Grade XI Physics (Application Based Questions) #GrowWithGreen
Questions 1. A child is running a fever of 102 F. He has taken the prescribed pill and his mother is using a wet cloth on his forehead in order to absorb heat from his body. After half an hour, the evaporation of sweat and the heat absorbed by the wet cloth cause his body to attain its normal temperature, i.e., 98 F. The child s mass is 36 kg. The average rate of evaporation caused by the pill is 2 g/min. Assume that the latent heat of evaporation of water is 580 Cal/g and that the specific heat of the human body is the same as the specific heat of water. What is the amount of heat absorbed during the process of putting the wet cloth on the child s forehead? 2. Two cylinders containing two ideal gases at temperatures 27 C and 17 C are allowed to mix together. The number of molecules in the gases is 6 10 24 and 2.3 10 23 respectively. The respective molecular masses are 28 g and 32 g. If there is no loss of energy, what will be the temperature of the mixture? 3. A lead bullet, having an initial temperature of 30 C, is fired with an initial velocity of 450 m/s. It penetrates a wall and melts. Assume that only 40% of the kinetic energy goes in heating the bullet and the rest dissipates in the wall. Latent heat of fusion of lead = 2.5 10 4 J/kg Specific heat of lead = 125 J/kg/K What is the melting point of lead? 4. The time period of a simple pendulum placed in an elevator increases when it moves down with a uniform acceleration. Justify the statement. 5. Tuning forks I and II are vibrating in air. It is observed that the difference between the wavelengths produced by the forks is 0.8 m. When these tuning forks are placed close to each other, 4 beats per second are produced. The speed of sound in air is 340 m/s. What are the respective wavelengths produced by tuning forks I and II? 2 marks 2 marks 3 marks 2 marks 3 marks
Solutions 1. Mass of the child, m = 36 kg Specific heat of the human body = Specific heat of water, S w = 1000 Cal/kg/ C Change in the temperature of the child s body is calculated as: Hence, the heat lost by the child body, This heat is lost in two ways by evaporation (Δ Q 1 ) and by absorption (Δ Q 2 ). Δ Q = Δ Q 1 + Δ Q 2 Mass of water that evaporates in 30 minutes = 2 30 = 60 g Latent heat of evaporation of water, L v = 580 Cal/g Hence, heat lost by the body, Δ Q 1 = ml v = 60 580 = 34800 Cal Heat absorbed during the wetting process, Δ Q 2 = Δ Q Δ Q 1 = 80000 34800 = 45200 Cal 2. Kinetic energy is given as: k = Boltzmann constant = 1.38 10 23 J/K n = number of molecules T = temperature of the system Before mixing: Kinetic energy of one gas,
Kinetic energy of another gas, Total Kinetic energy, E = E 1 + E 2 After mixing: T = temperature of the mixture Total kinetic energy of the mixture, Since there is no loss of energy, E = Given, n 1 = 6 10 24 T 1 = 27 C = 300 K n 2 = 2.3 10 23 T 2 = 17 C = 290 K 3. The heat gained by the bullet, which raises its temperature from 30 C to T C, is given
as: Q 1 = mcdt m = Mass of the bullet c = Specific heat of lead = 125 J/kg/K T = Melting point of lead dt = Rise in the temperature of the bullet = T 30 Q 1 = m 125 ( T 30) Cal ( i ) The heat required for melting the bullet is given as: Q 2 = ml L = Latent heat of fusion of lead = 2.5 10 4 J/kg Q 2 = m 2.5 10 4 Cal ( ii ) Kinetic energy of the bullet is given by the relation: u = Initial velocity of the bullet = 450 m/s The total amount of developed heat is given as: At thermal equilibrium, we can write: Hence, melting point of lead is 154 C.
4. Time period T,of a simple pendulum of length l, is given by the relation: As per Newton s second law of motion, the effective acceleration due to gravity acting on a pendulum placed in an elevator decreases by ( g a ) when the elevator moves downward with uniform acceleration a. Since time period is inversely proportional to the square root of effective acceleration, it increases when the elevator moves downward with uniform acceleration. This is given by the relation: It can be inferred from the two equations that. 5. Let the wavelengths produced by fork I and fork II be λ 1 and λ 2 respectively. It is given that: λ 2 λ 1 = 0.8 ( i ) Let the frequencies of the sounds produced by forks I and II be ν 1 and ν 2 respectively. Number of beats produced/sec by the forks = 4 Beats are produced when the frequency difference between the two sources is very small, given as: ν 1 ν 2 = m m = Number of beats/sec Speed of sound in air, v = 340 m/s [Using equation ( i )]
λ 1 = 8.66 m or 7.86 m A negative value of wavelength is unphysical. λ 1 = 7.86 m Substituting this value in equation ( i ), we get: λ 2 = 0.8 + 7.86 = 8.66 m.