
The resultant of two forces 3P and 2P is R. If the first force is doubled then the resultant is also doubled, then the angle between two forces is:
A). \[60{}^\circ \]
B). \[120{}^\circ \]
C). \[70{}^\circ \]
D). \[180{}^\circ \]
Answer
536.4k+ views
Hint: The question can be easily solved by using the formula for resultant vector. Here we can find the angle between the two forces by equating the two equations for resultant vectors R and 2R and 6P and 2P respectively with a constant angle between the forces.
Formula Used: The formula for resultant vector is given by:
\[R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\]
\[\theta \] is the angle between vectors A and B.
Complete step by step answer:
A resultant vector is a vector that produces the same effect as is produced by the individual vectors together.
Equation for resultant vector R when the given vector forces are 3P and 2P:
\[R=\sqrt{{{(3P)}^{2}}+{{(2P)}^{2}}+2(3P)(2P)\cos \theta }\]
\[R=\sqrt{9{{P}^{2}}+4{{P}^{2}}+12{{P}^{2}}\cos \theta }\]
\[R=\sqrt{13{{P}^{2}}+12{{P}^{2}}\cos \theta }\]
\[\Rightarrow {{R}^{2}}=13{{P}^{2}}+12{{P}^{2}}\cos \theta \] ………………..(1)
Now, the equations for resultant vector 2R when the given vectors are 6P and 2P:
\[2R=\sqrt{{{(6P)}^{2}}+{{(2P)}^{2}}+2(6P)(2P)\cos \theta }\]
\[2R=\sqrt{36{{P}^{2}}+4{{P}^{2}}+24{{P}^{2}}\cos \theta }\]
\[2R=\sqrt{40{{P}^{2}}+24{{P}^{2}}\cos \theta }\]
\[{{(2R)}^{2}}=40{{P}^{2}}+24{{P}^{2}}\cos \theta \]
\[4{{R}^{2}}=40{{P}^{2}}+24{{P}^{2}}\cos \theta \] …………………………..(2)
On substituting the value of \[{{R}^{2}}\] from equation (1) in (2) we get,
\[4(13{{P}^{2}}+12{{P}^{2}}\cos \theta )=40{{P}^{2}}+24{{P}^{2}}\cos \theta \]
\[52{{P}^{2}}+48{{P}^{2}}\cos \theta =40{{P}^{2}}+24{{P}^{2}}\cos \theta \]
\[24{{P}^{2}}\cos \theta =-12{{P}^{2}}\]
\[\cos \theta =-\dfrac{1}{2}\]
\[\Rightarrow \theta =120{}^\circ \]
Hence, the correct answer is option B. \[120{}^\circ \].
Note: The magnitude of resultant of two vectors is maximum when the vectors are in the same direction and it is minimum when the two vectors act in the opposite direction. When in the same direction the vectors are added to get the resultant and when they are in the opposite direction they are subtracted.
Formula Used: The formula for resultant vector is given by:
\[R=\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \theta }\]
\[\theta \] is the angle between vectors A and B.
Complete step by step answer:
A resultant vector is a vector that produces the same effect as is produced by the individual vectors together.
Equation for resultant vector R when the given vector forces are 3P and 2P:
\[R=\sqrt{{{(3P)}^{2}}+{{(2P)}^{2}}+2(3P)(2P)\cos \theta }\]
\[R=\sqrt{9{{P}^{2}}+4{{P}^{2}}+12{{P}^{2}}\cos \theta }\]
\[R=\sqrt{13{{P}^{2}}+12{{P}^{2}}\cos \theta }\]
\[\Rightarrow {{R}^{2}}=13{{P}^{2}}+12{{P}^{2}}\cos \theta \] ………………..(1)
Now, the equations for resultant vector 2R when the given vectors are 6P and 2P:
\[2R=\sqrt{{{(6P)}^{2}}+{{(2P)}^{2}}+2(6P)(2P)\cos \theta }\]
\[2R=\sqrt{36{{P}^{2}}+4{{P}^{2}}+24{{P}^{2}}\cos \theta }\]
\[2R=\sqrt{40{{P}^{2}}+24{{P}^{2}}\cos \theta }\]
\[{{(2R)}^{2}}=40{{P}^{2}}+24{{P}^{2}}\cos \theta \]
\[4{{R}^{2}}=40{{P}^{2}}+24{{P}^{2}}\cos \theta \] …………………………..(2)
On substituting the value of \[{{R}^{2}}\] from equation (1) in (2) we get,
\[4(13{{P}^{2}}+12{{P}^{2}}\cos \theta )=40{{P}^{2}}+24{{P}^{2}}\cos \theta \]
\[52{{P}^{2}}+48{{P}^{2}}\cos \theta =40{{P}^{2}}+24{{P}^{2}}\cos \theta \]
\[24{{P}^{2}}\cos \theta =-12{{P}^{2}}\]
\[\cos \theta =-\dfrac{1}{2}\]
\[\Rightarrow \theta =120{}^\circ \]
Hence, the correct answer is option B. \[120{}^\circ \].
Note: The magnitude of resultant of two vectors is maximum when the vectors are in the same direction and it is minimum when the two vectors act in the opposite direction. When in the same direction the vectors are added to get the resultant and when they are in the opposite direction they are subtracted.
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