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

Explanation:

Given:
A spring mass system with:
Spring constant ( k ) = 120 N/m
Damping coefficient ( c ) = 40 N·s/m
Displaced by 0.5 m and released.

The condition for critical damping is:
     
 \( c_{\text{critical}} = 2 \sqrt{mk}\)
     
    Given damping coefficient ( c = 40 N·s/m ) implies the system is critically damped if \(c = c_{\text{critical}}\) .

    Since c matches the critical condition, the system is critically damped .
    Option 1 is correct .

2. Trajectory of the Motion :
    For a critically damped system, the displacement ( y(t) ) takes the form:
     
     \(y(t) = (A + Bt) e^{-\alpha t}\)
     
     where A and B are constants depending on initial conditions.

    Given the expression \(y(t) = 0.25 (1 + 2t) e^{-5t} \) , it matches the critically damped form.
3. Overdamped Condition :
    Since the system is critically damped , it is not overdamped .
 

4. Trajectory :
    The given form in Option 4 does not match the correct constants for the critically damped solution.

Correct option is 1)

Explanation:

The magnetic field inside a solenoid is given by:

\( B = \mu_0 n I\)

where n is the number of turns per unit length, and I is the current flowing through the solenoid. The magnetic flux through a single turn of the solenoid is:

\(\Phi_1 = B \cdot A = \mu_0 n I \pi R^2\)

where \(A = \pi R^2\) is the cross-sectional area of the solenoid.

For a solenoid of length l , there are n l turns, so the total flux \Phi is:

\( \Phi = \mu_0 n I \pi R^2 \cdot n l = \mu_0 n^2 l \pi R^2 I\)

The self-inductance L is related to the flux by the formula \(\Phi = L I\) . Therefore, the self-inductance per unit length is:

\(\mathcal{L} = \frac{L}{l} = \mu_0 n^2 \pi R^2\)

Thus, the self-inductance per unit length of the solenoid is:

 The correct solution is 2):   \( \boxed{\mathcal{L} = \mu_0 n^2 \pi R^2}\)

Concept:

The magnetic flux (Φ) through a loop is given by the product of the magnetic field (B) and the area (A) of the loop: Φ = B ⋅ A.

For a circular loop, the area is given by: A = πr², where r is the radius of the loop.

Calculation:

Given,

Magnetic field, B = 0.125 T

Rate of change of radius, dr/dt = -2 × 10⁻³ m/s

Radius, r = 4 cm = 0.04 m

The rate of change of area is:

dA/dt = π × 2r × dr/dt

dA/dt = π × 2 × 0.04 × (-2 × 10⁻³)

dA/dt = -1.6 × 10⁻⁴ π m²/s

The induced emf is:

ε = -B × dA/dt

ε = -0.125 × (-1.6 × 10⁻⁴ π)

ε = 2 × 10⁻⁵ π V

Converting to microvolts (μV), we get:

ε = 20π μV

∴ The induced emf is 20π μV.
Hence, the correct option is 2) 20π μV.

Calculation:

An ideal transformer operates based on the principle that the ratio of the primary voltage (V_p) to the secondary voltage (V_s) is equal to the ratio of the number of turns in the primary winding (N_p) to the number of turns in the secondary winding (N_s)
Mathematically, this is expressed as:

For ideal transformer  V s /V p = N s /N p

\(\frac{V_s}{V_p} = \frac{Ns}{Np} =\frac{2}{1}\)

∴ The correct option is 2)

Calculation: 

Motional emf is given by, \(\int_0^L\)Bωx dx

⇒ ε = \(B \omega \int_0^L x d x\)

⇒ ε = \(\mathrm{B} \omega\left[\frac{x^2}{2}\right]_0^{\mathrm{L}}=\frac{1}{2} \mathrm{~B} \omega \mathrm{L}^2\)

∴ The induced emf will be \(\frac{1}{2} \mathrm{BL}^2 \omega\)

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.