Physics MCQ Quiz - Objective Question with Answer for Physics - Download Free PDF

Concept:

Bohr's Model and Wavelength Calculation:

In Bohr's model, the wavelength of the radiation emitted during an electron transition is related to the energy difference between two states.

The energy difference between two states A and B (or C) is given by:

E = hc / λ,

where:

E: Energy difference between the states (Joules)

h: Planck’s constant (6.626 × 10⁻³⁴ J·s)

c: Speed of light (3 × 10⁸ m/s)

λ: Wavelength of the radiation (meters)

Calculation:

Given,

Wavelength for transition from state A to C: λ_AC = 2000 Å

Wavelength for transition from state B to C: λ_BC = 6000 Å

Now, we calculate the energy difference for both transitions using the relation E = hc / λ:

For transition A to C, we have:

⇒ E_A-C = hc / 2000

For transition B to C, we have:

⇒ E_B-C = hc / 6000

Now, the energy difference between states A and B is:

⇒ E_A-B = E_A-C - E_B-C

⇒ E_A-B = (hc / 2000) - (hc / 6000)

⇒ E_A-B = hc × (1 / 2000 - 1 / 6000)

⇒ E_A-B = hc × (1 / 3000)

Thus, the wavelength of radiation emitted during the transition from state A to state B is:

⇒ λ_AB = 3000 Å

∴ The wavelength of the radiation emitted during the transition from state A to state B is 3000 Å.

Calculation:

N₂ = N₀e^–3λt 

N1 = N0e–λt 

\(\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}\) = e–2λt 

t_halflifeofN1 t = \(\frac{\ln 2}{\lambda}{ }_{n} 1\)

\(\frac{\mathrm{N}_{0}}{\mathrm{2}}\)= N0e–λt 

λt = ln2

t = \(\frac{\ln 2}{\lambda}\)

= \(\mathrm{e}^{-2 \lambda \frac{\ln 2}{\lambda}}\)

\(\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}}=\frac{1}{4}\)

Concept:

Radioactive decay follows the equation:

N(t) = N₀ e^-λt

Where:

• N(t): Remaining activity after time t
• N₀: Initial activity
• λ: Decay constant
• t: Time elapsed

The relationship between the half-life t_1/2 and the decay constant λ is:

λ = ln 2 / t_1/2

Calculation:

Given:

• Half-life t_1/2 = 2.5 years
• Remaining activity = 1.5625%, or 0.015625 of initial activity

First, calculate λ:

λ = ln 2 / 2.5 = 0.277 years^-1

Now, calculate the time required for the activity to reduce to 1.5625% of its initial value:

0.015625 = e^-λt

Taking the natural logarithm:

ln(0.015625) = -λt

t = ln(0.015625) / -λ = ln(0.015625) / -0.277 ≈ 10 years

Hence, the time required for the activity to reduce to 1.5625% is 10 years.

Concept Used:

Energy Released During Alpha Decay:

During alpha decay, an atom of Uranium-238 (U) loses an alpha particle (helium nucleus) and transforms into Thorium-234 (Th). The energy released in the process can be calculated using the mass defect.

The energy released (E) is given by the formula:

E = Δm × c²

Where Δm is the mass defect (difference in mass before and after the decay) and c is the speed of light.

We can also use the equivalent mass-energy relation in atomic mass units (u), where 1 u = 931.5 MeV/c².

The mass defect is:

Δm = (mass of U) - (mass of Th + mass of He)

Calculation:

Given,

Mass of U (238U) = 238.05079 u

Mass of He (4He) = 4.00260 u

Mass of Th (234Th) = 234.04363 u

1 u = 931.5 MeV/c²

The mass defect Δm is:

Δm = (Mass of U) - (Mass of Th + Mass of He)

⇒ Δm = 238.05079 u - (234.04363 u + 4.00260 u)

⇒ Δm = 0.00456 u

Now, to find the energy released:

E = Δm × 931.5 MeV/c²

⇒ E = 0.00456 u × 931.5 MeV/c²

⇒ E ≈ 4.25 MeV

∴ The energy released during the alpha decay of U-238 is 4.25 MeV.

Concept Used:

Bohr’s Model and Quantum Number:

In Bohr's model, the angular momentum of an electron moving in a circular orbit is quantized and is given by:

m × v × r = n × h / 2π

Where:

m = Mass of the object (kg)

v = Orbital speed (m/s)

r = Radius of the orbit (m)

n = Quantum number (dimensionless)

h = Planck’s constant = 6.625 × 10^-34 J·s

For the Earth's revolution around the sun, we can treat the Earth as an electron and apply this formula to find the quantum number.

SI Units: Mass (kg), Speed (m/s), Radius (m), Planck's constant (J·s)

Calculation:

Given,

Mass of Earth, m = 6 × 10²⁴ kg

Orbital speed, v = 3 × 10⁴ m/s

Radius of orbit, r = 1.5 × 10¹¹ m

Planck’s constant, h = 6.625 × 10^-34 J·s

Using Bohr’s quantization condition for angular momentum:

m × v × r = n × h / 2π

⇒ n = (m × v × r × 2π) / h

⇒ n = (6 × 10²⁴ × 3 × 10⁴ × 1.5 × 10¹¹ × 2π) / (6.625 × 10^-34)

⇒ n ≈ 2.6 × 10⁷⁴

∴ The quantum number is 2.6 × 10⁷⁴.

CONCEPT:

• Electric Power: The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

\(P = \frac{W}{t} = VI = {I^2}R = \frac{{{V^2}}}{R}\)

Where V = Potential difference, R = Resistance and I = current.

CALCULATION:

Given - Potential difference (V) = 220 V, power of the bulb (P) = 100 W and actual voltage (V') = 110 V

• The resistance of the bulb can be calculated as,

​\(\Rightarrow R=\frac{V^2}{P}=\frac{(220)^2}{100}=484 \,\Omega\)

• The power consumed by the bulb.

\(\Rightarrow P=\frac{V^2}{R}=\frac{(110)^2}{484}=25 \,W\)

CONCEPT:

Galvanometer:

• A galvanometer is used for detecting current in an electric circuit.
• The galvanometer is the device used for detecting the presence of small currents and voltage or for measuring their magnitude.
• The galvanometer is mainly used in the bridges and potentiometer where they indicate the null deflection or zero current.
• The potentiometer is based on the premise that the current sustaining coil is kept between the magnetic field experiences a torque.

EXPLANATION:

• From the above, it is clear that the galvanometer is the instrument used for detecting the presence of electric current in a circuit. Therefore option 1^st is correct.

​Additional Information 

Difference between Ammeter and Galvanometer:

• The ammeter shows only the magnitude of the current.
• The galvanometer shows both the direction and magnitude of the current.

The correct answer is Newton's first law.

CONCEPT:

• Newton’s first law of motion: It is also called the law of inertia. Inertia is the ability of a body by virtue of which it opposes a change.
• According to Newton’s first law of motion, an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
• The inertia of rest: When a body is in rest, it will remain at rest until we apply an external force to move it. This property is called inertia of rest.
• The inertia of motion: When a body is in a uniform motion, it will remain in motion until we apply an external force to stop it. This property is called inertia of motion.

EXPLANATION:

• When a bus suddenly starts moving, the passengers fall backward due to the law of inertia of rest or 1st law of Newton.
• Because the body was in the state of rest and when the bus suddenly starts moving the lower body tends to be in motion, but the upper body still remains in a state of rest due to which it feels a jerk and falls backward. Hence option 1 is correct.

Additional Information

Laws of Motion given by Newton are as follows:

Option 4 is correct

CONCEPT:

• Electric potential (V): The amount of work done to move a unit charge from a reference point (or infinity) to a specific point in an electric field without producing an acceleration is called electric potential at that point.

\({\rm{Electric\;potential\;}}\left( {\rm{V}} \right) = \frac{{{\rm{Work\;done\;}}\left( {\rm{W}} \right)}}{{{\rm{Charge\;}}\left( {\rm{q}} \right)}}\)

• Electrostatic Potential Energy: The amount of work done to move a charged particle from infinity to a point in an electric field is known as the potential energy of that charged particle.

CALCULATION:

Given that:

Electric charge (q) = 5 C

Potential difference (V) = 16 V

Work done (W) = charge (q) × potential difference (V)

Work done (W) = 5 × 16 = 80 J

The correct answer is 2 kJ.

CONCEPT:

• Potential energy: The energy of any object due to its position with respect to a reference point is called potential energy. It is denoted by PE.

Potential energy is given by:

PE = m g h.

Here, PE is the Potential Energy, m is the mass, g is the acceleration due to gravity, and h is the height at which the object is placed

CALCULATION:

Given that: 

Mass (m) = 10 Kg

Height (h) = 20 m

P.E. = 10 x 10 x 20

P.E.= 2000 J

 P.E. = 2 kJ

• Kinetic energy: The energy due to the motion of the object is called kinetic energy. 
  • Kinetic energy (KE) = 1/2 (mv²)
  • Where m is mass and v is velocity. 
• Since the object is stationary (at rest) so the velocity is zero. Hence the kinetic energy of the object will be zero.
• Only the potential energy of the object will be there at the height.

The correct answer is 490J

CONCEPT:

• Work-energy theorem: It states that the sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body i.e.,

Work done by all the forces = Kf - Ki

\(W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} = {\bf{Δ }}K\)

Where v = final velocity, u = initial velocity and m = mass of the body

CALCULATION:

It is given that,

Mass (m) = 20 kg

Final Velocity (v) = 7 m/s and initial velocity (u) = 0 m/s

According to the work-energy theorem,

⇒  Work done = Change in K.E

⇒  W = Δ K.E

Since initial speed is zero so the initial KE will also be zero.

⇒  Work done (W) = Final K.E = 1/2 mv²

⇒  W = 1/2 × 20 × 7²

⇒  W = 10 × 49

⇒  W = 490J

CONCEPT:

Coulomb's law in Electrostatics –

• Coulomb's law state’s that force of interaction between two stationary point charges is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them and acts along the straight line joining the two charges.

Force (F) ∝ q₁ × q₂

\(F \propto \;\frac{1}{{{r^2}}}\)

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)

Where K is a constant = 9 × 10⁹ Nm²/C²

EXPLANATION:

Given – q₁ = 2 × 10^-7 C, q₂ = 3 × 10^-7 C and r = 30 cm = 30 × 10^-2 m

Force is equal to

\(F = \left( {9{\rm{\;}} \times {\rm{\;}}{{10}^9}} \right)\times \frac{{2 \times {{10}^{ - 7}} \times 3 \times {{10}^{ - 7}}}}{{{{\left( {30 \times {{10}^{ - 2}}} \right)}^2}}}\)

\( \Rightarrow F = \frac{{54 \times {{10}^{ - 5}}}}{{900 \times {{10}^{ - 4}}}} = 6 \times {10^{ - 3}}N\)

The correct answer is Kudankulam.

• Kudankulam Nuclear Power Plant is the largest nuclear power station in India by capacity.

Key Points

• Kudankulam Nuclear Power Plant is located 650 km south of Chennai, in the Tirunelveli district of Tamilnadu, India.
• The power plant will have a combined capacity of 6000 Mega Watt upon completion.
• The Atomic Energy Commission was established in 1948 by the efforts of Dr. Homi Jahangir Bhabha, the father of Atomic Energy Research in India.
• India's first atomic research reactor 'Apsara' started working in Trombay (near Mumbai) but India's first Nuclear Power reactor was established at Tarapur in 1969.
• Production of nuclear energy requires uranium, thorium, and heavy water, Uranium is found in Jharkhand, Rajasthan, and Meghalaya.

CONCEPT:

• Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
• These equations are only valid when the acceleration of the body is constant and they move on a straight line.

There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

\({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

Where, V = final velocity, u = initial velocity, s = distance traveled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

EXPLANATION:

Given that:

Initial velocity (u) = 0

Distance (S) = 20 m

Time (t) = 4 sec

Use \({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

20 = 0 + \(\frac{1}{2} \times a \times 4^2\)

acceleration = a = 20/8 = 2.5 m/s²

CONCEPT:

• When moving through a magnetic field, the charged particle experiences a force.
• When the direction of the velocity of the charged particle is perpendicular to the magnetic field:
  • Magnetic force is always perpendicular to velocity and the field by the Right-Hand Rule.
  • And the particle starts to follow a curved path.
  • The particle continuously follows this curved path until it forms a complete circle.
  • This magnetic force works as the centripetal force.
• Centripetal force (F_C) = Magnetic force (F_B)

​⇒ qvB = mv²/R

​⇒ R = mv/qB

where q is the charge on the particle, v is the velocity of it, m is the mass of the particle, B is the magnetic field in space where it circles, and R is the radius of the circle in which it moves.

EXPLANATION:

Given that particle has charge e; mass = m; and moves with a velocity v in a magnetic field B. So

• Centripetal force (FC) = Magnetic force (FB)

​​⇒ qvB = mv2/R

\(\Rightarrow R=\frac{mv}{qB}\)

\(\Rightarrow r=\frac{mv}{Be}\)

So the correct answer is option 1.