Home
Class 12
PHYSICS
Assuming the radius of the earth to be 6...

Assuming the radius of the earth to be `6.38 xx 10^(8) cm`, the gravitational constant to be `6.67 xx 10^(-8) cm^(3) g^(-1) s^(-2)`, acceleration due to gravity on the surface to be `980 cm / s^(2)`, find the mean density (in `(g) / (cm)^(3)` ) of the earth.

Text Solution

Verified by Experts

The correct Answer is:
5.49
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • GEOMETRICAL OPTICS

    CENGAGE PHYSICS|Exercise Integer Type|4 Videos

  • HEATING EFFECT OF CURRENT

    CENGAGE PHYSICS|Exercise Thermal Power in Resistance Connected in Circuit|28 Videos

Similar Questions

Explore conceptually related problems

The radius of the moom is 1.7 xx 10^(6) m and its mass is 7.4 xx 10^(22) kg . If G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) , find the value of acceleration due to gravity on the surface of moon.

The radius of the earth is 6.37 xx 10^(6) m and its mean density is 5.5 xx 10^(3) "kg m"^(-3) and G = 6.67 xx 10^(-11) "N-m"^(2) kg^(-2) Find the gravitational potential on the surface of the earth.

Knowledge Check

  • The value of acceleration due to gravity 'g' on the surface of the moon with radius 1/2 that of the earth and same mean density as that of the earth

    A
    g/4
    B
    g/2
    C
    4 g
    D
    8 g

  • If R is the radius of the earth and g the acceleration due to gravity on the earth’s surface, the mean density of the earth is

    A
    `4pi G//3gR`
    B
    `3pi R//4gG`
    C
    `3g//4pi RG`
    D
    `pi Rg//12G`

  • If R is the radius of the earth and g the acceleration due to gravity on the earth's surface, the mean density of the earth is

    A
    `(4piG)/(3gR)`
    B
    `(3piR)/(4gG)`
    C
    `(3g)/(4piRG)`
    D
    `(piR)/(12G)`

    • Similar Questions

    Explore conceptually related problems

    The radius of the earth is 6.37xx10^(6)m , its mean density is 5.5xx10^(3)Kg m^(-3) and G= 6.66xx10^(-11)Nm^(2)Kg^(-2) . Determine the gravitational potential on the surface of the earth.

    The value of universal gravitational constant is 6.67xx10^(-8) "dyne" g^(-2) cm^2. What is its value in Mks system?

    The value of universal gravitational constant G = 6.67 xx 10^(-11) N m^(2) kg^(-2) . The value of G in units of g^(-1) cm (^(3)s^(-2) is

    If R is the radius of the Earth and g the acceleration due to gravity on the Earts surface, the mean density of the Earth is

    Assuming the earth to be a homogeneous sphere of radius R , its density in terms of G (constant of gravitation) and g (acceleration due to gravity on the surface of the earth)

    Home
    Profile