Verified Solution • 2025–26 Session moderate

QJEEM18Ph9731: JEE Main 2018 Physics - Class 11 JEEM18Ph9731 Solution: 5 beats / econd are heard when a tuning fork is sounded with a…

5 beats / econd are heard when a tuning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be :

Multiple Choice Options

  • A 195 Hz
  • B 150 Hz
  • C 300 Hz
  • D 251 Hz

Step-by-Step Verified Solution

Length of wire is L1 = 0.95 m; L2 = 1 m. Number of beats per second heard = 5

Let frequency of fork be f. Therefore,

v2L1−f=5⇒v2L1=5+f{v \over {2{L_1}}} - f = 5 \Rightarrow {v \over {2{L_1}}} = 5 + f ...... (1)

and f−v2L2=5⇒v2L2=f−5f - {v \over {2{L_2}}} = 5 \Rightarrow {v \over {2{L_2}}} = f - 5 .... (2)

Dividing Eq. (1) by Eq. (2), we get

v2L1v2L2=5+ff−5⇒L2L1=f+5f−5{{{v \over {2{L_1}}}} \over {{v \over {2{L_2}}}}} = {{5 + f} \over {f - 5}} \Rightarrow {{{L_2}} \over {{L_1}}} = {{f + 5} \over {f - 5}}

⇒10.95=f+5f−5⇒f−5=0.95f+4.75\Rightarrow {1 \over {0.95}} = {{f + 5} \over {f - 5}} \Rightarrow f - 5 = 0.95f + 4.75

⇒f−0.95f=5+4.75⇒0.05f=9.75⇒f=9.750.05\Rightarrow f - 0.95f = 5 + 4.75 \Rightarrow 0.05f = 9.75 \Rightarrow f = {{9.75} \over {0.05}}

⇒f=195\Rightarrow f = 195 Hz

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