Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } } $
${ F } _{ 2 }=4{ F } _{ 1 }$

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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. $\dfrac{\sqrt{308}}{3} \dfrac{Gm}{l^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The gravitational field at the midpoint of BC is the vector sum of the fields from the three masses. Using the geometry of the equilateral triangle, the components cancel or add up to the calculated resultant.

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The escape velocity is related to temperature by the formula v_rms = sqrt(3RT/M). Setting v_rms equal to the escape velocity (11.2 km/s) and solving for T gives approximately 10075 K.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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. $-x \hat{i} - y \hat{j} - z \hat{k}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $30.53 \ minutes$

  2. $50.38 \ minutes$

  3. $52.68 \ minutes$

  4. $84.75 \ minutes$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.