Two horizontal forces of magnitudes of `10N` and `P N` act on a particle. The force of magnitude `10N` acts due west and the force of magnitude `P N` acts on a bearing of `30^(@)` east of north as shown in figure. The resultant of these two force acts due north. Find the magnitude of the resultant.

Options

A. `10 sqrt(2) N`
B. `15 sqrt(3) N`
C. `12sqrt(5) N`
D. `10 sqrt(3) N`

Answer

D

Explanation

Resultant acts along y-axis
`implies P_(x)= P cos 60^(@)= 10 N or P/2= 10 or P =20`
Resultant `P_(y)= P sin 60^(@)= 20xx(sqrt(3))/2= 10sqrt(3) N`