In Carius method, halogen containing organic compound
Question: In Carius method, halogen containing organic compound is heated with fuming nitric acid in the presence of :$\mathrm{HNO}_{3}$$\mathrm{AgNO}_{3}$$\mathrm{CuSO}_{4}$$\mathrm{BaSO}_{4}$Correct Option: , 2 Solution: Organic compound is heated with fuming nitric acid in the presence of silver nitrate in carius method. Lunar caustic $\left(\mathrm{AgNO}_{3}\right)$ is used as reagent hare to distinguish $\mathrm{Cl}^{-}, \mathrm{Br}$ and $\mathrm{I}^{-}$respectively as follows. $\mathrm{Cl}...
Read More →In a typical combustion engine the work don
Question: In a typical combustion engine the work don by a gas molecule is given $W=\alpha^{2} \beta e^{\frac{-\beta x^{2}}{k T}}$ where $x$ is the displacement, $k$ is the Boltzmann constant and $\mathrm{T}$ is the temperature. If $\alpha$ and $\beta$ are constants, dimensions of $\alpha$ will be : $\left[\mathrm{MLT}^{-2}\right]$$\left[\mathrm{M}^{0} \mathrm{LT}^{0}\right]$$\left[\mathrm{M}^{2} \mathrm{LT}^{-2}\right]$$\left[\mathrm{MLT}^{-1}\right]$Correct Option: , 2 Solution: kT has dimensi...
Read More →Cu^2+ salt reacts with potassium
Question: $\mathrm{Cu}^{2+}$ salt reacts with potassium iodide to give$\mathrm{Cu}_{2} \mathrm{I}_{2}$$\mathrm{Cu}_{2} \mathrm{I}_{3}$Cul$\mathrm{Cu}\left(\mathrm{I}_{3}\right)_{2}$Correct Option: 1 Solution: $2 \mathrm{Cu}^{+2}+4 \mathrm{I}^{-} \longrightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}(\mathrm{~s})+\mathrm{I}_{2}$ $2 \mathrm{Cu}^{+2}+3 \mathrm{I}^{-} \longrightarrow 2 \mathrm{CuI}+\mathrm{I}_{2}$...
Read More →A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle T as shown in figure.
Question: . A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle T as shown in figure. The coefficient of kinetic friction is K. Then, the block's acceleration 'a' is given by : (g is acceleration due to gravity) $-\frac{F}{m} \cos \theta-\mu_{K}\left(g-\frac{F}{m} \sin \theta\right)$$\frac{\mathrm{F}}{\mathrm{m}} \cos \theta-\mu_{\mathrm{K}}\left(\mathrm{g}-\frac{\mathrm{F}}{\mathrm{m}} \sin \theta\right)$$\frac{F}{m} \cos \theta-\mu_{K}\left(g+\frac{...
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Question: Major product P of above reaction, is :Correct Option: , 4 Solution:...
Read More →Consider the combination of 2 capacitors
Question: Consider the combination of 2 capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$, with $\mathrm{C}_{2}\mathrm{C}_{1}$, when connected in parallel, the equivalent capacitance is $\frac{15}{4}$ time the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$. $\frac{15}{11}$$\frac{111}{80}$$\frac{29}{15}$$\frac{15}{4}$Correct Option: 2, Solution: When connected in parallel $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\m...
Read More →For changing the capacitance of a given parallel plate capacitor,
Question: For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} \mathrm{~d}$, where ' $\mathrm{d}$ ' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0 ) is given by the following relation :$\mathrm{C}^{\prime}=\frac{3+\mathrm{K}}{4 \mathrm{~K}} ...
Read More →Find the gravitational force of attraction between the ring and sphere as shown in the diagram,
Question: Find the gravitational force of attraction between the ring and sphere as shown in the diagram, where the plane of the ring is perpendicular to the line joining the centres. If $\sqrt{8} \mathrm{R}$ is the distance between the centres of a ring (of mass ' $m$ ') and a sphere (mass 'M') where both have equal radius 'R'. $\frac{\sqrt{8}}{9} \cdot \frac{\mathrm{GmM}}{\mathrm{R}}$$\frac{2 \sqrt{2}}{3} \cdot \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$$\frac{1}{3 \sqrt{8}} \cdot \frac{\mathrm{GMm}}...
Read More →One main scale division of a vernier callipers is 'a' cm and nth division of the vernier scale coincide with (n – 1)th division of the main scale.
Question: One main scale division of a vernier callipers is ' $\mathrm{a}$ ' $\mathrm{cm}$ and $\mathrm{n}^{\text {th }}$ division of the vernier scale coincide with $(\mathrm{n}-1)^{\text {th }}$ division of the main scale. The least count of the callipers in $\mathrm{mm}$ is :$\frac{10 n a}{(n-1)}$$\frac{10 a}{(n-1)}$$\left(\frac{\mathrm{n}-1}{10 \mathrm{n}}\right) \mathrm{a}$$\frac{10 a}{n}$Correct Option: , 4 Solution: $(\mathrm{n}-1) \mathrm{a}=\mathrm{n}\left(\mathrm{a}^{\prime}\right)$ $a...
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Question: Consider the above reaction, compound B is :Correct Option: , 3 Solution:...
Read More →Benzene on nitration gives nitrobenzene in presence
Question: Benzene on nitration gives nitrobenzene in presence of $\mathrm{HNO}_{3}$ and $\mathrm{H}_{2} \mathrm{SO}_{4}$ mixture, where :both $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{HNO}_{3}$ act as a bases$\mathrm{HNO}_{3}$ acts as an acid and $\mathrm{H}_{2} \mathrm{SO}_{4}$ acts as a baseboth $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{HNO}_{3}$ act as an acids$\mathrm{HNO}_{3}$ acts as a base and $\mathrm{H}_{2} \mathrm{SO}_{4}$ acts as an acidCorrect Option: , 4 Solution: Reagent fo...
Read More →Bakelite is a cross-linked polymer
Question: Bakelite is a cross-linked polymer of formaldehyde and :PHBVBuna-SNovolacDacronCorrect Option: , 3 Solution: Novolac (phenol formaldehyde Resin) $\rightarrow$ Bakelite...
Read More →The hybridisations of the atomic orbitals of nitrogen
Question: The hybridisations of the atomic orbitals of nitrogen in $\mathrm{NO}_{2}^{-}, \mathrm{NO}_{2}^{+}$and $\mathrm{NH}_{4}^{+}$respectively are.$\mathrm{sp}^{3}, \mathrm{sp}^{2}$ and $\mathrm{sp}$$\mathrm{sp}, \mathrm{sp}^{2}$ and $\mathrm{sp}^{3}$$\mathrm{sp}^{3}, \mathrm{sp}$ and $\mathrm{sp}^{2}$$\mathrm{sp}^{2}, \mathrm{sp}$ and $\mathrm{sp}^{3}$Correct Option: , 4 Solution:...
Read More →The transformation occurring in Duma's method is given below :
Question: The transformation occurring in Duma's method is given below : $\mathrm{C}_{2} \mathrm{H}_{7} \mathrm{~N}+\left(2 \mathrm{x}+\frac{\mathrm{y}}{2}\right) \mathrm{CuO} \rightarrow \mathrm{xCO}_{2}+\frac{y}{2} \mathrm{H}_{2} \mathrm{O}+\frac{\mathrm{z}}{2} \mathrm{~N}_{2}+\left(2 \mathrm{x}+\frac{\mathrm{y}}{2}\right) \mathrm{Cu}$ The value of $y$ is (Integer answer) Solution: $\mathrm{C}_{2} \mathrm{H}_{7} \mathrm{~N}+\left(2 \mathrm{x}+\frac{\mathrm{y}}{2}\right) \mathrm{CuO} \rightarro...
Read More →Data given for the following reaction is as follows:
Question: Data given for the following reaction is as follows: $\mathrm{FeO}_{(\mathrm{s})}+\mathrm{C}_{\text {(graplite) }} \longrightarrow \mathrm{Fe}_{(\mathrm{s})}+\mathrm{CO}_{(\mathrm{g})}$ The minimum temperature in K at which the reaction becomes spontaneous is ________ . (Integer answer) Solution: $\mathrm{T}_{\min }=\left(\frac{\Delta^{0} \mathrm{H}}{\Delta^{0} \mathrm{~S}}\right)$ $\Delta^{0} \mathrm{H}_{\mathrm{rxn}}=\left[\Delta_{\mathrm{f}}^{0} \mathrm{H}(\mathrm{Fe})+\Delta_{\math...
Read More →Which one of the following statements
Question: Which one of the following statements is not true about enzymes ?Enzymes are non-specific for a reaction and substrate.Almost all enzymes are proteins.Enzymes work as catalysts by lowering the activation energy of a biochemical reaction.The action of enzymes is temperature and pH specificCorrect Option: 1 Solution: Fact...
Read More →According to molecular orbital theory, the number of unpaired
Question: According to molecular orbital theory, the number of unpaired electron(s) in $\mathrm{O}_{2}^{2-}$ is : Solution: Molecular orbital configuration of $\mathrm{O}_{2}^{2-}$ is $\sigma_{1 \mathrm{~s}}^{2} \sigma_{1 \mathrm{~s}}^{* 2} \sigma_{2 \mathrm{~s}}^{2} \sigma_{2 \mathrm{~s}}^{* 2}\left(\pi 2 \mathrm{p}_{\mathrm{x}}^{2}=\pi 2 \mathrm{p}_{\mathrm{y}}^{2}\right)\left(\pi_{2 \mathrm{px}}^{* 2}=\pi_{2 \mathrm{py}}^{* 2}\right)$ Zero unpaired electron...
Read More →The number of species having non–pyramidal shape among the following is_______.
Question: The number of species having nonpyramidal shape among the following is_______. (A) $\mathrm{SO}_{3}$ (B) $\mathrm{NO}_{3}^{-}$ (C) $\mathrm{PCl}_{3}$ (D) $\mathrm{CO}_{3}^{2-}$ Solution: Hence non-pyramidal species are $\mathrm{SO}_{3}, \mathrm{NO}_{3}^{-}$and $\mathrm{CO}_{3}^{2-}$...
Read More →Which one of the following species doesn't have a magnetic
Question: Which one of the following species doesn't have a magnetic moment of 1.73 BM, (spin only value) ?$\mathrm{O}_{2}^{+}$$\mathrm{CuI}$$\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2}$$\mathrm{O}_{2}^{-}$Correct Option: , 2 Solution: Species must not contain single unpaired (1) $\mathrm{O}_{2}^{+} \rightarrow \sigma_{1 \mathrm{~s}}^{2}\sigma_{1 \mathrm{~s}}^{* 2}\sigma_{2 \mathrm{~s}}^{2}\sigma_{2 \mathrm{~s}}^{* 2}\sigma_{2 \mathrm{pz}}^{2}\pi_{2 \mathrm{px}}^{2}=...
Read More →Sodium oxide reacts with water to produce sodium hydroxide.
Question: Sodium oxide reacts with water to produce sodium hydroxide. $20.0 \mathrm{~g}$ of sodium oxide is dissolved in $500 \mathrm{~mL}$ of water. Neglecting the change in volume, the concentration of the resulting $\mathrm{NaOH}$ solution is $\times 10^{-1}$ M. (Nearest integer) $[$ Atomic mass : $\mathrm{Na}=23.0, \mathrm{O}=16.0, \mathrm{H}=1.0]$ Solution: $\mathrm{Na}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaOH}$ $\frac{20}{62}$ moles Moles of $\mathrm{NaOH}$ forme...
Read More →When 5.1 g of solid NH4 HS is introduced into
Question: When $5.1 \mathrm{~g}$ of solid $\mathrm{NH}_{4} \mathrm{HS}$ is introduced into a two litre evacuated flask at $27^{\circ} \mathrm{C}, 20 \%$ of the solid decomposes into gaseous ammonia and hydrogen sulphide. The $\mathrm{K}_{\mathrm{p}}$ for the reaction at $27^{\circ} \mathrm{C}$ is $\mathrm{x} \times 10^{-2}$. The value of $x$ is - (Integer answer) $\left[\right.$ Given $\left.\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$ Solution: moles of...
Read More →Solve this following
Question: For the reaction $\mathrm{A} \rightarrow \mathrm{B}$, the rate constant $\mathrm{k}\left(\right.$ in $\left.\mathrm{s}^{-1}\right)$ is given by $\log _{10} k=20.35-\frac{\left(2.47 \times 10^{3}\right)}{T}$ The energy of activation in $\mathrm{kJ} \mathrm{mol}^{-1}$ is__________________ (Nearest integer) [Given : $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ] Solution: Given $\log \mathrm{K}=20.35-\frac{2.47 \times 10^{3}}{\mathrm{~T}}$ We know $\log \mathrm{K}=\l...
Read More →Spin only magnetic moment of an octahedral complex
Question: Spin only magnetic moment of an octahedral complex of $\mathrm{Fe}^{2+}$ in the presence of a strong field ligand in BM is :4.892.8203.46Correct Option: , 3 Solution: In presence of SFL $\Delta_{0}\mathrm{P}$ means pairing occurs therefore $\therefore$ No of unpaired $\mathrm{e}^{-}(\mathrm{s})=0$ $\therefore \mu=\sqrt{n(n+2)} B M=0$ $\left[n=\right.$ No of unpaired $\left.e^{-}(s)\right]$ In $\mathrm{NiCl}_{2} \mathrm{Ni}^{+2}$ is having configuration $3 \mathrm{~d}^{8}$ $\therefore \...
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Question: The $\mathrm{pH}$ of a solution obtained by mixing $50 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{HCl}$ and $30 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{NaOH}$ is $\mathrm{x} \times 10^{-4}$. The value of $\mathrm{x}$ is (Nearest integer) $[\log 2.5=0.3979]$ Solution: $\mathrm{HCl}$ (aq.) $+\mathrm{NaOH}$ (aq.) $\rightarrow \mathrm{NaCl}$ (aq.) $+\mathrm{H}_{2} \mathrm{O}(\ell)$ $50 \mathrm{ml}, 1 \mathrm{M} \quad 30 \mathrm{ml}, 1 \mathrm{M}$ $\mathrm{t}=0 \quad 50 \mathrm{~mm} \quad 30 \ma...
Read More →The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm,
Question: The number of photons emitted by a monochromatic (single frequency) infrared range finder of power $1 \mathrm{~mW}$ and wavelength of $1000 \mathrm{~nm}$, in $0.1$ second is $x \times 10^{13}$. The value of $x$ is________ (Nearest integer) $\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{c}=3.00 \times 10^{8} \mathrm{~ms}^{-1}\right)$ Solution: Energy emitted in $0.1 \mathrm{sec}$. $=0.1 \mathrm{sec} \cdot \times 10^{-3} \frac{\mathrm{J}}{\mathrm{s}}$ $=10^{-4} \...
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