Find all points of discontinuity of f, where f is defined by
Question: Find all points of discontinuity off, wherefis defined by $f(x)=\left\{\begin{array}{l}\frac{|x|}{x} \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right.$ Solution: The given function $f$ is $f(x)=\left\{\begin{array}{l}\frac{|x|}{x} \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right.$ It is known that, $x0 \Rightarrow|x|=-x$ and $x0 \Rightarrow|x|=x$ Therefore, the given function can be rewritten as $f(x)=\left\{\begin{array}{l}\frac{|x|}{x}=\frac{-x}{x}=-1 \text ...
Read More →The Q value of a nuclear reaction A + b → C + d is defined by
Question: TheQvalue of a nuclear reactionA+bC+dis defined by Q= [mA+mbmCmd]c2where the masses refer to the respective nuclei. Determine from the given data theQ-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) ${ }_{1}^{1} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H}$ (ii) ${ }_{6}^{12} \mathrm{C}+{ }_{6}^{12} \mathrm{C} \rightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_{2}^{4} \mathrm{He}$ Atomic masses are ...
Read More →Find the sum of the following series up to n terms:
Question: Find the sum of the following series up tonterms:$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots$ Solution: The $n^{\text {th }}$ term of the given series is $\frac{1^{3}+2^{3}+3^{3}+\ldots+n^{3}}{1+3+5+\ldots+(2 n-1)}=\frac{\left[\frac{n(n+1)}{2}\right]^{2}}{1+3+5+\ldots+(2 n-1)}$ Here, $1,3,5, \ldots(2 n-1)$ is an A.P. with first term a, last term $(2 n-1)$ and number of terms as $n$ $\therefore 1+3+5+\ldots .+(2 n-1)=\frac{n}{2}[2 \times 1+(n-1) 2]=n^...
Read More →What are the reactions involved for ozone layer depletion in the stratosphere?
Question: What are the reactions involved for ozone layer depletion in the stratosphere? Solution: In the stratosphere, ozone is a product of the action of UV radiations on dioxygen as: (i) $\mathrm{O}_{2(g)} \stackrel{\mathrm{UV}}{\longrightarrow} \mathrm{O}_{(g)}+\mathrm{O}_{(g)}$ (ii) $\mathrm{O}_{2(g)}+\mathrm{O}_{(g)} \stackrel{\mathrm{UV}}{\longleftrightarrow} \mathrm{O}_{3(g)}$ Reaction (ii) indicates the dynamic equilibrium existing between the production and decomposition of ozone molec...
Read More →The nucleus
Question: The nucleus ${ }_{10}^{23}$ Ne decays by $\beta^{-}$emission. Write down the $\beta$ decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: $m\left({ }_{10}^{23} \mathrm{Ne}\right)=22.994466 \mathrm{u}$ $m\left({ }_{11}^{23} \mathrm{Na}\right)=22.989770 \mathrm{u} .$ Solution: In $\beta^{-}$emission, the number of protons increases by 1 , and one electron and an antineutrino are emitted from the parent nucleus. $\beta^{-}$emission of the nucleus $...
Read More →If S1, S2, S3 are the sum of first n natural numbers,
Question: If $\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}$ are the sum of first $n$ natural numbers, their squares and their cubes, respectively, show that $9 \mathrm{~S}_{2}^{2}=\mathrm{S}_{3}\left(1+8 \mathrm{~S}_{1}\right)$ Solution: From the given information, $S_{1}=\frac{n(n+1)}{2}$ $\mathrm{S}_{3}=\frac{\mathrm{n}^{2}(\mathrm{n}+1)^{2}}{4}$ Here, $S_{3}\left(1+8 S_{1}\right)=\frac{n^{2}(n+1)^{2}}{4}\left[1+\frac{8 n(n+1)}{2}\right]$ $=\frac{n^{2}(n+1)^{2}}{4}\left[1+4 n^{2}+4 n\right...
Read More →Find all points of discontinuity of f, where f is defined by
Question: Find all points of discontinuity off, wherefis defined by $f(x)=\left\{\begin{array}{l}|x|+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3x3 \\ 6 x+2, \text { if } x \geq 3\end{array}\right.$ Solution: The given function $f$ is $f(x)=\left\{\begin{array}{l}|x|+3=-x+3, \text { if } x \leq-3 \\ -2 x, \text { if }-3x3 \\ 6 x+2, \text { if } x \geq 3\end{array}\right.$ The given functionfis defined at all the points of the real line. Letcbe a point on the real line. Case I: If $c-3$, then...
Read More →The radionuclide
Question: The radionuclide11C decays according to ${ }_{6}^{11} \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{~B}+e^{+}+v: \quad \mathrm{T}_{1 / 2}=20.3 \mathrm{~min}$ The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: $m\left({ }_{6}^{11} \mathrm{C}\right)=11.011434 \mathrm{u}$ and $m\left({ }_{6}^{11} \mathrm{~B}\right)=11.009305 \mathrm{u}$, calculateQand compare it with the maximum energy of the positron emitted Solution: The givennuclear reaction is: ${ }_{6}^{11} ...
Read More →Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …
Question: Find the sum of the firstnterms of the series: 3 + 7 + 13 + 21 + 31 + Solution: The given series is 3 + 7 + 13 + 21 + 31 + $S=3+7+13+21+31+\ldots+a_{n-1}+a_{n}$ $S=3+7+13+21+\ldots+a_{n-2}+a_{n-1}+a_{n}$ On subtracting both the equations, we obtain $S-S=\left[3+\left(7+13+21+31+\ldots+a_{n-1}+a_{n}\right)\right]-\left[\left(3+7+13+21+31+\ldots+a_{n-1}\right)+a_{n}\right]$ $S-S=3+\left[(7-3)+(13-7)+(21-13)+\ldots+\left(a_{n}-a_{n-1}\right)\right]-a_{n}$ $0=3+[4+6+8+\ldots(n-1)$ terms $]...
Read More →The radionuclide
Question: The radionuclide11C decays according to ${ }_{6}^{11} \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{~B}+e^{+}+v: \quad \mathrm{T}_{1 / 2}=20.3 \mathrm{~min}$ The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: $m\left({ }_{6}^{11} \mathrm{C}\right)=11.011434 \mathrm{u}$ and $m\left({ }_{6}^{11} \mathrm{~B}\right)=11.009305 \mathrm{u}$, calculateQand compare it with the maximum energy of the positron emitted Solution: The givennuclear reaction is: ${ }_{6}^{11} ...
Read More →The radionuclide
Question: The radionuclide11C decays according to ${ }_{6}^{11} \mathrm{C} \rightarrow{ }_{5}^{11} \mathrm{~B}+e^{+}+v: \quad \mathrm{T}_{1 / 2}=20.3 \mathrm{~min}$ The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: $m\left({ }_{6}^{11} \mathrm{C}\right)=11.011434 \mathrm{u}$ and $m\left({ }_{6}^{11} \mathrm{~B}\right)=11.009305 \mathrm{u}$, calculateQand compare it with the maximum energy of the positron emitted Solution: The givennuclear reaction is: ${ }_{6}^{11} ...
Read More →Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.
Question: Find the $20^{\text {th }}$ term of the series $2 \times 4+4 \times 6+6 \times 8+\ldots+n$ terms. Solution: The given series is $2 \times 4+4 \times 6+6 \times 8+\ldots n$ terms $\therefore n^{\text {th }}$ term $=a_{n}=2 n \times(2 n+2)=4 n^{2}+4 n$ $a_{20}=4(20)^{2}+4(20)=4(400)+80=1600+80=1680$ Thus, the $20^{\text {th }}$ term of the series is 1680 ....
Read More →Find the sum of the following series up to n terms:
Question: Find the sum of the following series up tonterms: (i) 5 + 55 + 555 + (ii) .6 +.66 +. 666 + Solution: (i) 5 + 55 + 555 + Let $S_{n}=5+55+555+\ldots . .$ to $n$ terms $=\frac{5}{9}[9+99+999+\ldots$ to $\mathrm{n}$ terms $]$ $=\frac{5}{9}\left[(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots\right.$ to $\mathrm{n}$ terms $]$ $=\frac{5}{9}\left[\left(10+10^{2}+10^{3}+\ldots \mathrm{n}\right.\right.$ terms $)-(1+1+\ldots \mathrm{n}$ terms $\left.)\right]$ $=\frac{5}{9}\left[\frac{1...
Read More →Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of
Question: Find the Q-value and the kinetic energy of the emitted $\alpha$-particle in the $\alpha$-decay of (a) ${ }_{88}^{226} \mathrm{Ra}$ and (b) ${ }_{86}^{220} \mathrm{Rn}$. Given $m\left({ }_{88}^{226} \mathrm{Ra}\right)=226.02540 \mathrm{u}, m\left(\frac{222}{86} \mathrm{Rn}\right)=222.01750 \mathrm{u}$, $m\left({ }_{86}^{220} \mathrm{Rn}\right)=220.01137 \mathrm{u}, m\left({ }_{84}^{216} \mathrm{Po}\right)=216.00189 \mathrm{u} .$ Solution: (a) Alpha particle decay of ${ }_{88}^{226} \mat...
Read More →Find all points of discontinuity of f, where f is defined by
Question: Find all points of discontinuity off, wherefis defined by $f(x)=\left\{\begin{array}{l}2 x+3, \text { if } x \leq 2 \\ 2 x-3, \text { if } x2\end{array}\right.$ Solution: The given function $f$ is $f(x)=\left\{\begin{array}{l}2 x+3, \text { if } x \leq 2 \\ 2 x-3, \text { if } x2\end{array}\right.$ It isevident that the given functionfis defined at all the points of the real line. Letcbe a point on the real line. Then, three cases arise. (i) $c2$ (ii) $c2$ (iii) $c=2$ Case (i) $c2$ The...
Read More →Obtain approximately the ratio of the nuclear radii of the gold isotope
Question: Obtain approximately the ratio of the nuclear radii of the gold isotope ${ }_{79}^{197} \mathrm{Au}$ and the silver isotope ${ }_{47}^{107} \mathrm{Ag}$. Solution: Nuclear radius of the gold isotope $_{79} \mathrm{Au}^{197}=R_{\mathrm{Au}}$ Nuclear radius of the silver isotope ${ }_{47} \mathrm{Ag}^{107}=R_{\mathrm{Ag}}$ Mass number of gold,AAu= 197 Mass number of silver,AAg= 107 The ratio ofthe radii of the two nuclei is related with their mass numbers as: $\frac{R_{\mathrm{Au}}}{R_{\...
Read More →The half-life of
Question: The half-life of ${ }_{38}^{90} \mathrm{Sr}$ is 28 years. What is the disintegration rate of $15 \mathrm{mg}$ of this isotope? Solution: Half life of ${ }_{38}^{90} \mathrm{Sr}, t_{1 / 2}=28$ years = 28 365 24 60 60 = 8.83 108s Mass of theisotope,m= 15 mg $90 \mathrm{~g}$ of ${ }_{38}^{90} \mathrm{Sr}$ atom contains $6.023 \times 10^{23}$ (Avogadro's number) atoms. Therefore, $15 \mathrm{mg}$ of ${ }_{38}^{90} \mathrm{Sr}$ contains: $\frac{6.023 \times 10^{23} \times 15 \times 10^{-3}}...
Read More →Obtain the amount of
Question: Obtain the amount of ${ }_{27}^{60}$ Co necessary to provide a radioactive source of $8.0 \mathrm{mCi}$ strength. The half-life of ${ }_{27}^{60}$ Co is $5.3$ years. Solution: The strength of the radioactive source is given as: $\frac{d N}{d t}=8.0 \mathrm{mCi}$ $=8 \times 10^{-3} \times 3.7 \times 10^{10}$ $=29.6 \times 10^{7}$ decay $/ \mathrm{s}$ Where, N= Required number of atoms Half-life of ${ }_{27}^{60} \mathrm{Co}, T_{1 / 2}=5.3$ years = 5.3 365 24 60 60 = 1.67 108s For decay ...
Read More →The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon.
Question: The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive ${ }_{6}^{14} \mathrm{C}$ present with the stable carbon isotope ${ }_{6}^{12} \mathrm{C}$. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life ( 5730 years) of ${ }_{6}^{14} \math...
Read More →A radioactive isotope has a half-life of T years.
Question: A radioactive isotope has a half-life ofTyears. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value? Solution: Half-life ofthe radioactive isotope =Tyears Original amount of the radioactive isotope =N0 (a)After decay, the amount of the radioactive isotope =N It is given that only 3.125% ofN0remains after decay. Hence, we can write: $\frac{N}{N_{0}}=3.125 \%=\frac{3.125}{100}=\frac{1}{32}$ But $\frac{N}{N_{0}}=e^{-\lambda t}$ Where, = Decay constant t=...
Read More →A radioactive isotope has a half-life of T years.
Question: A radioactive isotope has a half-life ofTyears. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value? Solution: Half-life ofthe radioactive isotope =Tyears Original amount of the radioactive isotope =N0 (a)After decay, the amount of the radioactive isotope =N It is given that only 3.125% ofN0remains after decay. Hence, we can write: $\frac{N}{N_{0}}=3.125 \%=\frac{3.125}{100}=\frac{1}{32}$ But $\frac{N}{N_{0}}=e^{-\lambda t}$ Where, = Decay constant t=...
Read More →A radioactive isotope has a half-life of T years.
Question: A radioactive isotope has a half-life ofTyears. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value? Solution: Half-life ofthe radioactive isotope =Tyears Original amount of the radioactive isotope =N0 (a)After decay, the amount of the radioactive isotope =N It is given that only 3.125% ofN0remains after decay. Hence, we can write: $\frac{N}{N_{0}}=3.125 \%=\frac{3.125}{100}=\frac{1}{32}$ But $\frac{N}{N_{0}}=e^{-\lambda t}$ Where, = Decay constant t=...
Read More →Is the function f defined by
Question: Is the functionfdefined by $f(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 5, \text { if } x1\end{array}\right.$ continuous at $x=0$ ? At $x=1$ ? At $x=2$ ? Solution: The given function $f$ is $f(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 5, \text { if } x1\end{array}\right.$ At $x=0$ It isevident thatfis defined at 0 and its value at 0 is 0. Then, $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x=0$ $\therefore \lim _{x \rightarrow 0} f(x)=f(0)$ Therefore,fis...
Read More →Prove that the function is continuous at x = n, where n is a positive integer.
Question: Prove that the function $f(x)=x^{n}$ is continuous at $x=n$, where $n$ is a positive integer. Solution: The given function is $f(x)=x^{n}$ It is evident that $f$ is defined at all positive integers, $n$, and its value at $n$ is $n^{n}$. Then, $\lim _{x \rightarrow n} f(n)=\lim _{x \rightarrow n}\left(x^{n}\right)=n^{n}$ $\therefore \lim _{x \rightarrow n} f(x)=f(n)$ Therefore,fis continuous atn, wherenis a positive integer...
Read More →What are the harmful effects of photochemical
Question: What are the harmful effects of photochemical smog and how can they be controlled? Solution: EFFECTS OF PHOTOCHEMICAL SMOG: Photochemical smog isoxidizing smog owing to the presence of NO2and O3,causing corrosion of metals, stones, rubber, and painted surfaces. The other major components of photochemical smog are PAN, acrolein, and formaldehyde. Both PAN and ozone are eye irritants, while nitric oxide (formed from NO2) causes nose and throat irritation. At higher concentrations, photoc...
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