Two radioactive elements R and S disintegrate as
`RtoP+alpha,lamda_(R)=4.5xx10^(-3)"years"^(-1)`
`StoQ+betal,lamda_(S)=3xx10^(-3)"years"^(-1)`
Starting with number of atoms of R and S in the ratio of 2:1 this ratio afte4r the lapse of three half lives of R will be

Two radioactive elements R and S disintegrate as R rarr P+alpha , lambda_(R)=4.5 xx 10^(-3) years ^(-1) S rarr P+beta , lambda_(s)=3 xx 10^(-3) years ^(-1) Starting with number of atoms of R and S in the ratio of 2: 1 , this ratio after the lapse of three half lives of R will be

If in a Delta the ex-reddii r_(1) r_(2) r_(3) are in the ratio 1:2:3 then their sides are in the ratio :-

An aluminium (alpha_(Al)=4xx10^(-3)//^(@)C) wire resistance 'R_(1)' and carbon wire (alpha_(c)=0.5xx10^(-3)//^(@)C) resistance 'R_(2)' are connected in series to have a resultant resistance of 18 ohm at all temperatures The values of R_(1) and R_(2) in ohms

Some energy levels of a molecule are shown in the figure. The ratio of the wavelength r=lamda_(1)lamda_(2) ,is given by:

Some energy levels of a molecule are shown in the figure. The ratio of the wavelength r=lamda_(1)lamda_(2) ,is given by:

(1)/(2)r-(1)/(3)s=8 (5)/(8)r-(1)/(4)s=29 For the system of equations above, what is the value of r+s?