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A body of mass `10kg` is acted upon by two per pendicular forces `6N` and `8N`. The resultant ac-celeration of the body is .
A. `1ms^(-2)" at angle of tan"^(-1)((3)/(4))w.r.t. 8N "force"`
B. `0.2ms^(-2)" at angle of tan"^(-1)((3)/(4))w.r.t. 8N "force"`
C. `1ms^(-2)" at angle of tan"^(-1)((4)/(3))w.r.t. 8N "force"`
D. `0.2ms^(-2)" at angle of tan"^(-1)((4)/(3))w.r.t. 8N "force"`

1 Answer

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Best answer
Correct Answer - A
Here, m=10kg
The resultant force acting on the body is
Let the resultant force F makes an angle `theta` w.r.t. 8N force,
From figure, `tan theta =(6)/(8)(N)/(N)=(3)/(4)`
The resultant acceleration of the body is, `a=(F)/(m)=(10)/(10)(N)/(kg)=1ms^(-2)`
The resultant acceleration, is along the direction of the resultant force. Hence, the resultant acceleration of the body is `1m s^(-2)` at an angle `tan ^(-1) ((3)/(4))` w.r.t. 8N force.
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