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Standard XII

Physics

Springs

The elongatio...

Question

The elongation of a spring of length $L$ and of negligible mass due to a force is $x$ . The spring is cut into two pieces of lengths in ratio $1 : n .$ The ratio of the respective spring constants is

n : 1

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1 : n

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n^{2}

: 1

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1 :

n^{2}

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Solution

The correct option is A $n : 1$
let the force constant be $k$
$k = \frac{E A}{l}$
$l_{1} : l_{2} = 1 : n; l_{1} + l_{2} = l$
$(1 + n) l_{1} = l$

$l_{1} = \frac{l}{1 + n}$
$l_{2} = \frac{n l}{1 + n}$
$k$ is inversely proportional to $l$
Thus ratio is $n / 1$

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The elongation of a spring of length L and of negligible mass due to a force is x. The spring is cut into two pieces of lengths in ratio 1:n. The ratio of the respective spring constants is

The correct option is A n:1let the force constant be kk=EAll1:l2=1:n;l1+l2=l(1+n)l1=l l1=l1+n l2=nl1+nk is inversely proportional to lThus ratio is n/1

The elongation of a spring of length $L$ and of negligible mass due to a force is $x$ . The spring is cut into two pieces of lengths in ratio $1 : n .$ The ratio of the respective spring constants is

The correct option is A $n : 1$
let the force constant be $k$
$k = \frac{E A}{l}$
$l_{1} : l_{2} = 1 : n; l_{1} + l_{2} = l$
$(1 + n) l_{1} = l$

$l_{1} = \frac{l}{1 + n}$
$l_{2} = \frac{n l}{1 + n}$
$k$ is inversely proportional to $l$
Thus ratio is $n / 1$