Chemistry, asked by pikkigaming1234, 4 days ago

1. Copper is drawn into wires to make electric circuits. This is because it is
(i) Malleable and good conductor of electricity
(ii) Ductile is good conductor of electricity
(iii) Sonorous and good conductor of electricity
(iv) Brittle and good conductor of electricity

Answers

Answered by MAAJJA

$Let the assumed mean be A = 50 and h = 20Calculation of mean\begin{gathered}\begin{array}{ |c|c|c|c|c|} \hline \rm Class& \rm Frequency&\rm Mid-values&u_i= \dfrac{x_i-A}{h} &f_iu_i \\\\ \hline \\ \: \: \: 0-20&17&10& - 1& - 34\\20 - 40&f _1&30& - 2& - f_ i\\40 - 60&32&50& \: \: \: \: 0& \: \: \: \: \: \: \: 0\\60 - 80&f_ 2&70& \: \: \: \: 1& \: \: \: \: f_2 \\\!\!\!80 - 100 &19&90& \: \: \: \: 2& \: \: \: \: \: 38 \\ \hline &N=\sum \!f_i=68+f_1+f_2&&&\sum\! f_iu_i=4-f_1+f_2\\ \hline\end{array}\end{gathered} Class0−2020−4040−6060−8080−100 Frequency17f 1 32f 2 19N=∑f i =68+f 1 +f 2 Mid−values1030507090 u i = hx i −A −1−2012 f i u i −34−f i 0f 2 38∑f i u i =4−f 1 +f 2 We have,\begin{gathered}N = \sum\!f_i = 120\\\end{gathered} N=∑f i =120 \begin{gathered} \implies68 + f_1 + f_2 = 120 \\ \end{gathered} ⟹68+f 1 +f 2 =120 \begin{gathered} \implies f_1 + f_2 = 120 - 68 \\ \end{gathered} ⟹f 1 +f 2 =120−68 \begin{gathered} \implies f_1 + f_2 = 52 \xrightarrow{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: }(1) \\ \end{gathered} ⟹f 1 +f 2 =52 (1) Now,\rm{ \: \: \: \: \: \: \: \: \: \: Mean = 50}Mean=50\begin{gathered} \implies \: \: \: \: A + h \bigg \{ \dfrac{1}{N} \sum \!f _ iu_i \bigg \} = 50 \\\\ \end{gathered} ⟹A+h{ N1 ∑f i u i }=50 \begin{gathered} \implies \: 50 + 20 \bigg \{ \dfrac{4 - f _1 + f _ 2}{120} \bigg \} = 50 \\\\ \end{gathered} ⟹50+20{ 1204−f 1 +f 2 }=50 \begin{gathered} \implies \: 50 + \cancel{ 20} \bigg \{ \dfrac{4 - f _1 + f _ 2}{ \cancel{120}} \bigg \} = 50 \\\\ \end{gathered} ⟹50+ 20 { 120 4−f 1 +f 2 }=50 \begin{gathered} \implies \: \: \: \: \: \: 50 + \bigg \{ \dfrac{4 - f _1 + f _ 2}{ 6 } \bigg \} = 50 \\\\ \end{gathered} ⟹50+{ 64−f 1 +f 2 }=50 \begin{gathered} \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg \{ \dfrac{4 - f _1 + f _ 2}{ 6} \bigg \} = 50 - 50\\\\ \end{gathered} ⟹{ 64−f 1 +f 2 }=50−50 \begin{gathered} \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{4 - f _1 + f _ 2}{ 6 } = 0 \\\\ \end{gathered} ⟹ 64−f 1 +f 2 =0 \begin{gathered} \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 4 - f _1 + f _ 2 = 0 \\\\ \end{gathered} ⟹4−f 1 +f 2 =0 \begin{gathered}\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: f _1 - f _ 2 = 4 \xrightarrow{ \: \: \: \: \: \: \: \: \: \: }(2)\\\\ \end{gathered} ⟹f 1 −f 2 =4 (2) Solving 1 and 2\begin{gathered}:\implies \: \: \: \: f _1 - f _ 2 = 4 \xrightarrow{ \: \: \: \: \: \: \: \: \: \: \: \: \: }(1)\\ \end{gathered} :⟹f 1 −f 2 =4 (1) \begin{gathered}:\implies \: \: \: \: f_1 + f_2 = 52\xrightarrow{ \: \: \: \: \: \: \: \: \: \: \: \: \: }(2) \\ \\ \end{gathered} :⟹f 1 +f 2 =52 (2) \begin{gathered}:\implies \: \: \: \: f _1 = 4 + f _ 2 \\ \end{gathered} :⟹f 1 =4+f 2 \begin{gathered}:\implies \: \: \: \: f_1 + f_2 = 52 \\ \end{gathered} :⟹f 1 +f 2 =52 \begin{gathered}:\implies \: \: \: \: 4 + f_2 + f_2 = 52 \\ \end{gathered} :⟹4+f 2 +f 2 =52 \begin{gathered}:\implies \: \: \: \: 2f_2 = 48 \\ \end{gathered} :⟹2f 2 =48 \begin{gathered}:\implies \: \: \: \: f_2 = \ \dfrac{48 }{2}\\ \end{gathered} :⟹f 2 = 248 \begin{gathered}:\implies \: \: \: \: f_2 = \cancel\dfrac{48 }{2}\\ \end{gathered} :⟹f 2 = 248 \begin{gathered}:\implies \: \: \: \: f_2 = 24\\ \\ \end{gathered} :⟹f 2 =24 \begin{gathered}:\implies \: \: \: \: f_1 + f_2 = 52 \\ \end{gathered} :⟹f 1 +f 2 =52 \begin{gathered}:\implies \: \: \: \: f_1 + 24 = 52 \\ \end{gathered} :⟹f 1 +24=52 \begin{gathered}:\implies \: \: \: \: f_1 = 28\\ \\ \end{gathered} :⟹f 1 =28 \therefore f_1 = 28 \: \mathrm{and} \: f_2 = 24∴f 1 =28andf 2 =24$

Answered by kallunniexpress

The property of metal to be drawn to wires are called ductility and the best conductors of electricity are - silver,gold,copper etc
Since copper is more cheaper it is used as electric wires - therefore ans - option (ii)

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