Question

19. The original length of a wire is 153.7 40.6 cm. It is stretched to 155.3 0.2 cm. Calculate the
elongation in the wire with error limits.​

Answer 1

frequency of fundamental note is given by

\bold{\nu=\frac{v}{2l}\sqrt{\frac{T}{\mu}}}

Here, ν is the frequency of fundamental note , v is speed of wave , l is the length of string , T is tension in the string and μ is mass density { mass per unit length}

in first case ,

\bold{256Hz=\frac{v}{2l}\sqrt{\frac{T}{\mu}}}---(1)

in second case,

\bold{320Hz=\frac{v}{2(l-10)}\sqrt{\frac{T}{\mu}}} ----(2)

From equations (1) and (2)

256/320 = 2(l-10)/2l

⇒ 4/5 = ( l - 10)/l

⇒ 4l = 5( l - 10)

⇒ 4l = 5l - 50 ⇒ l = 50 cm

Hence, length of string is 50 cm

Answer 2

Answer:

the answer is ( 1.6 +- 0.8 ) cm

Explanation:

original length  = x = ( 153.7 +- 0.6 ) cm

new length  = y = ( 155.3 +- 0.2 ) cm

elongation  = y - x

                   = ( 155.3 - 153.7 ) +- ( 0.6 + 0.2 )

                  = ( 1.6 +- 0.8 ) cm

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