A spring of force constant 15 N/m is cut into two pieces. If the ratio of their length is 1 : 3, then the force constant of smaller piece is ____ N/m.

A spring of force constant 15 N/m is cut into two pieces. If the ratio of their length is 1 : 3, then the force constant of smaller piece is ____ N/m.

Q. A spring of force constant 15 N/m is cut into two pieces. If the ratio of their length is 1 : 3, then the force constant of smaller piece is ____ N/m.

(A) 20

(B) 45

(C) 60

(D) 15

Correct Answer: 60

Explanation (Conceptual and Mathematical)

For a spring, the force constant is inversely proportional to its length.

\[ k \propto \frac{1}{L} \]

Let the original length of the spring be \(L\) and its force constant be \(k = 15 \, \text{N/m}\).

The spring is cut into two parts in the ratio \(1 : 3\). Hence, the length of the smaller piece is:

\[ L_1 = \frac{L}{1+3} = \frac{L}{4} \]

Since force constant is inversely proportional to length:

\[ k_1 = k \times \frac{L}{L_1} \]

Substitute the values:

\[ k_1 = 15 \times \frac{L}{L/4} \]

\[ k_1 = 15 \times 4 \]

\[ k_1 = 60 \, \text{N/m} \]

Thus, the force constant of the smaller piece is 60 N/m.

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