Let m be the midpoint and u be the upper class limit of a class in a continuous frequency distribution.
Question: Letmbe the midpoint andube the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is(a) 2mu(b) 2m+u(c)mu(d)m+u Solution: (a) 2m-uGiven:Mid value =mUpper limit =u We know: $\frac{\text { Lower limit+Upper limit }}{2}=$ Mid value $\Rightarrow \frac{\text { Lower limit }+u}{2}=m$ $\Rightarrow$ Lower limit $+u=2 m$ $\Rightarrow$ Lower limit $=2 m-u$...
Read More →The graph between
Question: $3 s$ orbital$2 s$ orbital$1 s$ orbital$2 p$ orbitalCorrect Option: , 2 Solution: The given probability density curve is for $2 s$ orbitaı due to the presence of only one radial node. $1 s$ and $2 p$ orbital do not have any radial node and $3 s$ orbital has two radial nodes. Hence, option (2) is correct....
Read More →A parallel plate capacitor whose capacitance
Question: A parallel plate capacitor whose capacitance $\mathrm{C}$ is $14 \mathrm{pF}$ is charged by a battery to a potential difference $\mathrm{V}=12 \mathrm{~V}$ between its plates. The charging battery is now disconnected and a porcelin plate with $\mathrm{k}=7$ is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of ___ pJ (Assume no friction) Solution: $\mathrm{U}_{\mathrm{i}}=\frac{1}{2} \times 14 \times 12 \ti...
Read More →A parallel plate capacitor whose capacitance
Question: A parallel plate capacitor whose capacitance $\mathrm{C}$ is $14 \mathrm{pF}$ is charged by a battery to a potential difference $\mathrm{V}=12 \mathrm{~V}$ between its plates. The charging battery is now disconnected and a porcelin plate with $\mathrm{k}=7$ is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of ___ pJ (Assume no friction) Solution: $\mathrm{U}_{\mathrm{i}}=\frac{1}{2} \times 14 \times 12 \ti...
Read More →The mid-value of a class interval is 42 and the class size is 10.
Question: The mid-value of a class interval is 42 and the class size is 10. The lower and upper limits are(a) 3747(b) 37.547.5(c) 36.547.5(d) 36.546.5 Solution: (a) 3747Let the lower limit bex.Here,Class size = 10 Upper limit = Class size + Lower limitUpper limit = (x+ 10)Mid value of the class interval = 42 $\therefore \frac{x+x+10}{2}=42$ $\Rightarrow \frac{2 x+10}{2}=42$ $\Rightarrow 2 x+10=84$ $\Rightarrow 2 x=74$ $\Rightarrow x=37$ Thus, we have : Lower limit $=37$ Upper limit $=37+10=47$...
Read More →The sum of the real values of $x$ for which the middle term in
Question: The sum of the real values of $x$ for which the middle term in the binomial expansion of $\left(\frac{x^{3}}{3}+\frac{3}{x}\right)^{8}$ equals 5670 is :(1) 0(2) 6(3) 4(4) 8Correct Option: 1 Solution: Middle Term, $\left(\frac{n}{2}+1\right)^{\text {th }}$ term in the binomial expansion of $\left(\frac{x^{3}}{3}+\frac{3}{x}\right)^{8}$ is, $T_{4+1}={ }^{8} C_{4}\left(\frac{x^{3}}{3}\right)^{4}\left(\frac{3}{x}\right)^{4}=5670$ $\Rightarrow \quad \frac{8 \times 7 \times 6 \times 5}{4 \ti...
Read More →In a frequency distribution, the mid-value of a class is 10 and width of each class is 6. The lower limit of the class is
Question: In a frequency distribution, the mid-value of a class is 10 and width of each class is 6. The lower limit of the class is(a) 6(b) 7(c) 8(d) 12 Solution: (b) 7Given:Mid value of the class = 10Width of each class = 6Now,Let the lower limit bex.We know:Upper limit = Lower limit + Class size =x+ 6 Also, Mid value $=\frac{x+x+6}{2}=\frac{2 x+6}{2}=x+3$ $\Rightarrow x+3=10$ $\Rightarrow x=7$ Thus, the lower limit is 7....
Read More →Which one of the following about an electron occupying the 1 s orbital in a hydrogen atom is incorrect?
Question: Which one of the following about an electron occupying the $1 s$ orbital in a hydrogen atom is incorrect? (The Bohr radius is represented by $a_{0}$ ).The probability density of finding the electron is maximum at the nucleus.The electron can be found at a distance $2 a_{0}$ from the nucleus.The magnitude of the potential energy is double that of its kinetic energy on an average.The total energy of the electron is maximum when it is at a distance $a_{0}$ from the nucleus.Correct Option:...
Read More →The value of r for which
Question: The value of $\mathrm{r}$ for which ${ }^{20} C_{r}{ }^{20} C_{0}+{ }^{20} C_{r-1}{ }^{20} C_{1}+{ }^{20} C_{r-2}{ }^{20} C_{2}+\ldots+{ }^{20} C_{0}{ }^{20} C_{r}$ is maximum, is :(1) 15(2) 20(3) 11(4) 10Correct Option: , 2 Solution: Consider the expression ${ }^{20} C_{r}{ }^{20} C_{0}+{ }^{20} C_{r}-1{ }^{20} C_{1}$ $+{ }^{20} C_{r-2}{ }^{20} C_{2}+\ldots+{ }^{20} C_{0} \cdot{ }^{20} C_{r}$ For maximum value of above expression $r$ should be equal to 20 . as ${ }^{20} C_{20} \cdot{ ...
Read More →The class marks of a frequency distribution are 15, 20, 25, 30, .. .
Question: The class marks of a frequency distribution are 15, 20, 25, 30, .. . The class corresponding to the class marks 20 is(a) 12.517.5(b) 17.522.5(c) 18.521.5(d) 19.520.5 Solution: (b) 17.5-22.5We are given frequency distribution 15, 20, 25, 30,...Class size = 20-15 = 5Class marks = 20Now, Lower limit $=\left(20-\frac{5}{2}\right)=\frac{35}{2}=17.5$ Upper limit $=\left(20+\frac{5}{2}\right)=\frac{45}{2}=22.5$ Thus, the required class is 17.5-22.5....
Read More →In the class intervals 10−20, 20−30, the number 20 is included in
Question: In the class intervals 1020, 2030, the number 20 is included in(a) 1020(b) 2030(c) in each of 1020and 2030(d) in none of 1020and 2030 Solution: (b) 2030This is the continuous form of frequency distribution. Here, the upper limit of each class is excluded, while the lower limit is included. So, the number 20 is included in the class interval 2030....
Read More →For any given series of spectral lines of atomic hydrogen,
Question: For any given series of spectral lines of atomic hydrogen, let $\Delta \bar{v}=\bar{v}_{\max } \mathrm{fl}^{\bar{v}_{\min } \text { be the difference in maximum and }}$ minimum frequencies in $\mathrm{cm}^{-1}$. The ratio $\Delta \bar{v}_{\text {Lyman }} / \Delta \bar{v}$ Balmer is :$4: 1$$9: 4$$5: 4$$27: 5$Correct Option: , 2 Solution: $\bar{v} \propto \Delta \mathrm{E}$ For H-atom $\bar{v}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ For Lyman ser...
Read More →The class mark of the class 100−120 is
Question: The class mark of the class 100120 is(a) 100(b) 110(c) 115(d) 120 Solution: (b) 110 Class mark $=\frac{\text { Upper limit+Lower limit }}{2}=\frac{120+100}{2}=110$...
Read More →If
Question: If $\sum_{r=0}^{25}\left\{{ }^{50} C_{r} \cdot{ }^{50-r} C_{25-r}\right\}=K\left({ }^{50} C_{25}\right)$, then $K$ is equal to:(1) $(25)^{2}$(2) $2^{25}-1$(3) $2^{24}$(4) $2^{25}$Correct Option: , 4 Solution: $\sum_{r=0}^{25}\left({ }^{50} C_{r} \cdot{ }^{50-r} C_{25-r}\right)=\sum_{r=0}^{25}\left(\frac{\lfloor 50}{|50-r| r} \frac{\mid 50-r}{|25| 25-r}\right)$ $=\sum_{r=0}^{25}\left(\frac{\lfloor 50}{\lfloor 25} \times \frac{1}{\lfloor 25} \times\left(\frac{\lfloor 25}{\lfloor 25-r \mi...
Read More →If
Question: If $\sum_{r=0}^{25}\left\{{ }^{50} C_{r} \cdot{ }^{50-r} C_{25-r}\right\}=K\left({ }^{50} C_{25}\right)$, then $K$ is equal to:(1) $(25)^{2}$(2) $2^{25}-1$(3) $2^{24}$(4) $2^{25}$Correct Option: , 4 Solution: $\sum_{r=0}^{25}\left({ }^{50} C_{r} \cdot{ }^{50-r} C_{25-r}\right)=\sum_{r=0}^{25}\left(\frac{\lfloor 50}{|50-r| r} \frac{\mid 50-r}{|25| 25-r}\right)$ $=\sum_{r=0}^{25}\left(\frac{\lfloor 50}{\lfloor 25} \times \frac{1}{\lfloor 25} \times\left(\frac{\lfloor 25}{\lfloor 25-r \mi...
Read More →The range of the data
Question: The range of the data12, 25, 15, 18, 17, 20, 22, 6, 16, 11, 8, 19, 10, 30, 20, 32 is(a) 10(b) 15(c) 18(d) 26 Solution: (d) 26We have:Maximum value = 32Minimum value = 6We know:Range = Maximum value-Minimum value=32-6=26...
Read More →Four identical rectangular plates with length,
Question: Four identical rectangular plates with length, $l=2 \mathrm{~cm}$ and breadth, $\mathrm{b}=\frac{3}{2} \mathrm{~cm}$ are arranged as shown in figure. The equivalent capacitance between $\mathrm{A}$ and $\mathrm{C}$ is $\frac{\mathrm{x} \varepsilon_{0}}{\mathrm{~d}}$. The value of $\mathrm{x}$ is (Round off to the Nearest Integer) Solution: $\mathrm{C}_{\mathrm{eq}}=\frac{2 \mathrm{C}_{0}}{3}=\frac{2}{3} \frac{\mathrm{E}_{0} \mathrm{~A}}{\mathrm{~d}}$ $\mathrm{C}_{\mathrm{eq}}=\frac{2 \...
Read More →The positive value of
Question: The positive value of $\lambda$ for which the co-efficient of $x^{2}$ in the expression $x^{2}\left(\sqrt{x}+\frac{\lambda}{x^{2}}\right)^{10}$ is 720 , is:(1) 4(2) $2 \sqrt{2}$(3) $\sqrt{5}$(4) 3Correct Option: 1 Solution: Since, coefficient of $x^{2}$ in the expression $x^{2}\left(\sqrt{x}+\frac{\lambda}{x^{2}}\right)$ is a constant term, then Coefficient of $x^{2}$ in $x^{2}\left(\sqrt{x}+\frac{\lambda}{x^{2}}\right)^{10}$ $=$ co-efficient of constant term in $\left(\sqrt{x}+\frac{\...
Read More →If p is the momentum of the fastest electron ejected from
Question: If $p$ is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength $\lambda$, then for $1.5 p$ momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function):$\frac{3}{4} \lambda$$\frac{1}{2} \lambda$$\frac{2}{3} \lambda$$\frac{4}{9} \lambda$Correct Option: , 4 Solution: In photoelectric effect, $\frac{h c}{\lambda}=w+\mathrm{K...
Read More →Find the values of a, b, c, d, e, f, g from the following frequency distribution of the heights of 50 students in a class:
Question: Find the values ofa, b, c, d, e, f, gfrom the following frequency distribution of the heights of 50 students in a class: Solution: The complete table will be...
Read More →If the third term in the binomial expansion of
Question: If the third term in the binomial expansion of $\left(1+x^{\log _{2} x}\right)^{5}$ equals 2560 , then a possible value of $x$ is:(1) $\frac{1}{4}$(2) $4 \sqrt{2}$(3) $\frac{1}{8}$(4) $2 \sqrt{2}$Correct Option: 1 Solution: Third term of $\left(1+x^{\log _{2} x}\right)^{5}={ }^{5} C_{2}\left(x^{\log _{2} x}\right)^{5-3}$ $={ }^{5} C_{2}\left(x^{\log _{2} x}\right)^{2}$ Given, ${ }^{5} \mathrm{C}_{2}\left(x^{\log _{2} x}\right)^{2}=2560$ $\Rightarrow\left(x^{\log _{2} x}\right)^{2}=256=...
Read More →The quantum number of four electrons are given below :
Question: The quantum number of four electrons are given below : I. $\quad n=4, l=2, m_{l}=-2, m_{s}=-1 / 2$ II. $n=3, l=2, m_{l}=1, m_{s}=+1 / 2$ III. $n=4, l=1, m_{l}=0, m_{s}=+1 / 2$ IV. $n=3, l=1, m_{l}=1, m_{\mathrm{s}}=-1 / 2$ The correct order of their increasing energies will be :$\mathrm{IV}\mathrm{III}\mathrm{II}\mathrm{I}$ $\mathrm{I}\mathrm{II}\mathrm{III}\mathrm{IV}$$\mathrm{IV}\mathrm{II}\mathrm{III}\mathrm{I}$$\mathrm{I}\mathrm{III}\mathrm{II}\mathrm{IV}$Correct Option: , 3 Soluti...
Read More →If
Question: If $\sum_{i=1}^{20}\left(\frac{{ }^{20} \mathrm{C}_{i-1}}{{ }^{20} \mathrm{C}_{i}+{ }^{20} \mathrm{C}_{i-1}}\right)^{3}=\frac{k}{21}$, then $k$ equals:(1) 400(2) 50(3) 200(4) 100Correct Option: , 4 Solution: Consider the expression, $\frac{{ }^{20} C_{i-1}}{{ }^{20} C_{i}+{ }^{20} C_{i-1}}=\frac{{ }^{20} C_{i-1}}{{ }^{21} C_{1}}$ $=\frac{20 !}{(i-1) !(21-i) !} \times \frac{i !(21-i) !}{21 !}=\frac{i}{21}$ $\therefore \quad \sum_{i=1}^{20}\left(\frac{{ }^{20} C_{i-1}}{{ }^{20} C_{i}+{ }...
Read More →The coefficient of
Question: The coefficient of $t^{4}$ in the expansion of $\left(\frac{1-t^{6}}{1-t}\right)^{3}$(1) 14(2) 15(3) 10(4) 12Correct Option: , 2 Solution: Consider the expression $\left(\frac{1-t^{6}}{1-t}\right)^{3}=\left(1-t^{6}\right)^{3}(1-t)^{-3}$ $\begin{aligned}=\left(1-3 t^{6}+3 t^{12}-t^{18}\right) \left(1+3 t+\frac{3 \cdot 4}{2 !} t^{2}\right.\\ \left.+\frac{3 \cdot 4 \cdot 5}{3 !} t^{3}+\frac{3 \cdot 4 \cdot 5 \cdot 6}{4 !} t^{4}+\mathrm{K} \infty\right) \end{aligned}$ Hence, the coefficien...
Read More →The de Broglie wavelength of an electron in the
Question: The de Broglie wavelength of an electron in the $4^{\text {th }}$ Bohr orbit is:$2 \pi \mathrm{a}_{0}$$4 \pi \mathrm{a}_{0}$$6 \pi \mathrm{a}_{0}$$8 \pi \mathrm{a}_{0}$Correct Option: , 4 Solution: $2 \pi r=n \lambda$ $r=\frac{n^{2} a_{0}}{Z}$ $2 \pi \times \frac{4^{2}}{1} a_{0}=4 \lambda$ $\lambda=2 \pi \times \frac{4}{1} a_{0}$ $\lambda=8 \pi a_{0}$...
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