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Hint: Use triangle law of vector addition, according to which if two vectors acting on a particle at the same time are represented in magnitude and direction by two sides of a triangle taken in one order, then their resultant vector is represented in magnitude and direction by the third side of triangle taken in opposite order.

Then use the Given condition, and find the value of $\theta $ .

$R=\sqrt{{{\left( {{F}_{1}} \right)}^{2}}+{{\left( {{F}_{2}} \right)}^{2}}+2{{F}_{1}}{{F}_{2}}\cos \theta }$

$\begin{align}

& {{F}_{1}}\text{ is first force} \\

& {{\text{F}}_{2}}\text{ is second force} \\

& \text{and }\theta \text{ is angle between the force}\text{.} \\

\end{align}$

We have $\begin{align}

& {{F}_{1}}=3N \\

& {{F}_{2}}=2N \\

\end{align}$

The resultant force is,

\[\begin{align}

& R=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 2 \right)}^{2}}+\left( 3 \right)\left( 2 \right)\cos \theta } \\

& {{R}^{2}}=9+4+12\cos \theta \\

\end{align}\]

\[{{R}^{2}}=13+12\cos \theta \]……. (1)

Now force ${{F}_{1}}$ is increased to $6N$ and resultant become $2R.$

$\begin{align}

& 2R=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 2 \right)}^{2}}+2\left( 6 \right)\left( 2 \right)\text{ }\cos \theta } \\

& 4{{R}^{2}}=36+4+24\text{ }\cos \theta \\

\end{align}$

$4{{R}^{2}}=40+24\text{ cos}\theta $…….. (2)

Put the value of ${{R}^{2}}\text{ from }$ equation (1) into equation (2)

\[\begin{align}

& 4\left( 13+12\text{ cos}\theta \right)=40+24\text{ cos}\theta \\

& \text{52+48 cos}\theta =40+24\text{ cos}\theta \\

& \text{12+24 cos}\theta \text{=0} \\

\end{align}\]

\[\begin{align}

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

& \text{ cos}\theta \text{=180}{}^\circ -60{}^\circ \\

& \text{ }\theta =120{}^\circ \\

& \text{ The value of }\theta \text{ is }120{}^\circ \\

\end{align}\]

Also read,

Parallelogram law of vectors and polygon law of vectors.

By Graphical and Analytical method both.

Then use the Given condition, and find the value of $\theta $ .

**Formula used**The resultant Force is given by$R=\sqrt{{{\left( {{F}_{1}} \right)}^{2}}+{{\left( {{F}_{2}} \right)}^{2}}+2{{F}_{1}}{{F}_{2}}\cos \theta }$

$\begin{align}

& {{F}_{1}}\text{ is first force} \\

& {{\text{F}}_{2}}\text{ is second force} \\

& \text{and }\theta \text{ is angle between the force}\text{.} \\

\end{align}$

**Complete step by step solution**We have $\begin{align}

& {{F}_{1}}=3N \\

& {{F}_{2}}=2N \\

\end{align}$

The resultant force is,

\[\begin{align}

& R=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 2 \right)}^{2}}+\left( 3 \right)\left( 2 \right)\cos \theta } \\

& {{R}^{2}}=9+4+12\cos \theta \\

\end{align}\]

\[{{R}^{2}}=13+12\cos \theta \]……. (1)

Now force ${{F}_{1}}$ is increased to $6N$ and resultant become $2R.$

$\begin{align}

& 2R=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 2 \right)}^{2}}+2\left( 6 \right)\left( 2 \right)\text{ }\cos \theta } \\

& 4{{R}^{2}}=36+4+24\text{ }\cos \theta \\

\end{align}$

$4{{R}^{2}}=40+24\text{ cos}\theta $…….. (2)

Put the value of ${{R}^{2}}\text{ from }$ equation (1) into equation (2)

\[\begin{align}

& 4\left( 13+12\text{ cos}\theta \right)=40+24\text{ cos}\theta \\

& \text{52+48 cos}\theta =40+24\text{ cos}\theta \\

& \text{12+24 cos}\theta \text{=0} \\

\end{align}\]

\[\begin{align}

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

& \text{ cos}\theta \text{=180}{}^\circ -60{}^\circ \\

& \text{ }\theta =120{}^\circ \\

& \text{ The value of }\theta \text{ is }120{}^\circ \\

\end{align}\]

**Note:**To find the resultant of the two vectors, must read triangle law of vector addition.Also read,

Parallelogram law of vectors and polygon law of vectors.

By Graphical and Analytical method both.

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