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JEE Main 201912 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Evening Shift
Subject: Physics
Chapter: Laws of Motion

JEE Main 2019 · 12 April · Evening Shift · Physics · Laws of Motion

A spring whose unstrentches length is l has a force constant k ....

A spring whose unstrentches length is

l

has a force constant

k .

The spring is cut into two pieces of unstretches 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:

Options

A $\dfrac{1}{{n}^{2}}$
B ${n}^{2}$
C $n$
D $\dfrac{1}{n}$

Correct answer

D. $\dfrac{1}{n}$

Step-by-step solution

Spring constants of the spring will be inversely proportional to length.

\dfrac{{k}_{1}}{{k}_{2}} = \dfrac{{l}_{2}}{{l}_{1}} = \dfrac{1}{n}

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