CBSE Class 11 Physics Chapter 15: Waves NCERT Solutions

We are given the following information:

• Mass of the string, $M\; =\; 2.50\; \backslash text\{\; kg\}$
• Tension in the string, $T\; =\; 200\; \backslash text\{\; N\}$
• Length 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:

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

Substituting the values:

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

Now, we can find the time ($t$) taken for the disturbance to travel the length of the string using the formula:

$t\; =\; \backslash frac\{\backslash text\{distance\}\}\{\backslash text\{speed\}\}\; =\; \backslash frac\{l\}\{v\}$

Substituting the values:

$t\; =\; \backslash frac\{20.0\; \backslash text\{\; m\}\}\{40\; \backslash text\{\; m/s\}\}\; =\; 0.50\; \backslash text\{\; s\}$

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

We are given:

• Height of the tower, $s\; =\; 300\; \backslash text\{\; m\}$
• Initial velocity of the stone, $u\; =\; 0\; \backslash text\{\; m/s\}$ (since it is dropped)
• Acceleration due to gravity, $g\; =\; 9.8\; \backslash text\{\; m/s\}^2$
• Speed of sound in air, $v\_\{sound\}\; =\; 340\; \backslash text\{\; m/s\}$

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

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

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

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

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

Next, we calculate the time ($t\_2$) taken for the sound of the splash to travel from the water surface back up to the top of the tower. We use the formula: $time\; =\; \backslash frac\{\backslash text\{distance\}\}\{\backslash text\{speed\}\}$

$$t\_2\; =\; \backslash frac\{s\}\{v\_\{sound\}\}\; =\; \backslash frac\{300\; \backslash text\{\; m\}\}\{340\; \backslash text\{\; m/s\}\}\; \backslash approx\; 0.88\; \backslash text\{\; s\}$$

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

$t\_\{total\}\; =\; t\_1\; +\; t\_2$

$t\_\{total\}\; \backslash approx\; 7.82\; \backslash text\{\; s\}\; +\; 0.88\; \backslash text\{\; s\}\; =\; 8.70\; \backslash text\{\; s\}$

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

We are given the following information:

• Length of the steel wire, $l\; =\; 12.0\; \backslash text\{\; m\}$
• Mass of the steel wire, $m\; =\; 2.10\; \backslash text\{\; kg\}$
• Desired velocity of the transverse wave, $v\; =\; 343\; \backslash text\{\; m/s\}$ (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. We need to find the tension $T$. We can rearrange the formula to solve for $T$:

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

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

Now, substitute the given values:

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

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

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

Rounding to a reasonable number of significant figures (based on the input values), the tension should be approximately 20589 N.

Therefore, the tension in the wire should be approximately 20589 N for the transverse wave speed to equal the speed of sound in dry air at 20 °C.