Question

f the perimeter of a sector of a circle of radius 6.5 cm is 29 cm, then its area isa) 58 sqcm b) 52 sqcm c) 25 sqcm d) 56 sqcm *

1 point

a

b

c

d

​

Answer 1

Answer:

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Answer 2

Answer:

To Find :

Change in gravitational force when masses of both objects are tripled

Solution :

Gravitational force between two maases m₁ and m₂ when the distance of seperation between them is r is given by ,

\begin{gathered} \\ \star \: {\boxed{\purple{\sf{F = \dfrac{Gm_1m_2}{ {r}^{2} } }}}} \\ \\ \end{gathered}

⋆

F=

r

2

Gm

1

m

2

Where ,

G is Gravitational constant

Let the original masses be m₁ and m₂ . Then the orginal gravitational force is ,

\begin{gathered} \\ : \implies \sf \:F_1 = \dfrac{Gm_1m_2}{ {r}^{2} } \: \: .........(1)\\ \\ \end{gathered}

:⟹F

1

=

r

2

Gm

1

m

2

.........(1)

Now , masses of objects when they are tripled is 3m₁ and 3m₂ and distance is r (since no change in distance is mentioned). Substituting the new masses ;

\begin{gathered} \\ : \implies \sf \:F_2= \dfrac{G \times 3m_1 \times 3m_2}{ {r}^{2} } \\ \\ \end{gathered}

:⟹F

2

=

r

2

G×3m

1

×3m

2

\begin{gathered} \\ : \implies \sf \: F_2 = \dfrac{ 9Gm_1m_2}{ {r}^{2} } \\ \\ \end{gathered}

:⟹F

2

=

r

2

9Gm

1

m

2

\begin{gathered} \\ : \implies \sf \: F_2 = 9 \bigg( \dfrac{Gm_1m_2}{ {r}^{2} } \bigg) \\ \\ \end{gathered}

:⟹F

2

=9(

r

2

Gm

1

m

2

)

\begin{gathered} \\ : \implies{\underline{\boxed{\pink{\mathfrak{F_2 = 9F_1}}}}} \bigstar \\ \\ \end{gathered}

:⟹

F

2

=9F

1

★

Hence ,

When the masses of objects are tripled keeping the distance between them constant the gravitational force becomes 9 times the original one.