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(D) Given: Speed of sound in air,$v = 350 \, m/s$. Frequency of tuning fork,$f = 1750 \, Hz$.For a pipe closed at one end,the resonant frequencies are

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(D) Given: Speed of sound in air,$v = 350 \, m/s$. Frequency of tuning fork,$f = 1750 \, Hz$.For a pipe closed at one end,the resonant frequencies are given by $f_n = \frac{(2n-1)v}{4L}$,where $n = 1, 2, 3, \dots$ and $L$ is the length of the air column.Initially,$L_1 = 25 \, cm = 0.25 \, m$. Substituting the values: $1750 = \frac{(2n-1) 	imes 350}{4 	imes 0.25}$.$1750 = \frac{(2n-1) 	imes 350}{1} \implies 2n-1 = \frac{1750}{350} = 5 \implies 2n = 6 \implies n = 3$.To resonate again with the same tuning fork,the next resonant length $L_2$ corresponds to $n = 4$.$1750 = \frac{(2 	imes 4 - 1) 	imes 350}{4 	imes L_2} \implies 1750 = \frac{7 	imes 350}{4 	imes L_2}$.$L_2 = \frac{7 	imes 350}{4 	imes 1750} = \frac{2450}{7000} = 0.35 \, m = 35 \, cm$.The minimum distance the pipe should be raised is $\Delta L = L_2 - L_1 = 35 \, cm - 25 \, cm = 10 \, cm$.

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