Question

The tension in a wire is decreased by 19%.the percentage decrease in frequency will be

Answer 1

Answer is 10%

Frequency is directly proportional to √T

(f1/f2) =√(T1 / T2)

T2 = 81% T1

T1 is decreased by 19%

Formula= (f1 / f2)=√(T1) / (0.81 T1)

(f2 / f1)=(9 / 10)

f1 – (f2 / f1)=1 – (9 / 10)

(f1 – f2) / f1=(1 / 10)

Calculating percentage

frequency=(1 / 10)× 100 = 10%

Answer 2

Explanation:

f ∝ √T (f1/f2) = √(T1/T2)

given: T2 = 81% T1 as T1 is decreased by 19%

∴ (f1/f2)  = √{(T1)/(0.81 T1)}

∴ (f1/f2)   = √{(100)/81} = (10/9) (f2/f1) = (9/10)

∴ f1 – (f2/f1) = 1 – (9/10)

∴ {(f1 – f2)/f1} = (1/10)

∴ % decrease in frequency = (1/10) × 100 = 10%

peace out