Solve the following
Question: The volume (in $\mathrm{mL}$ ) of $0.1 \mathrm{~N} \mathrm{NaOH}$ require to neutralise $10 \mathrm{~mL}$ of $0.1 \mathrm{~N}$ phosphinic acid is_________________ . Solution: (10)...
Read More →The radius R of a nucleus of mass number A can be estimated
Question: The radius $R$ of a nucleus of mass number $A$ can be estimated by the formula $R=\left(1.3 \times 10^{-15}\right) A^{1 / 3} \mathrm{~m}$. It follows that the mass density of a nucleus is of the order of : $\left(M_{\text {prot. }} \cong M_{\text {neut. }} \simeq 1.67 \times 10^{-27} \mathrm{~kg}\right)$ (1) $10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$(2) $10^{10} \mathrm{~kg} \mathrm{~m}^{-3}$(3) $10^{24} \mathrm{~kg} \mathrm{~m}^{-3}$(4) $10^{17} \mathrm{~kg} \mathrm{~m}^{-3}$Correct Optio...
Read More →In a molecule of pyrophosphoric acid,
Question: In a molecule of pyrophosphoric acid, the number of $P$ $\mathrm{OH}, \mathrm{P}=\mathrm{O}$ and $\mathrm{P}-\mathrm{O}-\mathrm{P}$ bonds/ moiety(ies) respectivel are:2,4 and 13,3 and 34,2 and 04,2 and 1Correct Option: , 4 Solution:...
Read More →In a radioactive material, fraction of active material remaining after time t is 9 / 16. The fraction that was remaining after t / 2 is :
Question: In a radioactive material, fraction of active material remaining after time $t$ is $9 / 16$. The fraction that was remaining after $t / 2$ is :(1) $\frac{4}{5}$(2) $\frac{3}{5}$(3) $\frac{3}{4}$(4) $\frac{7}{8}$Correct Option: , 3 Solution: (3) As we know, for first order decay, $N(t)=N_{0} e^{-\lambda t}$ According to question, $\frac{N(t)}{N_{0}}=\frac{9}{16}=e^{-\lambda t}$ After time, $t / 2 ;$ $N(t / 2)=N_{0} e^{-\lambda(t / 2)}$ $\frac{N(t / 2)}{N_{0}}=\sqrt{e^{-\lambda t}}=\sqrt...
Read More →Aqua regia is used for dissolving noble metals (Au, Pt, etc.).
Question: The gas evolved in this process is :NO$\mathrm{N}_{2} \mathrm{O}_{5}$$\mathrm{N}_{2}$$\mathrm{N}_{2} \mathrm{O}_{3}$Correct Option: 1 Solution:...
Read More →Prove the following
Question: All the pairs $(\mathrm{x}, \mathrm{y})$ that satisfy the inequality $2 \sqrt{\sin ^{2} x-2 \sin x+5} \cdot \frac{1}{4 \sin ^{2} y} \leq 1$ also satisfy the equation:(1) $2|\sin x|=3 \sin y$(2) $2 \sin x=\sin y$(3) $\sin x=2 \sin y$(4) $\sin x=|\sin y|$Correct Option: , 4 Solution: Given inequality is, $2 \sqrt{\sin ^{2} x-2 \sin x+5} \leq 2^{2 \sin ^{2} y}$ $\Rightarrow \sqrt{\sin ^{2} x-2 \sin x+5} \leq 2 \sin ^{2} y$ $\Rightarrow \sqrt{(\sin x-1)^{2}+4} \leq 2 \sin ^{2} y$ It is tru...
Read More →In a reactor,
Question: In a reactor, $2 \mathrm{~kg}$ of ${ }_{92} \mathrm{U}^{235}$ fuel is fully used up in 30 days. The energy released per fission is $200 \mathrm{MeV}$. Given that the Avogadro number, $\mathrm{N}=6.023 \times 10^{26}$ per kilo mole and $1 \mathrm{eV}$ $=1.6 \times 10^{-19} \mathrm{~J}$. The power output of the reactor is close to:(1) $35 \mathrm{MW}$(2) $60 \mathrm{MW}$(3) $125 \mathrm{MW}$(4) $54 \mathrm{MW}$Correct Option: , 2 Solution: (2) Power output of the reactor, $P=\frac{\text ...
Read More →If the boiling point of
Question: If the boiling point of $\mathrm{H}_{2} \mathrm{O}$ is $373 \mathrm{~K}$, the boiling point of $\mathrm{H}_{2} \mathrm{~S}$ will be :less than $300 \mathrm{~K}$equal to $373 \mathrm{~K}$more than $373 \mathrm{~K}$greater than $300 \mathrm{~K}$ but less than $373 \mathrm{~K}$Correct Option: 1 Solution: At room temperature, water is liquid and has boiling point $373 \mathrm{~K}$ due to hydrogen bonding. Whereas $\mathrm{H}_{2} \mathrm{~S}$ is a gas and it has no hydrogen bonding. Hence b...
Read More →Prove the following
Question: Let $S=\left\{\theta \in[-2 \pi, 2 \pi]: 2 \cos ^{2} \theta+3 \sin \theta=0\right\}$. Then the sum of the elements of $S$ is:(1) $\frac{13 \pi}{6}$(2) $\frac{5 \pi}{3}$(3) $2 \pi$(4) $\pi$Correct Option: , 3 Solution: $2 \cos ^{2} \theta+3 \sin \theta=0$ $\Rightarrow(2 \sin \theta+1)(\sin \theta-2)=0$ $\Rightarrow \sin \theta=-\frac{1}{2}$ or $\sin \theta=2 \rightarrow$ Not possibe The required sum of all solutions in $[-2 \pi, 2 \pi]$ is $=\left(\pi+\frac{\pi}{6}\right)+\left(2 \pi-\f...
Read More →A radioactive sample is undergoing
Question: A radioactive sample is undergoing $\alpha$ decay. At any time $t_{1}$, its activity is $A$ and another time $t_{2}$ the activity is $\frac{A}{5} .$ What is the average life time for the sample?(1) $\frac{t_{2}-t_{1}}{\ln 5}$(2) $\frac{1 n\left(t_{2}+t_{1}\right)}{2}$(3) $\frac{t_{1}-t_{2}}{1 n 5}$(4) $\frac{1 n 5}{t_{2}-t_{1}}$Correct Option: 1 Solution: (1) For activity of radioactivesample $A=A_{0} e^{-\alpha t_{1}} \ldots(1)$ $\frac{A}{5} A_{0} e^{-\alpha t_{2}} \ldots(2)$ From $(1...
Read More →On heating compound (A) gives a gas (B) which is a constituent of air.
Question: On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with $\mathrm{H}_{2}$ in the presence of a catalyst gives another gas (C) which is basic in nature. (A) should not be :$\mathrm{NaN}_{3}$$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$$\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}$$\mathrm{NH}_{4} \mathrm{NO}_{2}$Correct Option: , 3 Solution: $\mathrm{Cu}^{2+}$ ions get precipitated every quickely due to low $K_{\mathrm{sp}}$ va...
Read More →The wavelength of an X-ray beam is 10 A.
Question: The wavelength of an X-ray beam is $10 \AA$. The mass of a fictitious particle having the same energy as that of the $X$ - ray photons is $\frac{x}{3} h \mathrm{~kg}$. The value of $x$ is Solution: (10) Given wavelength of an $x$-ray beam =10 A $\because E=\frac{h c}{\lambda}=m c^{2}$\ $m=\frac{h}{c \lambda}$ The mass of a fictitious particle having the same energy as that of the $x$-ray photons $=\frac{x}{3}$ hkg $\frac{x}{3} h=\frac{h}{c \lambda}$ $x=\frac{3}{c \lambda}$ $=\frac{3}{3...
Read More →The number of distinct solutions
Question: The number of distinct solutions of the equation, $\log _{1 / 2}|\sin x|=2-\log _{1 / 2}|\cos x|$ in the interval $[0,2 \pi]$, is_________. Solution: $\log _{1 / 2}|\sin x|=2-\log _{1 / 2}|\cos x|$ $\Rightarrow \log _{1 / 2}|\sin x \cos x|=2$ $\Rightarrow|\sin x \cos x|=\frac{1}{4}$ $\Rightarrow \quad \sin 2 x=\pm \frac{1}{2}$...
Read More →Two radioactive substances X and Y
Question: Two radioactive substances $X$ and $Y$ originally have $N_{1}$ and $N_{2}$ nuclei respectively. Half life of $X$ is half of the half life of $Y$. After there half lives of $Y$, number of nuclei of both are equal. The ratio $\frac{N_{1}}{N_{2}}$ will be equal to:(1) $\frac{8}{1}$(2) $\frac{1}{8}$(3) $\frac{3}{1}$(4) $\frac{1}{3}$Correct Option: 1 Solution: (1) After $\mathbf{n}$ half life no of nuclei undecayed $=\frac{N_{o}}{2^{\mathrm{n}}}$ given $\mathrm{T}_{\frac{1}{2} \mathrm{x}}=\...
Read More →Match List-I with List-II.
Question: Match List-I with List-II. Choose the correct answer from the option given below:$(\mathrm{a})-(\mathrm{iii}),(\mathrm{b})-(\mathrm{ii}),(\mathrm{c})-(\mathrm{i}),(\mathrm{d})-(\mathrm{i} v)$(a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)$(\mathrm{a})-(\mathrm{i} v),(\mathrm{b})-(\mathrm{i}),(\mathrm{c})-(\mathrm{ii}),(\mathrm{d})-(\mathrm{iii})$(a) $-(\mathrm{i}),(\mathrm{b})-(\mathrm{iii}),(\mathrm{c})-(\mathrm{i} v),(\mathrm{d})-(\mathrm{ii})$Correct Option: , 2 Solution: Sodium carb...
Read More →Prove the following
Question: If $\theta_{1}$ and $\theta_{2}$ be respectively the smallest and the largest values of $\theta$ in $(0,2 \pi)-\{\pi\}$ which satisfy the equation,(1) $\frac{\pi}{3}$(2) $\frac{2 \pi}{3}$(3) $\frac{\pi}{3}+\frac{1}{6}$(4) $\frac{\pi}{9}$Correct Option: 1 Solution: $2 \cot ^{2} \theta-\frac{5}{\sin \theta}+4=0$ $\frac{2 \cos ^{2} \theta}{\sin ^{2} \theta}-\frac{5}{\sin \theta}+4=0$ $\Rightarrow \quad 2 \cos ^{2} \theta-5 \sin \theta+4 \sin ^{2} \theta=0, \sin \theta \neq 0$ $\Rightarrow...
Read More →An X-ray tube is operated at 1.24 million volt. The shortest wavelength of the produced photon will be :
Question: An X-ray tube is operated at $1.24$ million volt. The shortest wavelength of the produced photon will be :(1) $10^{-2} \mathrm{~nm}$(2) $10^{-3} \mathrm{~nm}$(3) $10^{-4} \mathrm{~nm}$(4) $10^{-1} \mathrm{~nm}$Correct Option: , 2 Solution: (2) $\lambda_{\min }=\frac{h c}{\mathrm{eV}}$ $\lambda_{\min }=\frac{1240 \mathrm{~nm}-\mathrm{eV}}{1.24 \times 10^{6}}$ $\lambda_{\min }=10^{-3} \mathrm{~nm}$...
Read More →Given below are two statements :
Question: Given below are two statements : Statement-I: Two photons having equal linear momenta have equal wavelengths. Statement-II: If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease. In the light of the above statements, choose the correct answer from the options given below.(1) Statement-I is false but Statement-II is true(2) Both Statement-I and Statement-II are true(3) Both Statement-I and Statement-II are false(4) Statement-I is true but ...
Read More →Prove the following
Question: Let $\alpha$ and $\beta$ be two real roots of the equation $(k+1) \tan ^{2} x-\sqrt{2} \cdot \lambda \tan x=(1-k)$, where $k(\neq-1)$ and $\lambda$ are real numbers. If $\tan ^{2}(\alpha+\beta)=50$, then a value of $\lambda$ is:(1) $10 \sqrt{2}$(2) 10(3) 5(4) $5 \sqrt{2}$Correct Option: , 2 Solution: $(k+1) \tan ^{2} x-\sqrt{2} \lambda \tan x+(k-1)=0$ $\tan \alpha+\tan \beta=\frac{\sqrt{2} \lambda}{k+1}$ [Sum of roots] $\tan \alpha \cdot \tan \beta=\frac{k-1}{k+1}$ [Product of roots] $...
Read More →Which pair of oxides is acidic in nature?
Question: Which pair of oxides is acidic in nature?$\mathrm{N}_{2} \mathrm{O}, \mathrm{BaO}$$\mathrm{CaO}, \mathrm{SiO}_{2}$$\mathrm{B}_{2} \mathrm{O}_{3}, \mathrm{CaO}$$\mathrm{B}_{2} \mathrm{O}_{3}, \mathrm{SiO}_{2}$Correct Option: , 4 Solution: $\mathrm{B}_{2} \mathrm{O}_{3}$ and $\mathrm{SiO}_{2}$ both are oxides of non-metal and hence are acidic in nature....
Read More →The decay of a proton to neutron is :
Question: The decay of a proton to neutron is :(1) not possible as proton mass is less than the neutron mass(2) possible only inside the nucleus(3) not possible but neutron to proton conversion is possible(4) always possible as it is associated only with $\beta^{+}$decayCorrect Option: , 2 Solution: (2) It is possible only inside the nucleus and not otherwise....
Read More →Prove the following
Question: If $\sqrt{3}\left(\cos ^{2} x\right)=(\sqrt{3}-1) \cos x+1$, the number of solutions of the given equation when $\mathrm{x} \in\left[0, \frac{\pi}{2}\right]$ is Solution: $\sqrt{3} t^{2}-(\sqrt{3}-1) t-1=0$ (where $t=\cos x$ ) Now, $t=\frac{(\sqrt{3}-1) \pm \sqrt{4+2 \sqrt{3}}}{2 \sqrt{3}}$ $t=\cos x=1$ or $-\frac{1}{\sqrt{3}} \rightarrow$ rejected as $x \in\left[0, \frac{\pi}{2}\right]$ $\Rightarrow \cos x=1$ $\Rightarrow$ No. of solution $=1$...
Read More →Compound A used as a strong oxidizing agent is amphoteric in nature.
Question: Compound $\mathrm{A}$ used as a strong oxidizing agent is amphoteric in nature. It is the part of lead storage batteries. Compound $\mathrm{A}$ is :$\mathrm{Pb}_{3} \mathrm{O}_{4}$$\mathrm{PbO}_{2}$$\mathrm{PbSO}_{4}$$\mathrm{PbO}$Correct Option: , 2 Solution: lead storage batteries $\mathrm{PbO}_{2}$ is used. In this $\mathrm{O} . \mathrm{S}$. of $\mathrm{Pb}$ is $+4$ so it is always reduced and behaves as oxidizing agent...
Read More →A radioactive sample disintegrates via two independent
Question: A radioactive sample disintegrates via two independent decay processes having half lives $\mathrm{T}_{1 / 2}^{(1)}$ and $\mathrm{T}_{1 / 2}^{(2)}$ respectively. The effective half- life $\mathrm{T}_{1 / 2}$ of the nuclei is:(1) None of the above(2) $\mathrm{T}_{1 / 2}=\mathrm{T}_{1 / 2}^{(1)}+\mathrm{T}_{1 / 2}^{(2)}$(3) $\mathrm{T}_{1 / 2}=\frac{\mathrm{T}_{1 / 2}^{(1)} \mathrm{T}_{1 / 2}^{(2)}}{\mathrm{T}_{1 / 2}^{(1)}+\mathrm{T}_{1 / 2}^{(2)}}$(4) $\mathrm{T}_{1 / 2}=\frac{\mathrm{T...
Read More →The number of integral values of k
Question: The number of integral values of ' $k$ ' for which the equation $3 \sin \mathrm{x}+4 \cos \mathrm{x}=\mathrm{k}+1$ has a solution, $k \in R$ is Solution: $3 \sin x+4 \cos x=k+1$ $-5 \leq \mathrm{k}+1 \leq 5$ $-6 \leq \mathrm{k} \leq 4$ $-6,-5,-4,-3,-2,-1,0,1,2,3,4 \mid \Rightarrow 11$ integral values...
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