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

Calculation:

The x-component of OA = (OA) cos 90° = 0

The x-component of OB = (OB) cos 0° = OB

The x-component of OC = (OC) cos 135° = – (1/√2) OC

Hence, the x-component of the resultant:

⇒ OB – (1/√2) OC    ... (i)

It is given that the resultant is zero, and hence its x-component is also zero. From (i):

⇒ OB = (1/√2) OC     ... (ii)

The y-component of OA = OA cos 180° = – OA

The y-component of OB = OB cos 90° = 0

The y-component of OC = (OC) cos 45° = (1/√2) OC

Hence, the y-component of the resultant:

⇒ (1/√2) OC – OA     ... (iii)

As the resultant is zero, so is its y-component. From (iii):

(1/√2) OC = OA

⇒ OC = √2 OA = 5√2 m

From (ii), OB = (1/√2) OC = 5 m

Thus the OB / OC = 1/ √2 = sin(π/4) 

Calculation:

We can easily add two or more vectors if we know their components along the rectangular coordinate axes. Let us have:

a = a_x î + a_y ĵ + a_z k̂

b = b_x î + b_y ĵ + b_z k̂

c = c_x î + c_y ĵ + c_z k̂

Then, a + b + c = (a_x + b_x + c_x) î + (a_y + b_y + c_y) ĵ + (a_z + b_z + c_z) k̂

If all the vectors are in the X-Y plane, then all the z components are zero, and the resultant is simply:

a + b + c = (a_x + b_x + c_x) î + (a_y + b_y + c_y) ĵ

This is the sum of two mutually perpendicular vectors of magnitude (a_x + b_x + c_x) and (a_y + b_y + c_y).

The resultant can easily be found to have a magnitude: √[(a_x + b_x + c_x)² + (a_y + b_y + c_y)²]

making an angle α with the X-axis where: tan α = (a_y + b_y + c_y) / (a_x + b_x + c_x)

Thus, The angle made is tanα = 4/3 ⇒ α = 53^∘ .

Calculation:

The magnitude of the resultant vector:

R' = √(x² + y² + 2xy cos θ)

Here, x = y = a

Then, R' = √(a² + a² + 2a² cos θ)

⇒ R' = a√2 √(1 + cos π ) = 0

∴ The correct answer is Option (1)

Calculation:

The point A is (a, 0, a) and point B is (0, a, a).

The vector AF will be:

AF = a î − a ĵ

The unit vector will be:

n̂ = (1 / √2) (î − ĵ)

Concept:

Let the magnitude of (A + B) be R and the magnitude of (A − B) be R'.

Given, the magnitude of R is √3 times the magnitude of R'.

Thus, R = √3 R'

Now,

R = A + B

R² = A² + B² + 2AB cos θ

⇒ R² = A² + A² + 2AA cos θ

R² = 2A² + 2A² cos θ

Again,

R' = A − B

R'² = A² + B² − 2AB cos θ

⇒ R'² = A² + A² − 2AA cos θ

R'² = 2A² − 2A² cos θ

(R / R')² = 3

3 = (1 + cos θ) / (1 − cos θ)

⇒ 4 cos θ = 2

⇒ θ = cos⁻¹(1 / 2)

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,

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

• The power consumed by the bulb.

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

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.

$\mathrm{E}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\left(\mathrm{V}\right)=\frac{\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{d}\mathrm{o}\mathrm{n}\mathrm{e}\left(\mathrm{W}\right)}{\mathrm{C}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\left(\mathrm{q}\right)}$

• 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

$W=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=\mathbf{\Delta }K$

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

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:

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

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

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

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

The correct answer is 1.2 × 108 m/s.

Key Points

CONCEPT:

• Refractive index (μ): The ratio of the velocity of light in vacuum to the velocity of light in the medium is called refractive index of that medium.

$\text{The refractive index of a substance/medium}=\frac{\text{Velocity of light in vacuum}}{\text{Velocity of light in the medium}}$

So μ = c/v

Where c is the speed of light in vacuum and v is the speed of light in the medium.

CALCULATION:

Given that:

Refractive index of the diamond (µ_d)= 2.5

We know

The velocity of light in vacuum (c) = 3 × 10⁸ m/s

To find the velocity of light in diamond (v)

Now,

$$\begin{gathered} {\mu }_d=\frac{c}{v} \\ or,2.5=\frac{3\times {10}^8}{v} \\ or,v=\frac{3\times {10}^8}{2.5}=1.2\times {10}^{8}m/s \end{gathered}$$     

Hence option 1 is correct.