The work done in an isothermal expansion is W = nRT ln(Vf/Vi). Assuming 1 mole of gas, W = 1 * 8.31 * 300 * ln(2) ≈ 600 J.

Using the formula v_rms = sqrt((3RT)/M), we can rearrange to find T = (v_rms^2 * M) / (3R). Substituting v_rms = 500 m/s and M = 0.02 kg/mol gives T = 500 K.

In an adiabatic compression, the temperature of the gas increases due to work done on it.

Using Boyle's Law, P1V1 = P2V2, we find P2 = P1 * (V1/V2) = 1 atm * (4 L / 1 L) = 4 atm.

Total duration = (4 sessions * 2 hours) + (3 breaks * 0.5 hours) = 8 hours + 1.5 hours = 9 hours.

Profit Margin = ((Revenue - Cost) / Revenue) * 100 = ((600 - 500) / 600) * 100 = 16.67%.

20% of 1,000,000 = 0.20 * 1000000 = 200,000 trees.

Total expenditure = ₹2000 * 1500 families * 12 months = ₹36,000,000 or ₹36 lakhs.

Each family will receive ₹1,20,000 / 8 = ₹15,000.

Fraction = ₹50,000 / ₹2,00,000 = 1/4.

Amount reserved for administrative costs = 10% of ₹5,00,000 = ₹50,000.

Using the inclusion-exclusion principle: Total liking at least one sport = 6 + 4 - 2 = 8. Therefore, those liking neither = 10 - 8 = 2.

Using the principle of inclusion-exclusion: Total = Football + Basketball - Both. Thus, 12 = 7 + 5 - Both, leading to Both = 0.

Let B be the number of basketball players. Total = Football + Basketball - Both. 12 = 7 + B - 5. Thus, B = 10, and only basketball = 10 - 5 = 5.

To find the number of people who play only football, we subtract those who play both from those who play football: 12 - 5 = 7.

To find the number of friends who play only football, we subtract those who play both from those who play football: 12 - 5 = 7.

To find the number of friends who like only football, we subtract those who like both from those who like football: 5 - 2 = 3.

To find the number of friends who play only one sport, we calculate: (Football only + Basketball only) = (5 - 2) + (4 - 2) = 3 + 2 = 5.

The number of friends who like only football is calculated as: Only Football = Total Football - Both = 5 - 2 = 3.

B is the youngest as A is older than B and C is younger than D.

Total cost = Average cost per person × Number of people = 20 × 5 = $100.

Let the initial number of students be x. The work done is inversely proportional to the number of days. Setting up the equation gives x = 10.

Let the initial number of students be x. The work done is constant, so x * 12 = (x + 4) * 8. Solving gives x = 8.

The total age of the original group is 20 * 15 = 300. Adding the new student gives 300 + 22 = 322. The new average age is 322 / 16 = 20.125, which rounds to 20.6.

Let the number of students be n. Total height = 150n. After one leaves, total height = 148(n - 1). Setting the equations gives 150n - 148(n - 1) = height of student.

Total height = 160 * 10 = 1600 cm. New total = 1600 - 170 = 1430 cm. New average = 1430 / 9 = 158.89 cm, approximately 159 cm.

Total score = 75 * 10 = 750. New total = 750 - 90 = 660. New average = 660 / 9 = 73.33.

The mode is 2, as it appears four times in the data set.

The mode is 3, as it appears 3 times, which is more than any other number.

The mode is 50, which appears 3 times in the scores.