Question

A thin wire of mass m and length ℓ is bent to form a circular loop. The gravitational potential at the centre of the loop is

Answer 1

Answer:

A thin wire of mass M and length L is bent to from a circular ring. The moment of inertia of this ring about its axis is :

A .

4π

2

1

ML

2

B .

12

1

ML

2

C .

3π

2

1

ML

2

D .

π

2

1

ML

2

December 27,

Answer 2

Given:

• A thin wire of mass m and length l is bent to form a circular loop.

To find:

• Gravitational potential at centre ?

Calculation:

Let's assume an elemental mass dm from the loop having a length of dl .

Now, we know :

$2\pi r = l$

$\implies \: r = \dfrac{l}{2 \pi }$

Now, field potential at centre due to elemental mass:

$dV = \dfrac{G(dm)}{ {r}^{2} }$

Now, integrating both sides :

$\displaystyle \implies \int dV = \int\dfrac{G(dm)}{ {r}^{2} }$

$\displaystyle \implies \int dV = \dfrac{G}{ {r}^{2} } \int \: dm$

$\displaystyle \implies V = \dfrac{G}{ {r}^{2} } \times m$

$\displaystyle \implies V = \dfrac{4 {\pi}^{2} Gm}{ {l}^{2} }$

So, final answer is :

$\boxed{ \bf V = \dfrac{4 {\pi}^{2} Gm}{ {l}^{2} } }$