Answer:
option D
Explanation:
given,
length of the pipe, L = 0.96 m
Speed of sound,v = 345 m/s
Resonating frequency when both the end is open
n is the Harmonic number
2nd overtone = 3rd harmonic
so, here n = 3
now,
f = 540 Hz
The common resonant frequency of the string and the pipe is closest to 540 Hz.
the correct answer is option D
Answer:
L = 0.109 H
Explanation:
Given that,
Number of loops in the solenoid, N = 1500
Radius of the wire, r = 4 cm = 0.04 m
Length of the rod, l = 13 cm = 0.13 m
To find,
Self inductance in the solenoid
Solution,
The expression for the self inductance of the solenoid is given by :
L = 0.109 H
So, the self inductance of the solenoid is 0.109 henries.