Gravitation

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New answer posted•a year ago

Gravitation physics ncert exemplar solutions class 12th chapter two

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

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$g' = g - ω^{2} R s i n θ$

$g' (a t θ = 0 p o l e) = g$

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

g decreases as we move bole to equator

So, A is false statement

But ‘R’ statement is true

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New answer posted•a year ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

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:

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

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

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

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New answer posted•a year ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

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

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year 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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New answer posted•a year ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be:

(Radius of earth will, be:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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New answer posted•a year ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

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:

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Using conservation of Mechanical Energy

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

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

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New answer posted•a year ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Assume there are two identical simple pendulum clocks. Clock – 1 is placed on the earth and Clock – 2 is placed on a space station located at a height h above the earth surface. Clock – 1 and Clock – 2 operate at time periods 4s and 6s respectively. Then the value of h is – (consider radius of earth R_E = 6400km and g on earth 10 m/s²)

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\frac{T_{1}}{T_{2}} = \frac{4}{6} = \sqrt[]{\frac{g_{2}}{g_{1}}}$

$\Rightarrow {(1 + \frac{h}{R})}^{2} = \frac{9}{4} \Rightarrow 1 + \frac{h}{R} = \frac{3}{2} \Rightarrow h = \frac{R}{2} = 3200 k m .$

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New answer posted•a year ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

On the $x$ -axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$ 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:

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

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

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

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

$\Rightarrow x d x = t d t$

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

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

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New answer posted•a year ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

On the $x$ -axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$ 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:

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

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

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

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

$\Rightarrow x d x = t d t$

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

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

Upvote

New answer posted•a year ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

Two planets A and B of equal mass are having their period of revolutions T_A and T_B such that T_A = 2T_B. These planets are revolving in the circular orbits or radii r_A and r_B respectively. Which out of the following would be the correct relationship of their orbits?

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

According to kepler’s third law of time period –

${(\frac{T_{A}}{T_{B}})}^{2} = {(\frac{r_{A}}{r_{B}})}^{3} \Rightarrow \frac{r_{A}}{r_{B}} = 4^{1 / 3}$

$\Rightarrow r_{4}^{3} = 4$ $r$ $B$ $3$

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New answer posted•a year ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

A body is moving in a low circular orbit about a planet of mass $M$ and radius $R$ . The radius of the orbit can be taken to be $R$ itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is :

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Sol. $\overset{?}{U} = 5 \hat{j}$

$\overset{?}{a} = 10 \hat{i} + 4 \hat{j}$

$\Rightarrow \overset{?}{S} = \overset{?}{U} t + \frac{1}{2} (\overset{?}{a}) t^{2}$

$\Rightarrow 20 \hat{i} + y_{0} \hat{j} = (5 t^{2}) \hat{i} + (5 t + 2 t^{2}) \hat{j}$

$\begin{array}{r} 20 = 5 t^{2} y_{0} = 5 t + 2 t^{2} \\ t = 2 \Rightarrow 18 m . \end{array}$

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