IIT JEE Physics Exam Quiz!

Explanation
The change in internal energy of a gas can be calculated using the equation ΔU = nCvΔT, where ΔU is the change in internal energy, n is the number of moles of gas, Cv is the molar specific heat at constant volume, and ΔT is the change in temperature. In this case, the volume changes from V to 2V under constant pressure, so the change in temperature is zero. Therefore, ΔU = nCv(0) = 0. Since the change in internal energy is zero, the correct answer is pV I (γ-1), as it represents the product of pressure, volume, and the ratio of specific heat at constant pressure to that at constant volume.

Explanation
In the steady-state, heat flows from the region with higher temperature to the region with lower temperature. The rate of heat transfer through each wall depends on its thermal conductivity and thickness. The temperature at the outer wall can be found using the formula (k1T1d2+ k2T2d1 ) / ( k1d2+ k2d1), which takes into account the thermal conductivities and thicknesses of both walls, as well as the temperatures at the outer surfaces. This formula calculates the weighted average of the temperatures at the outer surfaces, where the weights are determined by the thermal conductivities and thicknesses of the walls.

Explanation
The given answer is correct because:
1. The equation E=(3/4) * (mv^2/qa) represents the electric field E in terms of the charge q, velocity v, and magnetic field B. This equation is applicable in this scenario.
2. The rate of work done by the electric field at point P is given by (3/4) * (mv^3/a), which is also correct.
3. The rate of work done by both the electric and magnetic fields at point Q is zero because the velocity is perpendicular to both the electric and magnetic fields, resulting in no work being done.

Explanation
The ratio l1/(l1+l2) represents the proportion of the initial length of the aluminum rod to the total length of the combined rod. The fact that the length of each rod increases by the same amount when their temperatures are raised by tC indicates that the expansion of the aluminum and steel rods is equal. Therefore, the expansion coefficient of the aluminum rod (αa) is equal to the expansion coefficient of the steel rod (αs). The correct answer, αs/( αa + αs), represents the ratio of the expansion coefficient of the steel rod to the sum of the expansion coefficients of both rods.

Explanation
In order for a single process to be both isothermal and isobaric, a change of phase is essential. This means that the substance undergoing the process must transition from one phase to another, such as from a solid to a liquid or from a liquid to a gas. This change in phase allows for the temperature to remain constant (isothermal) and the pressure to remain constant (isobaric) throughout the process. Without a change of phase, it would not be possible to maintain both isothermal and isobaric conditions simultaneously.

Explanation
In the given question, the displacement of the wire in the first experiment is y1 = A sin(πx/L) sinωt, and in the second experiment, it is y2 = A sin (2πx/L) sin2ωt. The energy of the wire is directly proportional to the square of the amplitude of the displacement. Comparing the two equations, we can see that the amplitude of y2 is twice that of y1, and the square of the amplitude is four times that of y1. Therefore, the energy in the second experiment (E2) is four times the energy in the first experiment (E1), leading to the answer E2=4E1.

Explanation
The maximum speed of the block can be determined using the principle of conservation of mechanical energy. Initially, the block has potential energy stored in the spring, given by 0.5*k*x^2, where k is the spring constant and x is the elongation of the spring. The initial kinetic energy of the block is given by 0.5*m*v^2, where m is the mass of the block and v is its initial velocity. As the block oscillates, the potential energy is converted into kinetic energy and vice versa. At the maximum speed, all the potential energy is converted into kinetic energy. Therefore, equating the potential energy to the kinetic energy, we have 0.5*k*x^2 = 0.5*m*v^2. Solving for v, we get v = sqrt((k*x^2)/m) = sqrt((200*0.15^2)/0.25) = 5.2 m/s.

Explanation
When a thin film is introduced in the path of the upper beam in a double-slit experiment, it causes a phase shift in the light waves passing through it. This phase shift leads to a change in the interference pattern observed on the screen. In this case, the film has a thickness of 2 μm, which is comparable to the wavelength of the light (500 nm). As a result, the central maximum will shift upward by nearly two fringes. This is because the film introduces a path difference between the upper and lower beams, causing a shift in the interference pattern.

Explanation
The size of the image formed by a concave mirror can be determined using the magnification formula, which is given by the ratio of the height of the image to the height of the object. In this case, the formula b*[f / (u – f)]^2 represents the magnification of the image. The term f / (u – f) represents the linear magnification, which determines how much the image is enlarged or reduced compared to the object. Squaring this term ensures that the magnification is positive and gives the correct size of the image. Therefore, the correct answer is b*[f / (u – f)]^2.

Explanation
When an object is at infinity, the image formed by a convex lens is focused at the focal point of the lens. In this case, the image size is 2 cm.

When a concave lens is placed between the convex lens and the image, it will diverge the light rays and create a virtual image. The image formed by the concave lens will be located at a distance of 26 cm from the convex lens.

To calculate the new size of the image, we can use the lens formula:

1/f = 1/v - 1/u

Where f is the focal length of the convex lens, v is the image distance from the convex lens, and u is the object distance from the convex lens.

By substituting the given values into the formula, we can find the new image distance v. Then, using the magnification formula:

m = -v/u

We can find the magnification of the concave lens. Finally, we can multiply the magnification by the original image size to find the new size of the image.

Therefore, the new size of the image is 2.5 cm.