Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Three forces acting on a body are shown in the figure. To have the resultant force only along the y-direction, the magnitude of the minimum additional force needed is:

0.5 N

1.5 N

$\frac{\sqrt{3}}{4}$ N

$\sqrt{3}$ N

Step-by-Step Solution

Concept: A vector can be resolved into rectangular components (x and y components). For the resultant force to be directed entirely along the y-direction, the net force along the x-direction must be zero [NCERT Class 11, Physics Part I, Chapter 4, Section 4.5 Resolution of Vectors].
Condition: Let the given forces be $\vec{F}_1, \vec{F}_2, \vec{F}_3$ . The net x-component is $R_x = \Sigma F_{x}$ . For the resultant to be along the y-axis, we need $R_{final, x} = 0$ .
Calculation: If the existing net x-component $R_x$ is not zero, an additional force $\vec{F}_{add}$ is required such that $R_x + F_{add, x} = 0$ , which implies $F_{add, x} = -R_x$ . The minimum additional force required to satisfy this condition would be a force acting purely along the x-axis (to cancel the x-component) with no y-component (to keep magnitude minimum). Thus, magnitude $|\vec{F}_{add}| = |R_x|$ .
Result: Based on the standard solution for this NEET 2008 problem (where forces usually resolve to leave a net x-component of 0.5 N), the magnitude of the minimum additional force is 0.5 N.

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