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Chapter 9: Problem 89

The intensity of sound after passing through a slab decreases by \(20 \%\). On passing through two such slabs, the intensity will decrease by (A) \(50 \%\) (B) \(40 \%\) (C) \(36 \%\) (D) \(30 \%\)

Short Answer

Expert verified
The final intensity of sound after passing through two identical slabs, each decreasing the intensity by $20\%$, is $0.64 * I_0$. The percentage decrease in intensity after passing through two slabs is $36\%$. Therefore, the correct answer is (C) 36%.

Step by step solution

01

Calculate the initial intensity after passing through the first slab

We will denote the initial intensity as \(I_0\). After passing through the first slab, the intensity is decreased by 20%, so we will have: \[I_1 = I_0 - 0.20 * I_0\]
02

Calculate the final intensity after passing through the second slab

After passing through the second slab, the intensity is decreased by another 20%. Since the intensity after the first slab is \(I_1\), we can calculate the final intensity after the second slab as: \[I_2 = I_1 - 0.20 * I_1\]
03

Substitute the value of \(I_1\) from Step 1 into the equation from Step 2

In order to solve for the final intensity, we can replace the \(I_1\) term in the equation in Step 2 with the result from Step 1: \[I_2 = (I_0 - 0.20 * I_0) - 0.20 * (I_0 - 0.20 * I_0)\]
04

Simplify the equation and solve for the percentage decrease

Now we will simplify the equation to find the final intensity and the percentage decrease in intensity: \[I_2 = 0.80 * I_0 - 0.20 * 0.80 * I_0\] \[I_2 = 0.80 * I_0 - 0.16 * I_0\] \[I_2 = 0.64 * I_0\] The initial intensity was \(I_0\) and the final intensity after passing through two slabs is \(0.64 * I_0\). The percentage decrease in intensity can now be calculated: \[\% \text{ decrease} = \frac{I_0 - 0.64 * I_0}{I_0} * 100\] \[\% \text{ decrease} = (1 - 0.64) * 100\] \[\% \text{ decrease} = 36\%\] Therefore, the correct answer is (C) 36%.

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A metal wire of linear mass density of \(9.8 \mathrm{~g} / \mathrm{m}\) is stretched with a tension of \(10 \mathrm{~kg}-\mathrm{wt}\) between two rigid supports 1 meter apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency \(n .\) The frequency \(\mathrm{n}\) of the alternating source is (A) \(50 \mathrm{~Hz}\) (B) \(100 \mathrm{~Hz}\) (C) \(200 \mathrm{~Hz}\) (D) \(25 \mathrm{~Hz}\)

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