All questions of Physics for JEE Exam

A circle around a point charge represents an equipotential surface (V= Kq/r throughout the perimeter), and work done on an equipotential surface is zero.

2 minutes 20 seconds=140 seconds or 40+40+40+40/2 seconds One round takes 40 second so it is clear that 3 and1/2 rounds are done. after three round final and initial status are same after 1/2 more round athlete completes half of the circle So final and inial status are on the edges of diameter. Displacement is diatance between final and initial status So it is equal to diameter or 2R.

Correct answer is A because when we connect a connecting wire across any capacitor then it is short circuited...so net capacitance is just 2C....and the 3rd capacitor is simply a waste one.

Each part resistance is 10 ohm as R is directly proportional to length now 10 ohms 5 resistance are connected in parallel. Therefore 2 is equivalent resistance.

KE=I w²=Iv²/r² ( since omega =v/r)
This implies that KE is inversely proportional to r²
So when radius is decreased , KE will increase due to which PE will decrease
Hence option D is correct.

The density of ice(about 0.9kg/cc) as u know is less than water(1kg/cc),therefore when ice melts it converts into denser water so the volume of the system(water+ice) contracts resulting in decrease in water level. At 4 degree C..the density of water is maximum and hence when tem falls below 4 degree its density decrease and so volume increases again.

The potential energy of a stretched spring can be calculated using the formula U = (1/2)kx^2, where U is the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

Given that the spring is stretched by 2 cm and its potential energy is U, we can write the equation as U = (1/2)k(0.02)^2.

To find the potential energy when the spring is stretched by 10 cm, we can use the same formula. Let's call this potential energy U'.

U' = (1/2)k(0.10)^2

Now, let's compare U and U' to determine the relationship between them.

U' = (1/2)k(0.10)^2
= (1/2)(0.01)k
= 0.005k

We can see that U' is equal to 0.005 times the original potential energy U.

Therefore, the potential energy when the spring is stretched by 10 cm is 0.005 times the potential energy when it is stretched by 2 cm. In other words, U' = 0.005U.

To express this relationship in terms of fractions, we can write U' = U/200.

However, we need to find the correct answer option. Let's check the options given:

a) U/25 - This is not the correct answer as U' is not equal to U/25.
b) U/5 - This is not the correct answer as U' is not equal to U/5.
c) 5U - This is not the correct answer as U' is not equal to 5U.
d) 25U - This is the correct answer as U' is equal to 25U.

Therefore, the correct answer is option 'D' - 25U.

-degree incline with an initial velocity of 30 m/s. The coefficient of kinetic friction between the particle and the incline is 0.2. Find the distance traveled up the incline before the particle comes to rest.

To find the distance traveled up the incline before the particle comes to rest, we need to consider the forces acting on the particle.

First, let's break down the forces acting on the particle:

1. Gravitational force (Fg): This force acts vertically downward and is given by the equation Fg = m * g, where m is the mass of the particle and g is the acceleration due to gravity (approximately 9.8 m/s^2).

2. Normal force (Fn): This force acts perpendicular to the incline and counteracts the component of the gravitational force that is parallel to the incline. It is equal to the weight of the particle projected onto the incline, which can be found using the equation Fn = m * g * cos(theta), where theta is the angle of the incline (45 degrees).

3. Frictional force (Ff): This force acts parallel to the incline and opposes the motion of the particle. It can be calculated using the equation Ff = coefficient of kinetic friction * Fn.

4. Net force (Fnet): This force is the sum of the forces acting on the particle and is equal to the mass of the particle times its acceleration. Since the particle eventually comes to rest, the net force is zero.

Now, let's analyze the forces acting on the particle along the incline:

- The component of the gravitational force parallel to the incline is Fg_parallel = m * g * sin(theta).

- The net force along the incline is Fnet = Ff - Fg_parallel.

Since the particle eventually comes to rest, we can set Fnet to zero:

0 = Ff - Fg_parallel

We can substitute the formulas for Ff and Fg_parallel to get:

0 = (coefficient of kinetic friction * Fn) - (m * g * sin(theta))

Since we are given the coefficient of kinetic friction and the angle of the incline, we can substitute those values:

0 = (0.2 * m * g * cos(theta)) - (m * g * sin(theta))

Dividing both sides of the equation by m * g gives:

0 = 0.2 * cos(theta) - sin(theta)

Now, let's solve this equation for theta:

0.2 * cos(theta) = sin(theta)

Dividing both sides of the equation by cos(theta) gives:

0.2 = tan(theta)

Taking the inverse tangent of both sides gives:

theta = arctan(0.2)

Using a calculator, we find that theta is approximately 11.31 degrees.

Now, we can find the distance traveled up the incline before the particle comes to rest. We'll use the formula for displacement along an inclined plane:

s = (v^2 - u^2) / (2 * a),

where s is the displacement, v is the final velocity (which is zero in this case), u is the initial velocity (30 m/s), and a is the acceleration.

Since the net force is zero, the acceleration along the incline is also zero (since there is no force to oppose the motion). Therefore, the displacement s is zero.

So, the distance traveled

Collision between two carom coins is an example of an inelastic collision. In this collision, the kinetic energy of the system is not conserved, and some energy is lost due to deformation or other forms of energy transfer. Let's discuss this in more detail.

Definition of inelastic collision
An inelastic collision is defined as a collision between two bodies where the kinetic energy of the system is not conserved. In an inelastic collision, some of the kinetic energy is converted into other forms of energy, such as deformation, sound, or heat. The total momentum of the system is conserved, but not the total kinetic energy.

Characteristics of inelastic collision
- The colliding objects stick together after the collision.
- The kinetic energy of the system decreases.
- The momentum of the system is conserved.

Explanation of collision between two carom coins
When two carom coins collide with each other, their surfaces come in contact, and some deformation occurs due to the force of the impact. This deformation results in the loss of some of the kinetic energy of the system. The two coins will stick together after the collision, and their combined velocity will be less than the initial velocity of the first coin. Therefore, the collision between two carom coins is an example of an inelastic collision.

Conclusion
Inelastic collisions are common in everyday life, and they play an important role in understanding the behavior of physical systems. The collision between two carom coins is a simple example of an inelastic collision, where some of the kinetic energy of the system is lost due to deformation.

Vector a +vector PR = vector c
vector b +vector QR = vector c
where, vector PR = -vector QR
add both and get answer

Given:
- Kinetic energy of the electron, KE = 6.6 * 10^-14 J
- Magnetic field, B = 4 * 10^-3 T
- The electron enters the magnetic field at a right angle to it.

Formula:
The radius of the circular path of a charged particle moving in a magnetic field is given by the formula:

r = (mv)/(qB)

where,
- r is the radius of the circular path
- m is the mass of the particle
- v is the velocity of the particle
- q is the charge of the particle
- B is the magnetic field strength

Calculating the velocity:
Since the kinetic energy is given, we can use the formula for kinetic energy:

KE = (1/2)mv^2

Simplifying the equation, we get:

v^2 = (2KE)/m

v = sqrt((2KE)/m)

Calculating the radius:
Using the formula for the radius of the circular path, we substitute the values:

r = (mv)/(qB)

Substituting the value of v from the previous calculation:

r = (sqrt((2KE)/m) * m)/(qB)

Simplifying the equation, we get:

r = (sqrt(2KE) * sqrt(m))/(qB)

Substituting the values:
- KE = 6.6 * 10^-14 J
- m = mass of an electron = 9.1 * 10^-31 kg
- q = charge of an electron = 1.6 * 10^-19 C
- B = 4 * 10^-3 T

Substituting these values into the equation:

r = (sqrt(2 * 6.6 * 10^-14) * sqrt(9.1 * 10^-31))/(1.6 * 10^-19 * 4 * 10^-3)

Simplifying the equation, we get:

r ≈ 4.7 * 10^-2 m

Converting the radius to centimeters:

r ≈ 4.7 * 10^-2 * 100 cm

r ≈ 4.7 cm

The nearest option to the calculated radius is 50 cm (option D).

Motion of the Particle Dropped from a Tunnel Dug Along a Diameter of the Earth

The correct answer is option 'C' - the motion of the particle is on a straight line. Let's explore why this is the case.

Gravity and Free Fall
- When a particle is dropped from a height above the surface of the Earth, it falls freely under the influence of gravity.
- In free fall, the only force acting on the particle is the force of gravity, which causes the particle to accelerate towards the Earth.
- The acceleration due to gravity is approximately constant near the Earth's surface and is denoted by 'g'.

Path of the Particle
- As the particle falls, it follows a curved path due to the gravitational force acting on it.
- However, if the particle is dropped from a point directly above the tunnel dug along the Earth's diameter, its path will align with the straight line passing through the tunnel.
- This is because the gravitational force acts along this line, causing the particle to fall directly towards the tunnel.

Motion as Seen from the Earth
- When we observe the motion of the particle from the Earth's frame of reference, it appears to move in a straight line.
- This is because the Earth's rotation and the curvature of the Earth's surface are much larger in scale compared to the motion of the particle.
- Therefore, the motion of the particle appears as a straight line to an observer on the Earth.

Characteristics of the Motion
- The motion of the particle is not simple harmonic because simple harmonic motion requires a restoring force that is proportional to the displacement from the equilibrium position.
- The motion is not parabolic because parabolic motion occurs when an object is projected with an initial velocity and follows a curved path.
- The motion is not periodic because the particle falls towards the tunnel once and does not repeat the same motion in a periodic manner.

Conclusion
In conclusion, when a particle is dropped from a point directly above a tunnel dug along the Earth's diameter, its motion, as seen from the Earth, appears to be on a straight line. This is because the gravitational force acts along this line, causing the particle to fall directly towards the tunnel.

Avg E= L(∆i/∆t) 220=L (10/0.5) 220=L(20) L=11H

Explanation : Boyle's law is only hold good when temperature is high and pressure is low. Angle of repose is equal to angle of friction.

(11.05×2+0.5×2)

Horizontal dist. travelled=nb Vertical dist. travelled=nh nb=ut nh=0+(1/2)gt^2 => n=(2hu^2)/(gb^2)

Since a dipole is made up of 2 charges of opposite sign and equal magnitude, the net force on it is always zero, whatever may be the orientation.

The frequency of a wave depends on the source by which the waves are produced. The frequency of a wave is fixed and does not change even when it passes through different substances. However, the speed of sound changes when it travel to one medium to anothe. As v=f�lambda, where v is the speed of sound,, f is the frequency and lambda is The wavelength of sound.