Three particles, each of mass m, are placed at the vertices of an equilateral triangle of side a. The gravitational field intensity at the cenrtroid of the triangle is then

z e r o

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{g m^{2}}{a^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 G m^{2}}{a^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3 G m^{2}}{a^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $z e r o$
$r = A O = B O = C O = \frac{l}{\sqrt{3}}$

$E_{A} = \frac{G m}{r^{2}}$

$= \frac{3 G m}{l^{2}}$ along $O A$
$E_{B} = \frac{3 G M}{l}$

$E_{C} = \frac{3 G m}{l}$

They are co planer and equal in magnitude and make an angle of $120^{\circ}$
The resultant field is zero.

Suggest Corrections

Similar questions

Three particles, each of mass m, are placed at the vertices of an equilateral triangle of side a. The gravitational field intensity at the centroid of the triangle is

Three particles, each of mass m, are placed at the vertices of an equilateral triangle of side $a$ . The net gravitational force on a body of mass $M$ at the centroid of the triangle is

Q. Three particles, each of mass

M

are placed at the three corners of an equilateral triangle of side

l

. What is the gravitational field strength at the location of

m

Consider a system of 4 charged particles each of mass m placed at the vertices and centre of an equilateral triangle of side a as shown in figure. Under mutually electrostatic interaction charges at vertices of the triangle are revolving in a circle. Find speed of charges. [ $k = \frac{1}{4 π ϵ_{0}}$ and neglect gravitational effect]

Q. Three mass points each of mass m are placed at the vertices of an equilateral triangle of side l. What is the gravitational field and potential due to three masses at the centroid of the triangle?

Join BYJU'S Learning Program

Three particles, each of mass m, are placed at the vertices of an equilateral triangle of side a. The gravitational field intensity at the cenrtroid of the triangle is then

The correct option is A zeror=AO=BO=CO=l√3 EA=Gmr2 =3Gml2 along OA EB=3GMl EC=3Gml They are co planer and equal in magnitude and make an angle of 120∘ The resultant field is zero.