A body of mass $10 k g$ is acted upon by two perpendicular forces, $6 N$ and $8 N$ . The resultant acceleration of the body is

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$1 m s^{- 2}$ at an angle of $\tan^{- 1} (\frac{3}{4})$ w.r.t $8 N$ force

$0.2 m s^{- 2}$ at an angle of $\tan^{- 1} (\frac{3}{4})$ w.r.t $8 N$ force

$1 m s^{- 2}$ at an angle of $\tan^{- 1} (\frac{4}{3})$ w.r.t $8 N$ force

$0.2 m s^{- 2}$ at an angle of $\tan^{- 1} (\frac{4}{3})$ w.r.t $8 N$ force

answer is A.

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Detailed Solution

Here, $m = 10 k g$ . The resultant force acting on the body is $F = \sqrt{{(8 N)}^{2} + {(6 N)}^{2}} = 10 N$ Let the resultant force $F$ makes an angle $θ$ w.r.t. $8 N$ force. From figure, $\tan θ = \frac{6 N}{8 N} = \frac{3}{4}$
The resultant acceleration of the body is $a = \frac{F}{M} = \frac{10 N}{10 k g} = 1 m s^{- 2}$
The resultant acceleration is along the direction of the resultant force. Hence, the resultant acceleration of the body is $1 m s^{- 2}$ at an angle of $\tan^{- 1} (\frac{3}{4})$ w.r.t $8 N$ force.