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QJEEM17Ph1269: JEE Main 2017 Physics - Class 11 JEEM17Ph1269 Solution: Two wires W 1 and W 2 have the same radius r and respective…

Two wires W1 and W2 have the same radius r and respective densities ρ\rho 1 and ρ\rho 2 such that ρ2 = 4 ρ\rho 1 . They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in W1 to W2 is :

JEE Main 2017 (Online) 8th April Morning Slot Physics - Waves Question 105 English

Multiple Choice Options

  • A 1:01
  • B 1:02
  • C 1:03
  • D 4:01

Step-by-Step Verified Solution

Fundamental frequency of a stretched string is ,

    f = n2LTμ{n \over {2L}}\sqrt {{T \over \mu }}

Here, n = number of antinodes

μ\mu = mass per unit length.

As, O is the midpoint of two bridegs, hence length of two wires are equal.

∴   \therefore\,\,\, L1 = L2 = L

As frequency of both wires same

f1 = f2

⇒\Rightarrow    \,\,\, n12LTπr2ρ1{{{n_1}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _1}}}} = n22LTπr2ρ2{{{n_2}} \over {2L}}\sqrt {{T \over {\pi {r^2}{\rho _2}}}}

⇒\Rightarrow    \,\,\, n1n2{{{n_1}} \over {{n_2}}} = ρ1ρ2\sqrt {{{{\rho _1}} \over {{\rho _2}}}}

⇒\Rightarrow    \,\,\, n1n2{{{n_1}} \over {{n_2}}} = ρ14ρ1\sqrt {{{{\rho _1}} \over {4{\rho _1}}}} = 12{1 \over 2}

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