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KEAM Paper 1 Mock Test - 1 - JEE MCQ


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30 Questions MCQ Test - KEAM Paper 1 Mock Test - 1

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KEAM Paper 1 Mock Test - 1 - Question 1

The energy required to remove the electron from a singly ionized Helium atom is 2.2 times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is:

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

On removing an electron, the Helium atom becomes a single electron atom.
Applying Bohr's model, the energy of the single electron is given by 
(where Z is atomic number and n is number of the shell in which the electron resides).
Thus, the energy required is:

Energy required to remove the electron from singly ionized Helium atom = 54.4 eV

Energy required to remove the electron from Helium atom = x eV

So, 54.4 eV = 2.2x

x = 24.73 eV

Total energy required to ionize the Helium atom completely = 54.4 + 24.73 = 79.13 eV

Hence, the nearest option is 'C'.

KEAM Paper 1 Mock Test - 1 - Question 2

The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the volume of the sheet to correct significant figures.

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

Length , l = 4,234 m
Breadth, b = 1.005 m
Thickness , t = 2.01 x 10-2 m
Volume = l x b x t
⇒ V = 4.234 x 1.005 x 0.0201 = 0.0855289 = 0.0855 m3 (significant figure = 3)

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KEAM Paper 1 Mock Test - 1 - Question 3

A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure. When the string is cut, the initial angular acceleration of the rod is:
[2013]

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

Weight of the rod will produce the torque,
τ = mg x L / 2 = I α = mL2 / 3 α (∵ Irod = ML2 / 3)
Hence, angular acceleration, α = 3g / 2L

KEAM Paper 1 Mock Test - 1 - Question 4

A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is   

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

KEAM Paper 1 Mock Test - 1 - Question 5

An object is said to be in uniform motion in a straight line if its displacement

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

Explanation:Uniform motion is the kind of motion in which a body covers equal displacement in equal intervals of time. It does not matter how small the time intervals are, as long as the displacements covered are equal.

If a body is involved in rectilinear motion and the motion is uniform, then the acceleration of the body must be zero.

KEAM Paper 1 Mock Test - 1 - Question 6

Two cars are moving in same direction with speed of 30 kmph. They are separated by a distance of 5 km. What is the speed of a car moving in opposite direction if it meets the two cars at an interval of 4 min?   

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

Lets say the speed of second car be v. Now with respect to car 1, speed of car 2 is v - 30. Now if the cross each other at just after 4 min then the distance traveled by car 2 relative to car 1 is (v -30) x 4/60 and we know the distance travelled relatively is 5km
Thus we get 
(v -30) x 4/60 = 5
V = 300/4 - 30
= 45 km/h

KEAM Paper 1 Mock Test - 1 - Question 7

A body starts from rest, the ratio of distances travelled by the body during 3rd and 4thseconds is :

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

The velocity after 2 sec = 2a and velocity after 3 sec = 3a
For some constant acceleration a,
Now distance travelled in third second, s3 = 2a + ½ a
= 5/2 a
Similarly distance travelled in fourth second, s4= 3a + ½ a
= 7/2 a
Hence the reqiuired ratio is 5/7

KEAM Paper 1 Mock Test - 1 - Question 8

The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 × 107)ct + sin(6.28 × 107)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
(c = 3 × 10ms-1, h = 6.6 × 10-34 J-s)

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

B = B0sin(π × 107C)t + B0sin (2π × 107C)t
Since there are two EM waves with different frequency, to get maximum kinetic energy, we take the photon with higher frequency.
B1 = B0sin(π × 107C)t  v1 = 107/2 x c 
B2 = B0sin(2π × 107C)t v2 = 107C
where C is speed of light C = 3 × 108 m/s
v2 > v1
so KE of photoelectron will be maximum for photon of higher energy.
v= 107C Hz
hv = φ + KEmax
energy of photon
Eph = hv = 6.6 × 10-34 × 107 × 3 × 109
Eph = 6.6 × 3 × 10-19J

KEmax = Eph - φ
= 12.375 - 4.7 = 7.675 eV  7.72 eV

KEAM Paper 1 Mock Test - 1 - Question 9

One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27°C. The work done on the gas will be:

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

Work done in isothermal process on the gas,

KEAM Paper 1 Mock Test - 1 - Question 10

The electric field at a point associated with a light wave is E = (100 Vm −1) sin [(3.0 ×1015s−1)t] sin [(6.0×1015s−1)t]. If this light falls on a metal surface having a work function of 2.0 eV, what will be the maximum kinetic energy of the photo electrons ?

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

KEAM Paper 1 Mock Test - 1 - Question 11

A ray of light is incident at an angle of 60° on one face of a prism of angle 30°. The emergent ray of light makes an angle of 30° with incident ray. The angle made by the emergent ray with second face of prism will be:

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

For the prism, angle of deviation is given by:

Hence, emergent ray will be perpendicular to the second face or we say that the angle made by the emergent ray is 90°.

KEAM Paper 1 Mock Test - 1 - Question 12

Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths λN and λA, respectively. The ratio λNis closest to:

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

Energy of the emitted photon,

where EA and EN are energies of photons from atom and nucleus, respectively.

EA is order of eV and EN is order of MeV

KEAM Paper 1 Mock Test - 1 - Question 13

A gravity meter can detect change in acceleration due to gravity (g) of the order of 10-9%. Calculate the smallest change in altitude near the surface of the earth that results in a detectable change in g. Radius of the earth R = 6.4 x 106m.

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

KEAM Paper 1 Mock Test - 1 - Question 14

The diagram of a logic circuit is given below. The output F of the circuit is given by:

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

By law of distribution of Boolean Algebra, we have A+ (B. C) = (A+B). (A + C)
Step 1: Write outputs for the OR gates with inputs W and X, and W and Y.
Input for first OR gate are W and X. The output for this OR gate is, Y1 = W + X
Input for second OR gate are W and Y. The output for this OR gate is, Y2 = W + Y
Step 2: Write output for AND gate whose inputs are the output of OR gate.
F = Y1 • Y⇒ F = (W + X). (W + Y)
By using Law of distribution of Boolean Algebra, A + (B. C) = (A+B). (A + C)
Therefore, F = W + (X. Y)

KEAM Paper 1 Mock Test - 1 - Question 15

In the circuit shown in the Figure, cell is ideal and R2 = 100Ω. A voltmeter of internal resistance 200Ω reads V12 = 4 V and V23 = 6 V between the pair of points 1 - 2 and 2 - 3 respectively. What will be the reading of the voltmeter between the points 1 - 3. 

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


Let emf of the cell be E. Current through the voltmeter (when connected between 1-2) is 
Current through R1 = 4/R1
∴ Current through 
∴ Potential difference across 

When the voltmeter is connected between 2 - 3
Current through voltmeter = 6/200 A
Current through R2 = 6/100 A
∴ Current through


From (1) and (2) 
Put this in 
When connected across 1-3, the voltmeter will read E = 12 V.

KEAM Paper 1 Mock Test - 1 - Question 16

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

Given, Z 50, A - Z = 120 - 50 = 70
Δm = Z.mp + (A - Z) mn - M= [50 × 1.007825 +70 × 1.008665 - 119.902199] = 1.095601 MeV
E = 1.095601 × 931.478 MeV = 1020.53 MeV ≈ 1021 MeV

KEAM Paper 1 Mock Test - 1 - Question 17

The astronomical phenomenon when the planet Venus passes directly between the Sun and the earth is known as Venus transit. For two separate persons standing on the earth at points M and N, the Venus appears as black dots at points M' and N' on the Sun. The orbital period of Venus is close to 220 days. Assuming that both earth and Venus revolve on circular paths and taking distance MN = 1000 km, calculate the distance M'N' on the surface of the Sun.
[Take (2.75)1/3 = 1.4]

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

Let radius of circular orbit of the Earth and Venus be re and rv respectively (rc/rv) = (365/220)2 [Kepler's third law] 

From the drawing given in the problem M'N'/MN = N'V/NV

KEAM Paper 1 Mock Test - 1 - Question 18

Consider the following figure:

Which of the following labelled points in the figure given above indicate an unstable state of an object?

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

Explanation: 

  • From the given graph we can see that the potential energy is minimum at point B and D
  • Whereas it has maximum potential energy at point A and C respectively
  • Hence point A and C will be an unstable state since they have maximum potential energy compared to B and D.
KEAM Paper 1 Mock Test - 1 - Question 19

A straight wire of length L and radius a has a current I. A particle of mass m and charge q approaches the wire moving at a velocity v in a direction anti parallel to the current. The line of motion of the particle is at a distance r from the axis of the wire. Assume that r is slightly larger than a so that the magnetic field seen by the particle is similar to that caused by a long wire. Neglect end effects and assume that speed of the particle is high so that it crosses the wire quickly and suffers a small deflection θ in its path. Calculate θ.

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




Force on the particle is 
This force is always perpendicular to the velocity. Since deflection is small, the force is nearly in (↑) direction always.
Impulse is: 

KEAM Paper 1 Mock Test - 1 - Question 20

A water barrel having water up to depth 'd' is placed on a table of height 'h'. A small hole is made on the wall of the barrel at its bottom. If the stream of water coming out of the hole falls on the ground at a horizontal distance 'R' from the barrel, then the value of 'd' is:

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


By Bernoulli's equation: mgd = 1/2 mv2
v = √2gd ........... (1)
By Projectile motion equation: 1/2 gt2 = h

Therefore, 
So, d = R2/4h

KEAM Paper 1 Mock Test - 1 - Question 21

A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let P2 be the pressure inside the inner bubble and P0 be the pressure outside the outer bubble. Radius of another bubble with pressure difference P2 - P0 between its inside and outside would be

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


KEAM Paper 1 Mock Test - 1 - Question 22

A planet of mass M is revolving around sun in an elliptical orbit. If dA is the area swept in a time dt, angular momentum can be expressed as

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

KEAM Paper 1 Mock Test - 1 - Question 23

A particle is projected upward from the surface of earth (radius = R) with a speed equal to the orbital speed of a satellite near the earth’s surface. The height to which it would rise is

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

  (orbital speed υ0 of a satellite)

Near the earth’s surface is equal to 1/√2
times the escape velocity of a particle on earth’s surface) Now from conservation of mechanical energy: Decrease in kinetic energy = increase in potential energy

KEAM Paper 1 Mock Test - 1 - Question 24

A spherical hole is made in a solid sphere of radius R. The mass of the original sphere was M.The gravitational field at the centre of the hole due to the remaining mass is

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

By the principle of superposition of fields

Here  = net field at the centre of hole due to entire mass

 = field due to remaining mass

 = field due to mass in hole = 0

KEAM Paper 1 Mock Test - 1 - Question 25

Three particles P, Q and R placed as per given figure. Masses of P, Q and R are √3 m, √3 m and m respectively. The gravitational force on a fourth particle ‘S’ of mass m is equal to

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

in horizontal direction  

in vertical direction

KEAM Paper 1 Mock Test - 1 - Question 26

A cylindrical tank has a hole of diameter 2r in its bottom. The hole is covered wooden cylindrical block of diameter 4r, height h and density ρ/3.

Situation I : Initially, the tank is filled with water of density ρ to a height such that the height of water above the top of the block is h1 (measured from the top of the block).

Situation II : The water is removed from the tank to a height h2 (measured from the bottom of the block), as shown in the figure.
The height h2 is smaller than h (height of the block) and thus the block is exposed to the atmosphere.

Q. Find the minimum value of height h1 (in situation 1), for which the block just starts to move up?    

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

Consider the equilibrium of wooden block.
Forces acting in the downward direction are

(a) Weight of wooden cylinder

(b) Force due to pressure (P1) created by liquid of height h1 above the wooden block is
= P1 × π (2r)2 = [P0 + h1rg] × π (2r)2
Force acting on the upward direction due to pressure P2 exerted from below the wooden block and atmospheric pressure is

At the verge of rising

KEAM Paper 1 Mock Test - 1 - Question 27

If the elastic limit of copper is 1.5 × 108 N/ m2, determine the minimum diameter a copper wire can have under a load of 10.0 kg if its elastic limit is not to be exceeded.

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

KEAM Paper 1 Mock Test - 1 - Question 28

A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 400 N is applied to each end of the wire. What minimum diameter is required for the wire?

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

*Multiple options can be correct
KEAM Paper 1 Mock Test - 1 - Question 29

At what temperature is the r.m.s velocity of a hydrogen molecule equal to that of an oxygen molecule at 47°C?

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

KEAM Paper 1 Mock Test - 1 - Question 30

Heat generated through a resistive wire will increase _______ times in unit time, if current in the wire becomes twice. 

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

So heat is given by 

H = I2 Rt

time is fixed, unit time, Resistance of wire will remain the same. 

H ∝ I

So,

H = k I2    ---(1)

if the current becomes twice'

H' = k (2I)   ---(2)

(2) divided by (1) we get

So, H' : H = 4 : 1

So we see, the heat generated in unit time becomes 4 times of its initial value.

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