A spring whose unstretched length is $\ell$ has a force constant k . The spring is cut into two pieces of unstretched lengths $\ell_{1}$ and $\ell_{2}$ where, $\ell_{1}=\mathrm{n} \ell_{2}$ and n is an integer. The ratio $k_{1} / k_{2}$ of the corresponding force constants, $k_{1}$ and $k_{2}$ will be:

(B) $\frac{1}{\mathrm{n}^{2}}$

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Elasticity & SHM

A spring whose unstretched length is

ℓ

has a force constant k . The spring is cut into two pieces of unstretched lengths

ℓ_{1}

and

ℓ_{2}

where,

ℓ_{1} = n ℓ_{2}

and n is an integer. The ratio

k_{1} / k_{2}

of the corresponding force constants,

k_{1}

and

k_{2}

will be:

Answer: (D)

A spring's force constant is inversely proportional to its length, so the product

k ℓ

is the same for the whole spring and for each cut piece. Hence

k_{1} ℓ_{1} = k_{2} ℓ_{2}

, giving

\frac{k _{1}}{k _{2}} = \frac{ℓ _{2}}{ℓ _{1}}

. With

ℓ_{1} = n ℓ_{2}

this is

\frac{k _{1}}{k _{2}} = \frac{1}{n}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option