Question

If c is small in comparision with l then ${\left(\frac{l}{l+c}\right)}^{\frac{1}{2}}+{\left(\frac{l}{l-c}\right)}^{\frac{1}{2}}=$

• A. $2+\frac{3c}{4l}$
• B. $2+\frac{3c^2}{4l^2}$
• C. $l+\frac{3c^2}{4l^2}$
• D. $l+\frac{3c}{4l}$

Answer 1

Answer: (B)

$2+\frac{3c^2}{4l^2}$

${\left(\frac{l}{l+c}\right)}^{1/2}+{\left(\frac{l}{l-c}\right)}^{1/2}$

$={\left(\frac{1}{1+c/l}\right)}^{1/2}+{\left(\frac{1}{1-c/l}\right)}^{1/2}$

$={\left(1+\frac{c}{l}\right)}^{-\frac{1}{2}}+{\left(1-\frac{c}{l}\right)}^{-\frac{1}{2}}$

$=\left[1-\left(\frac{1}{2}\right).\left(\frac{c}{l}\right)+\left(\frac{3}{8}\right).{\left(\frac{c}{l}\right)}^2-\left(\frac{5}{16}\right).{\left(\frac{c}{l}\right)}^3+.....\right]$

$+\left[1+\left(\frac{1}{2}\right).\left(\frac{c}{l}\right)+\left(\frac{3}{8}\right).{\left(\frac{c}{l}\right)}^2+\left(\frac{5}{16}\right).{\left(\frac{c}{l}\right)}^3+.....\right]$

$\text{{}\because {\left(\text{1 + x}\right)}^{\text{n}}\text{= 1 + nx +}\frac{\text{n}\left(\text{n + 1}\right)}{\text{2!}}{\text{x}}^{\text{2}}\text{+}\frac{\text{n}\left(\text{n - 1}\right)\left(\text{n - 2}\right)}{\text{3!}}\text{,}\text{put}\text{n}\text{=}\text{-}\frac{\text{1}}{\text{2}}\text{and}\text{x}\text{=}\pm \frac{\text{c}}{\text{l}}$

$=2+\frac{6c^2}{8l^2}+....$

$=2+\frac{3c^2}{4l^2}$

[ingnoring higher powers, since c<l.]