A mass M is divided into two parts xM and (1 | vedclass

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

Question

(A) The gravitational force $F$ between two masses $m_1 = xM$ and $m_2 = (1 - x)M$ separated by a distance $r$ is given by $F = \frac{G(xM)((1 - x)M)}

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(A) The gravitational force $F$ between two masses $m_1 = xM$ and $m_2 = (1 - x)M$ separated by a distance $r$ is given by $F = \frac{G(xM)((1 - x)M)}{r^2}$.Since $G$,$M$,and $r$ are constants,the force $F$ is proportional to $x(1 - x) = x - x^2$.To find the value of $x$ for which the force is maximum,we differentiate $F$ with respect to $x$ and set it to zero:$\frac{d}{dx}(x - x^2) = 1 - 2x = 0$.Solving for $x$,we get $1 = 2x$,which implies $x = 0.5$.

$A$ mass $M$ is divided into two parts $xM$ and $(1 - x)M$. For a given separation,the value of $x$ for which the gravitational attraction between the two pieces becomes maximum is

Similar Questions

The gravitational force between two stones of mass $1 \, kg$ each,separated by a distance of $1 \, m$ in a vacuum,is:

The gravitational constant depends upon .........

Two stars of masses $m_1$ and $m_2$ are parts of a binary star system. The radii of their orbits are $r_1$ and $r_2$ respectively,measured from the centre of mass of the system. The magnitude of the gravitational force that $m_1$ exerts on $m_2$ is

Two identical particles each of mass $m$ go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be

The gravitational force $F_g$ between two objects does not depend on

Vedclass Test Series

Exam Paper Generator

Online Exam Module