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Transverse waves of same frequency are generated in two steel wires A and B. The diameter of A is twice of B and the tension in A is half that in B. The ratio of velocities of wave in A and B is
- a)1 : 3√2
- b)1 : 2√2
- c)1 : 2
- d)√2 : 1
Correct answer is option 'B'. Can you explain this answer?
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Transverse waves of same frequency are generated in two steel wires A ...
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B)3 : 1
c)1 : 2
d)2 : 1
Solution:
The velocity of a wave in a string is given by the formula:
v = √(T/μ)
where T is the tension in the string and μ is the linear density (mass per unit length) of the string.
Let the linear density of wire A be μA and that of wire B be μB. Let the tension in wire B be TB.
Given that the diameter of A is twice of B, we have:
diameter of A = 2 × diameter of B
or, radius of A = 2 × radius of B
or, cross-sectional area of A = π(2rB)^2 = 4πrB^2
and cross-sectional area of B = πrB^2
Since the volume of the wires is the same, we have:
length of A × cross-sectional area of A = length of B × cross-sectional area of B
or, length of A × 4πrB^2 = length of B × πrB^2
or, length of A = length of B × 4
So, the linear density of wire A is:
μA = mass of A / length of A
= (πrA^2 × length of A × density of steel) / length of A
= πrA^2 × density of steel
= 4πrB^2 × density of steel
= 4μB
Given that the tension in A is half that in B, we have:
TA = TB / 2
The velocity of the wave in wire A is:
vA = √(TA / μA) = √((TB/2) / (4μB)) = √(TB / 8μB)
The velocity of the wave in wire B is:
vB = √(TB / μB)
So, the ratio of velocities is:
vA / vB = √(TB / 8μB) / √(TB / μB)
= √(μB / 8μB)
= 1 / √8
= √2 / 4
Therefore, the answer is (d) 2 : 1.
c)1 : 2
d)2 : 1
Solution:
The velocity of a wave in a string is given by the formula:
v = √(T/μ)
where T is the tension in the string and μ is the linear density (mass per unit length) of the string.
Let the linear density of wire A be μA and that of wire B be μB. Let the tension in wire B be TB.
Given that the diameter of A is twice of B, we have:
diameter of A = 2 × diameter of B
or, radius of A = 2 × radius of B
or, cross-sectional area of A = π(2rB)^2 = 4πrB^2
and cross-sectional area of B = πrB^2
Since the volume of the wires is the same, we have:
length of A × cross-sectional area of A = length of B × cross-sectional area of B
or, length of A × 4πrB^2 = length of B × πrB^2
or, length of A = length of B × 4
So, the linear density of wire A is:
μA = mass of A / length of A
= (πrA^2 × length of A × density of steel) / length of A
= πrA^2 × density of steel
= 4πrB^2 × density of steel
= 4μB
Given that the tension in A is half that in B, we have:
TA = TB / 2
The velocity of the wave in wire A is:
vA = √(TA / μA) = √((TB/2) / (4μB)) = √(TB / 8μB)
The velocity of the wave in wire B is:
vB = √(TB / μB)
So, the ratio of velocities is:
vA / vB = √(TB / 8μB) / √(TB / μB)
= √(μB / 8μB)
= 1 / √8
= √2 / 4
Therefore, the answer is (d) 2 : 1.
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Transverse waves of same frequency are generated in two steel wires A and B. The diameter of A is twice of B and the tension in A is half that in B. The ratio of velocities of wave in A and B isa)1 : 3√2b)1 : 2√2c)1 : 2d)√2 : 1Correct answer is option 'B'. Can you explain this answer?
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