Answer
Verified
96k+ views
Hint: Try to recall the concept of dividing current in branches having resistance \[{R_1}\] and \[{R_2}\] which are firstly connected in series and secondly in parallel. From that concept we will get the idea of adding spring constants, either springs are connected in series or connected in parallel. And then simply take the ratio of and we will get the answer.
Complete step by step answer:
First we will see the concept of adding spring constants when they are connected in
1. Series
2. Parallel
Case:1 When Springs are connected in series.
When two spring are connected in series having spring constants \[{K_1}\] and \[{K_2}\] as shown in below figure
Then the resulting Spring constant ${K_r}$ is given by the formula:
\[\dfrac{1}{{{K_{}}}} = \dfrac{1}{{{K_1}}} + \dfrac{1}{{{K_2}}}\]
Now from the question
\[{K_1} = K\]
\[{K_2} = 2K\]
Putting values in formula
\[\dfrac{1}{{{K_r}}} = \dfrac{1}{K} + \dfrac{1}{{2K}}\]
\[\dfrac{1}{{{K_r}}} = \dfrac{{2K + K}}{{2{K^2}}}\]
\[\dfrac{1}{{{K_r}}} = \dfrac{3}{{2K}}\]
\[{K_r} = \dfrac{{2K}}{3}\]
Case 2: When springs are connected in parallel
When two spring are connected in parallel having spring constants \[{K_1}\] and \[{K_2}\] as shown in below figure
Then the resulting Spring constant \[{K_p}\] is given by the formula:
\[{K_p} = {K_1} + {K_2}\]
Given values of \[{K_1}\] and \[{K_2}\]from question:
\[{K_1} = K\]
\[{K_2} = 2K\]
Putting values in above equation:
\[{K_p} = K + 2K\]
\[{K_p} = 3K\]
Now we have both spring constant
\[\dfrac{{{K_i}}}{{{K_p}}} = \dfrac{{{K_r}}}{{{K_p}}} = \dfrac{{\dfrac{{2K}}{3}}}{{3K}}\]
\[\dfrac{{{K_i}}}{{{K_p}}} = \dfrac{2}{9}\]
Hence, option C is correct.
Note: This concept can be used to combine two springs and replace it with a new one. Not only we are bound to combine two springs we can also combine many strings. If the spring constant of a spring is high then it is difficult to stretch it more while spring having less or low spring constant can be elongated much more than the previous one.
Complete step by step answer:
First we will see the concept of adding spring constants when they are connected in
1. Series
2. Parallel
Case:1 When Springs are connected in series.
When two spring are connected in series having spring constants \[{K_1}\] and \[{K_2}\] as shown in below figure
Then the resulting Spring constant ${K_r}$ is given by the formula:
\[\dfrac{1}{{{K_{}}}} = \dfrac{1}{{{K_1}}} + \dfrac{1}{{{K_2}}}\]
Now from the question
\[{K_1} = K\]
\[{K_2} = 2K\]
Putting values in formula
\[\dfrac{1}{{{K_r}}} = \dfrac{1}{K} + \dfrac{1}{{2K}}\]
\[\dfrac{1}{{{K_r}}} = \dfrac{{2K + K}}{{2{K^2}}}\]
\[\dfrac{1}{{{K_r}}} = \dfrac{3}{{2K}}\]
\[{K_r} = \dfrac{{2K}}{3}\]
Case 2: When springs are connected in parallel
When two spring are connected in parallel having spring constants \[{K_1}\] and \[{K_2}\] as shown in below figure
Then the resulting Spring constant \[{K_p}\] is given by the formula:
\[{K_p} = {K_1} + {K_2}\]
Given values of \[{K_1}\] and \[{K_2}\]from question:
\[{K_1} = K\]
\[{K_2} = 2K\]
Putting values in above equation:
\[{K_p} = K + 2K\]
\[{K_p} = 3K\]
Now we have both spring constant
\[\dfrac{{{K_i}}}{{{K_p}}} = \dfrac{{{K_r}}}{{{K_p}}} = \dfrac{{\dfrac{{2K}}{3}}}{{3K}}\]
\[\dfrac{{{K_i}}}{{{K_p}}} = \dfrac{2}{9}\]
Hence, option C is correct.
Note: This concept can be used to combine two springs and replace it with a new one. Not only we are bound to combine two springs we can also combine many strings. If the spring constant of a spring is high then it is difficult to stretch it more while spring having less or low spring constant can be elongated much more than the previous one.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main