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Properties of Matter question

2021 · 20 Jul · Shift 2 · Q50
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  5. /2021 · 20 Jul · Shift 2 · Q50

Properties of Matter question

2021 · 20 Jul · Shift 2 · Q50

JEE MainPhysicsProperties of MatterMCQ+4 / −1
The length of a metal wire is l1, when the tension in it is T1 and is l2 when the tension is T2. The natural length of the wire is :
  1. A
    l1l2\sqrt {{l_1}{l_2}}l1​l2​​
  2. B
    l1T2−l2T1T2−T1{{{l_1}{T_2} - {l_2}{T_1}} \over {{T_2} - {T_1}}}T2​−T1​l1​T2​−l2​T1​​
  3. C
    l1T2+l2T1T2+T1{{{l_1}{T_2} + {l_2}{T_1}} \over {{T_2} + {T_1}}}T2​+T1​l1​T2​+l2​T1​​
  4. D
    l1+l22{{{l_1} + {l_2}} \over 2}2l1​+l2​​
View written solutionFree

Correct answer: B

  1. Use the relation between extension and tension

For a wire obeying Hooke’s law,

extension∝tension\text{extension} \propto \text{tension}extension∝tension

So if l0l_0l0​ is the natural length of the wire, then

l1−l0∝T1,l2−l0∝T2l_1 - l_0 \propto T_1, \qquad l_2 - l_0 \propto T_2l1​−l0​∝T1​,l2​−l0​∝T2​

Let the constant of proportionality be kkk. Then

l1−l0=kT1...(1)l_1 - l_0 = kT_1 \quad ...(1)l1​−l0​=kT1​...(1) l2−l0=kT2...(2)l_2 - l_0 = kT_2 \quad ...(2)l2​−l0​=kT2​...(2)
  1. Eliminate kkk

From (1),

k=l1−l0T1k = \frac{l_1 - l_0}{T_1}k=T1​l1​−l0​​

From (2),

k=l2−l0T2k = \frac{l_2 - l_0}{T_2}k=T2​l2​−l0​​

Equating,

l1−l0T1=l2−l0T2\frac{l_1 - l_0}{T_1} = \frac{l_2 - l_0}{T_2}T1​l1​−l0​​=T2​l2​−l0​​
  1. Cross-multiply and solve for l0l_0l0​
T2(l1−l0)=T1(l2−l0)T_2(l_1 - l_0) = T_1(l_2 - l_0)T2​(l1​−l0​)=T1​(l2​−l0​) T2l1−T2l0=T1l2−T1l0T_2l_1 - T_2l_0 = T_1l_2 - T_1l_0T2​l1​−T2​l0​=T1​l2​−T1​l0​

Bring terms containing l0l_0l0​ together:

T2l1−T1l2=l0(T2−T1)T_2l_1 - T_1l_2 = l_0(T_2 - T_1)T2​l1​−T1​l2​=l0​(T2​−T1​)

Hence,

l0=l1T2−l2T1T2−T1l_0 = \frac{l_1T_2 - l_2T_1}{T_2 - T_1}l0​=T2​−T1​l1​T2​−l2​T1​​
  1. Match with the given options

This matches Option B:

l1T2−l2T1T2−T1\boxed{\frac{l_1T_2 - l_2T_1}{T_2 - T_1}}T2​−T1​l1​T2​−l2​T1​​​

So the natural length of the wire is Option B.

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