Test: Electric Current & Measuring Instruments (December 1) - NEET MCQ

# Test: Electric Current & Measuring Instruments (December 1) - NEET MCQ

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## 10 Questions MCQ Test Daily Test for NEET Preparation - Test: Electric Current & Measuring Instruments (December 1)

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Test: Electric Current & Measuring Instruments (December 1) - Question 1

### When the position of cell and galvanometer in a Wheatstone bridge is inter-changed, its balanced condition

Detailed Solution for Test: Electric Current & Measuring Instruments (December 1) - Question 1

For balanced Wheatstone bridge which is shown in figure a, P/Q​=S/R​
If we interchange the cell and galvanometer then circuit becomes as shown in figure b.
and balanced condition, P/S​=Q​/R⇒P/Q​=S/R​
Thus, balanced point remains unchanged.

Test: Electric Current & Measuring Instruments (December 1) - Question 2

### Ten million electrons pass from point P to point Q in one micro second. The current and its direction is

Detailed Solution for Test: Electric Current & Measuring Instruments (December 1) - Question 2

Here,number of electron,
n = 10000000 = 107
Total charge on ten million electrons is
Q = ne [where e = 1.6 × 10−19C]
=10× 1.6 × 10−19C = 1.6 × 10−12C
Time taken by ten million electrons to pass from point P to point Q is
t = 1μs = 10−6s
The current
I =
Since the direction of the current is always opposite to the direction of flow of electrons. Therefore due to flow of electrons from point P to point Q the current will flow from point Q to point P.

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Test: Electric Current & Measuring Instruments (December 1) - Question 3

### A current in a wire is given by the equation, I =2t2 − 3t + 1, the charge through cross section of wire in time interval t = 3s to t = 5s is

Detailed Solution for Test: Electric Current & Measuring Instruments (December 1) - Question 3

As I = dQ/dt ;
dQ − Idt ;
dQ = (2t2 − 3t + 1)dt

Q.=
= [2/5(125 − 27) − 32(25 − 9) + 2]
= 43.34C

Test: Electric Current & Measuring Instruments (December 1) - Question 4

An electric bulb is rated 220 V and 100 W. Power consumed by it operated on 110 V is

Detailed Solution for Test: Electric Current & Measuring Instruments (December 1) - Question 4

Power, P= V2/R​
R= V2/P​
R=220×220/100 ​=484Ω
The current flow in the circuit,
i= V′/R​ (V′= supplied voltage)
i=110/484​=0.2272A
The voltage drop across the bulb is 110V
The power consumed by the bulb,
Pb​=V2/R​=110×110​/484
Pb​=25W
The correct option is C.

Test: Electric Current & Measuring Instruments (December 1) - Question 5

Meter Bridge or Slide Wire Bridge is a practical form of

Detailed Solution for Test: Electric Current & Measuring Instruments (December 1) - Question 5

The meter bridge principle is based on the Wheatstone bridge circuit which says that if at any point of(of a wire), the ratio of two resistances (say R1) is equal to the ratio of another two resistance (say R3 and R4 where R4is the unknown resistance),Then there shall be no flow of current at that point between those points and the edge containing the resistances (R1/R2 and R3/R4) therefore, applying it to the meter bridge, at such point, the galvanometer will show zero defection.

Test: Electric Current & Measuring Instruments (December 1) - Question 6

In a potentiometer a cell of e.m.f 2V gives a balance point at 30cm. If the cell is replaced by another cell and the balance point shifts to 60cm.What is the e.m.f of the other cell?​

Detailed Solution for Test: Electric Current & Measuring Instruments (December 1) - Question 6

Changes(if required):
Solution:
We can use the formula E1/E2=L1/L2
E1=2v,E2=?,l1=30,l2=60
2/E2=30/60
E2=120/30=4v

Test: Electric Current & Measuring Instruments (December 1) - Question 7

1  ampere current is equivalent to

Detailed Solution for Test: Electric Current & Measuring Instruments (December 1) - Question 7

Q = It also Q = ne [e = 1.6 × 10−19C]
∴ ne = It or
= 6.25 × 1018 electrons s−1

Test: Electric Current & Measuring Instruments (December 1) - Question 8

In an atom electrons revolves around the nucleus along a path of radius 0.72Å making 9.4×1018 revolution per second. The equivalent current is (e = 1.6×10−19C)

Detailed Solution for Test: Electric Current & Measuring Instruments (December 1) - Question 8

r = 0.72Å = 0.72 × 10−10m
Frequency of revolution of electron in orbit of given atom
v = 9.4 × 1018 rev/s
(where T is time period of revolution of electron in orbit)
∴ Then equivalent current is
I = e/T = ev = 1.6 × 10−19 × 9.4 × 1018
= 1.504A

Test: Electric Current & Measuring Instruments (December 1) - Question 9

Sensitivity of potentiometer can be increased by

Detailed Solution for Test: Electric Current & Measuring Instruments (December 1) - Question 9

Sensitivity of the potentiometer means the smallest potential difference it can measure. It can be increased by reducing the potential gradient. The same is possible by increasing the length of the potentiometer. Hence the correct option is option C.

Test: Electric Current & Measuring Instruments (December 1) - Question 10

The current in a wire varies with time according to the equation I = 4 + 2t, where I is in ampere and t is in second. The quantity of charge which has to be passed through a cross-section of the wire during the time t = 2s to t = 6s is?

Detailed Solution for Test: Electric Current & Measuring Instruments (December 1) - Question 10

Given:
I = 4 + 2t
Let dq be the charge which passes in a small interval of time dt. Then
dq = Idt
or dq = (4 + 2t) dt
On integrating, we get

= 48 C

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