Current Electricity MCQ Level - 2 - Physics MCQ

We know, Power delivered = i²R  where  i = current and  R = Resistance

Power delivered in both cases is equal then

The deflection of one division of galvanometer is given by the current = 4 × 10^–4 A
or the current sensitivity of galvanometer is
= 4 × 10^–4 A/div
i_g = full scale deflection
= current sensitivity × total number of turns
= 4 × 10^–4 × 25 = 10^–2 A
Resistance of galvanometer = 100Ω
The value of resistance required,

The correct answer is: 150

Let the resistance of the lamp filament be R. Then when the voltage drops expected power is 
Here R'  will be less than R,  because now the rise in temperature will be less. Therefore  P  is more than  

But it will not be 90% of earlier value, because fall in temperature is smaller.

The correct answers are: 81 W, between 81 and 90 W

R = R₀ + R₀α(T₁-T₀)
tan3α = R₀ + R₀α(T₁-T₀)...............(1)
tan2α = R₀ + R₀α(T₂-T₀)...............(2)
tan3α - tan2α = R₀α(T₁-T₂)
(T₁-T₂)  α  tan(3α) - tan(2α)
tan(3α-2α) = (tan3α - tan2α)/(1 + tan3αtan2α)
(T₁ - T₂)  α  tanα(1 + tan3αtan2α)
So, tanα is the right answer.

For a closed circuit cells supplies a constant current in the circuit.

Equation of cells E = V + ir
For V = 50 volt
E = 50 + 11r           ...(i)
Similarly for  V = 60V
E = 60 + r      ...(ii)
From Equation (i) and (ii), we get
r = 1Ω
Substituting the value of r  is equation (i), we get
E = 61V
The correct answer is: 61V, 1Ω

Ratio of masses of wires
(m₁, m₂, m₃) = 1 : 3 : 5
Ratio of length of wires

Electrical resistance is given by

The correct answer is: 125 : 15 : 1

Using the relation

Let  L₁  is the original length and  L₂  is the length after drawn out.
Let L₁ = l  and  L₂ – 3L₁ = 3l
Volume is constant

L is the latent heat of vaporisation of water, the heat required to produced 1g of steam.

L = 540 cal = 540 × 4. 2 = 2268J.
Energy supplied = 1080 Js^–1
Time required to boil 100g  of water

The correct answer is: 210 s