Test: Physics - 1 - MCAT MCQ

Torque is the turning effect of force applied on a body. It is given as t = r x F
Where
t is the torque vector;
r is the displacement vector;
F is the force vector;
Since dimensional formula of force is MLT^-2 and that of displacement is L hence, dimensional formula of torque becomes
MLT^-2* L
ML²T^-2
Therefore, C is the correct answer option.

The state of static equilibrium is characterized by a body either being at rest or moving with a constant velocity. In order to achieve static equilibrium, all forces acting on the body must be balanced, resulting in a net force of zero. Among the initial four options provided, only option D is correct. It is not possible for the body to accelerate since acceleration is the rate of change of velocity over time. When the velocity remains constant, the acceleration becomes zero. Therefore, option E is incorrect.

This problem can be solved by applying the kinematic equation of motion as shown below
v = u + at ---------------- (1)
Where
v is the final velocity of the car;
u is the initial velocity of the car = 25 m/s;
a is the acceleration the car is undergoing;
t is the time under consideration = 1.5 s;

Since the car finally stops, hence v = 0 m/s. Putting the values in equation 1 we get
0 = 25 + a *1.5
Or, -25 = 1.5a ---------------- (2)
Negative sign is coming because the car finally comes to rest and it is undergoing negative acceleration. We can neglect the negative sign here. Therefore, equation 2 becomes
a = 25/1.5
Or, a = 16.67 m/s²

Since, we know initial and final velocities and acceleration as well we can use kinematic equation
v² = u²- 2as ---------------- (3)
to solve for displacement.
Where
S is the displacement of the car; We have taken the negative sign for acceleration here due to the fact that acceleration is negative. Putting the values of v, u, a in equation 3 we get
0 = 25² - 2 × 16.67 × S
Or, 25² = 2 × 16.67 × S
Or, S = 25²/(2*16.67)
Or, S = 625/33.34
Or, S = 18.75m
Hence, B is the correct answer option.

Moment of inertia of thin-walled fluorescent lamp is given as
I = MR²  ---------------- (1)
Where
I is the moment of inertia;
M is the mass of the lamp = 50 g;
R is the radius of the lamp = 3 cm;
We have been given mass in grams so, we need to convert it into kg.
1 g = 10^-3kg
50g = 50 * 10^-3kg
50g = 0.05 kg
Also, radius is given in cm so we need to convert it into m.
1 cm = 10^-2m
3 cm = 3 * 10^-2m
3 cm = 0.03 m
Putting the values of M and R in equation 1 we get
I = 0.05 * 0.03²
Or, I = 4.5 * 10^-5 kg m²
Therefore, A is the correct answer option.

General expression for displacement of a body in SHM is given as
X(t) = A sin(ωt) ----------- (1)
Where
X(t) is the displacement of the particle;
A is the amplitude of the oscillation;
ω is the angular frequency;
Displacement of a particle as given in the question is
X(t) = 23 sin(314t) cm ------------- (2)
Comparing equation 1 with 2 we get
A = 2.3 cm and ω = 314 rad/s
Angular frequency is related to frequency as ω = 2Πf ----------- (3)
Where
f is the frequency of the oscillation;
Putting the value of in equation 3 we get
314 = 2Πf
Or, 314/(2*Π) = f
Or, 314/(2*3.14) = f
(Π = 3.14)
Or, f = 50 Hz
Hence, C is the correct answer option

Velocity of water coming out of a cylinder is given as
v² = 2gh ----------------- (1)
Where
v is the velocity of the water coming out of the cylinder;
h is the height to which water is filled= 2.5 m;
g is the acceleration due to gravity = 9.8 m/s²;
Putting the values of h and g in equation 1 we get
v² = 2 * 9.8 * 2.5
Or, v² = 49
Or, v = 7 m/s
Hence, E is the correct answer option.

Charge is given as
Q = ne --------------- (1)
Where
Q is one coulomb of charge;
n is the number of electronic charges;
e is the charge on electron = 1.6 * 10 ^-19 C;
Putting the values in equation 1 we get
1 = n * 1.6 * 10 ^-19
Or, n = 6.25 * 10¹⁸
Since n is a number therefore number of electronic charges that make one coulomb of charge is 6 *10¹⁸. Therefore, A is the correct answer option.

Magnetic moment is given as
M = IA ------------ (1)
Where
M is the magnetic moment;
I is the current flowing in the wire;
A is the area of the circular loop;
Since the wire is of circular shape hence, area of the circular loop is given as
A = πr² -------------- (2)
Where
r is the radius of the wire;
Let A₁ and r₁ be the area and radius of the initial wire. Therefore, expressing equation 2 in terms of A₁ and r₁ we get
A₁ = πr₁² ----------- (3)
Similarly, let A₂ and r₂ be the area and radius of the final wire. Therefore, expressing equation 2 in terms of A₂ and r₂ we get
A₂ = πr₂²------------ (4)
Since it is given in the question radius of the final wire is twice that of the initial wire hence, r₂ = 2r₁
Using the value of r₂ in equation 4 we get
A₂ = π(2r₁)² Or, A₂ = 4πr₁² ------------- (5)
Comparing equation 5 with 3 we find that final area can be expressed in terms of initial area and is given as
A₂ = 4πr₁²
A₂ = 4A₁ ------------------ (6)
Expressing initial and final magnetic moment in equation 1 in terms of initial area and final area we get
M₁ = IA₁ ---------------- (7)
And, M₂ = IA₂ ---------- (8)
Both initial and final current in the wire is same. Dividing equation 8 by equation 7 we get
M₂/ M₁ = (IA₂)/(IA₁) ---------------- (9)
Cancelling out the common term, in 9 we get
M₂/ M₁ = A₂/A₁ --------------- (10)
Using equation 6 in 10 we get
M₂/ M₁ = 4A₂/A₁
Or, M₂/ M₁ = 4
Therefore, M₂ = 4 M₁ -------------- (11)
Putting the value of M₁ in equation 11 we get M₂ = 4 * 3 * 10^-5 Or, M₂ = 1.2 * 10^-4 Am²
Hence, E is the correct answer option.

The nature of an electric field can be either negative or positive. If the electric field is generated by negative charges, it is negative, whereas if it is created by positive charges, it is positive. The deflection experienced by a charged particle passing through an electric field depends on the type of charge present. It can be either attractive or repulsive. Since gamma rays have no charge, they will continue moving further into the electric field without any deflection. Therefore, option A is incorrect. Infrared rays consist of uncharged particles, so they will not experience any deflection from their normal path. Hence, option B is also incorrect. Beta rays are essentially electrons, which means they carry the same charge as an electron. When beta rays pass through an electric field, they can be either attracted or repelled, depending on the nature of the electric field. Therefore, option C is the correct answer. Neutrons, like gamma rays and infrared rays, are neutral particles and will not exhibit any deflection while passing through an electric field. Thus, option D is incorrect. X-rays consist of neutral particles as well, and as a result, they undergo no deflection when passing through an electric field. Consequently, option E is also incorrect.

Since the table is frictionless, hence both the bodies will continue moving under the common force F and will have same acceleration. Since it is brought out in the question that the smaller block is not moving over the larger block hence, both the blocks will move as a combined body.
The situation above is shown below.

As both the bodies are moving together, hence, their combined mass becomes
M_total = M + 4M
Or, M_total = 5M
Common force acting on the bodies is F. Therefore, by Newton’s second law of motion we have
F = ma ------------- (1)
Where
F is the force applied;
m is the mass of the body;
a is the acceleration of both the blocks;
Mass of the body here is M_total = 5M. Putting the values in equation 1 we get
F = 5M * a
Or, F/5M = a
Hence, with this acceleration both the smaller and the larger mass will move under the effect of common force, F.
Therefore, C is the correct answer option.

Potential energy is the energy stored in a body by virtue of its height from the reference plane.
Potential energy is given as
PE = mgh --------------- (1)
Where
m is the mass of the body;
g is the acceleration due to gravity;
h is the height of the body above the reference plane;
The reference plane plays a vital role in defining the potential energy of a body. Most often reference plane is taken as the ground and hence plane B is the reference plane here. We will discuss potential energy of the body for all the planes individually
For plane A, the height of the body is zero as it is kept on the same plane A. Hence, putting the value of h in equation 1 we get
PE = 0
Therefore, potential energy of the body is zero with respect to the plane A. Thus, A is an incorrect option.
Both the planes A and C are at same height above the reference plane. Therefore, potential energy of the body with respect to the plane C is same as that in respect of A. Hence, B is incorrect as well. Since the plane A is at some height from the reference plane, h will have a non-zero value. Therefore, potential energy of the body with respect to the ground plane will have some non-zero value. Thus, C is the correct answer option. In the view of the above, D and E are also incorrect options.

Sound waves require a medium such as a solid, liquid, or gas to propagate. In the absence of a medium, sound waves cannot travel. Consequently, option A is incorrect. Seismic waves are generated as a result of faults in the Earth's tectonic plates. These waves originate from a specific point and travel through the various layers of the Earth. Without the presence of the Earth's layers, seismic waves cannot propagate. Therefore, option B is also incorrect. Tides and surface waves on water start from a point of disturbance and spread in all directions. Without a surrounding medium, such as liquid in the case of water waves, the waves will not propagate. Thus, options C and E are also incorrect. On the other hand, light waves do not require a medium to propagate. For instance, light waves from the Sun travel through the vast expanse of outer space, which is devoid of any medium. Hence, option D is the correct answer.

Heat always flows from a hot body to a cooler body. In other words, heat flows from a body at a higher temperature to a body at a lower temperature. If an object A, at 60⁰ C is kept in contact with another object B, at 65⁰ C, then this means that object A is at lower temperature than that of object B. Hence, heat will flow from B to A. Also, as the heat is transferred from B to A, temperature of B will reduce but temperature of A will increase to reach thermal equilibrium. Thus, B is the correct answer option and A, C, D and E are incorrect answer options.

Since the batteries are connected in series, equivalent potential of batteries can be found by adding
the individual potential. Let equivalent potential be V_eq. Therefore, V_eq = 25 + 25 + 25
Or, V_eq = 3 * 25 V
Or, V_eq = 75V
Since internal resistance of the batteries is zero, only resistance to which the equivalent potential is connected is 100 Ω. Hence, current flowing through the circuit can be found by Ohm’s law. Ohm’s law is given as
V = IR -------------- (1)
Where
V is the potential of the cell = 75 V;
I is the current flowing through the circuit;
R is the equivalent resistance of the circuit = 100 Ω
Putting the values of V and R in equation 1 we get
75 = I * 100
Or, I = 75/100
Or, I = 0.75 A
Therefore, B is the correct answer option.

In the figure given below, the brown arrow represents the incident ray, the green arrow represents the transmitted or the reflected ray, and the orange arrow represents the refracted ray. OA represents the normal to the interface between the air and the medium.

Since the incident light is travelling from one medium to another, some part of it will be reflected and some will be refracted. As we know from the property of reflection that, angle of incidence is same as angle of reflection, the green arrow, which is the reflected light will make the same angle with the normal as made by the incident beam with the normal. Therefore,
θ₁ = 30^o
θ₂ can be found by applying Snell’s law. According to Snell’s law we have
µ₁ sin θ_i = µ₂ + sin θ_r ---------------- (1)
Where
µ₁ is the refractive index of medium 1 (here air = 1);
θ_i is the angle of incidence = 30^o;
µ₂ is the refractive index of medium 2 = 1.5;
θ_r is the angle of refraction = θ₂ ;

Putting these values in equation 1 we get
Or, 1 * sin 30^o = 1.5 * sin θ_r
Or, 1/2 = 1.5 * sin θ_r
Or, 1/(2*1.5) = sin θ_r
Or, 1/3 = sin θ_r
Or, θ_r = sin^-1(1/3)
As θ_r is same as θ₂.
The angle of refraction is sin^-1(1/3).
Hence, E is the correct answer option.