Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A spring of force constant $k$ is cut into lengths of ratio $1:2:3$ . They are connected in series and the new force constant is $k'$ . If they are connected in parallel and force constant is $k''$ , then $k':k''$ is:

1:6

1:9

1:11

6:11

Step-by-Step Solution

Spring Constant and Length: The force constant $k$ of a spring is inversely proportional to its length $l$ ( $k \propto \frac{1}{l}$ ) [NCERT Class 11, Physics Part I, Sec 6.9].
Cutting the Spring: The spring of length $L$ is cut into ratio $1:2:3$ . The lengths of the segments are:

$l_1 = \frac{1}{6}L \implies k_1 = 6k$
$l_2 = \frac{2}{6}L = \frac{1}{3}L \implies k_2 = 3k$
$l_3 = \frac{3}{6}L = \frac{1}{2}L \implies k_3 = 2k$

Series Combination ( $k'$ ): When connected in series, the effective spring constant is given by: $\frac{1}{k'} = \frac{1}{k_1} + \frac{1}{k_2} + \frac{1}{k_3}$ $\frac{1}{k'} = \frac{1}{6k} + \frac{1}{3k} + \frac{1}{2k} = \frac{1+2+3}{6k} = \frac{6}{6k} = \frac{1}{k}$ $k' = k$ (This is expected as reconnecting parts in series reconstructs the original spring).
Parallel Combination ( $k''$ ): When connected in parallel, the effective spring constant is the sum of individual constants: $k'' = k_1 + k_2 + k_3$ $k'' = 6k + 3k + 2k = 11k$
Ratio: $\frac{k'}{k''} = \frac{k}{11k} = 1:11$

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started