A marine survey ship sends a sound wave straight to the seabed. It receives an echo 1.5 s later. The speed of sound in sea water is 150p ms⁻¹. Find the depth of the sea at this position.

Here are the solutions to the numerical problems:

Q:5 A marine survey ship sends a sound wave straight to the seabed. It receives an echo 1.5 s later. The speed of sound in sea water is 150p ms⁻¹. Find the depth of the sea at this position.

Step 1: Identify the given values and clarify the speed of sound.
Given:
Time for echo, $t = 1.5 \text{ s}$
Speed of sound, $v = 150 \text{p ms}^{-1}$. The unit "p" (pico) means $10^{-12}$. A speed of $150 \times 10^{-12} \text{ ms}^{-1}$ is extremely slow and unrealistic for sound in seawater. Sound in seawater is typically around $1500 \text{ ms}^{-1}$. Assuming "150p" is a typo and it should be $1500 \text{ ms}^{-1}$ (or possibly $150 \text{ ms}^{-1}$ if it's a very specific context, but $1500 \text{ ms}^{-1}$ is standard for seawater). We will proceed with $v = 1500 \text{ ms}^{-1}$.

Step 2: Calculate the depth of the sea.
The sound travels to the seabed and back, so the total distance covered is $2 \times \text{depth}$.
The formula for distance is $d = v \times t$.
So, $2 \times \text{depth} = v \times t$.

$$\text{depth} = \frac{v \times t}{2}$$

Substitute the values:

$$\text{depth} = \frac{1500 \text{ ms}^{-1} \times 1.5 \text{ s}}{2}$$ $$\text{depth} = \frac{2250 \text{ m}}{2}$$ $$\text{depth} = 1125 \text{ m}$$

The depth of the sea at this position is $\boxed{1125 \text{ m}}$.

Q:6 An electric bulb is marked with 220V, 100W. Find the resistance of the filament of the bulb. If the bulb is used 5 hours daily, find the energy in Kilowatt hour consumed by the bulb in one month ( 30 Days)

Step 1: Find the resistance of the filament.
Given:
Voltage, $V = 220 \text{ V}$
Power, $P = 100 \text{ W}$
The formula relating power, voltage, and resistance is $P = \frac{V^2}{R}$.
Rearrange to solve for $R$: $R = \frac{V^2}{P}$.

Substitute the values:

$$R = \frac{(220 \text{ V})^2}{100 \text{ W}}$$ $$R = \frac{48400 \text{ V}^2}{100 \text{ W}}$$ $$R = 484 \ \Omega$$

Step 2: Find the energy consumed in one month (30 days).
Given:
Power, $P = 100 \text{ W}$
Daily usage, $t_{daily} = 5 \text{ hours}$
Number of days, $N = 30 \text{ days}$
First, convert power to kilowatts: $P_{kW} = 100 \text{ W} = 0.1 \text{ kW}$.
Next, calculate the total time of usage in hours for one month:

$$T_{total} = t_{daily} \times N = 5 \text{ hours/day} \times 30 \text{ days} = 150 \text{ hours}$$

The formula for energy consumed is $E = P_{kW} \times T_{total}$.

Substitute the values:

$$E = 0.1 \text{ kW} \times 150 \text{ hours}$$ $$E = 15 \text{ kWh}$$

The resistance of the filament is $\boxed{484 \ \Omega}$.
The energy consumed by the bulb in one month is $\boxed{15 \text{ kWh}}$.

Q:7 A sound waves has frequency of 2 kHz and wavelength 35 cm. How long will it take to travel 1.5 km?

Step 1: Convert given values to standard units.
Given:
Frequency, $f = 2 \text{ kHz} = 2 \times 1000 \text{ Hz} = 2000 \text{ Hz}$
Wavelength, $\lambda = 35 \text{ cm} = 0.35 \text{ m}$
Distance, $d = 1.5 \text{ km} = 1.5 \times 1000 \text{ m} = 1500 \text{ m}$

Step 2: Calculate the speed of the sound wave.
The relationship between speed ($v$), frequency ($f$), and wavelength ($\lambda$) is $v = f \lambda$.

Substitute the values:

$$v = 2000 \text{ Hz} \times 0.35 \text{ m}$$ $$v = 700 \text{ ms}^{-1}$$

Step 3: Calculate the time taken to travel the given distance.
The formula for time ($t$) is $t = \frac{d}{v}$.

Substitute the values:

$$t = \frac{1500 \text{ m}}{700 \text{ ms}^{-1}}$$ $$t \approx 2.14 \text{ s}$$

It will take approximately $\boxed{2.14 \text{ s}}$ to travel 1.5 km.

Q:8 A current of 3mA is flowing through a wire for 1 minutes. What is the charge flowing through the wire?

Step 1: Convert given values to standard units.
Given:
Current, $I = 3 \text{ mA} = 3 \times 10^{-3} \text{ A}$
Time, $t = 1 \text{ minute} = 60 \text{ s}$

Step 2: Calculate the charge flowing through the wire.
The formula for charge ($Q$) is $Q = I \times t$.

Substitute the values:

$$Q = (3 \times 10^{-3} \text{ A}) \times (60 \text{ s})$$ $$Q = 180 \times 10^{-3} \text{ C}$$ $$Q = 0.18 \text{ C}$$

The charge flowing through the wire is $\boxed{0.18 \text{ C}}$.

Q:9 If at Anarkali Bazar Lahore, the intensity level of sound is 80 dB, what will be the intensity of sound there?

Step 1: Identify the given values and the formula for intensity.
Given:
Sound intensity level, $\beta = 80 \text{ dB}$
Reference intensity, $I_0 = 1.0 \times 10^{-12} \text{ Wm}^{-2}$ (standard value)
The formula for sound intensity level is $\beta = 10 \log_{10} \left(\frac{I}{I_0}\right)$.
To find $I$, rearrange the formula:

$$\frac{\beta}{10} = \log_{10} \left(\frac{I}{I_0}\right)$$ $$10^{\beta/10} = \frac{I}{I_0}$$ $$$$