KEAM Paper 1 Mock Test - 2 - JEE MCQ

# KEAM Paper 1 Mock Test - 2 - JEE MCQ

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## 120 Questions MCQ Test KEAM Mock Test Series 2024 - KEAM Paper 1 Mock Test - 2

KEAM Paper 1 Mock Test - 2 for JEE 2024 is part of KEAM Mock Test Series 2024 preparation. The KEAM Paper 1 Mock Test - 2 questions and answers have been prepared according to the JEE exam syllabus.The KEAM Paper 1 Mock Test - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for KEAM Paper 1 Mock Test - 2 below.
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KEAM Paper 1 Mock Test - 2 - Question 1

### Two radioactive substances A and B have decay constants 5λ and λ, respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become (1/e)2 will be:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 1

KEAM Paper 1 Mock Test - 2 - Question 2

### The magnetic field and number of turns of the coil of an electric generator is doubled then the magnetic flux of the coil will:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 2

Given: N = 2N1 and B = 2B1
The magnetic flux through the electric generator when the magnetic field is B, current flowing is A and the number of turns is N
φ = N B A cos θ ....(1)
The magnetic flux through the electric generator when the magnetic field and number of turns of the coil of an electric generator is doubled
φ1 = N1 B1 A cosθ   ...(2)
⇒ φ= (2N)(2B) A cos θ = 4 N B A cos θ = 4φ [∵φ = NBA cos θ]

KEAM Paper 1 Mock Test - 2 - Question 3

### If λ1 and λ2  are the wavelengths of the first members of Lyman and Paschen series, respectively, then λ1 : λ2 is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 3

For the wavelength of the first member of Lyman series:

For the wavelength of the first member of Paschen series:

KEAM Paper 1 Mock Test - 2 - Question 4

Three rods of Copper, Brass and Steel are welded together to from a Y –shaped structure. Area of cross – section of each rod = 4 cm2. End of copper rod is maintained at 100°C where as ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 4

KEAM Paper 1 Mock Test - 2 - Question 5

In an experiment for determination of refractive index of glass of a prism by i – δ plot, it was found that a ray incident at an angle of 35° suffers a deviation of 40° and that it emerges at an angle of 79°. In that case, which of the following is closest to the maximum possible value of the refractive index?

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 5

1.5 is the nearest option.

KEAM Paper 1 Mock Test - 2 - Question 6

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10-12 m, the minimum electron energy required is close to:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 6

KEAM Paper 1 Mock Test - 2 - Question 7

Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I/2. If a third polarizer C is placed between A and B, the intensity of emergent light is reduced to I/3. The angle between polarizers A and C is θ. Then,

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 7

For the system of polarizers A and B, the intensity of the emergent light is I/2. So, A and B have same alignment of transmission axis.
Let's assume that C is introduced at θ angle w.r.t. A.
Then,
Output Intensity = I/2
So, we have

KEAM Paper 1 Mock Test - 2 - Question 8

Assume that an electric field  exists in space. Then the potential difference VA – VO, where VO is the potential at the origin and VA the potential at x = 2 m is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 8

KEAM Paper 1 Mock Test - 2 - Question 9

In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 kΩ are used. The figure of merit of the galvanometer is 60 μA/division. In the absence of shunt resistance, the galvanometer produces a deflection of θ = 9 divisions, when current flows in the circuit. The value of the shunt resistance that can cause the deflection of θ/2 is closest to:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 9

For deflection of θ/2, current also reduces to I/2 with shunt resistance S.
Hence,

KEAM Paper 1 Mock Test - 2 - Question 10

A current of 1 A is flowing on the sides of an equilateral triangle of side 4.5 × 10-2 m. The magnetic field at the centroid of the triangle will be:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 10

Or 4 × 10-5 Wb/m2

KEAM Paper 1 Mock Test - 2 - Question 11

A rocket is fired from the earth towards the sun. At what distance from the earth's centre is the gravitational force on the rocket zero ? Mass of the sun = 2 × 1030 kg, mass of the earth 6 × 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 11

As the Rocket is fired from earth (E) towards the sun (s) at point (P) the distance from center of the earth.

The Force acts on the rocket are: These two gravitational forces are in opposite directions.

As we know gravitational force between two objects which are at distance R from each other.
G = universal gravitational constant, m1m2 = mass of the objects, R = Distance between them
By the gravitational Force
Force on Rocket due to Earth = Force on Rocket due to sun

By simplification:

KEAM Paper 1 Mock Test - 2 - Question 12

A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 12

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
As we know that, the value of the ratio of specific heat for standard gas = 1.40.

For an adiabatic process, we have:
The final volume is compressed to half of its initial volume.

KEAM Paper 1 Mock Test - 2 - Question 13

The figure shows a 2.0 V potentiometer used for the determination of the internal resistance of a 1.5 V cell. The balance point of the cell in the open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm of the potentiometer wire. Determine the internal resistance of the cell.

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 13

The internal resistance of the cell = r
Balance point of the cell in open circuit, I1 = 76.3cm
An external resistance (R.) is connected to the circuit with R = 9.5Ω
New balance point of the circuit, I2 = 64.8cm
Current flowing through the circuit = 1
Using the relation connecting resistance and emf is,

KEAM Paper 1 Mock Test - 2 - Question 14

What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm?

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 14

Given: Wavelength of sodium line (λ1) = 589nm Wavelength at which sodium line is observed (λ2) = 589.6nm
Change in wavelengths is given by Δλ = λ2 - λ
Substituting the values in above equation we get, Δλ 589.6 - 589 = 0.6nm
Velocity in terms of wavelength is given by,

Substituting the values in above equation we get,

KEAM Paper 1 Mock Test - 2 - Question 15

Light from a point source in the air falls on a spherical glass surface (μ=1.5 and radius of curvature 20cm). The distance of the light source from the glass surface is 100cm. The position where the image is formed is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 15

As per the given criteria, Refractive index of air, μ1 = 1, Refractive index of glass, μ2 = 1.5, Radius of curvature, R = 20 cm, Object distance, u = -100 cm
We know that,

KEAM Paper 1 Mock Test - 2 - Question 16

One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle ( directed towards the centre ) is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 16

When a particle connected to a string revolves in a circular path around a center, the centripetal force is provided by the tension produced in the string. Thus, in the given case, the net force on the particle is the tension T, i,e F = T = mv2/1 Where F is the net force acting on the particle.

KEAM Paper 1 Mock Test - 2 - Question 17

A metal block of area 0.10 m2 is connected to a 0.01 kg mass via a string that passes over a massless and frictionless 0.01 kg pulley as shown in the figure. A liquid with a film thickness of 0.3 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 ms-1. The coefficient of viscosity of the liquid is: (Take g = 10 ms-2)

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 17

Here, m = 0.01 kg, l = 0.3 mm = 0.3 × 10-3 m, g = 10 ms-2 V= 0.085 ms-1, A = 0.1 m2
The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m.
Thus, the shear force F is: F = T = mg = 0.010 kg x 9.8 ms-2 = 9.8 x 10-2 N
Shear stress on the fluid =

KEAM Paper 1 Mock Test - 2 - Question 18

Assume that the light of wavelength is 6000Å coming from a star. What is the time resolution of a telescope whose objective has a diameter of 100 inch?

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 18

Resolving power (R.P.) of the astronomical telescope is its ability to form separate images of two neighbouring astronomical objects like stars etc.

where D is diameter of objective lens and λ is wave length of light used.

KEAM Paper 1 Mock Test - 2 - Question 19

The magnetic field in a plane electromagnetic wave is given by By = (2 x 10-7) T Sin (0.5 x 103x + 1.5 x 1011t). This electromagnetic wave is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 19

The magnetic field in a plane of the electromagnetic wave is given by, B = 2 x 10-7 sin (0.5 × 103x + 1.5 × 1011t)
Comparing this equation with the standard wave equation: By = B0 sin[Kx + ωt]
From the above equation, K = 0.5 x 103 = propagation constant

The wavelength range of microwaves is 10-3 m to 0.3 m. The wavelength of this wave lies between 10-3m to 0.3 m, so the equation represents microwaves.

KEAM Paper 1 Mock Test - 2 - Question 20

A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 20

KEAM Paper 1 Mock Test - 2 - Question 21

A long infinite current-carrying wire is bent in the shape as shown in the figure. The magnetic induction at point O is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 21

An infinite wire, carrying current, is bent in the following way.
Two semi-infinite parts and a bent finite part. We have to find the induced magnetic field at an external point o, at a distance R. The corresponding figure is shown below.

Magnetic field induced by a finite current carrying conductor at a distance R is given by,
Where, θ1 and θ2 are the angles made by the endpoints of the wire to the external point.
And using right hand thumb rule, we get direction of magnetic field,
For the 1st semi-infinite part, θ1= 0 and θ2 = 90o = π/2
Substitute the above values in equation (i), we get
For the bent portion θ= 45o and θ2 = 135
Therefore, magnetic field will be,

For the second semi-infinite part,
θ= π/2 θ2 = 0
Therefore, magnetic field will be,
Total magnetic field at point o is, B = B1 + B2 + B3

KEAM Paper 1 Mock Test - 2 - Question 22

Two gases A and B have equal pressure P, temperature T, and volume V. The two gases are mixed together and the resulting mixture has the same temperature T and volume V as before. The ratio of pressure exerted by the mixture to either of the two gases is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 22

Let P1 and P2 be the partial pressures of gases A and B, respectively. Under equal conditions of temperature and volume, P1 = P2
Let P be the total pressure exerted by the mixture of two gases A and B. Using Dalton's law of partial pressures, P = P1 + P2
⇒ P = 2P1 ⇒ P/P1 = 2/1 ⇒ P : P1 = 2 : 1

KEAM Paper 1 Mock Test - 2 - Question 23

A current of 1.6 A is passed through CuSO4 solution. The number of Cu++ ions liberated per minute are:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 23

Current I = 1.6 Amps, Time t = 1 minute or 60 seconds
Two electrons are needed to liberate one ion of Cu++, i.e. the charge required is = 2 × 1.6 × 10-19 C = 3.2 × 10-19C
The total charge flow in 1 minute is Q = It = 1.6A × 60s = 96C
Thus, the number of ions liberated is n

KEAM Paper 1 Mock Test - 2 - Question 24

A book with many printing errors contains four different formulae for the displacement of a particle undergoing a certain periodic motion x. Which one is the wrong formula on dimensional grounds (A = amplitude, ω = angular velocity, T = time period of motion)?

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 24

The argument in the trigonometric function should be a dimensionless quantity.
The dimensions of time (t)  angular velocity (ω) and amplitude (A) are [t] = [T], [ω] = [T-1], [A] = [L]
A. The dimension of the argument in the equation:

Thus, option 'A' is not wrong on dimensional grounds.
B. In the equation: x = A / T sin (t/A), the dimension of the argument in the sine function is [t/A] = |T|/L

Thus, option 'B' is wrong on dimensional grounds.
C. The dimension of the argument in equation:

Thus, option 'C' is not wrong on dimensional grounds.
D. The dimension of the argument in equation: x = A sin (ωt) is [ωt] =  [T-1] [T] = dimensionless quantity
Thus, option 'D' is not wrong on dimensional ground. All the equations, except x = A/T sin (t/A) i.e. option 'B' have dimensionless argument.

KEAM Paper 1 Mock Test - 2 - Question 25

The figure below shows four plates each of area A and separated from one another by a distance d.

What is the capacitance between P and Q?

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 25

The given figure is shown below:

The above arrangement is equivalent to two parallel plate capacitors connected in parallel with the same separation d and area A.
Now, the capacitance of each capacitor is
The equivalent capacitance of C1 and C2 is C = C1 + C

KEAM Paper 1 Mock Test - 2 - Question 26

A spring of spring constant 5 × 103 is stretched initially by 5 cm from the unstretched position. The work required to stretch it further by another 5 cm is

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 26

Spring constant, k = 5 × 103 N/m
The work required to stretch the spring by the displacement x is W = 1/2 kx
When the spring is stretched by 5 cm, work done is W1 = 1/2 × (5 × 103N/m) × (5 × 10-2m)2
W1 = 6.25J
When the spring is again stretched further by 5 cm, the net displacement of the spring from its equilibrium position is
x' = 5 cm, +5 cm = 10 cm
Thus, the work required to stretch it further by an additional 5 cm is W2 = 1/2 x (5 × 103N/m) × (10 × 10-2m)
⇒ W2 = 25J
So, the extra work required to stretch the spring by an additional
5 cmis ΔW = W- W1 ⇒ ΔW = 25J - 6.25J ⇒ ΔW = 18.75J or 18.75 N - m

KEAM Paper 1 Mock Test - 2 - Question 27

Two sources of sound A and B produce progressive waves given by y1 = 6 cos(100πt) and y2 = 4 cos(102πt) ear the ears of an observer. It will hear

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 27

y1 = 6 cos(100πt) ...... (i)
y2 = 4 cos(102πt) ...... (ii)
Standard wave equation is given by y = Ao cos(ωt) ....... (iii)
From equation (i) and (iii), we get angular frequency ω1 = 100π = 2πf1 ⇒ f= 50 Hz
Amplitude, A01 = 6 unit From equation (ii) and (iii), we get ω=102π = 2πf2 ⇒ f2 =51 Hz
Amplitude, A02 = 4 unit
Beat frequency = difference in frequencies of the two given waves = f2 - f1 = 51 Hz - 50 Hz = 1 Hz, i.e 1 beat per sec
Resultant amplitude A of superposition of two waves of amplitudes A01 and A02 is A =
For maximum resultant amplitude, phase

For minimum resultant amplitude, phase

KEAM Paper 1 Mock Test - 2 - Question 28

Two beams of red and violet colors are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 28

Two beams of red and violet colors are made to pass separately through a prism (angle of the prism is 60°).
We have to find the angle of refraction at the position of minimum deviation.
In the position of minimum deviation, light passes symmetrically through the prism irrespective of the light wavelength used.

In the position of minimum deviation, r + r = 2r = A[ as deviation δ = 0]
for both the colour.

KEAM Paper 1 Mock Test - 2 - Question 29

A planet of radius R = 1/10 × ( radius of Earth) has the same mass density as Earth. Scientists dig a well of depth R/5 on it and lower a wire of the same length and of linear mass density 10-3 kg m-1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is: (take the radius of Earth = 6 × 106 m and the acceleration due to gravity on Earth is 10 ms-2)

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 29

Depth of the well = R/5 Linear mass density of the wire, μ = 10-3 kg m-1
We have to find the force applies at the top of the wire by a person holding it in.
Acceleration due to gravity on the planet will be given by g = GM/R2

(i) Acceleration due to gravity of earth
Dividing equation (i) by (ii), we get
Now, we will consider a mass element dm of width dx at depth x below the planet.
Mass of this element will be dm = μdx
Also, acceleration due to gravity at depth x below the planet will be
Force on the small segment of wire
Total force on the wire is obtained by integrating the above equation.

KEAM Paper 1 Mock Test - 2 - Question 30

A stone of mass m is attached to one end of a wire of cross-sectional area A and Young's Modulus Y. The stone is revolved in a horizontal circle at speed so that the wire makes an angle θ with the vertical, the strain produced in the wire is

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 30

The free-body diagram of the system is shown below:

[ac is centripetal acceleration]
Centripetal acceleration is in the horizontal direction as seen in the diagram. So, no acceleration is there in a vertical direction. So, the net force in a vertical direction is zero.
Thus

If L is the original length of the string, then the increase in length is given by ΔL = (T/A) (L/Y)
where, Y: Young's modulus of the wire, A: cross-sectional area of the wire

KEAM Paper 1 Mock Test - 2 - Question 31

When the temperature of a metal wire is increased from 0°C to 10°C, its length increases by 0.02%. The percentage change in its mass density will be closest to:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 31

KEAM Paper 1 Mock Test - 2 - Question 32

Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2 if the process is purely adiabatic and W3 if the process is purely isobaric. Then,

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 32

Area under the curve is work.
W2 < W1 < W3

KEAM Paper 1 Mock Test - 2 - Question 33

A leak proof cylinder of length 1 m, made of a metal which has very low coefficient of expansion, is floating vertically in water at 0°C such that its height above the water surface is 20 cm. When the temperature of water is increased to 4°C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4°C relative to the density at T = 0°C is close to: (Answer up to 2 decimal places)

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 33

KEAM Paper 1 Mock Test - 2 - Question 34

A monoatomic gas at pressure P and volume V is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 34

Constant entropy means the process is adiabatic.

KEAM Paper 1 Mock Test - 2 - Question 35

The efficiency of a Carnot's engine, working between steam point and ice point, will be:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 35

The ice point is 273 K and the steam point is 373 K.
Efficiency of the Carnot engine,

KEAM Paper 1 Mock Test - 2 - Question 36

A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. 'm' grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of 'm' is close to (Latent heat of water = 540 cal g-1, specific heat of water = 1 cal g-1 °C-1)

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 36

Heat lost by steam = heat gained by water
180 × 1 × (31 - 25) + 20 × (31-25) = m × 540 + m × 1 × (100-31)
180 × 6 +20 × 6 = 540m + 100 m - 31m
1080 + 120 = 640 m - 31m
1200 = 609m
m = 1200/609
= 1.97
m = 2

KEAM Paper 1 Mock Test - 2 - Question 37

The specific heat of water = 4200 J kg-1K-1 and the latent heat of ice = 3.4 × 105 J kg-1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (Answer upto 1 decimal place)

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 37

Here, the water will provide heat for ice to melt. Therefore,

= 0.0617 kg
= 61.7 grams
Remaining ice will remain unmelted.

KEAM Paper 1 Mock Test - 2 - Question 38

7 moles of certain monoatomic ideal gas undergoes a temperature increase of 40 K at constant pressure. The increase in the internal energy of the gas in this process is
(Given R = 8.3 JK-1 mol-1)

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 38

For a quasi-static process, the change in internal energy of an ideal gas is:
ΔU = nCV ΔT

[Molar heat capacity at constant volume for monoatomic gas = 3R/2]

KEAM Paper 1 Mock Test - 2 - Question 39

A Carnot's engine working between 400 K and 800 K has a work output of 1200 J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 39

KEAM Paper 1 Mock Test - 2 - Question 40

The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas from 0°C to 50°C when no work is done is ______. (R is the universal gas constant.

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 40

KEAM Paper 1 Mock Test - 2 - Question 41

The binding energy per nucleon of deuteron  and helium atom  is 1.1 MeV and 7MeV respectively. If two deuteron atoms react to form a single helium atom, the released energy, is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 41

Binding energy per nucleon of a deuteron

Binding energy per nucleon of helium

We have to find released energy if helium is formed from the fusion of two deuterons. Rest mass energy of deuteron,

KEAM Paper 1 Mock Test - 2 - Question 42

In an oscillating LC-circuit, the maximum charge on the capacitor is Q. The charge on this capacitor, when the energy is stored equally between the electric and magnetic fields, is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 42

An oscillating LC-circuit having a maximum charge on the capacitor as Q.
We have to find the charge on this capacitor when the energy is stored equally between the electric and magnetic fields.

Magnetic energy stored by the inductor at any instant is

Electrical energy stored by the capacitor at any instant is  [where, q is the charge at that instant]

Maximum electrical energy stored by the capacitor is

We are given that energy is equally stored between the electric and magnetic fields

By the principle of conservation of energy,

KEAM Paper 1 Mock Test - 2 - Question 43

Two particles A and B having charges 20 μC and -5 μC respectively are held fixed with a separation of 5 cm. At what position should a third charged particle be placed so that it does not experience a net electric force? (in integers)

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 43

KEAM Paper 1 Mock Test - 2 - Question 44

An annular ring with inner and outer radii R1 and R2  is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring F1/F is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 44

The annular ring is rolling with constant angular velocity ω. All the particles on the ring will have the same constant angular velocity ω. Since the two particles are moving in a circular orbit, the net force experienced by them is the centripetal force. The diagram given below represents the direction of forces acting on the two particles:

The centripetal force experienced by the particle on the inner circumference of radius R1
F1= mR1ωwhere m is the mass of each particle and ω the angular velocity of the ring.

Similarly, for the particle at the outer circumference of radius R2

Then, the ratio of the two forces is

KEAM Paper 1 Mock Test - 2 - Question 45

The I-V characteristic of a p-n junction diode in forward bias is shown in the figure. The ratio of dynamic resistance, corresponding to forward bias voltages of 2 V and 4 V respectively, is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 45

Dynamic resistance,

KEAM Paper 1 Mock Test - 2 - Question 46

Four identical particles of mass M are located at the corners of a square of side 'a'. What should be their speed if each of them revolves under the influence of other's gravitational field in a circular orbit circumscribing the square?

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 46

Net force on particle towards centre of circle is,

This force will act as centripetal force. Distance of particle from centre of circle is a/√2

KEAM Paper 1 Mock Test - 2 - Question 47

Out of a photon and an electron, the equation E=pc  is valid for:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 47

Total energy of a particle when it travels with a speed nearly equal to the speed of the light, the relativistic energy is given by

where, p: momentum of the particle, m : rest mass of the particle, c : speed of light
For photon, rest mass (m0) = 0
Therefore, equation (i) becomes
But for the electron rest mass is not zero, i.e., m0≠0
Therefore, E =pc for an electron.
thus, E = pc is valid for photon only.

KEAM Paper 1 Mock Test - 2 - Question 48

A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by  The corresponding electric field  is: (Speed of light, c = 3 x 108 ms-1)

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 48

KEAM Paper 1 Mock Test - 2 - Question 49

A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the inclined plane with horizontal is . The magnitude of the contact force will be:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 49

KEAM Paper 1 Mock Test - 2 - Question 50

In a resonance tube, using a tuning fork of frequency 325 Hz, two successive resonance lengths are observed at 25.4 cm and 77.4 cm respectively. The velocity of sound in air is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 50

For resonance, length of the tube is
Let first resonance occur at n, thus Length,  (given)
Successive resonance will occur at  n+1

length

Thus, the difference in lengths between the two resonances is given as

KEAM Paper 1 Mock Test - 2 - Question 51

An ideal gas at 27K is compressed adiabatically to 8/27 of its original volume. The rise in its temperature is:

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 51

Let initial volume V1 be V

Given: Final volume,

Initial temperature,

We have to find the rise in temperature, say T2.

KEAM Paper 1 Mock Test - 2 - Question 52

If gand gm are the accelerations due to gravity on the surface of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio to be;

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 52

The electric charge is a universal constant and thus, it has the same value on the Earth and on the Moon. It does not depend upon the value of acceleration due to gravity.

KEAM Paper 1 Mock Test - 2 - Question 53

Intensity of sunlight is observed as 0.092 Wm-2 at a point in free space. What will be the peak value of magnetic field at that point? (ε0 = 8.85 x 10-12C2N-1m-2)

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 53

Intensity of sunlight = I = 0.092 Wm-2

The peak value of electric and magnetic field are related as:

KEAM Paper 1 Mock Test - 2 - Question 54

In the arrangement shown in figure a1, a2, a3 and a4 are the accelerations of masses m1, m2, m3 and m4 respectively. Which of the following relation is true for this arrangement?

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 54

Using constraint

KEAM Paper 1 Mock Test - 2 - Question 55

Find the truth table for the function Y of A and B represented in the following figure.

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 55

KEAM Paper 1 Mock Test - 2 - Question 56

The truth table given below is for (A and B are the inputs, Y is the output). What type of GATE it will form?

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 56

Truth table with A and B as inputs and Y as output. We have to find which gate corresponds to the given truth table.
A. NOR gate can be made by using an OR gate followed by NOT gate , i.e. Given truth table does not satisfy this logic.

B. AND gate is a logic gate with the output as a product of the input, i.e.  The given truth table does not satisfy this logic.

C. XOR gate is a function of two logical variables A and B which gives the value 1 if one of the variables is zero and other is 1. If both the variables are 0 or both are 1, then the function is zero, i.e.,The given truth table does not satisfy this logic.

D. NAND gate is inverse of AND gate and is given by the expression

Truth table for NAND gate is

This satisfies the given truth table.

KEAM Paper 1 Mock Test - 2 - Question 57

An ideal gas is taken through the cycle  as shown in the figure. If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process  is

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 57

The given P-V graph of an ideal gas is reproduced below:

As the given process is cyclic, change in internal energy of the gas,

Using the first law of thermodynamics,

Thus, total work done by the gas is

Now, work done by the gas for any path is given as

From the figure, we get

is an isochoric process, thus
Therefore, total work done by the gas is given as

KEAM Paper 1 Mock Test - 2 - Question 58

The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit?

Detailed Solution for KEAM Paper 1 Mock Test - 2 - Question 58

KEAM Paper 1 Mock Test - 2 - Question 59

The value of acceleration due to gravity at Earth's surface is 9.8 ms-2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 ms-2, is close to:
(Radius of earth = 6.4 x 106 m)