Question

Maximum work done when electric dipole rotate in opposite direction that is ______________.​

Answer 1

Answer:

⇢

Acceleration is 2.22m/s

2

\large \dag† Step by step Explanation :-

Let for Earth ;

Mass = M

Radius = R

Now let for planet 'x' ;

Mass = M'

Radius = R'

As per the question :-

✧ Mass of planet 'x' is two times the mass of earth ;

\purple{ \large :\longmapsto \underline { \pmb{\boxed{{\sf M'=2M} }}}}----(1):⟼

M

′

=2M

M

′

=2M

−−−−(1)

✧ Radius of planet 'x' is three times the radius of earth ;

\purple{ \large :\longmapsto \underline { \pmb{\boxed{{\sf R'=3R} }}}}----(2):⟼

R

′

=3R

R

′

=3R

−−−−(2)

❒ We know that acceleration due to gravity is :-

\begin{gathered} \large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{ \blue{ g= \small\frac{GM}{R {}^{2} } }}}} \\ \end{gathered}

★

g=

R

2

GM

where,

G = Universal gravitational constant

M = Mass of planet

R = Radius of planet

Therefore for Earth ;

\begin{gathered}:\longmapsto \rm g= \frac{GM}{R {}^{2} } \\ \end{gathered}

:⟼g=

R

2

GM

and Given in question acceleration due to gravity on Earth is 10 m/s²,

\purple{ \large :\longmapsto \underline {\boxed{{\bf \small\frac{GM}{R {}^{2} } \large = 10 } }}}\small----(3):⟼

R

2

GM

=10

−−−−(3)

Now for planet 'x' ;

Acceleration due to gravity is :

\begin{gathered} \: \: \: \: \rm g'= \frac{GM'}{{(R')}^{2} } \\ \end{gathered}

g

′

=

(R

′

)

2

GM

′

⏩ Putting values of M' and R' from (1) and (2) :

\begin{gathered}:\longmapsto \rm g'= \frac{G(2M)}{ {(3R)^{2}}} \\ \end{gathered}

:⟼g

′

=

(3R)

2

G(2M)

\begin{gathered}:\longmapsto \rm g'= \frac{2GM}{9R {}^{2} } \\ \end{gathered}

:⟼g

′

=

9R

2

2GM

\begin{gathered}:\longmapsto \rm g'= \frac{GM}{R {}^{2} } \times \frac{2}{9} \\ \end{gathered}

:⟼g

′

=

R

2

GM

×

9

2

⏩ Using (3) :

\begin{gathered}:\longmapsto \rm g'=10 \times \frac{2}{9} \\ \end{gathered}

:⟼g

′

=10×

9

2

\begin{gathered}:\longmapsto \rm g'= \frac{20}{9} \\ \end{gathered}

:⟼g

′

=

9

20

\purple{ \large :\longmapsto \underline {\boxed{{\bf g' = 2.22 \: m/ {s}^{2} } }}}:⟼

g

′

=2.22m/s

2

Therefore acceleration due to gravity on the planet 'x' is

\large\underline{\pink{\underline{\frak{\pmb{\text Acceleration = 2.22 \: m/s^2 }}}}}

Acceleration=2.22m/s

2

Acceleration=2.22m/s

2