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3 months ago

Overview Class 11th

Four identical particles of equal masses 1kg made to move along the circumference of a circle of radius 1m under the actin of their own mutual gravitational attraction. The speed of each particle will be:

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alok kumar singh

Contributor-Level 10

$F_{1} c o s 45 ° + F_{2} c o s 45 ° + F_{3} = \frac{m v^{2}}{r}$

$\Rightarrow \frac{G m^{2}}{{(\sqrt[]{2} r)}^{2}} . \frac{1}{\sqrt[]{2}} + \frac{G m^{2}}{{(\sqrt[]{2} r)}^{2}} . \frac{1}{\sqrt[]{2}} + \frac{G m^{2}}{{(2 r)}^{2}} = \frac{m v^{2}}{r}$

$\Rightarrow v = \sqrt[]{\frac{G m}{4 r} (2 \sqrt[]{2} + 1)}$

Putting m = 1 kg and r = 1 m,

$v = \frac{1}{2} \sqrt[]{G (1 + 2 \sqrt[]{2})}$

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New answer posted

3 months ago

Overview Class 11th

Consider two satellites S₁ and S₂ with periods of revolution 1hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S₁ to the angular velocity of satellites S₂ is:

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alok kumar singh

Contributor-Level 10

$ω = \frac{2 π}{T} \Rightarrow \frac{ω_{1}}{ω_{2}} = \frac{T_{2}}{T_{1}} = 8$

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New answer posted

3 months ago

Overview Class 11th

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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Vishal Baghel

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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New answer posted

3 months ago

Overview Class 11th

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period

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Payal Gupta

Contributor-Level 10

F = m E cos θ

$= m (\frac{G M}{R^{3}} * r) * \frac{x}{r}$ $m a = \frac{m g}{R} * \Rightarrow a = \frac{g}{R} x$

$\Rightarrow T = 2 π \sqrt[]{\frac{R}{g}}$

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New answer posted

3 months ago

Overview Class 11th

Find the gravitational force of attraction between the ring and sphere as shown in the diagram where the plane of the ring is perpendicular to the joining the centres. If $\sqrt[]{8} R$ is the distance between the centres of a ring (of mass 'm') and a sphere (mass 'M') where both have equal radius 'R'.

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Payal Gupta

Contributor-Level 10

Method – I:

$F = M_{s p h e r e} E_{r i n g} = M * \frac{G m x}{{(R^{2} + x^{2})}^{\frac{3}{2}}} = M * \frac{G m x \sqrt[]{8} R}{{(R^{2} + 8 R^{2})}^{\frac{3}{2}}} = \frac{\sqrt[]{8} G M m}{27 R^{2}}$

Method – II:

$F = \int d F c o s θ = c o s θ \int (d m) \frac{G M}{(R^{2} + 8 R^{2})} = \frac{\sqrt[]{8} G M m}{27 R^{2}}$

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New answer posted

3 months ago

Overview Class 11th

A body weights 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?

[use g = $\frac{G m}{R^{2}} = 9.8 m s^{- 2}$ and radius of earth, R = 6400km].

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Vishal Baghel

Contributor-Level 10

As we go pole to equator, acceleration due to gravity decreases. So, weight of the body will also reduces. Thus, weight on equator will be slightly smaller than 49 N.

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New answer posted

3 months ago

Overview Class 11th

The same size images are formed by a convex lens by a convex lens when the object is placed at 20cm or at 10cm from the lens. The focal length of convex lens is ……………cm.

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Vishal Baghel

Contributor-Level 10

Image is virtual, when an object is placed at distance 10 cm from the lens and image is real when object is placed at 20 cm from the lens.

Thus, m₁ = -m₂ => $\frac{f}{f - 10} = \frac{- f}{f - 20} \Rightarrow f = 15 c m$

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New answer posted

3 months ago

Overview Class 11th

Two satellites A and B of masses 200kg and 400kg are revolving round the earth at height of 600km and 1600km respectively. If T_A and T_B are the time periods of A and B respectively then the value of T_B – T_A :

[Given : radius of earth = 6400km, mass of earth = 6 * 10²⁴kg]

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{r^{3}}{G M_{e}}}$

$T_{B} - T_{A} = \frac{2 π}{\sqrt[]{G M_{e}}} (r_{B}^{3 / 2} - r_{A}^{3 / 2}) = \frac{2 * 3.14}{\sqrt[]{6.67 * 1 0^{- 11} * 6 * 1 0^{24}}} [{(8 * 1 0^{6})}^{\frac{3}{2}} - {(7 * 1 0^{6})}^{\frac{3}{2}}]$

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New answer posted

3 months ago

Overview Class 11th

The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity v_e such that $v_{i} = \sqrt[]{\frac{x}{y}} * v_{e} .$ The value of x will be………

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Payal Gupta

Contributor-Level 10

Using conservation of energy :

$\frac{1}{2} m v_{i}^{2} - \frac{G m m}{R} = 0 - \frac{G m m}{11 R}$

$v_{i}^{2} = \frac{20}{11} \frac{G m}{R}$

Escape velocity is defined as $\sqrt[]{\frac{2 G M}{R}}$

$v_{i} = \sqrt[]{\frac{10}{11} v_{e}}$

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New answer posted

3 months ago

Overview Class 11th

If RE be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is:

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Vishal Baghel

Contributor-Level 10

Acceleration due to gravity at r distance above the surface = $\frac{G M}{{(R + r)}^{2}}$

Acceleration due to gravity at r distance below the surface = $\frac{G M}{R^{3}} (R - r)$

So, ratio = $\frac{(R - r) {(R + r)}^{2}}{R^{3}} = \frac{(R - r) (R^{2} + r^{2} + 2 R r)}{R^{3}} = 1 + \frac{r}{R} - \frac{r^{2}}{R^{2}} - \frac{r^{3}}{R^{3}}$

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