Question

answer karo iska...xD​

Answer 1

23. Given

• the mass of the Sun is $2.0*10^3^0$ kg.

• The radius of the Sun is $7.0*10^-^8$ m.

To find:-

The range where we can expect the Sun's mass density to be.

Solution:-

Let 'r' be the radius of the sun.

The volume of the Sun = $\frac{4}{3}\pi r^3$.

$\bold{\implies V=\frac{4}{3}\pi (7.0*10^1^8\ m)^3}\\\\\bold{\implies V=1.44*10^2^7\ m^3}$

The density of the sun is:-

$\bold{\rho = \frac{M}{V}}$

Now,

$\displaystyle{\bold{\rho=\frac{2.0*10^360\ kg}{1.44*10^2^7\ m^3} }}$

$\displaystyle{\bold{\implies \rho=1.38*10^3\ kg/m^3}}$

$\boxed{\displaystyle{\bold{\approxeq 1.4*10^3\ kg/m^3}}\\}$

The value of density lies in the range of densities of the solid and liquid.

The density of gas is very low. The density of sun is high because of the gravitational force of attraction between the outer and inner layer of the sun.

24. Distance of Jupiter from the Earth, S = $824.7*10^6$ km.

Angular diameter 35.72" = 35.72 * 4.874 * $10^-^6$ rad.

Diameter of Jupiter = s  

Using the relation,

θ = s/S

s = θS = $824.7*10^1^9*35.72*4.872*10^-^6=143520.76 *10^3$

$\boxed{s=1.435*10^5 \text{km}}$