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2 months 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]

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Payal Gupta

Contributor-Level 10

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

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

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

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

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

$= 48000 * 1 0^{3} J .$

$= 48 M J .$

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New answer posted

2 months 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:

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Vishal Baghel

Contributor-Level 10

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

2 months 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:

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Vishal Baghel

Contributor-Level 10

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

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

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New answer posted

2 months 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:

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alok kumar singh

Contributor-Level 10

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

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New answer posted

2 months 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:

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Vishal Baghel

Contributor-Level 10

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

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

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

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

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New answer posted

2 months 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 <e) from the earth surface. The value of a will be_______.

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Vishal Baghel

Contributor-Level 10

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

2 months 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

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Payal Gupta

Contributor-Level 10

$T^{2} α R^{3}$

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

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New answer posted

2 months ago

Overview Class 11th

The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three its previous value, then approximate new time period of the satellite will be

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Payal Gupta

Contributor-Level 10

According to question, we can write

$d = \sqrt[]{2 h R} \Rightarrow \frac{d_{2}}{d_{1}} = \sqrt[]{\frac{h_{2}}{h_{1}}} \Rightarrow h_{2} = {(\frac{d_{2}}{d_{1}})}^{2} h_{1} = {(\frac{2}{1})}^{2} * 125 = 500 m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

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