JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 13)
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A wire of length 2L, is made by joining two
wires A and B of same length but different radii
r and 2r and made of the same material. It is
vibrating at a frequency such that the joint of
the two wires forms a node. If the number of
antinodes in wire A is p and that in B is q then
the ratio p : q is :3 : 5
4 : 9
1 : 2
1 : 4
Explanation
Let mass per unit length of wires are $$\mu $$1 and $$\mu $$2
respectively
$$ \because $$ Materials are same, so density $$\rho $$ will be same.
$$ \therefore $$ $${\mu _1} = {{\rho \pi {r^2}L} \over L} = \mu $$ and $${\mu _2} = {{\rho 4\pi {r^2}L} \over L} = 4\mu $$
Tension in both are same = T, let speed of wave in wires are V1 and V2
$${V_1} = {{{V_1}} \over {2L}} = {V \over {2L}}\,\,\& \,{V_2} = {{{V_2}} \over {2L}} = {V \over {4L}}$$
Frequency at which both resonate is L.C.M. of both frequencies (i.e : $${V \over {2L}}$$ )
Hence number of loops in wires are 1 and 2 respectively
So, ratio of number of antinodes is 1 : 2.
$$ \because $$ Materials are same, so density $$\rho $$ will be same.
$$ \therefore $$ $${\mu _1} = {{\rho \pi {r^2}L} \over L} = \mu $$ and $${\mu _2} = {{\rho 4\pi {r^2}L} \over L} = 4\mu $$
Tension in both are same = T, let speed of wave in wires are V1 and V2
$${V_1} = {{{V_1}} \over {2L}} = {V \over {2L}}\,\,\& \,{V_2} = {{{V_2}} \over {2L}} = {V \over {4L}}$$
Frequency at which both resonate is L.C.M. of both frequencies (i.e : $${V \over {2L}}$$ )
Hence number of loops in wires are 1 and 2 respectively
So, ratio of number of antinodes is 1 : 2.
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