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Standard XII

Physics

Work Energy Theorem

Spring consta...

Question

Spring constants of two ideal springs A and B are $k_{1}$ and $k_{2}$ respectively such that $k_{1} > k_{2}$ . For same applied force, extension is more in:

Spring A

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Spring B

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Equal extension

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Depends on the mass of the spring

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Solution

The correct option is A Spring B
According to Hooke's Law for spring :
$F = - k x$
Now, two springs with spring constant $k_{1}$ and $k_{2}$ and same applied force. So,
$F = - k_{1} x_{1} = - k_{2} x_{2}$
$x_{1} = \frac{k_{2}}{k_{1}} x_{2}$
But $k_{1} > k_{2}$ ,
Therefore, $x_{1} < x_{2}$

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Spring constants of two ideal springs A and B are k1 and k2 respectively such that k1>k2. For same applied force, extension is more in:

The correct option is A Spring BAccording to Hooke's Law for spring : F=−kxNow, two springs with spring constant k1 and k2 and same applied force. So, F=−k1x1=−k2x2x1=k2k1x2But k1>k2, Therefore, x1<x2

Spring constants of two ideal springs A and B are $k_{1}$ and $k_{2}$ respectively such that $k_{1} > k_{2}$ . For same applied force, extension is more in:

The correct option is A Spring B
According to Hooke's Law for spring :
$F = - k x$
Now, two springs with spring constant $k_{1}$ and $k_{2}$ and same applied force. So,
$F = - k_{1} x_{1} = - k_{2} x_{2}$
$x_{1} = \frac{k_{2}}{k_{1}} x_{2}$
But $k_{1} > k_{2}$ ,
Therefore, $x_{1} < x_{2}$