A uniform spring of normal length l has a force constant k . It is cut into two pieces of length l1 and l2 such that l1=nl2 where n is an integer. Then the value of k1 (force constant of spring of length l1 ) is-

Question

A uniform spring of normal length l has a force constant k . It is cut into two pieces of length l1 and l2 such that l1=nl2 where n is an integer. Then the value of k1 (force constant of spring of length l1 ) is- (A) (k n/(n + 1)) (B) (k (n + 1)/n) (C) (k (n - 1)/n) (D) (k n/n - 1)

Tardigrade Admin · Accepted Answer

Using the relation kx = constant k1l1=k2l2=k (l1 + l2) k1=(k (l1 + l2)/l1) or k1=(k (n l2 + l2)/n l2) or k1=(k (n + 1)/n)

Q. A uniform spring of normal length $l$ has a force constant $k$ . It is cut into two pieces of length $l_{1}$ and $l_{2}$ such that $l_{1} = n l_{2}$ where $n$ is an integer. Then the value of $k_{1}$ (force constant of spring of length $l_{1}$ ) is-

$\frac{kn}{( n + 1 )}$

$\frac{k ( n + 1 )}{n}$

$\frac{k ( n - 1 )}{n}$

$\frac{kn}{n - 1}$

Using the relation kx = constant
$k_{1} l_{1} = k_{2} l_{2} = k (l_{1} + l_{2})$
$k_{1} = \frac{k ( l _{1} + l _{2} )}{l _{1}}$ or $k_{1} = \frac{k ( n l _{2} + l _{2} )}{n l _{2}}$
or $k_{1} = \frac{k ( n + 1 )}{n}$

Q. A uniform spring of normal length l has a force constant k . It is cut into two pieces of length l1​ and l2​ such that l1​=nl2​ where n is an integer. Then the value of k1​ (force constant of spring of length l1​ ) is-

(n+1)kn​

nk(n+1)​

nk(n−1)​

n−1kn​