Get Answers to all your Questions

header-bg qa

The total length of a sonometer wire  between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio 6 : 3 : 2. The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m. What is the minimum common frequency with which three parts can vibrate ?

 

  • Option 1)

    1100 Hz

  • Option 2)

    1000 Hz

  • Option 3)

    166 Hz

  • Option 4)

    100 Hz

 

Answers (1)

best_answer

Total length of sonometer wire l=110cm. = 1.1 m

length of wire is in ration, 6:3:2 v.e 60 cm., 30 cm., 20 cm

T=400N

w=0.01 kg

v=?

v=\frac{l}{2l}\sqrt{\frac{T}{w}}=\frac{1000}{11}Hz

v_{1}=\frac{1000}{6}Hz

v_{2}=\frac{1000}{3}Hz

v_{3}=\frac{1000}{2}Hz

hence common frequency = 1000Hz


Option 1)

1100 Hz

Incorrect Option

Option 2)

1000 Hz

Correct Option

Option 3)

166 Hz

Incorrect Option

Option 4)

100 Hz

Incorrect Option

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE