Verified answer

Answer:

Given :-

• The force of attraction (A) between two objects, whose masses are fixed, varies inversely with the square of their distance of separation (d).
• The force of attraction between them is 2 units when the distance is 5 cm.

To Find :-

• What is the distance between the two objects, if the attraction between them is 8 units.

Solution :-

[tex]\bigstar[/tex] The force is attaction (A) between two objects, whose masses are fixed, varies inversely with the square of their distance of separation (d).

So,

[tex]\implies \sf\boxed{\bold{A \propto \dfrac{1}{d^2}}}\\[/tex]

We can write as :

[tex]\implies \sf A =\: \dfrac{k}{d^2}[/tex]

where,

• A = Attraction between two objects
• k = Gravitational Constant
• d = Distance between two objects

Given :

• Force of Attraction (A) = 2 units
• Distance of Separation (d) = 5 cm

So, by putting those values we get,

[tex]\implies \sf 2 =\: \dfrac{k}{(5)^2}[/tex]

[tex]\implies \sf 2 =\: \dfrac{k}{(5 \times 5)}[/tex]

[tex]\implies \sf 2 =\: \dfrac{k}{(25)}[/tex]

[tex]\implies \sf 2 =\: \dfrac{k}{25}[/tex]

By doing cross multiplication we get,

[tex]\implies \sf k =\: 2(25)[/tex]

[tex]\implies \sf k =\: 2 \times 25[/tex]

[tex]\implies \sf\bold{k =\: 50}[/tex]

Hence, the value of gravitational constant is 50 .

Now,

[tex]\bigstar[/tex] The attraction between them is 8 units.

Given :

• Force of Attraction (A) = 8 units
• Gravitational Constant (k) = 50

According to the question by using the formula we get,

[tex]\implies \sf\bold{A =\: \dfrac{k}{d^2}}[/tex]

[tex]\implies \sf 8 =\: \dfrac{50}{d^2}[/tex]

By doing cross multiplication we get,

[tex]\implies \sf 8 \times d^2 =\: 50[/tex]

[tex]\implies \sf d^2 =\: \dfrac{50}{8}[/tex]

[tex]\implies \sf d =\: \sqrt{\dfrac{\cancel{50}}{\cancel{8}}}[/tex]

[tex]\implies \sf d =\: \sqrt{\dfrac{25}{4}}[/tex]

[tex]\implies \sf d =\: \sqrt{\dfrac{\underline{5 \times 5}}{\underline{2 \times 2}}}\\[/tex]

[tex]\implies \sf d =\: \dfrac{5}{2}[/tex]

[tex]\implies \sf\bold{\underline{d =\: 2.5\: cm}}[/tex]

[tex]\therefore[/tex] The distance between the two objects, if the attraction between them is 8 units is 2.5 cm .

[tex]\\[/tex]

VERIFICATION :-

Given :

• Distance between two objects (d) = 2.5 cm
• Gravitational Constant (k) = 50
• Force of Attraction (A) = 8 units

So, by putting those values according to the question we get,

[tex]\leadsto \bf A =\: \dfrac{k}{d^2}[/tex]

[tex]\implies \sf 8 =\: \dfrac{50}{(2.5)^2}[/tex]

[tex]\leadsto \sf 8 =\: \dfrac{50}{(2.5 \times 2.5)}[/tex]

[tex]\leadsto \sf 8 =\: \dfrac{50}{6.25}[/tex]

By doing cross multiplication we get,

[tex]\leadsto \sf 50 =\: 8(6.25)[/tex]

[tex]\leadsto \sf 50 =\: 8 \times 6.25[/tex]

[tex]\leadsto \sf\boxed{\bold{50 =\: 50}}[/tex]

[tex]\leadsto \bf L.H.S =\: R.H.S[/tex]

Hence, Verified !!

Henceforth, the above answer is correct.

Step-by-step explanation:

Solution:-

Given is that force of attraction (A) is inversely equal to d square of distance So, we can take k as constant

[tex]a = \frac{k}{ {d}^{2} } [/tex]

According to question if force of attraction (A) is 2 units distance is 5cm Find the distance between object if force of attraction is 8 units

So, first let's find the value of constant(k)

[tex]2 = \frac{k}{ {5}^{2} } \\ \\ 2 \times {5}^{2} = k \\ \\ 2 \times 25 = k \\ \\ k = 50[/tex]

So, Force of attraction is 8 units k is 50 and will remain constant we have to find distance between two objects

So, let's substitute the value in the equation

[tex]8 = \frac{50}{d {}^{2} } \\ \\ {d}^{2} = \frac{50}{8} \\ \\ {d }^{2} = \frac{25}{4} \\ \\ d = \sqrt{ \frac{25}{4} } \\ \\ d = \frac{5}{2} \\ \\ d = 2.5[/tex]

So, distance between two objects is 2.5cm

Hope this helps you dear friend

Have a great day ahead:D