IIT JEE Physics Exam Quiz!

2.

A block of mass m is attached to a spring of spring constant k and is oscillating on a frictionless horizontal surface. The amplitude of the oscillation is A. What is the speed of the block when its displacement from equilibrium is A/2?

0
√(k/m) * A
√(k/m) * A/2
√(3k/4m) * A

Correct Answer

A. √(3k/4m) * A

Explanation

When the block's displacement is A/2, its potential energy is (1/2)k(A/2)^2 = (1/8)kA^2. By conservation of energy, the kinetic energy of the block is (1/2)kA^2 - (1/8)kA^2 = (3/8)kA^2. If the speed of the block is v, then (1/2)mv^2 = (3/8)kA^2. Solving, we find v = √(3k/4m) * A.

Rate this question:

0

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?

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

Correct Answer(s)

A. E=(3/4) * (mv^2/qa)
A. Rate of work done by the electric field at P is (3/4) * (mv^3/a)
A. 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.

Rate this question:

1

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):

αs/ αa
αa/ αs
αs/( αa + αs)
αa/( αa + αs)

Correct Answer

A. 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.

Rate this question:

5.

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

One must use an ideal gas
Such a process is impossible
A change of phase is essential
One may use any real gas such as N2.

Correct Answer

A. 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.

Rate this question:

4

6.

A particle is moving in a circular path of radius r with constant speed v. What is the magnitude of the average acceleration of the particle over a half-revolution?

0
V^2 / r
2v^2 / (πr)
V^2 / (2r)

Correct Answer

A. 2v^2 / (πr)

Explanation

The average acceleration of a particle is defined as the change in velocity divided by the change in time. Since the particle is moving with constant speed, the change in speed is zero. However, the velocity is changing because the direction of motion is changing.

The change in velocity over a half-revolution is equal to 2v (the particle's speed changes from v in one direction to v in the opposite direction). The distance traveled over a half-revolution is πr (half the circumference of the circle). The time taken to complete a half-revolution is equal to the distance divided by the speed, which is πr / v.

Therefore, the average acceleration is:

(change in velocity) / (change in time) = 2v / (πr / v) = 2v^2 / (πr)

Rate this question:

1

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?

0.13 m/s
0.18 m/s
3.7 m/s
5.2 m/s

Correct Answer

A. 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.

Rate this question:

2

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:

Remain un-shifted.
Shift downwards by nearly two fringes.
Shift upward by nearly two fringes.
Shift downward by 10 fringes.

Correct Answer

A. 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.

Rate this question:

2

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:

B*[(u – f) / f]½
B*[f /(u – f)] ½
B*[(u – f)/f]
B*[f / (u – f)]^2

Correct Answer

A. 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.

Rate this question:

4

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.

1.25 cm
2.5 cm
1.05 cm
2 cm

Correct Answer

A. 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.

Rate this question:

3

IIT JEE Physics Exam Quiz!

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:

Quiz Preview

A block of mass m is attached to a spring of spring constant k and is oscillating on a frictionless horizontal surface. The amplitude of the oscillation is A. What is the speed of the block when its displacement from equilibrium is A/2?

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?

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

A particle is moving in a circular path of radius r with constant speed v. What is the magnitude of the average acceleration of the particle over a half-revolution?

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 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 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:

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.