A metal wire of uniform mass density having l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The length of the semicircular arc is $L = \pi R$,so the radius is $R = \frac{L}{\pi}$.Consider an infinitesimal element of the wire of length $dl

Vedclass · Accepted Answer

$\frac{2 GmM \pi}{L^2}$

Vedclass · Answer

(D) The length of the semicircular arc is $L = \pi R$,so the radius is $R = \frac{L}{\pi}$.Consider an infinitesimal element of the wire of length $dl = R d	heta$ at an angle $	heta$. The mass of this element is $dm = \lambda dl = \frac{M}{L} R d	heta$.The gravitational force exerted by this element on the particle of mass $m$ at the center is $dF = \frac{G m dm}{R^2} = \frac{G m (M/L) R d	heta}{R^2} = \frac{GMm}{LR} d	heta$.Due to symmetry,the horizontal components of the force cancel out,and the vertical components add up.The net force is $F = \int_{-\pi/2}^{\pi/2} dF \sin	heta = \frac{GMm}{LR} \int_{-\pi/2}^{\pi/2} \sin	heta d	heta$. Wait,the standard result for the field at the center of a semicircular arc is $E = \frac{2GM}{LR}$.Substituting $R = L/\pi$,we get $F = mE = m \left( \frac{2GM}{L(L/\pi)} ight) = \frac{2GMm\pi}{L^2}$.Thus,the correct option is $D$.

$A$ metal wire of uniform mass density having length $L$ and mass $M$ is bent to form a semicircular arc and a particle of mass $m$ is placed at the centre of the arc. The gravitational force on the particle by the wire is:

Similar Questions

Two bodies of mass $100 \ kg$ and $10^4 \ kg$ are lying one meter apart. At what distance from the $100 \ kg$ body will the intensity of the gravitational field be zero?

The gravitational intensity at the centre $O$ of a hemispherical shell of uniform mass density has the direction indicated by which arrow?

Two concentric spherical shells have masses $M_1$ and $M_2$ and radii $r_1$ and $r_2$ $(r_1 < r_2)$. What is the gravitational intensity at a distance $r$ from the center where $r_1 < r < r_2$?

The gravitational field in a region is given by $\vec{E} = (5\,N/kg)\,\hat{i} + (12\,N/kg)\,\hat{j}$. If the potential at the origin is taken to be zero,then the ratio of the potential at the points $(12\,m, 0)$ and $(0, 5\,m)$ is:

The mass density of a planet of radius $R$ varies with the distance $r$ from its centre as $\rho(r) = \rho_{0} \left(1 - \frac{r^{2}}{R^{2}}\right)$. Then the gravitational field is maximum at:

Vedclass Test Series

Exam Paper Generator

Online Exam Module