Mix Test

Result

Mix Test - 8

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Which of the following is NOT the property of equipotential surface?

A
They do not cross each other.

B
The rate of change of potential with distance on them is zero.

C
For a uniform electric field they are concentric spheres.

D
They can be imaginary spheres.

Solution

As all other statements are correct. In uniform electric field equipotential surfaces are never concentric spheres but are planes \(\bot\) to Electric field lines.
Question 2
1 / -0

Two point charges +8q and -2q are located at x = 0 and x = L respectively. The point on x axis at which net electric field is zero due to these charges is

A
8L

B
4L

C
2L

D
L

Solution

Let P is the observation point at a distance r from -2q and at (L+r) from +8q.

Given Now, Net EFI at P = 0

\(\therefore \vec{E_1}\)= EFI (Electric Field Intensity) at P due to +8q

\(\vec{E_2}\) = EFI (Electric Field Intensity) at P due to -2q

\(|\vec{E_1}|=|\vec{E_2}|\)

\(\therefore\frac{k(8q)}{(L+r)^2}=\frac{k(2q)}{r^2}\)

\(\therefore\frac{4}{(L+r)^2}=\frac{1}{(r)^2}\)

\(4r^2=(L+r)^2\)

2r = L+r

r = L

\(\therefore\) P is at x = L + L = 2L from origin.
Question 3
1 / -0

An electric dipole of moment p is placed parallel to the uniform electric field. The amount of work done in rotating the dipole by \(90^\circ\) is

A
2pE

B
pE

C
pE/2

D
Zero

Solution

\(W=pE(cos\theta_1–cos\theta_2)\)

\(\theta_1=0^\circ\)

\(\theta_2=90^\circ\)

\(W=pE(cos0^\circ–cos90^\circ)\)

= pE (1 – 0) = pE.
Question 4
1 / -0

Three capacitors \(2\mu F,3\mu F\,and\,6\mu F\) are joined in series with each other. The equivalent capacitance is

A
\(1/2\mu F\)

B
\(1\mu F\)

C
\(2\mu F\)

D
\(11\mu F\)

Solution

\(\frac{1}{C_{series}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\)

\(\frac{1}{C_{series}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

\(\frac{3+2+1}{6}=\frac{6}{6}\)

\(C_{series}=1\mu F.\)
Question 5
1 / -0

Which statement is true for Gauss law

A
All the charges whether inside or outside the gaussian surface contribute to the electric flux.

B
Electric flux depends upon the geometry of the gaussian surface.

C
Gauss theorem can be applied to non-uniform electric field.

D
The electric field over the gaussian surface remains continuous and uniform at every point.

Solution

The electric field over the Gaussian surface remains continuous and uniform at every point.
Question 6
1 / -0

The best instrument for accurate measurement of EMF of a cell is

A
Potentiometer

B
metre bridge

C
Voltmeter

D
ammeter and voltmeter

Solution

Potentiometer
Question 7
1 / -0

By increasing the temperature, the specific resistance of a conductor and a semiconductor

A
increases for both.

B
decreases for both.

C
increases for a conductor and decreases for a semiconductor.

D
decreases for a conductor and increases for a semiconductor.

Solution

Specific resistance of a conductor increases and for a semiconductor decreases with increase in temperature because for a conductor, a temperature.

coefficient of resistivity \(\alpha\) = + ve and for a semiconductor, \(\alpha\) = -ve.
Question 8
1 / -0

We use alloys for making standard resistors because they have

A
low temperature coefficient of resistivity and high specific resistance

B
high temperature coefficient of resistivity and low specific resistance

C
low temperature coefficient of resistivity and low specific resistance

D
high temperature coefficient of resistivity and high specific resistance

Solution

Alloys have low temperature coefficient of resistivity and high specific resistance. If \(\alpha\) = low, the value of ‘R’ with temperature will not change much and specific resistance is high then required length of the wire will be less.
Question 9
1 / -0

If the potential difference V applied across a conductor is increased to 2V with its temperature kept constant, the drift velocity of the free electrons in a conductor will

A
remain the same.

B
become half of its previous value.

C
be double of its initial value.

D
become zero.

Solution

We know

\(V_d=\frac{eE}{ml}\overline\tau\)

\(=e\frac{V}{ml}\overline\tau\)

If temperature is kept constant, relaxation time \(\overline\tau\) - will remain constant, and e, m are also constants.

\(V_d\propto V\)

\(V_d\propto 2V\)
Question 10
1 / -0

The SI unit of magnetic field intensity is

A
\(AmN^{-1}\)

B
\(NA^{-1}m^{-1}\)

C
\(NA^{-2}m^{-2}\)

D
\(NA^{-1}m^{-2}\)

Solution

We know

\(B=\frac{F}{Il\,sin\theta}\)

SI Unit of B \(=\frac{N}{Am}=NA^{-1}m^{-1}.\)
Question 11
1 / -0

The coil of a moving coil galvanometer is wound over a metal frame in order to

A
reduce hysteresis

B
increase sensitivity

C
increase moment of inertia

D
provide electromagnetic damping

Solution

The coil of a moving coil galvanometer is wound over metallic frame to provide electromagnetic damping so it becomes dead beat galvanometer.
Question 12
1 / -0

The horizontal component of earth’s magnetic field at a place is \(\sqrt3\) times the vertical component. The angle of dip at that place is

A
\(\pi/6\)

B
\(\pi/3\)

C
\(\pi/4\)

D
0

Solution

Tangent law \(B_v=B_H\,tan\delta\)

\(tan\delta=\frac{B_v}{B_H}\)

Given \(B_H=\sqrt3\,B_v\)

\(tan\delta=\frac{B_v}{\sqrt3\,Bv}=\frac{1}{\sqrt3}\)

\(\delta=30^\circ\,or\,\frac{\pi}{6}radians.\)
Question 13
1 / -0

The small angle between magnetic axis and geographic axis at a place is

A
Magnetic meridian

B
Geographic meridian

C
Magnetic inclination

D
Magnetic Declination

Solution

Correct option is Magnetic declination or Angle of declination. It is the small angle between geographic axis & magnetic axis.
Question 14
1 / -0

Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon the

A
rate at which current change in the two coils

B
relative position and orientation of the coils

C
rate at which voltage induced across two coils

D
currents in the two coils

Solution

Mutual inductance of a pair of two coils depends on the relative position and orientation of two coils, other statements are incorrect.
Question 15
1 / -0

The magnetic flux linked with the coil (in Weber) is given by the equation –
\(\phi=5t^2+3t+16\)

The induced EMF in the coil at time, t = 4 will be-

A
-27 V

B
-43 V

C
-108 V

D
210 V

Solution

\(\phi=5t^2+3t+16\)

\(|e|=\frac{d\phi}{dt}\)

\(=\frac{d}{dt}[5t^2+3t+16]\)

= 10t + 3

\(|e|^{t=4}\) = 10(4)+3 = 43V

e = -43Volts
Question 16
1 / -0

The instantaneous values of emf and the current in a series ac circuit are

\(E=Eo\,sin\,\omega t\,and\,I=Io\,sin(\omega t+\pi/3)\) respectively, then it is

A
Necessarily a RL circuit

B
Necessarily a RC circuit

C
Necessarily a LCR circuit

D
Can be RC or LCR circuit

Solution

\(E=E_0\,sin\,\omega t\)

\(I=I_0\,sin((\omega t+\frac{\pi}{3})\)

Correct option is (iv) as I can lead the Voltage in RC and LCR circuit, so it can be RC or LCR circuit.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Mix Test - 8

Result

Mix Test - 8

Score

-

Rank

-

SHARING IS CARING

Question 1

They do not cross each other.

The rate of change of potential with distance on them is zero.

For a uniform electric field they are concentric spheres.

They can be imaginary spheres.

Solution

Question 2

8L

4L

2L

L

Solution

Question 3

2pE

pE

pE/2

Zero

Solution

Question 4

\(1/2\mu F\)

\(1\mu F\)

\(2\mu F\)

\(11\mu F\)

Solution

Question 5

All the charges whether inside or outside the gaussian surface contribute to the electric flux.

Electric flux depends upon the geometry of the gaussian surface.

Gauss theorem can be applied to non-uniform electric field.

The electric field over the gaussian surface remains continuous and uniform at every point.

Solution

Question 6

Potentiometer

metre bridge

Voltmeter

ammeter and voltmeter

Solution

Question 7

increases for both.

decreases for both.

increases for a conductor and decreases for a semiconductor.

decreases for a conductor and increases for a semiconductor.

Solution

Question 8

low temperature coefficient of resistivity and high specific resistance

high temperature coefficient of resistivity and low specific resistance

low temperature coefficient of resistivity and low specific resistance

high temperature coefficient of resistivity and high specific resistance

Solution

Question 9

remain the same.

become half of its previous value.

be double of its initial value.

become zero.

Solution

Question 10

\(AmN^{-1}\)

\(NA^{-1}m^{-1}\)

\(NA^{-2}m^{-2}\)

\(NA^{-1}m^{-2}\)

Solution

Question 11

reduce hysteresis

increase sensitivity

increase moment of inertia

provide electromagnetic damping

Solution

Question 12

\(\pi/6\)

\(\pi/3\)

\(\pi/4\)

0

Solution