CBSE Class 11 Physics Chapter 15 Waves NCERT Solutions

We are given the following information:\nMass of the string, $M\; =\; 2.50\; \backslash text\{\; kg\}$\nTension in the string, $T\; =\; 200\; \backslash text\{\; N\}$\nLength of the string, $l\; =\; 20.0\; \backslash text\{\; m\}$

First, we need to calculate the mass per unit length ($\backslash mu$) of the string:

$$\backslash mu\; =\; \backslash frac\{M\}\{l\}\; =\; \backslash frac\{2.50\; \backslash text\{\; kg\}\}\{20.0\; \backslash text\{\; m\}\}\; =\; 0.125\; \backslash text\{\; kg\; m\}^\{-1\}$$

The velocity ($v$) of a transverse wave on a stretched string is given by the formula:

$$\backslash nv\; =\; \backslash sqrt\{\backslash frac\{T\}\{\backslash mu\}\}$$

Substituting the values:

$$\backslash nv\; =\; \backslash sqrt\{\backslash frac\{200\; \backslash text\{\; N\}\}\{0.125\; \backslash text\{\; kg\; m\}^\{-1\}\}\}\; =\; \backslash sqrt\{1600\; \backslash text\{\; m\}^2\backslash text\{s\}^\{-2\}\}\; =\; 40\; \backslash text\{\; m/s\}$$

Now, we can find the time ($t$) taken for the disturbance to travel the length of the string using the formula $t\; =\; \backslash frac\{\backslash text\{distance\}\}\{\backslash text\{velocity\}\}$:

$$\backslash nt\; =\; \backslash frac\{l\}\{v\}\; =\; \backslash frac\{20.0\; \backslash text\{\; m\}\}\{40\; \backslash text\{\; m/s\}\}\; =\; 0.50\; \backslash text\{\; s\}$$

Therefore, the disturbance takes 0.50 seconds to reach the other end of the string.

We are given:\nHeight of the tower, $s\; =\; 300\; \backslash text\{\; m\}$\nInitial velocity of the stone, $u\; =\; 0\; \backslash text\{\; m/s\}$ (since it is dropped)\nAcceleration due to gravity, $a\; =\; g\; =\; 9.8\; \backslash text\{\; m/s\}^2$\nSpeed of sound in air, $v\_\{sound\}\; =\; 340\; \backslash text\{\; m/s\}$

First, let's calculate the time ($t\_1$) taken by the stone to fall to the water surface. We use the second equation of motion: $s\; =\; ut\; +\; \backslash frac\{1\}\{2\}at^2$.

$$\backslash ns\; =\; ut\_1\; +\; \backslash frac\{1\}\{2\}gt\_1^2$$

Substituting the values:

$$300\; =\; (0\; \backslash times\; t\_1)\; +\; \backslash frac\{1\}\{2\}\; \backslash times\; 9.8\; \backslash times\; t\_1^2$$

$$300\; =\; 4.9\; \backslash times\; t\_1^2$$

Solving for $t\_1$:

$$\backslash nt\_1^2\; =\; \backslash frac\{300\}\{4.9\}\; \backslash approx\; 61.22$$

$$\backslash nt\_1\; =\; \backslash sqrt\{61.22\}\; \backslash approx\; 7.82\; \backslash text\{\; s\}$$

Next, we calculate the time ($t\_2$) taken by the sound of the splash to travel from the pond's surface to the top of the tower. The distance is the height of the tower (300 m), and the speed of sound is 340 m/s.

$$\backslash nt\_2\; =\; \backslash frac\{\backslash text\{distance\}\}\{\backslash text\{speed\; of\; sound\}\}\; =\; \backslash frac\{300\; \backslash text\{\; m\}\}\{340\; \backslash text\{\; m/s\}\}\; \backslash approx\; 0.88\; \backslash text\{\; s\}$$

The total time after which the splash is heard at the top of the tower is the sum of the time of fall and the time for the sound to travel back:

$$\backslash nt\; =\; t\_1\; +\; t\_2\; \backslash approx\; 7.82\; \backslash text\{\; s\}\; +\; 0.88\; \backslash text\{\; s\}\; =\; 8.70\; \backslash text\{\; s\}$$

Therefore, the splash is heard at the top of the tower approximately 8.70 seconds after the stone is dropped.

We are given:\nLength of the steel wire, $l\; =\; 12.0\; \backslash text\{\; m\}$\nMass of the steel wire, $m\; =\; 2.10\; \backslash text\{\; kg\}$\nDesired speed of the transverse wave, $v\; =\; 343\; \backslash text\{\; m/s\}$ (equal to the speed of sound in dry air at 20 °C).

First, we calculate the mass per unit length ($\backslash mu$) of the wire:

$$\backslash mu\; =\; \backslash frac\{m\}\{l\}\; =\; \backslash frac\{2.10\; \backslash text\{\; kg\}\}\{12.0\; \backslash text\{\; m\}\}\; =\; 0.175\; \backslash text\{\; kg\; m\}^\{-1\}$$

The formula for the speed of a transverse wave on a string is $v\; =\; \backslash sqrt\{\backslash frac\{T\}\{\backslash mu\}\}$, where $T$ is the tension in the wire.

To find the tension ($T$), we can rearrange the formula:

$$\backslash nv^2\; =\; \backslash frac\{T\}\{\backslash mu\}$$

$$\backslash nT\; =\; v^2\; \backslash mu$$

Now, substitute the known values:

$$\backslash nT\; =\; (343\; \backslash text\{\; m/s\})^2\; \backslash times\; (0.175\; \backslash text\{\; kg\; m\}^\{-1\})$$

$$\backslash nT\; =\; 117649\; \backslash text\{\; m\}^2\backslash text\{s\}^\{-2\}\; \backslash times\; 0.175\; \backslash text\{\; kg\; m\}^\{-1\}$$

$$\backslash nT\; \backslash approx\; 20588.575\; \backslash text\{\; N\}$$

Rounding to a reasonable number of significant figures based on the input values, the tension should be approximately 20600 N.

Therefore, the tension in the wire should be approximately 20600 N for the speed of the transverse wave to equal the speed of sound in dry air at 20 °C.