A cord attached to a vibrating tuning fork is divided into six segments under tension of 36 N. It will be divided into four segments, if the tension is

48N

81N

24N

16N

Solution

81N

String is vibrating in fundamental mode,it's frequency

'n' is given by ,

$\frac{1}{2 l} \sqrt{\frac{T}{m}} Therefore the frequency of the fork n will be N = 2 n = \frac{1}{2 l} \sqrt{\frac{T}{m}} If number of loops formed is P then N = \frac{P}{l} \sqrt{\frac{T}{m}} by squaring on both side N^{2} = \frac{p^{2}}{l^{2}} \times \frac{T}{m} \Rightarrow p^{2} T = N^{2} l^{2} m n, l and m are constant$

$P \sqrt{T} = constant hence, P_{1} \sqrt{T_{1}} = P_{2} \sqrt{T_{2}} 6 \times \sqrt{36} = 4 \sqrt{T_{2}} \sqrt{T_{2}} = 9 \Rightarrow T_{2} = 81 N$

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Mechanical Properties Of Fluids

A cord attached to a vibrating tuning fork is divided into six segments under tension of 36 N. It will be divided into four segments, if the tension is

48N

81N

24N

16N

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For Daily Free Study Material Join wiredfaculty

Mechanical Properties Of Fluids

A cord attached to a vibrating tuning fork is divided into six segments under tension of 36 N. It will be divided into four segments, if the tension is 48N 81N 24N 16N

Some More Questions From Mechanical Properties of Fluids Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

A cord attached to a vibrating tuning fork is divided into six segments under tension of 36 N. It will be divided into four segments, if the tension is

48N

81N

24N

16N