Two spherical planets P and Q have the same uniform density `rho,` masses `M_p and M_Q` and surface areas A and 4A respectively. A spherical planet R also has uniform density `rho` and its mass is `(M_P + M_Q).` The escape velocities from the plantes P,Q and R are `V_P V_Q and V_R` respectively. Then

Options

A. `V_(Q) gt V_(R )gtV_(P)`
B. `V_(R ) gt V_(R ) gtV_(P)`
C. `V_(R )//V_(P)=3`
D. `V_(P)//V_(Q)=1/2`

Answer

B

Explanation

Surface area of Q is four times therefore radius of Q is two times volume is eight times therefore mass of q is also eight times
so let `M_(P)-M` and `R_(P)-=r`
Then `M_(Q)=8M` and `R_(Q)=2r`
Now mass of R is `(M_(p)+M_(Q))` or 9 M Therefore radius of R is `(9)^(1//3)` r now escape velocity from the surface of a planet is given by
`V=sqrt(2(GM)/(r )`
`therefore V_(P)sqrt(2GM)/(r ) rarr V_(Q)=sqrt(2G(8M))//(2r)`
and `V_(R )=sqrt(2G(9M))/(9)`
from here we can see that `(V+(p))/(V_(Q))=1/2` and `V_(R) gt V_(Q) gt V_(P)`