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PCM - Mock Test (April 30) - JEE MCQ


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30 Questions MCQ Test Daily Test for JEE Preparation - PCM - Mock Test (April 30)

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PCM - Mock Test (April 30) - Question 1

If the position- time graph is a straight line parallel to the time axis

Detailed Solution for PCM - Mock Test (April 30) - Question 1

Explanation:Position-time graph of horizontal straight line parallel to time axis represents that the position of the body does not changes with the passage of time. So it represents the rest state of motion.It means velocity of object is zero.

PCM - Mock Test (April 30) - Question 2

A truck is coming towards you from a distance of 1000m on a straight road at time t=0. At time t=10 seconds it is at a distance of 900 m. The average velocity in m/sec is

Detailed Solution for PCM - Mock Test (April 30) - Question 2

PCM - Mock Test (April 30) - Question 3

A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final speed of 2.80 m/s. Find its original speed in m/s

Detailed Solution for PCM - Mock Test (April 30) - Question 3

PCM - Mock Test (April 30) - Question 4

For motion in a straight line

Detailed Solution for PCM - Mock Test (April 30) - Question 4

Explanation:To describe motion along a straight line, we can choose an axis, say X-axis, so that it coincides with the path of the object. We then measure the position of the object with reference to a conveniently chosen origin, say O,  Positions to the right of O are taken as positive and to the left of O, as negative.

PCM - Mock Test (April 30) - Question 5

The average acceleration 'a' over a time interval is defined as

Detailed Solution for PCM - Mock Test (April 30) - Question 5

Explanation:

Acceleration is the rate of change of velocity with time.It is a vector quantity.

Average acceleration is the change in velocity per unit time over an interval of time.

Let initial velocity at time interval t​​​​​​1 is v​​​​​​1 and final velocity at time interval t​​​​​2 is v​​​​​​2 Then

PCM - Mock Test (April 30) - Question 6

A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. Determine the time in seconds at which the stone reaches its maximum height.

g =9.8 m / sec2

Detailed Solution for PCM - Mock Test (April 30) - Question 6

PCM - Mock Test (April 30) - Question 7

A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final speed of 2.80 m/s. Find its acceleration in m/s2

Detailed Solution for PCM - Mock Test (April 30) - Question 7

PCM - Mock Test (April 30) - Question 8

Path length is defined as.

Detailed Solution for PCM - Mock Test (April 30) - Question 8

Explanation:Path length is defined as the total length of the path traversed by an object.

Unlike displacement, which is the total distance an object travels from a starting point, path length is the total distance travelled, regardless of where it travelled.

PCM - Mock Test (April 30) - Question 9

For motion with uniform acceleration, v-t graph is

Detailed Solution for PCM - Mock Test (April 30) - Question 9

Explanation:When velocity – time graph is plotted for an object moving with uniform acceleration, the slope of the graph is a straight line. 

PCM - Mock Test (April 30) - Question 10

If the kinetic energy of an electron is increased four times, the wavelength of the de-Broglie wave associated with it would become

Detailed Solution for PCM - Mock Test (April 30) - Question 10

The wavelength λ is inversely proportional to the square root of kinetic energy. So if KE is increased 4 times, the wavelength becomes half.

λ∝1/√KE

Hence Option A is the answer.

PCM - Mock Test (April 30) - Question 11

Calculate the wavelength (in nanometer) associated with a proton moving at 1.0×103ms-1 (Mass of proton = 1.67×10-27kg and h = 6.63×10-34Js)

Detailed Solution for PCM - Mock Test (April 30) - Question 11

Given mp = 1.67×10-27kg
h = 6.63×10-34Js
v = 1.0×103ms-1

We know wavelength λ = h/mv
∴λ = 6.63×10-34/(1.67×10-27 × 1.0×103)
= 0.40×10-10
≈ 0.40nm

Hence, Option (B) is the Correct Answer.

PCM - Mock Test (April 30) - Question 12

The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atom corresponding to which of the following

Detailed Solution for PCM - Mock Test (April 30) - Question 12

Correct Answer is A.

PCM - Mock Test (April 30) - Question 13

The maximum number of atomic orbitals associated with a principal quantum number 5 is

Detailed Solution for PCM - Mock Test (April 30) - Question 13

Number of orbitals in a shell = n2 = (5)= 25

PCM - Mock Test (April 30) - Question 14

A sub-shell with n = 6 , l = 2 can accommodate a maximum of

Detailed Solution for PCM - Mock Test (April 30) - Question 14

n = 6, l = 2 means 6d → will have 5 orbitals. 

∴ max 10 electrons can be accommodate as each orbital can have maximum of 2 electrons. 

PCM - Mock Test (April 30) - Question 15

Thomson’s plum pudding model explained:

Detailed Solution for PCM - Mock Test (April 30) - Question 15
  • An atom consists of a positively charged sphere with electrons filled into it. The negative and positive charges present inside an atom are equal and as a whole, an atom is electrically neutral.
  • Thomson’s model of the atom as compared to plum pudding and watermelon.
  • He compared the red edible part of the watermelon to a positively charged sphere whereas the seeds of watermelon to negatively charged particles.

electrical world

PCM - Mock Test (April 30) - Question 16

Which of the following conclusions could not be derived from Rutherford’s α -particle scattering experiment?

Detailed Solution for PCM - Mock Test (April 30) - Question 16
  • Concept of electrons moving in a circular path of fixed energy called orbits was put forward by Bohr and not derived from Rutherford's scattering experiment.
  • Out of a large number of circular orbits theoretically possible around the nucleus.
  • The electron revolves only in those orbits which have a tired value of energy Hence, these orbits are called energy level or stationary states.
PCM - Mock Test (April 30) - Question 17

The charge on electron was determined by:

Detailed Solution for PCM - Mock Test (April 30) - Question 17

The charge on electrons was determined by Milliken by using an oil drop experiment.

PCM - Mock Test (April 30) - Question 18

The nucleus of a tritium atom, 3H, contains

Detailed Solution for PCM - Mock Test (April 30) - Question 18
  • Tritium (3H) is the only radioactive isotope of hydrogen.
  • The nucleus of a tritium atom consists of a proton and two neutrons.
PCM - Mock Test (April 30) - Question 19

The wavelengths of two photons are 2000 Å and 4000 Å respectively. What is the ratio of their energies?

Detailed Solution for PCM - Mock Test (April 30) - Question 19



= 4000/2000
= 2

PCM - Mock Test (April 30) - Question 20

R = {(1, 1), (2, 2), (1, 2), (2, 1), (2, 3)} be a relation on A = {1, 2, 3}, then R is

Detailed Solution for PCM - Mock Test (April 30) - Question 20

A relation R on a non empty set A is said to be reflexive if fx Rx for all x ∈ R, Therefore , R is not reflexive.
A relation R on a non empty set A is said to be symmetric if fx Ry ⇔ yRx, for all x, y ∈ R Therefore, R is not symmetric.
A relation R on a non empty set A is said to be antisymmetric if fx Ry and y Rx ⇒ x = y, for all x, y ∈ R. Therefore, R is not antisymmetric.

PCM - Mock Test (April 30) - Question 21

If R is a relation from a set A to a set B and S is a relation from B to C, then the relation S  R.

Detailed Solution for PCM - Mock Test (April 30) - Question 21

If R is a relation from A→B and S is a relation fromB→C, then S ∘ R is a composite function from A to C.

PCM - Mock Test (April 30) - Question 22

The domain of definition of the function 

Detailed Solution for PCM - Mock Test (April 30) - Question 22

y is defined  If -x ≥ 0, i.e.

PCM - Mock Test (April 30) - Question 23

Detailed Solution for PCM - Mock Test (April 30) - Question 23

By definition, The Signum function =  

PCM - Mock Test (April 30) - Question 24

Let A = {1, 2, 3}. Which of the following is not an equivalence relation on A ?

Detailed Solution for PCM - Mock Test (April 30) - Question 24

A relation R on a non empty set A is said to be reflexive iff xRx for all x ∈ R . A relation R on a non empty set A is said to be symmetric iff xRy ⇔ yRx, for all x , y ∈ R .
A relation R on a non empty set A is said to be transitive iff xRy and yRz ⇒ xRz, for all x ∈ R. An equivalence relation satisfies all these three properties.
None of the given relations satisfies all three properties of equivalence relation.

PCM - Mock Test (April 30) - Question 25

The binary operation * defined on the set of integers as a∗b = |a−b|-1 is:

Detailed Solution for PCM - Mock Test (April 30) - Question 25

Here * is commutative as b*a = |b−a|−1 = |a−b|−1 = a∗b.
Because ,|−x| = |x| for all x ∈ R.

PCM - Mock Test (April 30) - Question 26

Let T be the set of all triangles in a plane with R a relation in T given by R = {(T1, T2) : T1 is congruent to T2}. Then R is

Detailed Solution for PCM - Mock Test (April 30) - Question 26

Let T be the set of all triangles in a plane with R a relation in T given by R = {(T1, T2): T1 is congruent to T2}. (T1T2) ∈ R iff T1 is congruent to T2
Reflexivity :T1≅T1 ⇒ (T1T1) ∈ R . Symmetry :(T1, T2)∈ R ⇒T1≅T2 ⇒ T2≅T1 ⇒ (T2, T1) ∈ R
Transitivity :(T1,T1) ∈ R and (T2, T3) ∈ R ⇒ T1≅T2 and T2≅T3 ⇒ T1≅T3 ⇒ (T1, T3) ∈ R .
Therefore, R is an equivalence relation on T.

PCM - Mock Test (April 30) - Question 27

The range of the function  

Detailed Solution for PCM - Mock Test (April 30) - Question 27

The given function is defined as Signum function. i.e.


domain is R and Range is {-1 , 0 , 1}.

PCM - Mock Test (April 30) - Question 28

The function f (x) = x2+ sin x is

Detailed Solution for PCM - Mock Test (April 30) - Question 28

For even function : f(-x) = f(x) , therefore, f(- x) = (− x)2+ sin (− x)
= x2−sin x ≠ f(x).
For an odd function : f(-x) = - f(x) , therefore, f(-x) = (− x)2+ sin (− x)
= x2− sin x
 ≠ − f(x).
Therefore f(x) is neither even nor odd.

PCM - Mock Test (April 30) - Question 29

R is a relation from { 11, 12, 13} to {8, 10, 12} defined by y = x – 3. The relation R−1

Detailed Solution for PCM - Mock Test (April 30) - Question 29

R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x – 3. The relation is given by x = y + 3,from{8, 10, 12} to {11, 12, 13} ⇒ relation = {(8,11),(10,13)}.

PCM - Mock Test (April 30) - Question 30

The range of the function f(x) = [sin x] is

Detailed Solution for PCM - Mock Test (April 30) - Question 30

The only possible integral values of sin x are {-1 ,0, 1}.

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