Question

need explaination to this!!!!​

Answer 1

$\underline{\blacksquare\:\:\:\footnotesize{\red{FigurE}}}$

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{0.3mm}\put(0,0){\vector(1,0){40}}\put(0,0){\vector(0,1){30}}\thicklines\put(15,14){\line(2,3){5.1}}\put(15,14){\vector(2,3){3}}\put(20,21.8){\line(2,-3){7}}\put(20,21.8){\vector(2,-3){4}}\put(15,14){\circle*{0.8}}\put(27.1,11.1){\circle*{0.8}}\linethickness{0.01mm}\put(20,21.8){\line(-1,0){20}}\put(15,14){\line(-1,0){15}}\put(27.1,11.1){\line(-1,0){27.1}}\put(15,14){\line(0,-1){14}}\put(27.1,11.1){\line(0,-1){11.1}}\put(20,21.8){\line(0,-1){21.8}}\footnotesize{\put(14.5,-2){V}}\footnotesize{\put(19,-2){2V}}\footnotesize{\put(26.1,-2){3V}}\footnotesize{\put(-2,11){p}}\footnotesize{\put(-3,14){2p}}\footnotesize{\put(-3,21.7){3p}}\footnotesize{\put(14,15){A}}\footnotesize{\put(19,22.8){B}}\footnotesize{\put(25,9.5){C}}\end{picture}$

$\underline{\blacksquare\:\:\:\footnotesize{\red{SolutioN}}}$

$\therefore\:\:\footnotesize{w=\dfrac{1}{2}(5p)v+\dfrac{1}{2}(4p)(4v)}$

$\footnotesize{\implies\:w=\dfrac{5pv}{2}+8pv}$

$\footnotesize{\implies\:w=\dfrac{5pv+16pv}{2}}$

$\footnotesize{\implies\:w=10.5pv}$

$\footnotesize{Now\:,\: let \:, \:\:T_A=T}$

$\footnotesize{\implies\:2pv=nRT}$

$\footnotesize{Now\:,\:T_c=T'}$

$\therefore\:\:\footnotesize{\triangle u=\dfrac{f}{2}nR(T'-T)}$

$\therefore\:\:\footnotesize{\triangle u=\dfrac{3}{2}(4pv)}$

$\footnotesize{\triangle u=6pv}$

$\therefore\:\:\footnotesize{\triangle Q=10.5pv+6pv}$

$\footnotesize{\implies\:\:\triangle Q=10.5pv+6pv}$

$\footnotesize{\implies\:\:\triangle Q=16.5pv}$

$\red{\footnotesize{\implies\:\:option \:(c)}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{2 mm}\put(1,1){\line(1,0){6.5}}\linethickness{1 mm}\put(1,1.5){\line(1,0){6.5}}\end{picture}$

Answer 2

Answer:

hope it helps you see the attachment for further information.....

inna easy to tha yaar..

2 min lga banane me