Test: Electric Current & Measuring Instruments(11 Nov) - JEE MCQ

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.

Here,number of electron,
n = 10000000 = 10⁷
Total charge on ten million electrons is
Q = ne [where e = 1.6 × 10⁻¹⁹C]
=10⁷ × 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.

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

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

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,
P_b​=V²/R​=110×110​/484
P_b​=25W
The correct option is C.
 

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.

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

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

Radius of electron orbit
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

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.

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