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Question

The sum of magnitudes of two forces acting at a point is $$16\,N$$. If the resultant force is $$8\,N$$ and its direction is perpendicular to smaller force, then the forces are:


A
6N and 10N
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B
8N and 8/N
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C
4N and 12N
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D
2N and 14N
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Solution

The correct option is A $$6\,N$$ and $$10\,N$$
The correct answer is A.

Given,

The magnitude of two n forces is $$16N$$

The resultant force is $$8N$$

so,

$$X+Y 16(X< Y)$$ and R is the resultant force perpendicular to X.

According to the Pythagoras theorem,

$$\Rightarrow R^2=Y^2-X^2=(Y-X)(Y+X)=16(Y-X)$$

$$\Rightarrow (Y-X)=\dfrac{R^2}{(Y+X)}=\dfrac{8^2}{16}$$

$$\Rightarrow\dfrac{64}{16}=4$$

$$(Y+X)=16$$ and $$(Y-X)=4$$

By solving a linear equation in two variable we get,

$$Y=10N$$  and $$X=6N$$

955921_1015886_ans_312ed766cbfe4a9f991e87f2c55e172f.png

Physics

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