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  • A stone hangs from the free end of a sonometer whose vibrating length, when tuned to a tuning fork is 40 cm. When the stone hangs wholly immersed in water, the resonant length is reduced to 30cm in

A stone hangs from the free end of a sonometer whose vibrating length, when tuned to a tuning fork is 40 cm. When the stone hangs wholly immersed in water, the resonant length is reduced to 30cm in same mode of oscillation. The relative density of the stone is-

Option: 1

\frac{16}{9}


Option: 2

\frac{16}{7}


Option: 3

\frac{16}{5}


Option: 4

\frac{16}{3}


Answers (1)

best_answer

f \alpha \frac{\sqrt{T}}{l}=f    remains same Here,

\begin{gathered} \frac{\sqrt{T_1}}{l_1}=\frac{\sqrt{T_2}}{l_2} \\ \sqrt{\frac{T_1}{T_2}}=\frac{l_1}{l_2}=\frac{40}{30}=\frac{4}{3} \\ or\ \sqrt{\frac{w}{w-f}}=\frac{4}{3} \\ \frac{v d g}{v d g-v \times 1 \times g}=\frac{4}{3} \\ \sqrt{\frac{d}{d-1}}=\frac{4}{3} \\ \frac{d}{d-1}=\frac{16}{9} \\ 9 d=16 d-16 \end{gathered}

d=\frac{16}{7}

Posted by

seema garhwal

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