Radius of earth is 6400 km. If mass of earth remains conserved and radius of earth would be 1cm
and then its density would be (approximately) :-
OPtion
1. 10^26 times the present value of density of earth
2. 10^24 times the present value of density of earth
3. 10^22 times the present value of density of earth
4. 10^20 times the present value of density of earth
5. 10^28 times the present value of density of earth
6. 10^30 times the present value of density of earth
7. 10^8 times the present value of density of earth
8. 10^16 times the present value of density of earth
9. 10^9 times the present value of density of earth
10. 10^18 times the present value of density of earth
Solution
Since mass is conserved hence
d2/d1 = V1/V2
= R13/R23
= 6400km*6400km*6400km/ 1cm*1cm*1cm
= 6400000m*6400000m*6400000m/0.01m*0.01m*0.01m
= 6.4*6.4*6.4*10^18/10^-6
= 262.144*10^24
approx 2.6*10^26
appro 10^26