Class 11th

$meqof{l}_2=meqofN{a}_2{S}_2{O}_3=20*0.002*1$

2 * mmol of l₂ = 0.4

mmol of l₂ = 0.2 mmol

mmol of Cu²⁺ = 0.2 * 2 * 10^-3

$\left[C{u}^{2+}\right]=\frac{0.4*10^{-3}}{10*10^{-3}}=0.04=4*10^{-2}M$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat{\sigma }_C=5\\ WeknowthatC=\frac{5}{9}\left(F-32\right)\Rightarrow F=\frac{9C}{5}+32\\ \therefore {\sigma }_F=\frac{9}{5}{\sigma }_C=\frac{9}{5}*5=9\\ \therefore {\sigma }_F^2={\left(9\right)}^2=81\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Here,wehaveC{V}_1=50,C{V}_2=60\\ \overline{x_1}=30and\overline{x_2}=25\\ \therefore C{V}_1=\frac{{\sigma }_1}{\overline{x_1}}*100\Rightarrow 50=\frac{{\sigma }_1}{30}*100\Rightarrow {\sigma }_1=\frac{50*30}{100}=15\\ andC{V}_1=\frac{{\sigma }_2}{\overline{x_2}}*100\Rightarrow 60=\frac{{\sigma }_2}{25}*100\Rightarrow {\sigma }_2=\frac{60*25}{100}=15\\ \therefore Difference{\sigma }_1-{\sigma }_2=15-15=0\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}WeknowthatVariance\left({\sigma }^2\right)=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{N}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{N}\right)}^2\\ =\frac{18000}{60}-{\left(\frac{960}{60}\right)}^2=300-256=44\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}First10positiveintegersare1,2,3,4,5,6,7,8,9,10\\ onmultiplyingeachnumberby-1,weget\\ -1,-2,-3,-4,-5,-6,-7,-8,-9,-10\\ onadding1toeachofthenumber,weget\\ 0,-1,-2,-3,-4,-5,-6,-7,-8,-9\\ \therefore {\displaystyle \sum _{}^{}{x}_i}=0-1-2-3-4-5-6-7-8-9\\ =-45\\ and{\displaystyle \sum _{}^{}{x}_i^2}=0^2+{\left(-1\right)}^2+{\left(-2\right)}^2+{\left(-3\right)}^2+{\left(-4\right)}^2+\dots +{\left(-9\right)}^2\\ =\frac{9*10*19}{6}=285\left[?{\displaystyle \sum _{}^{}{n}^2=}\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]\\ \therefore S.D=\sqrt[]{\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{N}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{N}\right)}^2}=\sqrt[]{\frac{285}{10}-{\left(\frac{-45}{10}\right)}^2}\\ =\sqrt[]{\frac{285}{10}-\frac{2025}{100}}=\sqrt[]{\frac{2850-2025}{100}}=\sqrt[]{8.25}\\ \therefore Variance={\left(SD\right)}^2={\left(\sqrt[]{8.25}\right)}^2=8.25\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Givennumbersare1,2,3,4,5,6,7,8,9,10\\ Numbersobtainedwhen1isaddedtotheabovenumbersis2,3,4,5,6,7,8,9,10and11.\\ \therefore {\displaystyle \sum _{}^{}{x}_i}=2+3+4+5+6+7+8+9+10+11\\ =\frac{10}{2}\left[2*2+\left(10-1\right).1\right]=5\left[4+9\right]=5*13=65\\ Now{\displaystyle \sum _{}^{}{x}_i^2}=2^2+3^2+4^2+\dots +11^2\\ =\left(1^2+2^2+3^2+4^2+\dots +11^2\right)-{\left(1\right)}^2\\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6}-1=\frac{11*12*23}{6}-1\\ =22*23-1=506-1=505\\ \therefore Variance\left({\sigma }^2\right)=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{n}\right)}^2=\frac{505}{10}-{\left(\frac{65}{10}\right)}^2\\ =50.5-{\left(6.5\right)}^2=50.5-42.25=8.25\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}WeknowthatSDoffirstnnaturalnumbers\sqrt[]{\frac{n^2-1}{12}}\\ Heren=10\\ \therefore SD=\sqrt[]{\frac{{\left(10\right)}^2-1}{12}}=\sqrt[]{\frac{99}{12}}=\sqrt[]{8.25}=2.87\\ Hence,thecorrectoptionis\left(d\right).\end{array}$