The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,

Options

A. the process during the path `ArarrB` is isothermal
B. heat flows out of the gas during the path `BrarrCrarrD`
C. work done during the path `ArarrB rarrC` is zero
D. positive work is done by the gas in the cycle ABCD

Answer

BD

Explanation

(a)`^nP_(r)=^(n-1)P_(r)+r.^(n-1)P_(r-1)`
`=((n-1)!)/((n-r-1)!)+(r(n-1)!)/((n-r)!)=((n-1)!)/((n-r-1)!)[(n-r+r)/((n-r))]=(n!)/((n-r)!)=^(n)P_(r)H.P.`
(b) `^20C_(r+2)=^(20)C_(2r-3) or r+2+2r-3=20,3r-1=20,3r=21,therefore r=7`
`r+2=2r-3`
`therefore" "r=5`
`^12C_r=^12C_5=^12C_7=(21xx11xx10xx8xx9)/(120)=72xx11=792 Ans`
(c)`^(n-1)C_3+^(n-1)c_4gt^(n)C_3`
`^nc_4gt^(n)c_3`
`(n!)/(4!(n-4)!)gt(n!)/(3!(n-3)!)`
`n-3gt4`
"therefore" `ngt7` Hence proved.(d)`^(15)C_(r+3)`
`3r=r+3`3r=r+3
`therefore" "2r=3 not possible`
`therefore" "3r+r+3=15`
`4r=12" " rArr r=3`Ans