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The escape velocity from the earth is about 11 km/second. The escape velocity from a planet having twice the radius and the same mean density as the earth, is

22 km/sec

11 km/sec

5.5 km/sec

15.5 km/sec

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Explanation

(a)

$v_{e} = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{2 G}{R} \times \frac{4}{3} {πR}^{3} ρ} = R \sqrt{\frac{8}{3} πG ρ} S o i f p = c o n s \tan t, v_{e} α R . S i n c e t h e p l a n e t h a v i n g d o u b l e r a d i u s i n c o m p a r i s o n t o e a r t h, T h e r e f o r e t h e e s c a p e v e l o c i t y b e c o m e s t w i c e i . e . 22 k m / s .$

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What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 km

$7.4 \times 10^{- 4}$ rad/sac

6.4 $\times 10^{- 4}$ rad/sac

$7.8 \times 10^{- 4}$ rad/sac

8.7 $\times 10^{- 4}$ rad/sac

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Explanation

$W e i g h t o f t h e b o d y a t e q u a t o r = \frac{3}{5} o f i n i t i a l w e i g h t s o g' = \frac{3}{5} g (b e c a u s e m a s s r e m a i n s c o n s \tan t) g' = g - ω^{2} R \cos^{2} θ \Rightarrow \frac{3}{5} g = g - ω^{2} R \cos^{2} (0^{0}) \Rightarrow ω^{2} = \frac{2 g}{5 R} \Rightarrow w = \sqrt{\frac{2 g}{5 R}} = \sqrt{\frac{2 \times 10}{5 \times 6400 \times 10^{3}}} = 7.8 \times 10^{- 4} \frac{r a d}{s a c}$

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If g is the acceleration due to gravity at the earth's surface and r is the radius of the earth, the escape velocity for the body to escape out of earth's gravitational field is

$\sqrt{2 g r}$

C.
g/r
D.

r/g

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Explanation

(b)

$K E + P E = 0 \frac{1}{2} m v^{2} - \frac{G M m}{r} = 0 \Rightarrow v = \sqrt{\frac{2 G M}{r}} O r, v = \sqrt{2 g r} (a s g = \frac{G M}{r^{2}})$

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The escape velocity of a projectile from the earth is approximately

11.2 m/sec

112 km/sec

11.2 km/sec

11200 km/sec

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Explanation

(c)

The escape velocity of a projectile from the earth is approximately 11.2 Km/s.

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The escape velocity of a particle of mass m varies as

$m^{2}$

$m^{0}$

$m^{- 1}$

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Explanation

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Acceleration due to gravity is ‘g’ on the surface of the earth. The value of acceleration due to gravity at a height of 32 km above earth’s surface is (Radius of the earth = 6400 km)

0.9 g

0.99g

0.8 g

1.01 g

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Explanation

(b) h=32 km, R=6400k, so h $< < R$

$g^{1} = g (1 - \frac{2 h}{R}) = g (1 \frac{2 \times 32}{6400}) \Rightarrow g^{1} = \frac{99}{100} g = 0.99 g$

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The time period of a simple pendulum on a freely moving artificial satellite is

Zero

2 sec

3 sec

Infinite

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Explanation

d) Time period of simple pendulum T=2 $π$ $\sqrt{\frac{L}{g_{e f f}}}$

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For the moon to cease to remain the earth's satellite, its orbital velocity has to increase by a factor of -

$\sqrt{2}$

1/ $\sqrt{2}$

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Explanation

(b) $v_{e} = \sqrt{2} v_{0,}$ i.e. if the orbital velocity of moon is increased by factor of $\sqrt{2}$ then it will escape out from the gravitational field of earth

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The height of the point vertically above the earth’s surface, at which acceleration due to gravity becomes 1% of its value at the surface is (Radius of the earth = R)

10r

20r

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Explanation

(b)

$g = \frac{G M}{R^{2}} g' = \frac{G M}{(R + h)^{2}} & g' = 1 % o f g' = \frac{1}{100} \times g \frac{g'}{g} = {(\frac{R}{R + h})}^{2} \Rightarrow \frac{1}{100} = {(\frac{R}{R + h})}^{2} \Rightarrow h = 9 r$

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Escape velocity on a planet is $v_{e}$ . If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes

$4 v_{d}$

$2 v_{d}$

$v_{d}$

$\frac{1}{2} v_{d}$

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Explanation

(b) $v_{e} = \sqrt{\frac{2 G M}{R}} \Rightarrow v_{e} \propto \sqrt{M} i f R = c o n s \tan t$

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Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.