IIT JEE Physics Exam Quiz!

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| By Meenujoshi87
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Meenujoshi87
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Quizzes Created: 5 | Total Attempts: 219,297
Questions: 10 | Attempts: 2,435

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IIT JEE Physics Exam Quiz! - Quiz

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Questions and Answers
  • 1. 

    If the ratio of specific heat of a gas at constant pressure to that at constant volume is γ, the change in internal energy of a mass of gas when the volume changes from V to 2V under constant pressure p is: 

    • A.

      RI (γ-1)

    • B.

      PV

    • C.

      PV I (γ-1)

    • D.

      γρV I(γ-1)

    Correct Answer
    C. PV I (γ-1)
    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.

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  • 2. 

    Two walls of thickness d1 and d2 and thermal conductivities k1 and k2 are in contact. In the steady-state, if the temperatures at the outer surfaces are T1 and T2, the temperature at the outer wall is:

    • A.

      (k1T1d2+ k2T2d1 ) / ( k1d2+ k2d1)

    • B.

      (k1T1+ k2d2 ) / ( d1+ d2)

    • C.

      [(k1d1+ k2d2 ) / (T1+ T2)] * T1T2

    • D.

      (k1d1T1+ k2d2T2 ) / ( k1d1+ k2d2)

    Correct Answer
    A. (k1T1d2+ k2T2d1 ) / ( k1d2+ k2d1)
    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.

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  • 3. 

    A particle of charge +q an Eî and uniform magnetic field Bk follows a trajectory from P to Q as shown in the figure. The velocities at P and Q are vî and -2vĵ. Which of the following statement(s) is/are correct? 

    • A.

      E=(3/4) * (mv^2/qa)

    • B.

      Rate of work done by the electric field at P is (3/4) * (mv^3/a)

    • C.

      Rate of work done by the electric field at P is zero.

    • D.

      Rate of work done by both the fields at Q is zero.

    Correct Answer(s)
    A. E=(3/4) * (mv^2/qa)
    B. Rate of work done by the electric field at P is (3/4) * (mv^3/a)
    D. Rate of work done by both the fields at Q is zero.
    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.

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  • 4. 

    Two rods, one of aluminum and the other made of steel, having initial lengths l1 and l2 are connected together to form a single rod of length l1+l2. The coefficients of linear expansion for aluminum and steel are αa and αs respectively. If the length of each rod increases by the same amount when their temperatures are raised by tC, then find ratio l1/(l1+l2):

    • A.

      αs/ αa

    • B.

      αa/ αs

    • C.

      αs/( αa + αs)

    • D.

      αa/( αa + αs)

    Correct Answer
    C. αs/( αa + αs)
    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.

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  • 5. 

    What is the condition for a single process to be both Isothermal and Isobaric?

    • A.

      One must use an ideal gas

    • B.

      Such a process is impossible

    • C.

      A change of phase is essential

    • D.

      One may use any real gas such as N2.

    Correct Answer
    C. A change of phase is essential
    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.

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  • 6. 

    The ends of a stretched wire of length L are fixed at x=0 and x=L. In one experiment the displacement of the wire is y1 = A sin(πx/L) sinωt and energy are E1 and in another experiment, its displacement is y2 = A sin (2πx/L) sin2ωt and energy is E2. Then:

    • A.

      E1=E2

    • B.

      E2=2E1

    • C.

      E2=4E1

    • D.

      E2=16E1

    Correct Answer
    C. E2=4E1
    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.

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  • 7. 

    A 0.25Kg block oscillates on the end of the spring with a spring constant of 200 Newton/meter. If the oscillation is started by elongating the spring 0.15meter and giving the block a speed of 3.0 meters/second, then what is the maximum speed of the block? 

    • A.

      0.13 m/s

    • B.

      0.18 m/s

    • C.

      3.7 m/s

    • D.

      5.2 m/s

    Correct Answer
    D. 5.2 m/s
    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.

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  • 8. 

    A double-slit experiment is performed with the light of wavelength 500 nm. A thin film of thickness 2 μm and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will:

    • A.

      Remain un-shifted.

    • B.

      Shift downwards by nearly two fringes.

    • C.

      Shift upward by nearly two fringes.

    • D.

      Shift downward by 10 fringes.

    Correct Answer
    C. Shift upward by nearly two fringes.
    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.

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  • 9. 

    A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror. The size of the image is approximately equal to:

    • A.

      B*[(u – f) / f]½

    • B.

      B*[f /(u – f)] ½

    • C.

      B*[(u – f)/f]

    • D.

      B*[f / (u – f)]^2

    Correct Answer
    D. B*[f / (u – f)]^2
    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.

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  • 10. 

    The size of the image of an object which is at infinity as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image.

    • A.

      1.25 cm

    • B.

      2.5 cm

    • C.

      1.05 cm

    • D.

      2 cm

    Correct Answer
    B. 2.5 cm
    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.

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  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Apr 09, 2012
    Quiz Created by
    Meenujoshi87
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