Laws of Motion

JEE Main level, single correct, medium

110 of 210

Question 110 of 210

A spring of force constant 15 N/m15 \mathrm{~N} / \mathrm{m} is cut into two pieces. If the ratio of their length is 1:31: 3, then the force constant of smaller piece is ____\_\_\_\_ N/m\mathrm{N} / \mathrm{m}.

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Answer: (A) 60

Spring force constant is inversely proportional to length: k∝1Lk \propto \frac{1}{L}
So kL=kL = constant for a given spring material and structure.
Original spring: k=15k = 15 N/m, length =L= L.
Smaller piece has length L4\frac{L}{4} (since length ratio is 1:3, smaller is 14\frac{1}{4} of total).
For smaller piece: k1×L4=15×Lk_1 \times \frac{L}{4} = 15 \times L
k1=15×4=60k_1 = 15 \times 4 = 60 N/m

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