Answer

Verified

18k+ views

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.

Recently Updated Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main

Find the points of intersection of the tangents at class 11 maths JEE_Main

For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main

The path difference between two waves for constructive class 11 physics JEE_MAIN

What is the difference between solvation and hydra class 11 chemistry JEE_Main

Other Pages

Derive an expression for maximum speed of a car on class 11 physics JEE_Main

Assertion Work and heat are two equivalent from of class 11 chemistry JEE_Main

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Formal charge on nitrogen and oxygen in NO3 ion are class 11 chemistry JEE_Main

Why does capacitor block DC and allow AC class 12 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main