Physics Gravitation

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].

0 Follower 2 Views

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.

0 0

New answer posted

3 months ago

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.

0 Follower 5 Views

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 => $\frac{f}{f - 10} = \frac{- f}{f - 20} \Rightarrow f = 15 c m$

0 0

New answer posted

3 months ago

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]

0 Follower 2 Views

Contributor-Level 10

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

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

0 0

New answer posted

3 months ago

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : The escape velocities of planet Ad and B are same. But A and B are of unequal mass.

Reason R : The product of their mass and radius must be same. M₁ R₁ = M₂R₂

In the light of the above statements, choose the most appropriate answer from the options given below:

0 Follower 3 Views

Contributor-Level 10

${(v_{e})}_{A} = {(v_{e})}_{B} \Rightarrow \frac{M_{1}}{R_{1}} = \frac{M_{2}}{R_{2}}$

R is not correct.

0 0

New answer posted

3 months ago

A solid sphere of radius R gravitationally attracts a particle placed at 3R form its centre with a force F₁,. Now a spherical cavity of radius $(\frac{R}{2})$ is made in the sphere (as shown in figure) and the force becomes F₂. The value of F₁: F₂ is:

0 Follower 3 Views

Contributor-Level 10

$F_{1} = \frac{G M m}{9 R^{2}}$

$F_{2} = \frac{G M m}{9 R^{2}} - \frac{G \frac{M}{8} m}{{(\frac{5 R}{2})}^{2}} = \frac{G M m}{R^{2}} (\frac{1}{9} - \frac{1}{50})$

$\frac{F_{1}}{F_{2}} = \frac{1}{1 - \frac{9}{50}} = \frac{50}{41}$

0 0

New answer posted

3 months ago

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………

0 Follower 4 Views

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

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

0 0

New answer posted

3 months ago

The minimum energy required to launch a satellite of mass $m$ from the surface of earth of mass $M$ and radius $R$ in a circular orbit at an altitude of $2 R$ from the surface of the earth is:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

Apply energy conservation,

$U_{i} + K_{i} = U_{f} + K_{f}$

$\Rightarrow - \frac{G M m}{R} + K_{i} = - \frac{G M m}{3 R} + \frac{1}{2} m v^{2}$

$\Rightarrow - \frac{G M m}{R} + K_{i} = - \frac{G M m}{3 R} + \frac{1}{2} * m * \frac{G M}{3 R}$

$\Rightarrow K_{i} = - \frac{1}{6} \frac{G M m}{R} + \frac{G M m}{R}$

$K_{i} = \frac{5}{6} \frac{G M m}{R}$

0 0

New answer posted

3 months ago

A parallel plate capacitor is charged by connecting it to a battery through a resistor. If $I$ is the current in the circuit, then in the gap between the plates:

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

According to modified Ampere's law

$? B . d I = μ_{0} (I_{C} + I_{D})$

For Loop $L_{1} I_{C} \neq 0$ and $I_{D} = 0$

For Loop $L_{2} I_{C} = 0$ and $I_{D} \neq 0$

Due to $KCL I_{C} = I_{D}$

0 0

New answer posted

3 months ago

The satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hour respectively. The radius of the orbit of nearer satellite is 2 * 10³ km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{π}{x} r a d h^{- 1}$ where x is__________.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

For A satellite T₁ = 1hour

$S o ω_{1} = 2 π r e d / h o u r$

for B satellite T₂ = 8 hour

given R₁ = 2 * 10³ Km

Relative $ω = \frac{V_{1} - V_{2}}{R_{2} - R_{1}} = \frac{2 π * 1 0^{3}}{6 * 1 0^{3}}$

$\Rightarrow \frac{π}{3}$ rad/hour

x = 3

0 0

New question posted

3 months ago