Story
NEET AITS
Story
Mock Tests
Story
Live Class
Story
JEE AITS

Questions  

A uniform rod AB of length 0 . 4 m and mass 1 .2 kg is suspended by two identical wires from a ceiling as shown in fig. (3). Where should a mass of 4.8 kg be 

suspended from the rod so that a tuning fork of frequency 256 Hz is in resonance with the first wire at Awhile its first overtone resonates with second wire at B?

Unlock the full solution & master the concept.

Get a detailed solution and exclusive access to our masterclass to ensure you never miss a concept
By Expert Faculty of Sri Chaitanya
a
5 cm
b
10 cm
c
15 cm
d
20 cm

Ready to Test Your Skills?

Check Your Performance Today with our Free Mock Tests used by Toppers!

detailed solution

Correct option is B

Here n1=12lT1m=256 and n2=12lT2m=2×256 From these equations, we have T1=4T2 Let the mass be suspended at a distance x ftom A T1+T2=4⋅8+1⋅2=6 Taking moments aboutA for equilibrium 1⋅2×0⋅2+4⋅8×x=T2×0⋅4  From eqs. (1) and (2), T2=1.2 1⋅2×0⋅2+4⋅8×x=1⋅2×0⋅4 x=5cm

ctaimg

Similar Questions

A sonometer wire vibrates with a frequency n. It is replaced by another wire of three times the diameter. If tension and other parameters remain the same, the frequency of vibration of the wire will be

Want to Improve your productivity
talk to our academic experts now !!

counselling
india
+91

whats app icon