Test Index

Oscillations Part 2

Section: Physics

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Question : 11 of 81

Marks: +1, -0

A spring of force constant

15 \mathrm{N/m}

is cut into two pieces. If the ratio of their length is

1:3

, then the force constant of smaller piece is

\_\_\_\_

\mathrm{N/m}

[24 jan 2026 Shift 1]

20
45
60
15

Solution:

Concept:

The spring constant

k

is inversely proportional to its length

L

.

Explanation:

Original spring constant

k_o = 15\ \mathrm{N/m}

and original length

L

.

The ratio of lengths after cutting is

1:3

.

Let the smaller piece length be

x

, then larger is

3x

, so total

x + 3x = L

, giving

x = L/4

.

Thus, smaller piece length

L_1 = L/4

.

Using

k \propto 1/L

, we have

k_o L = k_1 L_1

.

Substitute:

15 \times L = k_1 \times (L/4)

.

Cancel

L

and solve:

k_1 = 15 \times 4 = 60\ \mathrm{N/m}

.

Answer:

60\ \mathrm{N/m}

— Option (C).

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