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Properties of Matter question

2022 · 29 Jun · Shift 1 · Q46
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  5. /2022 · 29 Jun · Shift 1 · Q46

Properties of Matter question

2022 · 29 Jun · Shift 1 · Q46

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A wire of length L is hanging from a fixed support. The length changes to L1 and L2 when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to :
  1. A
    L1L2\sqrt {{L_1}{L_2}}L1​L2​​
  2. B
    L1+L22{{{L_1} + {L_2}} \over 2}2L1​+L2​​
  3. C
    2L1−L22{L_1} - {L_2}2L1​−L2​
  4. D
    3L1−2L23{L_1} - 2{L_2}3L1​−2L2​
View written solutionFree

Correct answer: C

  1. Use the relation for extension of a wire

For a wire obeying Hooke’s law,

ΔL∝F\Delta L \propto FΔL∝F

so the extension is directly proportional to the suspended load.

Let the original length of the wire be LLL.

Let the extension produced by a 1 kg1\,\text{kg}1kg mass be xxx.

Then, when a 1 kg1\,\text{kg}1kg mass is suspended:

L1=L+xL_1 = L + xL1​=L+x
  1. Case of 2 kg2\,\text{kg}2kg mass

Since force doubles, extension also doubles:

extension=2x\text{extension} = 2xextension=2x

So,

L2=L+2xL_2 = L + 2xL2​=L+2x
  1. Eliminate xxx to find LLL

From

L1=L+xL_1 = L + xL1​=L+x

we get

x=L1−Lx = L_1 - Lx=L1​−L

Substitute into

L2=L+2xL_2 = L + 2xL2​=L+2x

:

L2=L+2(L1−L)L_2 = L + 2(L_1 - L)L2​=L+2(L1​−L) L2=L+2L1−2LL_2 = L + 2L_1 - 2LL2​=L+2L1​−2L L2=2L1−LL_2 = 2L_1 - LL2​=2L1​−L

Hence,

L=2L1−L2L = 2L_1 - L_2L=2L1​−L2​
  1. Match with the options
L=2L1−L2L = 2L_1 - L_2L=2L1​−L2​

This corresponds to Option C.

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