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

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

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Payal Gupta | Contributor-Level 10

3 months ago

$\frac{G M}{{(3 R / 2)}^{2}} = \frac{G M}{R^{3}} \times r$

$\Rightarrow O A = \frac{4 R}{9} = r$

$A B = R - \frac{4 R}{9} = \frac{5 R}{9} \Rightarrow O A : A B = 4 : 5 = x : y \Rightarrow x = 4$

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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

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

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

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

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

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

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

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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

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$g' = g - ω^{2} R s i n θ$

$g' (a t θ = 0 p o l e) = g$

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

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:

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

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(Radius of earth will, be:

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

$g^{1} = \frac{99 g}{100} = 0.99 g$

% decrease is wt = $\frac{g - g^{1}}{g} \times 100 = 1 %$

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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

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