Physics Practice Test 5

Q.1

What would be the dimensional formula for the ratio between acceleration due to gravity and gravitational constant?

• A. M2L-2
• B. M2L-1
• C. ML-2
• D. ML-1
• E. ML
• Explanation: Acceleration due to gravity at Earth is expressed as
g = GM/R2 --------------- (1)
Since we have to find the dimensional formula for the ratio between g and G, equation 1 can be written as
(g/G) = (M/R2) ------------- (2)
Thus, dimensional formula g/G of can be found by calculating the dimensional formula for M/R2
Since, dimensional formula of M is M and R is L hence, equation 2 can be written as
(g/G) = (M/L2)
Or, (g/G) = ML-2
Thus, C is the correct answer option.

Q.2

A car speeds up from 15 m/s to 27 m/s in 45 s. What would be the acceleration of the car?

• A. 0.15 m/s2
• B. 0.32 m/s2
• C. 0.38 m/s2
• D. 0.27 m/s2
• E. 0.4 m/s2
• Explanation: Here the initial speed, the final speed and the time taken are given and so we can use the kinematic equation
v = u + at ----------- (1)
Putting these values in equation 1 we get
27 = 15 + a * 45
Or, a = 0.27 m/s2
Therefore, D is the correct answer option.

Q.3

Consider the following conclusions based on Newton’s first law.
1. Acceleration will be zero for a body obeying Newton’s first law
2. A balanced force is required to keep a body at rest or to keep it moving with uniform speed
3. An unbalanced force is required to keep a body at rest or to keep it moving with uniform speed
Based on the conclusions above, which of the following statement(s) is/are true?

• A. 1
• B. 2
• C. 3
• D. Both 1 and 2
• E. All three
• Explanation: If the net force acting on a body is zero then it means that the resultant of all the forces acting on the body is zero. Balanced forces produce zero net force acting on a body. Thus, if a body is acted upon by a balanced force, then net force acting on it is zero and hence, it will continue to retain its initial motion. Unbalanced forces have some non-zero value of force acting on a body and the body continues moving with non-uniform speed along the unbalanced force. Acceleration is time rate of change of velocity. It can also be concluded from Newton’s first law that as the velocity is uniform, change in velocity is zero and so is the acceleration of the body. Thus in view of the above discussion statements, 1 and 2 are correct and 3 is incorrect. Hence, D is the correct answer option.

Q.4

Consider the figure given below. A block of wood of mass 10 kg is kept on a rough surface. Which of the following statements is true in respect of the frictional force that will develop such that the block moves in the direction of a force of 100 N as shown in the figure below? • A. Frictional force < 100 N
• B. Frictional force > 100 N
• C. Frictional force = 100 N
• D. Frictional force = 0 N
• E. Not enough data is provided
• Explanation: Frictional force develops due to the actual motion of a block on a surface. As the surface is rough, some frictional force will develop against the force of 100 N. Hence, D is the incorrect option. If the applied force is greater than the frictional force then only will the block of wood move. Hence, A is the correct answer option and B is an incorrect option. Also, if the frictional force is the same as the applied force, then the block of wood is at the verge of moving but it will show no physical motion. Therefore, C is an incorrect option. In view of the above discussion, E is an incorrect option.

Q.5

A wheel is rotating with a constant acceleration of 5 rad/s2. After an angular displacement of radians, the angular velocity of the wheel becomes. What was the initial angular velocity of the wheel?

• Explanation: Here we will apply the equation of rotational motion, which is given as
ωf = ωi + 2 σ θ ----------- (1)
Putting the values in equation 1 we get
70 = ωi + (2*5*2 Π) (Π = 3.14)
70 = ωi + 62.8
Therefore, C is the correct answer option.

Q.6

A ball weighing 70 N falls from a height of 3 m. What would be the kinetic energy of the ball when it hits the ground?

• A. 200 J
• B. 195 J
• C. 210 J
• D. 180 J
• E. 22.5 J
• Explanation: Energy or work done is given as
E = Fd ---------- (1)
Putting the values in equation 1 we get
E = 70 * 30
Or, E = 210 J
Therefore, C is the correct answer option.

Q.7

If the amplitude of a body in SHM becomes one-half of the initial value, what would be the ratio between the final total energy and initial total energy?

• A. 1:2
• B. 1:1
• C. 2:1
• D. 1:4
• E. 4:1
• Explanation: Total energy of a body in SHM is given as
E = (1/2) kA2 ----------- (1)
Initial energy is given by
E1 = (1/2) kA12 ----------- (2)
Final energy is given by
E2 = (1/2) kA22 ------------ (3)
Since,
A2 = (A1/2),
E2 = (1/4)[(1/2) kA12] --------- (4)
Comparing equation 5 with 2 we get
E2 : E1 = 1:4
Therefore, D is the correct answer option.

Q.8

A seismic wave originates at point A and travels along every direction with a speed of 1.6 m/s. If the seismic wave reaches a point B which is at a distance of 2000 Km from A, what would be the time difference between the point A and B in minutes?

• A. 20.83 minutes
• B. 18.57 minutes
• C. 15.78 minutes
• D. 21.56 minutes
• E. 19.56 minutes
• Explanation: A seismic wave originates at point A, travels along every direction and reaches point B. As it originates from A, it will reach point B after some time and this is the time difference. Time difference can be given as
Time difference = (Distance / Speed) ------------ (1)
Distance between A and B is given as 2000 Km and speed of the seismic wave = 1.6 km/s. Therefore, putting these values in equation 1 we get
Time difference = (2000/1.6) seconds = 1250 seconds
1s = (1/60) minutes
1250s = 20.83 minutes
Therefore, A is the correct answer option.

Q.9

A cubical block of plastic of side length 0.1 m encloses a fluid of density 990 kg/m3. What is the weight of the block? (Assume weigh of the block is only due to the fluid inside it)

• A. 9.7 N
• B. 4.8 N
• C. 5.8 N
• D. 3.6 N
• E. 6.5 N
• Explanation: Weight is given as
W = mg ------------ (1)
Mass of the block can be given as
m = Vp ------------ (2)
Volume of a cube is given as
V = (side)3 ------------- (3)
Therefore, putting the value in equation 3 we get
V = (0.1)3
Or, V = 10-3 m3
Putting the values of V and density in equation 2 we get m = 0.99 kg.
Putting the value of m and g in equation 1 we get W = 9.7 N
Therefore, A is the correct answer option.

Q.10

A thermal power plant operates between the temperature range of 600 K and 1000 K. What is the efficiency of the plant?

• A. 30 %
• B. 40 %
• C. 45 %
• D. 34 %
• E. 100 %
• Explanation: Efficiency of a plant is given as
η = 1- (T2 / T1) ------------ (1)
Putting the values in equation 1 we get
η = 1- (600 / 1000)
Or, % η = 40%
Therefore, B is the correct answer option.

Q.11

A water drop of mass 10-9kg consists of free ions of charge 2.5*10-20 C. If the water drop is held at rest by an external field, what is the external field?

• A. 2.55*1010 N/C
• B. 3.92*1011 N/C
• C. 4.56*1010 N/C
• D. -4.56*1010 N/C
• E. -3.92*1011 N/C
• Explanation: Here, weight of the water drop is balanced by the electric force. Thus, we need to equate both kinds of forces to find the value of the electric field. Weight of the water drop is given as
W = mg --------- (1)
Putting the values in equation we get
W = 10-9 * 9.8
Or, W = 9.8 * 10-9 N
Electric force is given as F = 9E ---------- (2)
Putting the values in equation 2 we get
F = 2.5*10-20*E ------------- (3)
As discussed above, weight of the water drop is equal to the electric force and hence, equating weight with equation 3 we get
E = 3.92*1011 N/C
Hence, B is the correct answer option.

Q.12

The equivalent resistance of two wires connected in series is 50 Ω. If they are connected in parallel their resistance becomes 8 Ω, what is the resistance of both the conductors?

• A. 40 Ω and 10 Ω
• B. 20 Ω and 10 Ω
• C. 20 Ω and 20 Ω
• D. 40 Ω and 20 Ω
• E. Cannot be found
• Explanation: Let R1 and R2 be the resistances of the two wires connected in series and parallel respectively.
Rs = R1 + R2 = 50 Ω ------------ (1)
And,
RP = R1 ∥ R2 = [(R1*R2) / (R1+R2)] = 8 Ω ---------- (2)
From equation 1
R2 = 50- R1 ------------ (3)
Putting equation 3 in 2 we get
[(R1*50- R1) / (R1+50- R1)] = 8
Or, R1 = 10 Ω
Putting the value of R1 in equation 3 we get
R1 = 40 Ω
Therefore, A is the correct answer option.

Q.13

A beam of electrons enters normally in a region of magnetic field of 0.01 T with a velocity of 106m/s. What is the radius of the curved path traced by the beam of electrons? (Charge on electron = 1.6*10-19 C)

• A. 4.5 * 10-4 m
• B. 3.6 * 10-4 m
• C. 5.7 * 10-4 m
• D. 2.8 * 10-4 m
• E. Cannot be determined
• Explanation: If a charged particles enters a region of magnetic field. It experiences a force due to which the charged particles follow a curved path. The radius of the curved path is given as
r = mv/qB ----------- (1)
Putting the values in equations 1 we get
r = (9.1*10-31*106) / (1.6*10-19*0.01)
r = 5.4*10-4 m
Therefore, C is the correct answer option.

Q.14

If a ray of light is incident on ice from air at an angle of 300, which of the following is true? (Refractive index of ice = 1.309 and refractive index of air = 1)

• A. Angle of refraction = sin-1(0.382) and air is denser than ice
• B. Angle of refraction = sin-1(2.6) and ice is denser than air
• C. Angle of refraction = sin-1(0.382) and ice is denser than air
• D. Angle of refraction = cos-1(0.382) and air is denser than air
• E. Angle of refraction = tan-1(0.382) and ice is denser than air
• Explanation: Light entering from one medium to another medium of a different refractive index bends and this bending is governed by Snell’s law. It is given as
μ1 sin θi = μ2 sin θr --------------- (1)
Putting the values in equation 1 we get
1 * sin 300 = 1.309 * sin θr
Or, θr = sin-1 (0.382)
Therefore, angle of refraction is sin-1 (0.382). Where a medium is denser or rarer is determined by the refractive index of the medium. Higher the refractive index, denser will be the medium. Since, refractive index of ice (1.309) is more than that of air (1), ice is denser than air. Hence, C is the correct answer option.

Q.15

What is the momentum of a photon of energy 1 eV?

• A. 1.23 * 10-28 kgms-1
• B. 1.87 * 10-28 kgms-1
• C. 1.6 * 10-19 kgms-1
• D. 5.33 * 10-28 kgms-1
• E. 4.5 * 10-19 kgms-1