Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Gravitation

Question:

A system consists of a thin ring of radius r and of mass M and a straight wire of linear mass density λ of infinite length placed along the axis of the ring with one of its ends at the centre of the ring. What is the force of interaction between the wire and the ring.

Options:

$\frac{3 GM \lambda}{r}$

$\frac{2 GM \lambda}{r}$

$\frac{GM \lambda}{2 r}$

$\frac{GM \lambda}{r}$

Correct Answer:

$\frac{GM \lambda}{r}$

Explanation:

The force of interaction due to ring on the element dx is dF = $\frac{G M x \lambda d x}{\left(r^2+x^2\right)^{3 / 2}}$

⇒ F = GM $\lambda \int\limits_0^{\infty} \frac{x d x}{\left(r^2+x^2\right)^{3 / 2}}$

= $GM \lambda \int\limits_0^{\pi / 2} \frac{r^2 \tan \theta \sec ^2 \theta d \theta}{r^3 \sec ^3 \theta}$

$=\frac{GM \lambda}{r} \int\limits_0^{\pi / 2} \sin \theta d \theta=\frac{GM \lambda}{r}|-cos \theta|_0^{\pi / 2}=\frac{GM \lambda}{r}$.