Physics Practice Test 7
Q.1
Which of the following is a dimensionally correct representation for angular momentum?
- A. M^{2}L^{2}T^{-1}
- B. M^{2}L^{1}T^{-1}
- C. M^{2}L^{1}T^{-2}
- D. ML^{2}T^{-1}
- E. ML^{1}T^{-1}
- Answer: D
Explanation: Angular momentum is given as
L = I ω -------------- (1)
SI unit for I is kgm^{2}. Hence, dimensional formula of I becomes, Angular velocity is given as
ω = v/t
SI unit of v is m/s and v that of r is m. Hence, SI unit of angular velocity becomes
(m/s) / m
Or, 1/s
Therefore, dimensional formula of angular velocity is T^{-1}. Thus, dimensional formula of angular momentum from equation 1 becomes
(ML^{2}) / T Or, ML^{2}T^{-1}
Hence, D is the correct answer option.
Q.2
Consider the figure and the statements given below. (v is velocity and t is time)
1. The graph is for a body that is accelerating
2. The graph is for a body that is falling freely
3. The graph is for a body that has constant velocity
Which of the following statement(s) is true in accordance to the given figure?
- A. 2
- B. 1
- C. 3
- D. Both 1 and 2
- E. Both 2 and 3
- Answer: A
Explanation: The graph given in the equation is between velocity and time. Slope of the velocity time graph is linear here. Therefore, velocity increases with each instant but change in velocity per unit time is constant throughout the travel. Thus, 3 is an incorrect statement and so is option C. Acceleration of a body is defined as the rate of change of velocity per unit time. As said earlier, for a linear slope between velocity and time, change in velocity is constant per unit time and thus, acceleration is constant for the entire travel. Therefore, statement 1 is incorrect and also is option B. A body falls freely under gravity only. A free falling body always experiences uniform acceleration is constant for a free fall and the graph essentially represents the motion of free fall. Thus, statement 2 is correct and so A is the correct answer option. In view of the above discussion, options D and E are incorrect as well.
Q.3
If a block of wood weighs 1000 N, what would be its mass?
- A. 102.04 kg
- B. 10.24 kg
- C. 1020.44 kg
- D. 1.24 kg
- E. 110.24 kg
- Answer: A
Explanation: Weight of a body is given as
W = mg ---------- (1)
Putting the value of W and g in equation 1 we get
1000 = m * 9.8
m = 102.04 kg
Therefore, A is the correct answer option.
Q.4
Which of the following combinations of forces can lead to a resultant unbalanced force of 10 N, if both the forces are acting orthogonally?
- A. 4 N and 5 N
- B. 6 N and 8 N
- C. 5 N and 8 N
- D. 8 N and 8 N
- E. 5 N and 5 N
- Answer: B
Explanation: Resultant force is given as
R = √[F_{1}^{2}+F_{2}^{2}+F_{1}F_{2} cos θ]
As both the forces are acting orthogonally, θ = 90^{0}.
R = √[F_{1}^{2}+F_{2}^{2}+F_{1}F_{2} cos90^{0}]
Or, R = √[F_{1}^{2}+F_{2}^{2}+F_{1}F_{2}*0]
Or, R = √[F_{1}^{2}+F_{2}^{2}]
If we take F_{1} = 6 N and F_{2} = 8 N, we get
R = √62+82
Or, R = 10 N
Putting the values as mentioned in options A, C, D and E, we get the value of the resultant force as 6.4 N, 9.43 N, 11.3 N and 7.07 N. Hence B is the correct answer option.
Q.5
If a wheel is rotating with an angular velocity of 2.5 rad/s, what is the time taken by the wheel to complete one full rotation?
- A. 1.57 s
- B. 2.51 s
- C. 1.84 s
- D. 3.52 s
- E. 2.22 s
- Answer: B
Explanation: Time taken to complete one rotation is known as time period. Angular velocity is related to time period as
ω = (2 Π)/T -------------- (1)
Putting the values of (Π = 3.14) and ω in equation 1 we get
2.5 = (2*3.14) / T
T = 2.51 s
Therefore, B is the correct answer option.
Q.6
An object is to be lifted from the ground to a height of 5m in 25 seconds. If the energy consumed during the process is 2000 J, what is the power required?
- A. 75 J/s
- B. 72 J/s
- C. 80 J/s
- D. 100 J/s
- E. 125 J/s
- Answer: C
Explanation: Power is given as
P = W/t ------------- (1)
Putting the values in equation 1 we get
P = 2000/25
P = 80 J/s or Watt
Hence, C is the correct answer option.
Q.7
A body of mass 0.3 kg connected to a spring having constant of 20 N/m is executing SHM. What is the time period of the SHM?
- A. 0.65 s
- B. 0.56 s
- C. 0.75 s
- D. 0.87 s
- E. 0.14 s
- Answer: C
Explanation: Time period is given as
T = 2Π √(m/k) ------------- (1)
Putting the values in equation 1 we get
T = 2Π √(0.3/20)
Or, T = 2*3.14*0.12 (Π = 3.14)
T = 0.75 s
Therefore, C is the correct answer option.
Q.8
A radio FM is broadcasting signals at a frequency of 400 MHz. What is the wavelength of the signal, if it is travelling at a speed of 2.5*10^{8} m/s?
- A. 0.534 m
- B. 0.625 m
- C. 0.725 m
- D. 0.918 m
- E. 0.025 m
- Answer: B
Explanation: Speed of the wave is given as
v = f λ ------------ (1)
As the frequency is given in MHz, we need to convert it into Hz.
1 MHz = 10^{6} Hz
400 MHz = 400*10^{6} Hz
Putting the values in equation 1 we get
2.5 * 10^{8} = 400 * 10^{6} * λ
Or, λ = 0.625 m
Therefore, wavelength of the FM signal is 0.625 m and thus, B is the correct answer option.
Q.9
Consider the figure given below. Three vessels of different shapes are filled with water to the same height. Which of the following is a correct statement?
- A. Pressure at the bottom of A is more than that at the bottom of B
- B. Pressure at the bottom of B is more than that at the bottom of C
- C. Pressure at the bottom of C is more than that at the bottom of A
- D. All have the same volume
- E. All have the same pressure
- Answer: E
Explanation: All three vessels A, B and C are of different shapes. So, fluids inside the vessels will also have different shapes. Due to different shapes, volumes of the fluids enclosed inside the vessels for three different cases will be different as well. Hence, D is an incorrect option. Pressure at the bottom of a vessel due to the fluid it contains is given as
P = p gh
It can be concluded from the equation that pressure depends on density of the fluid and height to which a vessel is filled. Since, the fluid here is common for all the vessels and the height of the fluid is also the same; therefore, pressure exerted by the fluid at the bottom for all the vessels will be the same. Hence, E is the correct answer option.
Q.10
A sample of gas at 313 K occupies a volume of 10 m^{3}. What would be its volume at 373 K?
- A. 11.92 m^{3}
- B. 15.56 m^{3}
- C. 10.45 m^{3}
- D. 12.67 m^{3}
- E. Cannot be determined
- Answer: A
Explanation: This equation can be solved by applying Charles’s law, which is given as
(V_{1}/T_{1}) = (V_{2}/T_{2}) ------------ (1)
Putting the values in equation 1 we get
(10/313) = (V_{2}/372)
V_{2} = 11.92 m^{3}
Therefore, A is the correct answer option.
Q.11
What is the dipole moment between an electron and a proton separated by a distance of 1 nm?
- A. 3.6 * 10^{-28} Cm
- B. 4.5 * 10^{-28} Cm
- C. 4.5 * 10^{-26} Cm
- D. 1.6 * 10^{-28} Cm
- E. 3.2 * 10^{-28} Cm
- Answer: D
Explanation: Electric dipole moment between two charges of opposite polarity is given as
p = Qd ----------(1)
The distance is in nm so we need to convert it into m.
1 nm = 10^{-9} m
Putting the values in equation 1 we get
P = 1.6 * 10^{-19} * 10^{-9}
Or, 1.6 * 10^{-28} Cm
Hence, D is the correct answer option.
Q.12
What is the current density of electrons flowing through a steel block of length 1 m, if potential difference applied across the conductor is 200 V? (Conductivity of steel = 1.45*10^{6} Ω^{-1}m^{-1})
- A. 2.9*10^{8} A/m^{2}
- B. 1.4*10^{8} A/m^{2}
- C. 2.2*10^{8} A/m^{2}
- D. 3.4*10^{9} A/m^{2}
- E. 1.45*10^{4} A/m^{2}
- Answer: A
Explanation: Current density is given as
J = σE ------------- (1)
Electric field is given as
E = (V/I) ---------- (2)
Putting the values in equation 2 we get
E = (200/1) Or, E = 200 V/m
Putting the values in equation 1 we get
J = 1.45*10^{6}*200
Or, J = 2.9 * 10^{8} A/m^{2}
Hence, A is the correct answer option.
Q.13
A beta particle of mass is moving perpendicular to a magnetic field in a region of 0.01 T with a speed of 10^{4} m/s. What is the force experienced by the beta particle due to the magnetic field? (Charge on electron = 1.6*10^{-19} C)
- A. 3.2*10^{-17} N
- B. 1.6*10^{-17} N
- C. 4.8*10^{-17} N
- D. 2.4*10^{-17} N
- E. 10^{-17} N
- Answer: B
Explanation: Force experienced by the beta particle is given as
F = qΠB sin θ -------------- (1)
As the beta particle is moving perpendicularly to the magnetic field, θ=90^{0}. Charge on a beta particle is equal to the charge on an electron;
F = 1.6 * 10^{-19}*10^{4}*0.01*sin 90^{0}
F = 1.6 * 10^{-17} N
Hence, B is the correct answer option.
Q.14
What would be the minimum angle of incidence so that a ray of light travelling from medium 1 of refractive index 1.7 to medium 2 of refractive index 1.5 does not emerge out of medium 2?
- A. sin^{-1} (0.88)
- B. cos^{-1} (0.88)
- C. tan^{-1} (0.88)
- D. tan^{-1} (1.13)
- E. sin^{-1} (1.13)
- Answer: A
Explanation: For a ray of light incident on a medium to not emerge from it is known as limiting case of total internal reflection. For the minimum angle the refracted ray travels along the interface. For the incident ray making an angle more than the minimum angle the ray will reflect back and this is known as total internal reflection. From Snell’s law we have
μ_{1} sin θ_{i} = μ_{2} sin θ_{r} ------------ (1)
Putting the values in equation 1 we get
1.7*sin θ_{i} = 1.5* sin90^{0}
Θ_{i} =sin^{-1}(0.88)
Therefore, A is the correct answer option.
Q.15
A photon of frequency 350 GHz is incident on a metal surface. What is the energy of the photon in eV? (Planck’s constant = 6.6*10^{-34} Js)
- A. 2.34 * 10^{-3} eV
- B. 1.24 * 10^{-2} eV
- C. 1.44 * 10^{-3} eV
- D. 3.46 * 10^{-2} eV
- E. 0.45 eV
- Answer: C
Explanation: Energy possessed by a photon is given as
E = hv ------------ (1)
Frequency is given in GHz so, we need to convert it into Hz.
1 GHz = 10^{9} Hz
350 GHz = 3.5*10^{11} Hz
Putting the values in equation 1 we get
E = 6.6*10^{-34}*3.5*10^{11}
Or, E = 2.31 * 10^{-22} J
We need to convert energy in Joules into eV as given below
1.6*10^{-19} J = 1 eV
1 J = [1 / (1.6*10^{-19})]eV
Or, 2.31*10^{-22} J = 1.44*10^{-3} eV
Therefore, a photon with frequency of 350 GHz has energy of. Hence C is the correct answer option.