For each natural number K , let \[{{C}_{K}}\] denote the circle radius k centimetres and centre at origin O. On the circle \[{C_K}\]a particle moves K centimetres in the counter-clockwise direction. After completing its motion on\[{C_K}\] , the particle moves to \[{{C}_{K+1}}\] in the radial direction. The motion of the particle continues in this manner. The particle starts at \[(1,0)\] . If the particle crosses the positive direction of the x-axis for the first time on the circle \[{{C}_{n}}\], then find the value of \[n\].
VerifiedHint: O is the centre of the circle. The particle starts moving from . The radius of the first circle is 1cm. The second circle will be from the origin with 2 cm and it will go on until the \[{k^{th}}\]circle. As the particles move radially find the radius and the distance travelled radially in each circle and get their angular displacement. Equate them to the total angle of the circle \[2\pi \]and get the value of \[n\].
Complete step-by-step answer:
We have been told that \[{C_K}\]denotes the circle with radius K. Let us take the centre of the circle as O.
Now it is said that the particles start at (1,0). Thus the radius of the first circle formed will be of 1cm. The second circle will be 1cm from the first circle thus the radius of the second circle will be 2cm. Similarly, if we are drawing a third circle its radius will be 3cm from the origin O. Thus this can go up to \[{{k}^{th}}\] circle. You can understand more from the figure drawn below. We can represent the circles as \[{C_1},{C_2},{C_3}.....,{C_K}\].
Now after completing the motion on \[{{C}_{K}}\]. The particle starts moving in the radial direction i.e. \[{C_{K + 1}}\]. The particle moves 1cm in circle \[{C_1}\], then it moves radially to circle \[{{C}_{2}}\]and travels 2cm. Now the particle moves radially to the third circle \[{C_3}\]and travels 3cm. Now this can go until the circle \[{{C}_{K}}\], where it travels K cm. We can understand this from the figure we have drawn.
The path of the particle is shown by the bold line segment and arcs.
Now for circle \[{{C}_{1}}\], radius = 1cm and distance travelled by the particle= 1cm.
Similarly, for the circle \[{C_2}\], radius =2cm and distance travelled by the particle =2cm.
For circle \[{C_3}\], radius =3cm and the distance travelled by the particle =3cm.
Similarly, for the circle \[{{C}_{K}}\], radius = K cm and the distance travelled by the particle = K cm.
Now let us find the angular displacement on the \[{k^{th}}\] circle, it is given by,
\[\text{ }\!\!\theta\!\!\text{ =}\dfrac{\text{Distance travelled}}{\text{radius}}\]
Thus we need to find \[{\theta _1} + {\theta _2} + {\theta _3} + {\theta _4} + ...... + {\theta _K}\]. We know that the angle of a complete circle is \[2\pi \]. Thus the motion of the particle completes the full circle. Hence its angle will be \[2\pi \]. Thus we can write it as,
\[{{\theta }_{1}}+{{\theta }_{2}}+{{\theta }_{3}}+{{\theta }_{4}}+......+{{\theta }_{K}}=2\pi -\left( 1 \right)\]
Now let us find the value of \[{{\theta }_{1}},{{\theta }_{2}},{{\theta }_{3}},.....{{\theta }_{K}}\].
\[{\theta _1} = \dfrac{{\text{Distance travelled}}}{{\text{radius}}} = \dfrac{{1cm}}{{1cm}} = 1\]
\[{{\theta }_{2}}=\dfrac{2cm}{2cm}=1\]
\[{{\theta }_{3}}=\dfrac{3cm}{3cm}=1\]
Similarly, \[{\theta _K} = \dfrac{{kcm}}{{kcm}} = 1\]. Now let us substitute all this in \[\left( 1 \right)\].
\[{\theta _1} + {\theta _2} + {\theta _3} + {\theta _4} + ...... + {\theta _K} = 2\pi \], substitute the values. Put \[\pi =3.14\]
\[ 1+1+1+1+...+1=2\pi \]
\[ 1\times n=2\pi \]
\[ n=2\pi =2\times 3.14 \]
\[ n=6.28=7 \]
Thus we got the required value as \[n=7\].
Note: Here the total angular displacement \[n\] we got is greater than \[2\pi \] radians. Thus we take 6.28cm as approximately equal to 7cm. Here the particle crosses the positive direction of the x-axis for the first time on the \[{{n}^{th}}\] circle on \[{C_n}\].
