Tag: representing a gravitational field

Questions Related to representing a gravitational field

A gravitational field is

  1. a field of gravitons

  2. a field of massive particles

  3. the force field that exists in the space around every mass or group of masses.

  4. Force exerted on an unit charge


Correct Option: C
Explanation:

Gravitational field is the force field that exists in the space around every mass or group of masses. Mathematically, it is F/m

A gravitational field is a model used to explain the influence that a massive body extends into the space around itself, producing a force on another massive body. 

  1. True

  2. False


Correct Option: A
Explanation:
  • The gravitational field is the actually the force experimented by a unit mass due to a body of mass M.
  • So when a unit mass is kept at a distance 'r' from mass 'M' then it experiences a force which is equal to $\dfrac{GM}{r^2}$[as m=1],which is nothing but the gravitational field of mass M.
  • Its units are $\dfrac{N}{kg}$
  • Hence the above statement is true.

If the distance between two particles is reduced to half, the gravitational attraction between them will be

  1. Halved

  2. Quadrupled

  3. Doubled

  4. Reduced to a quarter


Correct Option: B
Explanation:

Gravitational force $=\dfrac { G{ m } _{ 1 }{ m } _{ 2 } }{ { r }^{ 2 } } ={ F } _{ 1 }$

If distance made $0$ half of original value
${ F } _{ 2 }=\dfrac { G{ m } _{ 1 }{ m } _{ 2 } }{ { \left( \dfrac { r }{ 2 }  \right)  }^{ 2 } } $
${ F } _{ 2 }=\dfrac { 4G{ m } _{ 1 }{ m } _{ 2 } }{ { r }^{ 2 } } $
${ F } _{ 2 }=4{ F } _{ 1 }$

Under the force of gravity, a heavy body falls quicker than a light body (neglect air resistance).

  1. True

  2. False


Correct Option: B
Explanation:

According to newton's laws the time taken to fall is not related to mass

$S=ut+\dfrac { 1 }{ 2 } { at }^{ 2 }$
${ V }^{ 2 }-{ u }^{ 2 }=2as$
It have not to do anything with mass so both lighter and heavier body will taken same time to fall from certain point.

Three particles of masses $2m, m$ and $2m$ are at the vertices $A, B$ and $C$ of an equilateral triangle $ABC$ of side length $'l'$. Then the intensity if gravitational field at the mid point of side $BC$ is:-

  1. $\dfrac{\sqrt{208}}{3} \dfrac{Gm}{l^2}$

  2. $\dfrac{\sqrt{59}}{3} \dfrac{Gm}{l^2}$

  3. $\dfrac{\sqrt{142}}{3} \dfrac{Gm}{l^2}$

  4. $\dfrac{\sqrt{308}}{3} \dfrac{Gm}{l^2}$


Correct Option: A

PRESSURE AND KINETIC INTERPRETATION OF TEMPERATURE
At what temperature the mean kinetic energy of hydrogen molecules increases to such that they will escape out of the gravitational field of earth for over?
take $({ v } _{ c }=11.2km/sec)$

  1. 12075 K

  2. 10000 K

  3. 20000 K

  4. 10075 K


Correct Option: D

The gravitational field in a region is given by $\vec {g} = 2\hat {i} + 3\hat {j} m/s^{2}$. The work done in moving a particle of mass $1\ kg$ from $(1, 1)$ to $\left (2, \dfrac {1}{3}\right )$ along the time $3y + 2x = 5$ is

  1. Zero

  2. $20\ J$

  3. $-15\ J$

  4. $18\ J$


Correct Option: C

At some planet gravitational acceleration is $1.96m/sec^{ -2 }$. If is safe to jump from a height of 2 m on earth, then what should be the corresponding safe height for jumping on the planet:

  1. 5 m

  2. 2 m

  3. 10 m

  4. 20 m


Correct Option: C
Explanation:

G on earth=9.8

ratio between earth and planet$=9.8:1.96$
= 5
so the safe height on the planet is 5 time greater than earth because
gravitational pull of that planet is 5 time less than earth
so required height = height on earth $\times$ 5
$2 \times 5$m
=10 m is the safe height on that planet.
Hence,
option $C$ is correct answer.

A body is acted upon by a constant force directed towards a fixed point. The magnitude of the force varies inversely as the square of the distance from the fixed point then path can be described by an equation similar to:

  1. $y=mx+c$

  2. ${ x }^{ 2 }+{ y }^{ 2 }={ r }^{ 2 }$

  3. $y=c{ x }^{ 2 }$

  4. none of these


Correct Option: D
Explanation:

The force field is similar to gravitational field.
So, the path is parabolic or elliptical or hyperbolic.
The exact trajectory will depend on the masses and  energy of the particle.
Option(A) is equation of a straight line, it is ruled out.
Option(B) is equation for a circle and is also eqn of an ellipse with length of major axis equal to minor axis.
Option(C) is eqn of a parabola.
Hence, (D) is the best option.

Potantial (V) at a point in space is given by $v = x^2 + y^2 + z^2$. Gravitational field at a point (x, y, z) is 

  1. $-2 x \hat{i} - 2 y \hat{j} - 2 z \hat{k}$

  2. $2 x \hat{i} + 2 y \hat{j} + 2 z \hat{k}$

  3. $x \hat{i} + y \hat{j} - z \hat{k}$

  4. $-x \hat{i} - y \hat{j} - z \hat{k}$


Correct Option: A
Explanation:

$v=x^2+y^2+z^2$

$E = \dfrac{{ - dv}}{{dx}}$

$E =  - \left[ {2\hat x + 2y\hat j + 2z\hat k} \right]$

$E =  - 2\hat x - 2y\hat j - 2z\hat k$
So, option $A$ is correct.