Radius of earth is 6400 km. If mass of earth remains conserved and radius of earth would be 1cm and

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26 Feb 2014 Solve

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)None of these

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

Correct Option: 1 Best Solution (9)

RAKESH   10 years AGO

(6400*1000*100)^3= 10^26

Like? Yes (14) | No (5)  

aviral bhargava   10 years AGO

density of earth with radius being 6400 km=
d=m/v
=m/(4/3)pie (r^3)
=m/m/(4/3)pie(6400*10^5)^3
density of earth with radius being 1 cm
d=m/v
=m/(4/3)pie(1^3)
density2/density2 =
2.6*(10^26)
Hence the first option is correct

Like? Yes (5) | No (1)  

S.SHAFI   10 years AGO

Density = mass/volume
volume = 4/3*pi*r^3
density of earth when radius is 6400km is (3/4)*(m/pi)*(1/6400km^3)
density of earth when radius is 1cm is (3/4)*(m/pi)*(1/1cm^3)
Calculating the value..
it is 10^26 times the present value of density of earth..
so answer is option 1.

Like? Yes (5) | No (2)  

rakesh   10 years AGO

density=mass/volume
since, mass is constant density is inversily proportion to volume.
d1*v1=d2*v2
volume=PI*r^3
v1=PI*(64*10^7)^3
v2=PI
d2=d1*2^18*10^21
d2 approx 10^26 times of d1

Like? Yes (3) | No (1)  

shilpi   10 years AGO

d= present density * 10^26*2.6

Like? Yes (1) | No   

Harsh Poddar   10 years AGO

mass remains constant
mass =x in both cases
case 1
radius=6400km=640000000cm
density=mass/ volume
x/(640000000)^3
case 2
new radius=1cm
density=x/1^3
so density of case 1=x/262144000000000000000000000
mass will be cancelled
after 2 there are 20 digits so earth will be 10^20 times density of current earth
option 4

Like? Yes (2) | No (1)  

anubhav verma   10 years AGO

(100*1000*100)^3=10^21
64^3=262144=2.6*10^5
total 10^26

Like? Yes (1) | No (1)  

ASHISH MITTAL   10 years AGO

(r1)^3/(r2)^3

Like? Yes (1) | No (1)  

ashish kumar   10 years AGO

D=M/V
Volume of sphere= (4*3.14*R^3)/3
if mass is conserved
so Dnew*Vnew =Dearth*volume of earth
Dnew=(6.4)^3*10^24*Dearth
Dnew=2.62*10^26 *Dearth

Like? Yes (1) | No (1)