Question

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

• 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

The correct option is B $2+\frac{3c^2}{4l^2}$

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

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

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

$\because \frac{c}{l}<<<1$

Applying binomial expansion, we have

$=(1-\frac{1}{2}\cdot \frac{c}{l}+\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{3}{2}\cdot {\left(\frac{c}{l}\right)}^2-.....)+(1+\frac{1}{2}\cdot \frac{c}{l}+\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{3}{2}\cdot {\left(\frac{c}{l}\right)}^2+.....)$

$=2+2\cdot \frac{3c^2}{8l^2}(\because \frac{c}{l}<<<1)$

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