8 Extremely hard Physics Questions that will outsmart you! (Solved)
1) An electron with velocity v =2 × 106 ms-1 enters a square region ABCD (of area a=1cm2 along one of its side AB. Inside the region there is a magnetic field B perpendicular to the area of the square. Find the minimum value of magnetic field for which the electron will come out of the square with a velocity parallel to its initial velocity(parallel doesn't necessarily mean in the same direction)
The magnetic force acting on the electron is given by F = q(v x B), where q is the charge of the electron and B is the magnetic field. Since the electron is entering the square region with velocity v parallel to side AB, the magnetic force will be perpendicular to the velocity of the electron, causing it to move in a circular path.
In order for the electron to exit the square with a velocity parallel to its initial velocity, the net force on the electron must be zero. This means that the magnetic force must be equal in magnitude but opposite in direction to the centripetal force, which is given by F = mv^2/r.
Here, v is the velocity of the electron, m is the mass of the electron, and r is the radius of the circular path. Now, if we know the value of v, m and a, we can calculate the magnetic field, B.
The minimum value of magnetic field for which the electron will come out of the square with a velocity parallel to its initial velocity is B = mv/(q*a)
Here, v = 210^6 m/s, m = 9.1110^-31 kg (mass of electron), q = 1.6*10^-19 C (charge of electron), a = 0.01 m^2
Plugging these values, we get
B = (9.1110^-31)(210^6)/(1.610^-190.01) = 1.410^-5 T
So the minimum value of magnetic field is 1.4*10^-5 T.
2) Two equal positive charge Q is placed 2 meter apart. Another charge q=0.001C of mass m=0.3gm is kept at a distance of 100m from both of the charges. The charge q is thrown with velocity v = 40ms-1 towards the center of the line segment joining the Q's. What is the minimum value of Q such that the charges will stop the charge from going through them?
The force experienced by the charge q due to the two charges Q is given by Coulomb's law as:
F = k(qQ/r^2)
where k is the Coulomb constant, q is the charge of the particle, Q is the charge of the two charges, and r is the distance between the charges.
The charge q is thrown towards the center of the line segment joining the Q's. Therefore, the net force on the charge q should be zero so that it stops at the center of the line segment joining the charges Q.
The net force on the charge q is given by
F = 2k(qQ/r^2) = ma
where m is the mass of the charge q and a is its acceleration.
As the charge q is thrown with velocity v = 40ms-1, the net force on it should be zero in order for it to stop at the center of the line segment joining the charges Q. Therefore,
ma = 0
As we know that F = ma, we can substitute the value of F from the Coulomb's law and get:
k*(qQ/r^2) = 0
We know that q = 0.001C, m = 0.310^-3 kg, v = 40 m/s, r = 100 m, and k = 8.9910^9 Nm^2/C^2.
The minimum value of Q can be found by substituting the known values into the equation.
Q = mv^2/(2kq/r^2)
Plugging in the values, we get:
Q = (0.310^-3)(40^2)/(28.9910^9*0.001/(100^2)) = 0.0001C
So the minimum value of Q is 0.0001 Coulombs.
3) In “Avengers: Infinity War” Thor has to hold open the star forge at Nidavellir to make his new weapon. During the process, an intense beam of light is falling on his back. Each second 5×1030photons are hitting him, each with a wavelength of 331 nm. What is the magnitude of force acting on Thor by the beam of light? (Assume that all the photons are being absorbed by Thor)
The magnitude of force acting on an object by a beam of light can be calculated using the equation:
F = P/c
where P is the power of the light and c is the speed of light.
4) In this case, the number of photons hitting Thor per second is given as 5x10^30. The wavelength of each photon is given as 331 nm. We can calculate the energy of each photon using the formula E=hc/λ , where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.
Plugging in the values, we get:
E = (6.626 x 10^-34 Js) * (3 x 10^8 m/s) / (331 x 10^-9 m) = 7.44 x 10^-19 J
We can calculate the total energy hitting Thor per second by multiplying the energy of each photon with the number of photons hitting him per second.
E = 7.44 x 10^-19 J * 5 x 10^30 = 3.72 x 10^11 J/s
Now, we can find the power by multiplying the energy with the number of photons per second:
P = E * N = 3.72 x 10^11 J/s
Finally, we can find the force experienced by Thor by dividing the power with the speed of light:
F = P/c = 3.72 x 10^11 J/s / 3 x 10^8 m/s = 1.24 x 10^3 N
So the magnitude of force acting on Thor by the beam of light is approximately 1240 N.
Note that this is a large force and would be enough to cause severe injury or death, but since this is a work of fiction, it is not meant to be taken as a realistic scenario.
5) A planar figure is formed from uniform wire that consists of two equal semi-circular arcs, each with its own closing diameter, joined so as to form a letter ‘B’. The figure is freely suspended from its top left-hand corner. What angle does the straight edge of the figure make with the vertical?
We can start by analyzing the forces acting on the planar figure when it is suspended from its top left-hand corner.
The force of gravity acts vertically downwards on the center of mass of the figure, creating a tension in the wire.
The tension in the wire creates an equal and opposite force on the top left-hand corner of the figure, which is where it is suspended.
Now, we can consider the two semi-circular arcs of the figure. The arc on the right-hand side of the figure creates a clockwise torque about the top left-hand corner of the figure, while the arc on the left-hand side creates a counterclockwise torque. These torques will be equal in magnitude but opposite in direction.
The net torque on the figure is zero, as the clockwise torque is equal to the counterclockwise torque. Therefore, the figure will rotate until the angle between the straight edge of the figure and the vertical is such that the clockwise torque about the top left-hand corner is equal to the counterclockwise torque.
The angle of the straight edge of the figure with the vertical is 90 degrees.
This is the angle at which the figure will be in equilibrium, meaning that the net torque will be zero and the angle will not change.
6) A spherical ball of volume V is moving at velocity v relative to an observer on the ground where v = 0.8c. What is the volume of the ball relative to the observer on the ground?
As the velocity of the ball relative to the observer on the ground is v = 0.8c, where c is the speed of light, we can use the Lorentz factor to calculate the volume of the ball relative to the observer on the ground.
The Lorentz factor is given by γ = 1/√(1 - (v^2/c^2))
Plugging in the value of v, we get:
γ = 1/√(1 - (0.8c)^2/c^2) = 1/√(1 - 0.64) = 1/√0.36 = 1/0.6 = 1.67
The volume of the ball relative to the observer on the ground is given by:
V' = γ*V
where V is the volume of the ball as observed by an observer traveling with it.
So, the volume of the ball relative to the observer on the ground is 1.67*V.
It is important to note that this is a consequence of special relativity and the contraction of length in the direction of motion, which makes the volume of the ball appear smaller for the observer on the ground.
7) An infinitely long hollow cylinder carries a uniform surface charge density σ. The radius of the cylinder is R. What is the potential at a distance r from the cylinder axis(r > R)? Take the potential at the surface of the cylinder to be V0.
The electric potential at a distance r from the cylinder axis can be calculated using the formula for the potential due to a uniform charge distribution:
V(r) = V0 + (σ/2πε) * ∫(ln(r/R) dθ)
Where V0 is the potential at the surface of the cylinder, σ is the uniform surface charge density, r is the distance from the cylinder axis, R is the radius of the cylinder, θ is the angle in radians and ε is the electric constant.
The integral can be solved to get:
∫(ln(r/R) dθ) = θ * ln(r/R)
So,
V(r) = V0 + (σ/2πε) * θ * ln(r/R)
As the cylinder is infinitely long, the potential is constant along the axis of the cylinder and independent of θ.
Therefore,
V(r) = V0 + (σ/2πε) * ln(r/R)
So, the potential at a distance r from the cylinder axis(r > R) is V0 + (σ/2πε) * ln(r/R)
It is important to notice that the potential is infinite at r = 0 as the logarithm of zero is not defined.
8) A disk of mass M and radius R is initially rotating at angular velocity of ω. While rotating, it is placed on a horizontal surface whose coefficient of friction is μ= 0.5. How long will it take for the disk to stop rotating? (Hint: use torque on the disk due to friction)
The time it takes for the disk to stop rotating can be calculated by finding the torque on the disk due to friction and then calculating the angular acceleration of the disk.
The torque on the disk due to friction is given by:
τ = μ * N * R
where τ is the torque, μ is the coefficient of friction, N is the normal force on the disk, and R is the radius of the disk.
As the disk is on a horizontal surface, the normal force on the disk is equal to the weight of the disk, which is given by:
N = M * g
where M is the mass of the disk and g is the acceleration due to gravity.
Plugging in these values, we get:
τ = μ * M * g * R
The angular acceleration of the disk is given by:
α = τ / I
where I is the moment of inertia of the disk.
The moment of inertia of a disk is given by:
I = (1/2) * M * R^2
Plugging in these values, we get:
α = μ * M * g * R / (1/2) * M * R^2
α = 2 * μ * g
The angular velocity of the disk is given by:
ω = α * t
where t is the time for which the disk stops rotating.
As the angular velocity is given in the problem, we can substitute it into the equation above and solve for t.
ω = 2 * μ * g * t
t = ω / (2 * μ * g)
Plugging in the given values, we get:
t = (ω) / (2 * 0.5 * g)
So the time it will take for the disk to stop rotating is inversely proportional to angular velocity and directly proportional to the coefficient of friction and gravity.
It's important to notice that this is a rough estimate as the torque on the disk due to friction is assumed to be constant and the disk is assumed to be a point mass, which in reality is not the case.
