Ask & Answer Question

Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eight Gravitation

A body is projected vertically upward from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be:

Option 1 -

800km

Option 2 -

1600km

Option 3 -

2133km

Option 4 -

4800km

2 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

Correct Option - 1

Detailed Solution:

$V_{C} = \sqrt[]{\frac{2 G M}{R}}$

Using conservation of Mechanical Energy

$\Rightarrow \frac{- G M m}{R} + \frac{1}{2} \times m (\frac{V e^{2}}{9}) = \frac{- G M m}{(R + h)}$

$\Rightarrow \frac{1}{R + h} = \frac{8}{9 R} \Rightarrow h = \frac{R}{8} = \frac{6400}{8} = 800 k m$

0 Upvotes 0 Downvotes

Similar Questions for you

On the $x$ -axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$ $\frac{}{{(^{}^{})}^{}}$ in the $x$ -direction. The magnitude of gravitational potential on the $x$ -axis at a distance $x$ , taking its value to be zero at infinity is:

Vishal Baghel

$g = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$ $\frac{}{{(^{}^{})}^{}}$

$\Rightarrow \int_{v}^{0} ? d V = - \int_{x}^{\infty} ? g d x$

$\Rightarrow O - V = - [\int \frac{A x}{{(a^{2} + x^{2})}^{3 / 2}}]$

Let, $a^{2} + x^{2} = t^{2}$ $^{}^{}^{}$

$\Rightarrow 2 x d x = 2 t d t$

$\Rightarrow x d x = t d t$

$\Rightarrow V = \int \frac{A t d t}{t^{3}} \Rightarrow - \frac{A}{t} \Rightarrow - {\frac{A}{\sqrt[]{a^{2} + x^{2}}}|}_{x}^{\infty}$

$\Rightarrow V = \frac{A}{\sqrt[]{a^{2} + x^{2}}}$

Two objects of equal masses placed at certain from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other, then the new force will be:

alok kumar singh

F = $\frac{G m_{2}}{d^{2}}$ - (i)

Now, $F' = \frac{G m_{1} m_{2}}{d^{2}} = \frac{G 2 m}{3} \times \frac{4 m}{3 d^{2}}$

$k' = \frac{8}{9} G \frac{m^{2}}{d^{2}}$

$∴ F' = \frac{8}{G} F$

Given below are two statements : One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : If we move form poles to equator, the direction of acceleration due to gravity of earth always points towards the centre of earth without any variation in its magnitude.

Reason R : At equator, the direction of acceleration due to the gravity is towards the centre of earth.

In the light of above statements, chose the correct answer from the options given below

Vishal Baghel

$g' = g - ω^{2} R s i n θ$

$g' (a t θ = 0 p o l e) = g$

$g' (a t θ = 90 °, e q u a t o r) = g - ω^{2} R$

g decreases as we move bole to equator

So, A is false statement

But ‘R’ statement is true

The escape velocity of a body on a planet ‘A’ is 12 kms^-1. The escape velocity of the body on another planet ‘B’, whose density is four times and radius is half of the planet ‘A’, is:

alok kumar singh

$M_{A} = ρ_{A} \times \frac{4}{3} π R_{A}^{3}$ $ρ_{B} = 4 ρ_{A}$

$M_{B} = ρ_{B} \times \frac{4}{3} π R_{B}^{3}$ $R_{B} = \frac{R_{A}}{2}$

$\frac{M_{A}}{M_{B}} = 2, \frac{R_{A}}{R_{B}} = 2$ $\frac{V_{E A}}{V_{E B}} = \sqrt[]{\frac{2 G_{1} M_{A}}{R_{A}} \times \frac{R_{B}}{2 G_{1} M_{B}}}$

$v_{E A} = v_{E B} = 12 k m s e c^{- 1}$

In the reported figure of earth, the value of acceleration due to gravity is same at point A and C but it is smaller than that of its value at point B (surface of the earth). The value of OA: AB will be x : y. The value of x is