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Properties of Matter question

2025 · 23 Jan · Shift 2 · Q54
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  5. /2025 · 23 Jan · Shift 2 · Q54

Properties of Matter question

2025 · 23 Jan · Shift 2 · Q54

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A massless spring gets elongated by amount x1x_1x1​ under a tension of 5 N . Its elongation is x2x_2x2​ under the tension of 7 N . For the elongation of (5x1−2x2)\left(5 x_1-2 x_2\right)(5x1​−2x2​), the tension in the spring will be,
  1. A
    20 N
  2. B
    39 N
  3. C
    11 N
  4. D
    15 N
View written solutionFree

Correct answer: C

  1. Use Hooke’s law

For a massless spring within elastic limit, T=kxT = kxT=kx where TTT is tension, kkk is spring constant, and xxx is elongation.

  1. Write expressions for x1x_1x1​ and x2x_2x2​

Given:

  • Under tension 5 N5\,\text{N}5N, elongation is x1x_1x1​
  • Under tension 7 N7\,\text{N}7N, elongation is x2x_2x2​

So, 5=kx1⇒x1=5k5 = kx_1 \Rightarrow x_1 = \frac{5}{k}5=kx1​⇒x1​=k5​ 7=kx2⇒x2=7k7 = kx_2 \Rightarrow x_2 = \frac{7}{k}7=kx2​⇒x2​=k7​

  1. Find the elongation 5x1−2x25x_1 - 2x_25x1​−2x2​

Substitute: 5x1−2x2=5(5k)−2(7k)5x_1 - 2x_2 = 5\left(\frac{5}{k}\right) - 2\left(\frac{7}{k}\right)5x1​−2x2​=5(k5​)−2(k7​) =25k−14k=11k= \frac{25}{k} - \frac{14}{k} = \frac{11}{k}=k25​−k14​=k11​

  1. Find the corresponding tension

Using T=kxT = kxT=kx for elongation 11k\dfrac{11}{k}k11​, T=k(11k)=11 NT = k\left(\frac{11}{k}\right) = 11\,\text{N}T=k(k11​)=11N

  1. Match with options

11 N\boxed{11\,\text{N}}11N​ So the correct option is C.

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