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**Q.1. Six similar bulbs are connected as shown in the figure with a DC source of emf E and zero internal resistance.The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section A and one from section B are glowing, will be : (2019)**

A: 4 : 9

B: 9 : 4

C: 1 : 2

D: 2 : 1

Ans:

Solution:

(i) All bulbs are glowing

(ii) Two from section A and one from section B are glowing.

B: V

C: V

D: V

For ideal voltmeter, resistance is infinite and for the ideal ammeter, resistance is zero.

A: 10

B: 11

C: 20

D: 9

Ans:

A:

B:

C:

D:

**Ans: ASolution:**

A: R/n

B: n

C: R/n

D: nR

Ans: B

Solution:

**Q.7. ****A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves :- (2017)A: Potential gradientsB: A condition of no current flow through the galvanometerC: A combination of cells, galvanometer and resistancesD: CellsAns: **B

Solution:

In zero deflection condition, potentiometer draws no current.

A: 3:2

B: 5:1

C: 5:4

D: 3:4

Ans:

Solution:

i.e., E∝ ℓ ... (i)

Now, in the first case, cells are connected in series

That is,

Net EMF = E

From equation (i),

E

Now, the cells are connected in series in the opposite direction,

Net emf = E

From equation (i)

E

From equation (ii) and (iii),

A: a

B: a

C: a

D: a

Ans: B

Solution:

Given,

Charge, Q = at - bt

We know that,

Current, I = dθ/dt

So, equation (i) can be written as,

For maximum value of t, the current is given by,

a-2bt = 0

Therefore, t = a/2b ..(iii)

Total heat produced (H) can be given as,

On solving the above equation, we get

Given,

Charge, Q = at - bt

We know that,

Current, I = dθ/dt

So, equation (i) can be written as,

For maximum value of t, the current is given by,

a-2bt = 0

Total heat produced (H) can be given as,

On solving the above equation, we get

Then :

A:

B:

C:

D:

Ans: B

Solution:**Q.11. ****A potentiometer wire has length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1 mV per cm on the wire is : (2015)A: 48 ΩB: 32 ΩC: 40 ΩD: 44 ΩAns: **B

Solution:

Figure alongside shows a potentiometer wire of length L = 4m and resistance RAB = 8Ω. Resistance connected in series is R.

When an accumulator of emf ε = 2V is used, we have current I given by,

The resistance per unit length of the potentiometer wire is given by,

**Q.12. ****Across a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is : (2015)A: electric filedB: current densityC: currentD: drift velocityAns: **C

As the cross-sectional area of the conductor is non-uniform so current density will be different.

As, I = JA

It is clear from Eq. (i) when an area increases the current density decreases so the number of the flow of electrons will be same and thus the current will be constant.

A: 1/500 G

B: 500/499 G

C: 1/499 G

D: 499/500 G

Ans:

Solution:

Let the shunt resistance be S.

**Q.14. ****The resistance in the two arms of the meter bridge are 5 Ω and R Ω, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6ℓ _{1}. The resistance ‘R’ is : (2014)**

A: 20 Ω

B: 25 Ω

C: 10 Ω

D: 15 Ω

Ans: B

Solution:

**Q.15. ****Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is 0.5 Ω. The power loss in the wire is : (2014)A: 19.2 JB: 12.2 kWC: 19.2 WD: 19.2 kWAns: **D

Solution:

= 1200 volt

Total resistance of wire

= 150 x 0.5 = 75 Ω

∴ current through wire

(i) infinity

(ii) 9.5 Ω, the ‘balancing lengths’ , on the potentiometer wire are found to be 3m and 2.85m,respectively.

The value of internal resistance of the cell is: (2014)

A: 0.5 Ω

B: 0.75 Ω

C: 0.25 Ω

D: 0.95 Ω

Ans:

Solution:

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