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Answer 1

Given :

Mass of a particle = m

Distance from the rod = d

Mass of rod = M.

Length of rod = L

To find :

The gravitational force between mass and rod.

Solution :

Here, mass of rod length = M/L

mass of small element dm = M/L dx

Gravitational force : $\sf dF = \dfrac{Gm(dm)}{x^2}$

$\dashrightarrow \sf dF = \dfrac{Gm \bigg(\dfrac{M}{L} dx \bigg)}{x^2}$

$\dashrightarrow \displaystyle \sf F = \int \limits_ {d}^{(d + L)}\dfrac{Gm \bigg(\dfrac{M}{L} dx \bigg)}{x^2}$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{L} \int \limits_ {d}^{(d + L)}\dfrac{ dx}{x^2}$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{L} \int \limits_ {d}^{(d + L)} {x}^{ - 2} dx$

$\dashrightarrow\displaystyle \sf F = \dfrac{GmM}{L} \bigg[ \dfrac{ {x}^{ - 1} }{ - 1} \bigg ]_ {d}^{(d + L)}$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{L} \bigg[ \dfrac{ - 1 }{x } \bigg ]_ {d}^{(d + L)}$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{L} \bigg[ \dfrac{ 1 }{d } - \dfrac{1}{d + L} \bigg ]$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{L} \bigg[ \dfrac{d + L - d}{d(d + L)} \bigg ]$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{ \not{L}} \bigg[ \dfrac{ \not{L}}{d(d + L)} \bigg ]$

$\dashrightarrow \underline{ \boxed{\sf F = \dfrac{GmM}{d(d + L)}}}$

Thus, the gravitational force between mass and rod is $\sf \dfrac{GmM}{d(d + L)}$

Answer 2

Given :

Mass of a particle = m

Distance from the rod = d

Mass of rod = M.

Length of rod = L

To find :

The gravitational force between mass and rod.

Solution :

Here, mass of rod length = M/L

mass of small element dm = M/L dx

Gravitational force : $\sf dF = \dfrac{Gm(dm)}{x^2}$

$\dashrightarrow \sf dF = \dfrac{Gm \bigg(\dfrac{M}{L} dx \bigg)}{x^2}$

$\dashrightarrow \displaystyle \sf F = \int \limits_ {d}^{(d + L)}\dfrac{Gm \bigg(\dfrac{M}{L} dx \bigg)}{x^2}$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{L} \int \limits_ {d}^{(d + L)}\dfrac{ dx}{x^2}$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{L} \int \limits_ {d}^{(d + L)} {x}^{ - 2} dx$

$\dashrightarrow\displaystyle \sf F = \dfrac{GmM}{L} \bigg[ \dfrac{ {x}^{ - 1} }{ - 1} \bigg ]_ {d}^{(d + L)}$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{L} \bigg[ \dfrac{ - 1 }{x } \bigg ]_ {d}^{(d + L)}$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{L} \bigg[ \dfrac{ 1 }{d } - \dfrac{1}{d + L} \bigg ]$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{L} \bigg[ \dfrac{d + L - d}{d(d + L)} \bigg ]$

$\dashrightarrow \displaystyle \sf F = \dfrac{GmM}{ \not{L}} \bigg[ \dfrac{ \not{L}}{d(d + L)} \bigg ]$

$\dashrightarrow \underline{ \boxed{\sf F = \dfrac{GmM}{d(d + L)}}}$

Thus, the gravitational force between mass and rod is $\sf \dfrac{GmM}{d(d + L)}$