Tag: physics

Questions Related to physics

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

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. Halved

  2. Quadrupled

  3. Doubled

  4. Reduced to a quarter

Show answer
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).

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. True

  2. False

Show answer
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:-

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

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

Show answer
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 $({ v } _{ c }=11.2km/sec)$

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. 12075 K

  2. 10000 K

  3. 20000 K

  4. 10075 K

Show answer
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

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. Zero

  2. $20\ J$

  3. $-15\ J$

  4. $18\ J$

Show answer
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:

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. 5 m

  2. 2 m

  3. 10 m

  4. 20 m

Show answer
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:

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. $y=mx+c$

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

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

  4. none of these

Show answer
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 

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

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

Show answer
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.

A large object is placed at exactly $65$% of the distance to the moon from the earth. Find out correct statement about the object ?

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. Fall to the sun

  2. Fall to the moon

  3. Fall to the earth

  4. Remain in the same place

  5. Drift out of the solar system

Show answer
Correct Option: C
Explanation:

The mass of moon is $1.2$% of that of earth. Hence $M _{moon}=0.012M _{earth}$

The acceleration due to gravity of the earth=$\dfrac{GM _{earth}}{(0.65R)^2}$
The acceleration due to gravity of the moon=$\dfrac{GM _{moon}}{(0.35R)^2}$
Hence $g _{earth}>g _{moon}$, and thus object falls towards the earth.

What is the time period satellite near the earth surface (neglect the height of orbit of satellite from the surface of ground)?

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. $30.53 \ minutes$

  2. $50.38 \ minutes$

  3. $52.68 \ minutes$

  4. $84.75 \ minutes$

Show answer
Correct Option: D
Explanation:

$T =$ Time period

$R=$ radius of earth $= 6400000m$
Gravitational constant $= G=$ $ 6.67\times {10}^{-11} Nm^2/kg^2$
$M=$ mass of earth $=$ $6\times{10}^{24}kg$

$T=$ $2\pi\sqrt{\dfrac {R^3}{GM}}$
Substitute and calculate
Whatever answer you get, divide it by $60$ so that it gets converted into minutes.