Practice Start Test

Current Electricity

JEE Physics · 159 questions · Page 1 of 16 · Click an option or "Show Solution" to reveal answer

Q1
If a wire is stretched to make it 0.1%0.1\% longer, its resistance will:
A increase by 0.2%0.2\%
B decrease by 0.2%0.2\%
C decrease by 0.05%0.05\%
D increase by 0.05%0.05\%
Correct Answer
Option A
Solution

Resistance of wire

R=ρlA=ρl2CR = {{\rho l} \over A} = {{\rho {l^2}} \over C}

(where

Al=CAl=C

) ∴\therefore Fractional charge in resistance

ΔRR=2Δll{{\Delta R} \over R} = 2{{\Delta l} \over l}

∴\therefore Resistance will increase by

0.2%0.2\%
Q2
The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5V and internal resistance of 20Ω\Omega , the null point on it is found to be a 1000 cm. The resistance of whole wire is :
A 80Ω\Omega
B 60Ω\Omega
C 120Ω\Omega
D 100Ω\Omega
Correct Answer
Option D
Solution

Let Resistance per unit length of potentiometer wire = λ\lambda 5 = λ\lambda ×\times 1000 ×\times 60 ×\times 10-3 ⇒\Rightarrow λ\lambda =

560{5 \over {60}}

Resistance of potentiometer wire = 1200 ×\times

560{5 \over {60}}

= 100

Ω\Omega
Q3
Consider a galvanometer shunted with 5Ω\Omega resistance and 2% of current passes through it. What is the resistance of the given galvanometer ?
A 300 Ω\Omega
B 344 Ω\Omega
C 245 Ω\Omega
D 226 Ω\Omega
Correct Answer
Option C
Solution

0.02i Rg = 0.98i ×\times 5 Rg = 245

Ω\Omega

Option (c)

Q4
Consider a moving coil galvanomenter (MCG): A. The torsional constant in moving coil galvanometer has dimensions [ML2 T−2]\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right] B. Increasing the current sensitivity may not necessarily increase the voltage sensitivity. C. If we increase number of turns (N)(\mathrm{N}) to its double (2 N)(2 \mathrm{~N}), then the voltage sensitivity doubles. D. MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel with galvanometer. E. Current sensitivity of MCG depends inversely on number of turns of coil. Choose the correct answer from the options given below:
A A, D Only
B A, B Only
C A, B, E Only
D B, D, E Only
Correct Answer
Option B
Solution

Let's analyze each statement step by step.

Statement A: “The torsional constant in moving coil galvanometer has dimensions

[ML2T−2][ML^2T^{-2}]

.” The torsional constant (often denoted by

kk

) relates the restoring torque to the angular displacement via

τ=kθ\tau = k \theta

. Since torque has dimensions of force × distance with dimensions

[MLT−2×L=ML2T−2][MLT^{-2} \times L = ML^2T^{-2}]

and the angular displacement (in radians) is dimensionless, the torsional constant indeed has the dimensions

[ML2T−2][ML^2T^{-2}]

.

Thus, Statement A is correct.

Statement B: “Increasing the current sensitivity may not necessarily increase the voltage sensitivity.”

The current sensitivity of a galvanometer is defined as its deflection per unit current (i.e.

θ/I\theta/I

).

The voltage sensitivity, on the other hand, is the deflection per unit applied voltage when the galvanometer is used as a voltmeter.

This sensitivity is proportional to the current sensitivity divided by the galvanometer’s internal resistance.

Often, if you try to increase current sensitivity (for instance, by increasing the number of turns or enhancing the magnetic field), the internal resistance may also change (typically increase when number of turns is increased).

Therefore, the improvement in current sensitivity does not automatically translate to a proportional increase in voltage sensitivity.

Hence, Statement B is true.

Statement C: “If we increase number of turns (

NN

) to its double (

2N2N

), then the voltage sensitivity doubles.” The deflection (current sensitivity) is proportional to

NN

since

θ∝NI\theta \propto N I

. However, when you double

NN

, the length of the wire increases, and so does the internal resistance (approximately doubling it, assuming the wire diameter and material remain the same).

Since voltage sensitivity is roughly given by

current sensitivityinternal resistance∝NRg\frac{\text{current sensitivity}}{\text{internal resistance}} \propto \frac{N}{R_g}

, if both

NN

and

RgR_g

double, the ratio remains essentially unchanged. Therefore, doubling

NN

does not double the voltage sensitivity, so Statement C is incorrect.

Statement D: “MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel with galvanometer.”

To convert a galvanometer into an ammeter, a shunt resistor is used to bypass most of the current so that only a small fraction (the full‐scale deflection current) passes through the galvanometer.

For this to work, the shunt resistance should be very low compared to the galvanometer’s resistance, not of “large value.”

Therefore, Statement D is false.

Statement E: “Current sensitivity of MCG depends inversely on number of turns of coil.”

As mentioned before, the electromagnetic torque in the coil is given by

τe∝NI\tau_e \propto N I

, so the deflection per unit current is

θ/I∝N\theta/I \propto N

. This shows that the current sensitivity increases with

NN

(i.e. it is directly proportional to

NN

), not inversely.

Hence, Statement E is false.

Only Statements A and B are correct.

Looking at the given options, the correct choice is: Option B: A, B Only.

Q5
An energy source will supply a constant current into the load if its internal resistance is
A very large as compared to the load resistance
B equal to the resistance of the load
C non-zero but less than the resistance of the load
D zero
Correct Answer
Option D
Solution
I=ER+r, I = {E \over {R + r}},\,

Internal resistance

(r)\left( r \right)

is zero,

I=ER=I = {E \over R} =

constant.

Q6
The mass of product liberated on anode in an electrochemical cell depends on (where tt is the time period for which the current is passed).
A (It)1/2{\left( {It} \right)^{1/2}}
B ItIt
C I/tI/t
D I2t{I^2}t
Correct Answer
Option B
Solution

According to Faraday's first law of electrolysis

m=ZIt⇒m∝Itm = ZIt \Rightarrow m \propto It
Q7
A 200 Ω\Omega resistor has a certain color code. If one replaces the red color by green in the code, the new resistance will be :
A 500 Ω\Omega
B 100 Ω\Omega
C 200 Ω\Omega
D 300 Ω\Omega
Correct Answer
Option A
Solution

200

Ω\Omega

= Red + Black + Brown Green = 5 So, Green + Black + Brown = 500

Ω\Omega
Q8
By what percentage will the illumination of the lamp decrease if the current drops by 20%?
A 26%
B 36%
C 46%
D 56%
Correct Answer
Option B
Solution
P=i2RPint =Iint 2RPfinal =(0.8Iint )2R\begin{aligned} & \mathrm{P}=\mathrm{i}^2 \mathrm{R} \\ & \mathrm{P}_{\text{int }}=\mathrm{I}_{\text{int }}^2 \mathrm{R} \\ & \mathrm{P}_{\text{final }}=\left(0.8 \mathrm{I}_{\text{int }}\right)^2 \mathrm{R} \end{aligned}

% change in power ==

Pfinal −Pint Pint ×100=(0.64−1)×100=−36%\frac{P_{\text{final }}-P_{\text{int }}}{P_{\text{int }}} \times 100=(0.64-1) \times 100=-36 \%
Q9
The number of electrons flowing per second in the filament of a 110 W110 \mathrm{~W} bulb operating at 220 V220 \mathrm{~V} is : (Given e=1.6×10−19C\mathrm{e}=1.6 \times 10^{-19} \mathrm{C})
A 1.25×10191.25 \times 10^{19}
B 31.25×101731.25 \times 10^{17}
C 6.25×10186.25 \times 10^{18}
D 6.25×10176.25 \times 10^{17}
Correct Answer
Option B
Solution
P=v×i⇒i=110220=12 Ai=ne⇒n=12×1.6×10−19=31.25×1017\begin{aligned} & P=v \times i \Rightarrow i=\frac{110}{220}=\frac{1}{2} \mathrm{~A} \\ & i=n e \Rightarrow n=\frac{1}{2 \times 1.6 \times 10^{-19}}=31.25 \times 10^{17} \end{aligned}
Q10
A circuit to verify Ohm's law uses ammeter and voltmeter in series or parallel connected correctly to the resistor. In the circuit :
A Ammeter is always connected in series and voltmeter in parallel
B Both ammeter and voltmeter must be connected in series
C Both ammeter and voltmeter must be connected in parallel
D Ammeter is always used in parallel and voltmeter is series
Correct Answer
Option A
Solution

Ammeter is always connected in series and voltmeter is connected in parallel.

Ready for a full JEE mock test? Timed · full syllabus · detailed solutions after submission
Take a Mock Test →