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New answer posted•a year ago

Overview Class 11th

A body of mass (2M) splits into four masses {m, M – m, m, M – m}, which are rearranged to form a square as shown in the figure. The ratio of $\frac{M}{m}$  for which, the gravitational potential energy of the system becomes maximum is x : 1. The value of x is

34 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$U = - \frac{G m (M - m)}{d} \times 4 + 2 x - (\frac{G m^{2}}{\sqrt[]{2} d})$

For Umax

$\frac{d U}{d m} = 0 \Rightarrow m = \frac{M}{2} \Rightarrow \frac{M}{m} = 2$

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New answer posted•a year ago

Overview Class 11th

Inside a uniform spherical shell:

(a) the gravitation field is zero (b) the gravitational potential is zero

(c) the gravitation field is same everywhere (d) the gravitation potential is same everywhere

(e) all the above

Choose the most appropriate answer from the options given below:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Inside a uniform spherical shell, electric field is zero every where & electric potential is constant but not zero, every where

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New answer posted•a year ago

Overview Class 11th

The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be: (Given g = acceleration due to gravity at the surface of earth.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$g^{1} = \frac{G M}{{(R + h)}^{2}}$

$g^{1} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

Given h = D = 2R

$g^{1} = \frac{g}{{(1 + \frac{2 R}{R})}^{2}}$

$g^{1} = \frac{g}{9}$

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New answer posted•a year ago

Overview Class 11th

An object of mass 1 kg is taken to a height form the surface of earth which is equal to three times that radius of earth. The gain in potential energy of the object will be

[If, g = 10ms^-2 and radius of earth = 6400km]

13 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

m = 1kg ; $Δ U = \frac{- G m_{e} m}{(R + h)} - (\frac{G m_{e} m}{R})$

$Δ U = \frac{G M m}{R} - \frac{G M m}{R + h}$

$= m g_{0} R - m g_{0} \frac{R}{(1 + \frac{h}{R})} = m g_{0} {1 - \frac{1}{(1 + \frac{3 R}{R})}}$

$= m g_{0} R (\frac{3}{4})$

$= \frac{3}{4} \times 1 \times 10 \times 6400 k m$

$= 48000 \times 1 0^{3} J .$

$= 48 M J .$

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New answer posted•a year ago

Overview Class 11th

Given below are two statements:

Statement I: The law of gravitation holds good for any pair of bodies in the universe.

Statement II: The weight of any person becomes zero when the person is at the centre of the earth.

In the light of the above statements, choose the correct answer from the option given below:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

The law of gravitation holds good for any pair of bodies in the universe.

This is correct statement.

The weight of any person becomes zero when the person is at the centre of the Earth.

Weight = mg

= m × 0 [g = 0 at the centre of the Earth]

= 0

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New answer posted•a year ago

Overview Class 11th

Three identical particles A, B and C of mass 100kg each are placed in a straight line with AB = BC = 13m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately:

11 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

F on P = $\frac{}{^{}} \frac{}{^{}}$ $\frac{G M M}{r^{2}} + \sqrt[]{2} \frac{G M M}{{(\sqrt[]{2} r)}^{2}}$

$F o n P = \frac{G \times 1 0^{4}}{1 3^{2}} (1 + \frac{1}{\sqrt[]{2}}) \approx 100 G$

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New answer posted•a year ago

Overview Class 11th

Two satellites A and B, having masses in the ration 4 : 3, are revolving in circular orbits of radii 3r and 4r respectively around the earth. The ratio of total mechanical energy of A to B is:

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$\frac{E_{A}}{E_{B}} = \frac{- \frac{G M m_{A}}{2 \times 3 r}}{\frac{- G M m_{B}}{2 \times 4 r}} = \frac{m_{A}}{m_{B}} \times \frac{4}{3} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$

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New answer posted•a year ago

Overview Class 11th

If the radius of earth shrinks by 2% while its mass remains same. The acceleration due to gravity on the earth’s surface will approximately:

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$g = \frac{G M}{r^{2}}$

$d g = - 2 \frac{G M}{r^{3}} d r$

$\Rightarrow d g \times 100 = - 2 (\frac{G M}{r^{2}}) \frac{d r}{r} \times 100$

$\Rightarrow \frac{d g}{g} \times 100 = \underset{}{\underset{?}{- 2 \times 2}} = - 4 %$

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New answer posted•a year ago

Overview Class 11th

If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a dept ah(h <<R_e) from the earth surface. The value of a will be_______.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$g_{(a t h)} = g_{(a t d e p t h α h)}$ h << R

$\Rightarrow g (1 - \frac{2 h}{R}) = g (1 - \frac{α h}{R})$

$\begin{array}{l} 1 - \frac{2 h}{R} = 1 - \frac{α h}{R} \Rightarrow \frac{2 h}{R} = α \frac{h}{R} \\ α = 2 \end{array}$

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New answer posted•a year ago

Overview Class 11th

The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes 3R will be

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$T^{2} α R^{3}$

$\begin{array}{l} \Rightarrow \frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{{R_{1}}^{3}}{{R_{2}}^{3}} \\ \Rightarrow \frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{R^{3}}{{(3 R)}^{3}} \\ T_{2} = 3 \sqrt[]{3} y e a r s \end{array}$

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