Question

16. There are two strings of equal
length and diameter but the
densities are in the ratio 1:2,
they are stretched by a tension T.
The ratio of frequencies will be​

Answer 1

A

2

:1

B

1:4

C

2:1

D

1:2

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Correct option is

A

2

:1

Given length of wire A = length of wire B =L and

diameter of wire A = diameter of wire B= D

Also Density ratio of wire A and wire B =1:2

and we know Density=

Volume

Mass

VolumeofwireB

MassofwireB

VolumeofwireA

MassofwireA

=

2

1

Volume of wire A = volume of wire B

Thus Mass of wire B= 2 Mass of wire A --------(1)

And ratio frequency of wire A to frequency of wire B

m

B

/L

T

m

A

/L

T

=

f

b

f

a

f

b

f

a

=

1

2