Question

23
31.
Two forces 3 N and 2 N are at an angle o such that the
resultant is R. The first force is now increased to 6 N and the
resultant become 2R. The value of is (HP PMT 2000]
(a) 30°
(b) 60°
(c) 90°
​

Answer 1

$\huge{{{\color{plum}{{ answer☃ :}}}}}$

\theta=120^{\circ}

The value of

Given that,

Explanation:

Angle=

\theta

First force

F= 3 N

Second force

F_{2}=2 N

The resultant force is

R= \sqrt{(F_{1})^2+(F_{2})^2+2F_{1}F_{2}cos\theta}

R= \sqrt{(3)^2+(2^2)+2\times3\times2\times cos\theta}

Now, if the first force increases to 6 n and R become 2 R.

Then,

2R= \sqrt{(6)^2+(2^2)+2\times6\times2\times cos\theta}

Put the value of R in equation in equation (II)

....(II)

2(\sqrt{(13+12\times cos\theta})= \sqrt{(40+24\times cos\theta}

On squaring both sides

4(13+12\ cos\theta)=40+24\ cos\theta

cos\theta=\dfrac{-1}{2}

\theta=cos^{-1}\dfrac{-1}{2}

\theta=120^{\circ}

Hence, The value of theta = 120°

$\huge{{{\color{plum}{{@ janvi ☃ :}}}}}$

Answer 2

$\large \tt\ red\ {answer}$

\theta=120^{\circ}

The value of

Given that,

Explanation:

Angle=

\theta

First force

F= 3 N

Second force

F_{2}=2 N

The resultant force is

R= \sqrt{(F_{1})^2+(F_{2})^2+2F_{1}F_{2}cos\theta}

R= \sqrt{(3)^2+(2^2)+2\times3\times2\times cos\theta}

Now, if the first force increases to 6 n and R become 2 R.

Then,

2R= \sqrt{(6)^2+(2^2)+2\times6\times2\times cos\theta}

Put the value of R in equation in equation (II)

....(II)

2(\sqrt{(13+12\times cos\theta})= \sqrt{(40+24\times cos\theta}

On squaring both sides

4(13+12\ cos\theta)=40+24\ cos\theta

cos\theta=\dfrac{-1}{2}

\theta=cos^{-1}\dfrac{-1}{2}

\theta=120^{\circ}

Hence, The value of theta = 120°