Test: Class 12 Physics: CBSE Sample Question Paper- Term I (2021-22) - Class 12 MCQ

As all other statements are correct. In uniform electric field, equipotential surfaces are never concentric spheres but are planes ⊥ to electric field lines.

Let P be the observation point at a distance r from –2q and at (L + r) from + 8q.
Given Now, Net EFI at P = 0
= EFI (Electric Field Intensity) at P due to + 8q
 = EFI (Electric Field Intensity) at P due to – 2q

∴ 
∴ 
4r² = (L + r)²
2r = L + r
r = L
∴ P is at x = L + L = 2L from origin
∴ Correct option is (C) 2L.

W = pE (cos θ₁ – cos θ₂)
θ₁ = 0°
θ₂ = 90°
W = pE (cos 0° – cos 90°)
= pE (1 – 0)
= PE

C_series = 1 µF

Q₁ _______Q₂
Force in the charges in the air is

= K F
= 5 F

All other statements except (iv) are incorrect.
The electric field over the Gaussian surface remains continuous and uniform at every point. 

Battery is disconnected.

 
Q = Charge remains context C′ = KC
Q′ = C′V′
Q = C′V′
Q = KCV′

V′ = 

Potentiometer

I₁ : I₂ = 3 : 2
r₁ : r₂ = 2 : 3
I₁ : I₂ = ?

= 
∴ 

Specific resistance of a conductor increases and for a semiconductor decreases with increase in temperature because for a conductor, a temperature:
coefficient of resistivity, a = + ve
and for a semiconductor, a = – ve

Alloys have low temperature coefficient of resistivity and high specific resistance. If a = low, the value of ‘R’ with temperature will not change much and specific resistance is high, then required length of the wire will be less.

∵ V = constant

= 
H′ /H = 2/1
H′ = 2H

We know

= 
If temperature is kept constant, relaxation time τ - will remain constant and e, m are also constants.
V_d ∝ V
V_d ∝ 2V

We know

SI Unit of B = N/Am = NA^–1m^–1

Redrawing the circuit, we get

3 Ω & 6 Ω are in parallel.
∴ R_{1 =}  3 x 6 / 3 + 6 = 18/9 = 2

Now R₁ and 8Ω in series
∴ R₂ = R₁ + 8 = 2 + 8 = 10 Ω
Now R₂ and 30Ωin parallel

= 300/40 = 30/4 = 15 /2
= 7.5 Ω

l = length of wire

Area of a Square = a²
Also here l = 4a
a = l /4
∴ Area = l² /16
A₁ = l²/16
Area of a circle = πr²
Also here, 2πr = l
r = l/2π
Now area = 

A₂ = l²/4π
Now Magnetic moment = IA
∴ M₁ = IA, & M₂ = I A₂
Since I (current) is same in both
∴  
 M₁M₂ = π : 4

The coil of a moving coil galvanometer is wound over metallic frame to provide electromagnetic damping so it becomes dead beat galvanometer.

Correct option is Magnetic declination or Angle of declination. It is the small angle between geographic axis and magnetic axis.

Target law, BV = BH tanδ
tanδ = BV / BH
Given BH = √3BV
tanδ = BV / √3BV = 1/√3
δ = 30° or π / 6 radians.

Current induced is I = |e/R|

Now |e| = dφ/dt
But there is no change of flux with time, as B',A' and θ all remain constant with time.
∴ No current is induced.

Mutual inductance of a pair of two coils depends on the relative position and orientation of two coils.

I = V / X_C  in pure capacitor

= 

φ = 5t² + 3t + 16
 |e| = dφ/dt
= d/dt(5t² + 3t + 16)
= 10t + 3
|e|_t=4 = 10(4) + 3 = 43 V
e = - 43 Volts

E = E_o sin ωt
I = 
as I can lead the voltage in RC and LCR circuit, so it can be RC or LCR circuit.

V_R = Effective voltage across R
∴ V_R = I_eff R
V_L = Effective voltage across L
V_L = I_eff × L
Net V = 
= 
20 = 
(20)² = (12)² + V_L²
400 = 144 + V_L²
V_L = √400 - 144 = √266 = 16 volts

Since, dielectric constant of metal is ∞. The electric force between the charges in any medium is given by 

⇒ F_m = F/∞
⇒ Fm = 0

Direction of area element is normal to X-Y plane, i.e., z axis. Therefore, . Now, using the relationship between electric flux f and electric field

= 40 units

The potential difference between two points of equipotential surface i.e.
V2 - V1 = 0 = W/q ⇒ W = 0
Therefore, no work done is done in moving a charge from one point to another on an equipotential surface.

We know that

Here x = 1 m
Hence, E = 6 V/m. 

Since, two copper wires are connected in series, current i flow through them is same. Both wires are made of same material, so n₁ = n₂
As, i = neAv_d
⇒ 
⇒ 
⇒