\right] d x$ $=\left[\frac{x^{3}}{3}-\frac{x^{2}}{2}+x\right]_{0}^{3}$ $=9-\frac{9}{2}+3=\frac{15}{2}$...
Read More →The diameter of the base of a cylinder is 6 cm and its height is 14 cm.
Question: The diameter of the base of a cylinder is 6 cm and its height is 14 cm. The volume of the cylinder is(a) 198 cm3(b) 396 cm3(c) 495 cm3(d) 297 cm3 Solution: (b) $396 \mathrm{~cm}^{3}$ Volume of the cylinder $=\pi r^{2} h$ $=\frac{22}{7} \times 3^{2} \times 14$ $=22 \times 9 \times 2$ $=396 \mathrm{~cm}^{3}$...
Read More →The area (in sq. units) in the first quadrant bounded by the parabola
Question: The area (in sq. units) in the first quadrant bounded by the parabola, $y=x^{2}+1$, the tangent to it at the point $(2,5)$ and the coordinate axes is :(1) $\frac{8}{3}$(2) $\frac{37}{24}$(3) $\frac{187}{24}$(4) $\frac{14}{3}$Correct Option: , 2 Solution: The equation of parabola $x^{2}=y-1$ The equation of tangent at $(2,5)$ to parabola is $y-5=\left(\frac{d y}{d x}\right)_{(2,5)}(x-2)$ $y-5=4(x-2)$ $4 x-y=3$ Then, the required area $=\int_{0}^{2}\left\{\left(x^{2}+1\right)-(4 x-3)\rig...
Read More →The area (in sq. units) of the region bounded by the curve
Question: The area (in sq. units) of the region bounded by the curve $x^{2}=4 y$ and the straight line $x=4 y-2$ is :(1) $\frac{5}{4}$(2) $\frac{9}{8}$(3) $\frac{7}{8}$(4) $\frac{3}{4}$Correct Option: , 2 Solution: Let points of intersection of the curve and the line be $P$ and $Q$ $x^{2}=4\left(\frac{x+2}{4}\right)$ $x^{2}-x-2=0$ Point are $(2,1)$ and $\left(-1, \frac{1}{4}\right)$ Area $=\int_{-1}^{2}\left[\left(\frac{x+2}{4}\right)-\left(\frac{x^{2}}{4}\right)\right] d x=\left[\frac{x^{2}}{8}...
Read More →If each side of a cube is doubled, then its volume
Question: If each side of a cube is doubled, then its volume(a) is doubled(b) becomes 4 times(c) becomes 6 times(d) becomes 8 times Solution: (d) becomes 8 timesSuppose that the side of the cube isa.When it is doubled, it becomes 2a. New volume of the cube $=(2 a)^{3}=8 a^{3}$ Hence, the volume becomes 8 times the original volume....
Read More →The first member of the Balmer series of hydrogen atom has a wavelength
Question: The first member of the Balmer series of hydrogen atom has a wavelength of $6561 \AA$. The wavelength of the second member of the Balmer series (in $\mathrm{nm}$ ) is_________ Solution: The wavelength of the spectral line of hydrogen spectrum is given by formula $\frac{1}{\lambda}=R\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)$ Where, $R=$ Rydberg constant For the first member of Balmer series $n_{\mathrm{F}}=2, n_{\mathrm{i}}=3$ $\therefore \frac{1}{\lambda}=\mathrm{R}\left(\fr...
Read More →If the area enclosed between the curves
Question: If the area enclosed between the curves $y=\mathrm{k} x^{2}$ and $x=\mathrm{k} y^{2},(\mathrm{k}0)$, is 1 square unit. Then $\mathrm{k}$ is:(1) $\frac{\sqrt{3}}{2}$(2) $\frac{1}{\sqrt{3}}$(3) $\sqrt{3}$(4) $\frac{2}{\sqrt{3}}$Correct Option: , 2 Solution: Two curves will intersect in the Ist quadrant at $A\left(\frac{1}{k}, \frac{1}{k}\right)$ $\because$ area of shaded region $=1 .$ $\therefore \int_{0}^{\frac{1}{k}}\left(\frac{\sqrt{x}}{\sqrt{k}}-k x^{2}\right) d x=1$ $\Rightarrow\lef...
Read More →The major product of the following reaction is:
Question: The major product of the following reaction is: Correct Option: , 4 Solution:...
Read More →The area of the region
Question: The area of the region $\mathrm{A}=\{(x, y): 0 \leq y \leq x|x|+1$ and $-1 \leq x \leq 1\}$ in sq. units is:(1) $\frac{2}{3}$(2) 2(3) $\frac{4}{3}$(4) $\frac{1}{3}$Correct Option: , 2 Solution: Given $A=\{(x, y): 0 \leq y \leq x|x|+1$ and $-1 \leq x \leq 1\}$ $\therefore$ Area of shaded region $=\int_{-1}^{0}\left(-x^{2}+1\right) d x+\int_{0}^{1}\left(x^{2}+1\right) d x$ $=\left(-\frac{x^{3}}{3}+x\right)_{-1}^{0}+\left(\frac{x^{3}}{3}+x\right)_{0}^{1}$ $=0-\left(\frac{1}{3}-1\right)+\l...
Read More →Among the following four aromatic compounds ,
Question: Among the following four aromatic compounds , which one will have the lowest melting point?Correct Option: , 4 Solution: The force of attraction between the molecules affects the melting point of a compound. Polarity increases the intermolecular force of attraction and as a result increases the melting point....
Read More →The major product of the following reaction is:
Question: The major product of the following reaction is: Correct Option: Solution: Reaction involved:...
Read More →Two cubes have their volumes in the ratio 1 : 27.
Question: Two cubes have their volumes in the ratio 1 : 27. The ratio of their surface areas is(a) 1 : 3(b) 1 : 8(c) 1 : 9(d) 1 : 18 Solution: (c) 1 : 9 Suppose that the edges of the cubes areaandb.We have: $\frac{a^{3}}{b^{3}}=\frac{1}{27}$ $\Rightarrow\left(\frac{a}{b}\right)^{3}=\frac{1}{27}$ $\Rightarrow \frac{a}{b}=\frac{1}{3}$ $\therefore$ Ratio of the surface areas $=\frac{6 a^{2}}{6 b^{2}}=\left(\frac{a}{b}\right)^{2}=\left(\frac{1}{3}\right)^{2}=\frac{1}{9}$...
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