Question

In a physics lab, we took two readings of dip angle by dip circle, in two readings planes of dip circle are mutually perpendicular, then ratio of vertical & horizontal component of field the to earth in lab is (reading 1 = ${45}^0$ reading 2 = ${60}^0$)

• A. $co{t}^{-1}\left\{\frac{4}{3}\right\}$
• B. $co{t}^{-1}\left\{\sqrt{\frac{7}{3}}\right\}$
• C. $co{t}^{-1}\left\{\sqrt{\frac{4}{3}}\right\}$
• D. $co{t}^{-1}\left\{\frac{7}{3}\right\}$

Answer 1

The correct option is C $co{t}^{-1}\left\{\sqrt{\frac{4}{3}}\right\}$

$co{t}^2\delta =co{t}^2{\delta }_1+co{t}^2{\delta }_2$
$co{t}^2\delta =co{t}^2{45}^o+co{t}^2{60}^o$
$=1+\frac{1}{3}=\frac{4}{3}$
$\delta =co{t}^{-1}\left\{\sqrt{\frac{4}{3}}\right\}$