A body of mass \(10\) kg is acted upon by two perpendicular forces, \(6\) N and \(8\) N. The resultant acceleration of the body is:

(a) \(1~\text{ms}^{-2}\) at an angle of \(tan^{-1}\Big(\frac43\Big)\) w.r.t. \(6\) N force
(b) \(0.2~\text{ms}^{-2}\) at an angle of \(tan^{-1}\Big(\frac34\Big)\) w.r.t. \(8\) N force
(c) \(1~\text{ms}^{-2}\) at an angle of \(tan^{-1}\Big(\frac34\Big)\) w.r.t. \(8\) N force
(d) \(0.2~\text{ms}^{-2}\) at an angle of \(tan^{-1}\Big(\frac34\Big)\) w.r.t. \(6\) N force


Choose the correct option:

1. (a), (c)
2. (b), (c)
3. (c), (d)
4. (a), (b), (c)

(1) Hint: The resultant acceleration will be in the direction of the resultant force.
Step 1: Find the resultant force and the magnitude of the acceleration.
Consider the adjacent diagram
Given,     mass = m = 10 kg
 Resultant force =F=F12+F22=36+64                             =10N                           a=Fm=1010=1m/s2; along R
Step 2: Find the direction of the acceleration.
 Let θ1 be angle between R and F1tan θ1=86=43θ1=tan1 (4/3)w.r.t.F1=6N
Let θ2 beanglebetweenF and F2
tan θ2=68=34
θ2=tan1 (34)w.r.t.F2=8N