Calculate acceleration due to gravity at a height of $2000 k m$ above the Earth's surface. Mass of Earth = $6 \times 10^{24} k g$ , radius of Earth = $6.4 \times 10^{6}$ m and (Gravitational constant, G $= 6.7 \times 10^{- 11} N m^{2} k g^{- 2}$ )

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When a star of mass 9×1030 kg ends as a black hole, the Schwarzschild radius of the star is (G=6.7×10−11Nm2kg−2)

The correct option is C 13.4 kmSchwarzschild radius is given byR=2GMC2⇒R=2×6.7×10−11×9×1030(3×108)2=13.4×103m=13.4kmOption C.

When a star of mass $9 \times 10^{30}$ kg ends as a black hole, the Schwarzschild radius of the star is $(G = 6.7 \times 10^{- 11}$ Nm $^{2}$ kg $^{- 2})$

The correct option is C 13.4 km
Schwarzschild radius is given by
$R = \frac{2 G M}{C^{2}}$
$\Rightarrow R = \frac{2 \times 6.7 \times 10^{- 11} \times 9 \times 10^{30}}{(3 \times 10^{8})^{2}} = 13.4 \times 10^{3}$ m $= 13.4$ km
Option C.