The speed of a wave on a string is 150 `"ms"^(-1)` when the tension is 120 N. The percentage increase in the tension in orderr to raise the wave speed
The speed of a wave on a string is 150 `"ms"^(-1)` when the
tension is 120 N. The percentage increase in the
tension in orderr to raise the wave speed by `20%` is
A. `44%`
B. `40%`
C. `20%`
D. `10%`
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1 Answers
Correct Answer - a
Speed of wave on a string
`v=sqrt(T/m)rArr v prop sqrt(T)`
`T_(1)/T_(2)=(v_(1)/v_(2))^(2)rArr (T_(2)-T_(1))/T_(1)=(v_(2)^(2)-v_(1)^(2))/v_(1)^(2)` ...(i)
Given, `T_(1) =120 m and v_(1)=150" ms"^(-1)`
`v_(2)=v_(1) + 20/100 v_(1)=120/100 v_(1)=6/5v_(1)=6/5 xx150`
`= 180" ms"^(-1)`
Substituting the values in Eq. (i) we get
`(T_(2)-T_(1))/T_(1)=((180)^(2)-(150)^(2))/((150)^(2))= (30xx330)/(150xx150)=0.44`
Percent increase in tension
`(T_(2)-T_(1))/T_(1)=xx100`
`=0.44 xx 100 = 44%`
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