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3 months 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

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Vishal Baghel

Contributor-Level 10

$g' = g - ω^{2} R s i n θ$

$g' (a t θ = 0 p o l e) = g$

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

g decreases as we move bole to equator

So, A is false statement

But 'R' statement is true

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3 months 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:

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alok kumar singh

Contributor-Level 10

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

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

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

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

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3 months 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

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Payal Gupta

Contributor-Level 10

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

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

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

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3 months 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:

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Vishal Baghel

Contributor-Level 10

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

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

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

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3 months 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:

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alok kumar singh

Contributor-Level 10

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

Using conservation of Mechanical Energy

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

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

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3 months 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)

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Vishal Baghel

Contributor-Level 10

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

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

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3 months 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:

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alok kumar singh

Contributor-Level 10

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

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

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

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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3 months ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

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alok kumar singh

Contributor-Level 10

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

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

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

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

3 months 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?

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Payal Gupta

Contributor-Level 10

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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3 months 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 :

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alok kumar singh

Contributor-Level 10

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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