Physics Practice Test 6

Q.1

Addition of vector A = 2 î + 3 ĵ + k̂ with vector B, result in unit vector along z-axis. What is the vector B?

  • A. -2 î + 3 ĵ
  • B. 2 î + 3 ĵ
  • C. 2 î - 3 ĵ
  • D. -2 î - 3 ĵ
  • E. 2 î
  • Answer: D
  • Explanation: Unit vector has the magnitude of 1 and points along z-axis here. Let C be the vector resulting from the addition of A and B. Therefore,
    C = A+B ------------- (1)
    Putting the value of in equation 1 we get
    C = 2 î + 3 ĵ + k̂B ------------ (2)
    Vector C is the unit vector along z-axis. Thus, it has the magnitude of 1 and points along z-axis and can be written as
    C = K ------------- (3)
    Using equation 3 in 2 we get
    B = -2 î - 3 ĵ
    Thus, D is the correct answer option.

Q.2

Consider the figure give below. A man starts his journey from position A travels in a circle to position B, just opposite to A. What are values of displacement and distance?(OB is the radius)
t6-q2-img1

  • A. 6.28 m and 4 m
  • B. 4.28 m and 4 m
  • C. 4 m and 6.28 m
  • D. 4 m and 4.28 m
  • E. 6.28 m and 4 m
  • Answer: C
  • Explanation: Displacement = Length (AB)
    Length (AB) = OA + OB
    Therefore, displacement = 4m
    Circumference of a circle is given as = 2Πr ------------ (1)
    Since, we need half of the circumference of the circle, dividing equation 1 by 2 we get distance travelled as
    Distance = Πr --------------- (2)
    Putting the value of r in equation 2 we get
    Distance = 3.14 * 2
    Or, Distance = 6.28 m
    Therefore, C is the correct answer option.

Q.3

Consider the figure given below. Two trusses are connected at common point O. What would be the force F required to keep the system in static equilibrium?
t6-q3-img1

  • A. 129.9 N
  • B. -129.9 N
  • C. 134.6 N
  • D. -134.6 N
  • E. -125.6 N
  • Answer: B
  • Explanation: Since the forces are in equilibrium, we get
    Σ Fx = 0 and Σ Fy = 0
    Resolving the force (150 N) along x and y axis we get the orange arrow represents the vertical component of 150 N and the green arrow represents the horizontal component of 150 N. As the system of trusses is in equilibrium, equating the forces along x-axis we get
    F + 150(cos 300)
    Or, F = -129.9 N
    The direction of force F is assumed to be towards the left of the screen as indicated in the figure. However, the negative sign signifies that the force is acting towards the right of the screen.
    Hence, B is the correct answer option.

Q.4

The weight of a man is measured to be 600 N inside an elevator at rest. If his weight is measured inside the elevator when it is accelerating downwards, which of the following statements is true?

  • A. Weight of the man will be more than measured before
  • B. Weight of the man will be less than measured before
  • C. There will be no change in the weight
  • D. Weight of the man inside the elevator will be zero
  • E. Not enough data is provided to predict the result
  • Answer: B
  • Explanation: In most of the cases normal reaction is same as the weight of any object. When the lift was at rest,
    N = mg ------------ (1)
    As the elevator is accelerating downwards with some value of acceleration, an unbalanced force is acting along the downward direction. So, we can write the equation of motion as
    ma + mg = N Or, mg = N – ma -------------- (2)
    mg is the actual weight of the man. When the elevator was not accelerating down then, the weight of the man should be equal to the normal reaction. But, as the elevator accelerates down the actual weight is given by difference between the normal reaction and. Thus, the weight of the man inside the elevator that is accelerating downwards will be less than the weight of the man measured when it was at rest.
    Therefore, B is the correct answer option.

Q.5

A disc rotating about its axis completes one complete rotation. What is the angle formed by the center of the disc with its radius after it completes one rotation?

  • A. 1800
  • B. 450
  • C. 900
  • D. 3600
  • E. 7200
  • Answer: D
  • Explanation: A disc appears to be a circle when seen from above. Here OA is the radius of the disc or the circle. As the disc is rotating on its axis, we will assume any point, here A to be rotating along with the disc. If A moves to a point just opposite (B) to its initial point, center of the disc forms an angle of 1800 with the diameter BA. Again if the point B moves from its initial position to the position A (one complete rotation), it forms an angle of Π. Therefore, the total angle formed between the center and the radius OA for one complete rotation by the disc along its axis is 2Π. Thus, D is the correct answer option.

Q.6

If a vehicle has twice the velocity of an identical (same mass) vehicle, what is the ratio of the kinetic energies between the faster moving vehicle and the slower moving vehicle?

  • A. 1 : 4
  • B. 1 : 2
  • C. 1 : 1
  • D. 2 : 1
  • E. 4 : 1
  • Answer: E
  • Explanation: Let the kinetic energy of the slower and faster moving vehicles be
    KE1 = (1/2) m1v12 and KE2 = (1/2) m2v22 respectively.
    Dividing the equations, we get
    (KE1 / KE2) = {[(1/2)m1v12] / [(1/2)m2v22]} ------------- (4)
    Since, msub > 1 = msub > 2 and, vsub > 2 = 2 vsub > 1,
    Equation 4 becomes
    (KE1 / KE2) = 4
    Or, KE1 : KE2 = 4 : 1
    Thus, E is the correct answer option.

Q.7

A pendulum of unknown length is executing SHM with a time period of 2s. What is the length of the pendulum?

  • A. 1.56 m
  • B. 0.52 m
  • C. 0.67 m
  • D. 0.79 m
  • E. 0.99 m
  • Answer: E
  • Explanation: Time period of pendulum is given as
    T = 2Π √(I/g) ------------- (1)
    Putting the values in equation 1 we get
    T = 2Π √(I/9.8)
    Or, 2 = Π √(I/9.8) ------------ (2)
    Squaring both the sides of equation 2 we get
    4 = 4 Πr2*(I/9.8)
    Or, I = 0.99 m
    Therefore, E is the correct answer option.

Q.8

Two boys are holding the ends of a string and one of them hits the string in such a way that waves start forming on the string. If it was observed that 4 troughs pass through a point in 10 seconds, what is the time period and frequency of the wave?

  • A. 1.5 s and 0.1 s-1
  • B. 15 s and 1 s-1
  • C. 0.4 s and 2.5 s-1
  • D. 2.5 s and 0.4 s-1
  • E. 4 s and 0.25 s-1
  • Answer: D
  • Explanation: Duration between consecutive troughs = Time period
    Or, One oscillation = Time period ------------- (1)
    Time taken to complete 4 oscillations = 10 s
    Time taken to complete 1 oscillations = 2.5 seconds
    Therefore, from equation 1 we can conclude that time period of the wave is 2.5 seconds. Frequency is related to time period as f = 1/T --------- (2)
    Putting the value in equation 2 we get
    f = 1/2.5 or f = 0.4 s-1.
    Therefore, D is the correct answer option.

Q.9

If a force of 1000 N is applied on a piston of surface area 100 cm2, what is the pressure exerted on the surface area of the piston?

  • A. 104 N
  • B. 107 N
  • C. 103 N
  • D. 105 N
  • E. 102 N
  • Answer: D
  • Explanation: Pressure exerted on the surface of the piston is given as
    P = F/A --------- (1)
    We are given area in cm2 so we need to convert it into m2.
    Putting the values in equation 1 we get
    P = 1000/10-2
    Or, P = 105 N/m2
    Therefore, D is the correct answer option.

Q.10

What would be the change in entropy for 2 kg of a substance heated from 400 K to 500 K? (Specific heat of the substance = 3700 J/Kg K)

  • A. 1634.4 J/K
  • B. 1510.5 J/K
  • C. 1200.9 J/K
  • D. 1623.3 J/K
  • E. 1644.4 J/K
  • Answer: E
  • Explanation: The amount of heat transferred is given by
    dQ = mcΔT ------------ (1)
    Change is temperature is given as ΔT = T2 – T1 ------------- (2)
    Putting the values in equation 2 we get ΔT = 100 K
    Therefore, dQ = 740000 J
    According to second law of thermodynamics dS = dQ/T
    Since, T = [(T2 + T1) / 2]
    T = 450 K
    Putting the values in equation 3 we get dS = 1644.4 J/K
    Therefore, E is the correct answer option.

Q.11

A charged particle (10-9 C) of mass 10-5 is projected with a speed of 40 m/s. If the electric field of 4*106 N/C in the region is making the charged particle stop, what is the distance that the charged particle will travel before coming to rest?

  • A. 1 m
  • B. 1.5 m
  • C. 2.8 m
  • D. 4 m
  • E. 2 m
  • Answer: E
  • Explanation: Kinematic equation of motion is
    v2 = u2 + 2as --------- (1)
    Electric force experienced by the charged particle in the region is
    F = qE ------------ (2)
    Putting the values in equation 2 we get
    F = 4 * 10-3 N
    Force due to motion of the charged particles is given as
    F = ma ----------- (3)
    Ma = 4*10-3
    10-5 * a = 4*10-3
    Or, a = 400 m/s2
    Putting the values in equation 1 we get
    0 = 402 – 2 * 400 * S ----------- (4)
    Negative sign is used because the charged particle is finally stopping i.e. charged particle is under retardation. Further solving equation 4 we get
    2 * 400 * S = 402
    Or, S = 2m
    Hence, E is the correct answer option.

Q.12

Consider the electrical circuit given below
t6-q12-img1

  • A. 33.35 V
  • B. 23.45 V
  • C. 35.65 V
  • D. 23.56 V
  • E. 12.45 V
  • Answer: A
  • Explanation: Let the equivalent resistance of the series connection between the arms xz and zy be Req1. Therefore, Req1 = 20 Ω
    Now, Req1 is in parallel with the resistance in the arm xy (10 Ω).
    Therefore, Req = R1 ∥ R2
    Req = (R1 * R2) / (R1 + R2)
    Let the equivalent resistance of the parallel connection between 20 Ω and 10 Ω be Req2. Therefore,
    Req2 = 20 ∥ 10
    Or, Req2 [(20*10)/(20+10)]
    Req2 = 6.67 Ω
    By Ohm’s law E = IR Or, E = 33.35 V
    Therefore, A is the correct answer option.

Q.13

What is the dimensional formula for magnetic field?

  • A. MA-2T-2
  • B. M-1A-2T-2
  • C. M-1A-2T-1
  • D. MA-2T-1
  • E. MA-1T-2
  • Answer: E
  • Explanation: Force experienced by a changed particle in a magnetic field is given as
    F = qvB sin θ Or, B = F/(qv sin θ) ------------ (1)
    Dimensional formula of force is given as
    MLT-2 ------------ (2)
    Dimensional formula of velocity becomes
    LT-1 ------------- (3)
    Charge is given as q = it ------------ (4)
    Dimensional formula of q from equation 4 becomes AT ----------- (5)
    Putting equation 2,3 and 5 in 1 we get
    B = MA-1T-2
    Therefore, E is the correct answer option.

Q.14

Consider the figure given below.
t6-q14-img1

  • A. Refractive index of medium 1 is greater than the refractive index of medium 2
  • B. Refractive index of medium 2 is higher than the refractive index of medium 1
  • C. Refractive index of medium 1 is equal to the refractive index of medium 2
  • D. Refractive index of medium 1 is zero
  • E. Not enough data is provided to predict the answer
  • Answer: B
  • Explanation: The figure given in the question represents refraction of light. The bending of light is related to the refractive indices, when a ray of light is incident from medium 1 to medium 2 as follows
    1. If the refractive index of medium 1 is lesser than the refractive index of medium 2, then the rays of light will bend towards the normal.
    2. If the refractive index of medium 1 is greater than the refractive index of medium 2, then the ray of light will bend away from the normal.
    It can be concluded from the above statements that the ray of light from medium 1 is bending towards the normal in medium 2. Hence, B is the correct answer option.

Q.15

If the momentum of a photon is 2.5*10-25 kg ms, what is the frequency of the photon? (Planck’s constant = 6.6*10-34 Js)

  • A. 0.88*1017 Hz
  • B. 1.14*1017 Hz
  • C. 2.76*1017 Hz
  • D. 3.5*1017 Hz
  • E. 0.88*1015 Hz
  • Answer: B
  • Explanation: Energy of a photon is
    E = mc ---------- (1)
    Momentum of a photon is
    p = mc ---------- (2)
    Using equation 2 in 1 we get
    E = pc -------------- (3)
    Energy of a photon is also expressed as
    E = hv ----------- (4)
    Equating equation 4 and 3 we get,
    hv = pc --------- (5)
    Putting the values in equation 5 we get
    6.6*10-34*v = 2.5*10-25*3*108
    Or, v = 1.14*1017 Hz
    Hence, B is the correct answer option.

Score: 0/10