EmSAT Physics Mock Test- 1 - EmSAT Achieve MCQ

Density of the rod will be calculated as

The percentage error will be calculated as

Substituting the values in the above equation

= (0.001 + 0.002 + 0.001) × 100 = 0.4%
Therefore, it is our required solution.

From the free body diagram of the central block:


Hence,


For the 5 kg block,

From the FBD below

Acceleration of the blocks in the given problem

From the equation of motion

From FBD
Let the tension in the string is T

Work done by tension

Kinetic energy of a rolling body is given as:

This is our required solution.

By the law of gravitation,

This is our required solution.

Acceleration of the chain,


We have to find the tension at point P.
Both the parts OP and PQ will move with the same acceleration.
Equation of motion for OP
T = ma

This is our required solution.

W = F.s
F = 20 N
s = ?
By the equation of motion

As the object is stationary, u = 0
Therefore,

Therefore, it is a correct option.

We know that moment of inertia of a thin circular ring, I = MR²

Where ρ is the linear mass density , is same as materials of both rings is same
So, we have 

Let the minimum velocity be v.
By the conservation of energy law,

Total potential at centre is

This is our required solution.

Let R be the radius of the bigger droplet and 'r' be the radius of the smaller droplet.
Equating the volume of the droplets:

Energy of droplet with radius R = 4πR²T
Energy of N droplets with radius r = N × 4πr²T
Ratio of final surface energy to the initial surface energy

This is our required solution.

When a wire of irregular shape turns into a circular loop, area of the loop tends to increase. Therefore, magnetic flux linked with the loop increases. According to Lenz’s law, the direction of induced current must oppose the change in magnetic flux, for which induced current should flow along adcba.

Let the radius of bigger coil be R and radius of smaller coil be r. Let I be the current passing through the bigger coil in anti-clockwise direction.
Magnetic field produced due to circular coil at its centre is vertically upwards.

Let M be the mutual induction of the coils. Then, we can write,

Here, ϕ = magnetic flux.
Since, ϕ = BAcos(θ), we can write,

In case of (i): Magnetic field due to bigger coil and surface vector of smaller coil lies in same direction, i.e., θ = 0°

In case of (ii): Magnetic field due to bigger coil and surface vector of smaller coil perpendicular to each other, i.e., θ = 90°

In case of (iii): No field line passes through the smaller coil since the smaller coil is placed beside bigger coil and B is vertically upwards. Hence, the field at centre of smaller coil will be zero.

By comparing equation (1), (2) and (3), we can write,

Hence, mutual inductance is maximum in situation (i).

Consider the larger loop to be made up of four rods each of length L, the field at the centre, i.e. at a distance of (L/2) from each rod, will be


So, the flux in the smaller loop,

Efficiency of transformer is given by,

In a transformer,

1. Iron losses in actual iron cores, inspite of lamination, eddy current are produced. The magnitude of eddy current may however be small and a part of energy is lost as the heat produced in the iron core.
2. Copper losse in practice, the coils of the transformer possess resistance. So, a part of the energy is lost due to the heat produced in the resistance of the coil.
3. Flux leakage the coupling between the coils is seldom perfect. So, whole of the magnetic flux produced by the primary coil is not linked up with the secondary coil.

And hysteresis loss, humming losses also occur in the transformer.

N  pole moves closer to coil CD and S pole moves closer to coil AB on rotation of magnet.
According to Lenz's law, N pole should develop at the end corresponding to C.
∴ Induced current flows from C to D
Also S pole should develop at the end corresponding to B.
∴ Induced current flows from A to B
According to an observer E₁ current will be clockwise in coil−1 and for an observer E^₂ current will be anti-clockwise in coil−2 as shown in figure.

When the iron is switched off, then the current starts dropping to zero in circuit, and thus, the magnetic field in the circuit drops. This causes the magnetic flux linked with loop to decrease. Due to self-induction, the current still tries to continue to flow. This causes the spark.
So, both Assertion and Reason are true and Reason is the correct explanation for the Assertion.

JUST AFTER CLOSING THE KEY
Let t = 0 be the time when we need to find the current, i.e., when the switch is just turned on. The inductor opposes the flow of current and behaves as an open circuit. So the current will flow only in the 2 Ω and 3 Ω resistors.
The equivalent resistance of the circuit is,
R_s = 2 + 3 = 5 Ω
So, 

LONG TIME AFTER CLOSING THE KEY
After a long time, the inductor does not oppose the flow of current and hence acts as a short circuit.
Now, the equivalent resistance of the circuit is,

The potential difference across the inductor is


The slope of the curve is positive and constant for the first half of the time and then negative and a constant value.

The efficiency of transformer

 For electron, Where, me is the mass of an electron. m = the mass of a proton. For alpha particles,  charge on proton. For proton, 

Work done by magnetic field is given by 

∵ Potential energy of needle is given by

Torque experienced by the needle in final position is 

Hence, the torque needed to maintain the needle in this position is √3W.

When θis the angle made by the magnetic field with the BA cosθ plane of the coil, the field component parallel to the coil is Deflecting couple = torque on the loop = Bi A cosθ.

The force acting on a charged particle in magnetic field is given by  when angle between v and B is 180^o , F = 0.

In figure II and III field due to three wires add up. Hence, the magnitude is same. In figure I and IV field due to two wires cancel and the field is only due to third wire.
∴ B_I = B_IV < B_II = B_III

Given, Volume of iron rod is, V = 10⁻³ m³
Relative permeability, μ_r = 1000
No of turns per unit length of solenoid is, n = N/l = 10 turns cm⁻¹ = 1000
Current through the solenoid is, i = 0.5 A
Magnetic moment of an iron core solenoid is given by,

As shown in figure,
mass M = 50mg = 50 × 10⁻³g
Strength of each pole,
m = 98.1 amp cm , g = 981 cm s⁻², V = ?

In equilibrium, mV × 2l = Mg × l