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Properties of Matter question

2023 · 11 Apr · Shift 1 · Q73
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Properties of Matter question

2023 · 11 Apr · Shift 1 · Q73

JEE MainPhysicsProperties of MatterNumerical+4 / −1
The length of a wire becomes l1l_{1}l1​ and l2l_{2}l2​ when 100 N100 \mathrm{~N}100 N and 120 N120 \mathrm{~N}120 N tensions are applied respectively. If 10 l2=11 l110 ~l_{2}=11~ l_{1}10 l2​=11 l1​, the natural length of wire will be 1x l1\frac{1}{x} ~l_{1}x1​ l1​. Here the value of xxx is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 2

  1. Use Hooke’s law for extension

For a wire within elastic limit, l=L+Δll = L + \Delta ll=L+Δl where LLL is the natural length and extension Δl∝T\Delta l \propto TΔl∝T (tension).

So we can write l=L+kTl = L + kTl=L+kT for some constant kkk.

Thus,

  • for 100 N100\,\text{N}100N tension: l1=L+100kl_1 = L + 100kl1​=L+100k
  • for 120 N120\,\text{N}120N tension: l2=L+120kl_2 = L + 120kl2​=L+120k
  1. Use the given relation

Given: 10l2=11l110l_2 = 11l_110l2​=11l1​ Substitute the expressions: 10(L+120k)=11(L+100k)10(L+120k)=11(L+100k)10(L+120k)=11(L+100k)

Expand: 10L+1200k=11L+1100k10L+1200k=11L+1100k10L+1200k=11L+1100k

Rearrange: 1200k−1100k=11L−10L1200k-1100k=11L-10L1200k−1100k=11L−10L 100k=L100k=L100k=L

  1. Find l1l_1l1​ in terms of LLL

From l1=L+100kl_1=L+100kl1​=L+100k और since L=100kL=100kL=100k, l1=L+L=2Ll_1=L+L=2Ll1​=L+L=2L

So, L=l12L=\frac{l_1}{2}L=2l1​​

  1. Match with the required form

Natural length is given as 1xl1\frac{1}{x}l_1x1​l1​ Comparing with L=l12L=\frac{l_1}{2}L=2l1​​ we get x=2x=2x=2

Final Answer

2\boxed{2}2​

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