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UGEE SUPR Mock Test-2 - JEE MCQ


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30 Questions MCQ Test - UGEE SUPR Mock Test-2

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UGEE SUPR Mock Test-2 - Question 1

A metal surface is illuminated by the light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum KE of the emitted photoelectrons would be

Detailed Solution for UGEE SUPR Mock Test-2 - Question 1

The maximum kinetic energy of photootecirons is given by KEmax = h(v-v0) …(i)

Where, h = Planck's constant,

v = frequency of radiation

and v0 = threshold frequency.

It can be seen from Eq.(i), that the maximum KE of the emitted photoelectron is proportional to the frequency of the radiation and is independent of the intensity of radiation, so it remains unchanged.

UGEE SUPR Mock Test-2 - Question 2

A force (F) = acting on a particle causes a displacement (s) = in its own direction. If the work done is 14 J, then the value of ‘a’ is

Detailed Solution for UGEE SUPR Mock Test-2 - Question 2
Given,F=

S=

and W = 14 J

The work done by a force in displacing a particle through a distance is given by W = F.s …(i)

Substituting the above values in Eq. (i), we get

14 =

⇒ 14 = --15 + 14+ 3a

⇒ a = 15/3 = 5

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UGEE SUPR Mock Test-2 - Question 3

Light of wavelength ‘λ‘ is incident on a single slit of width 'a', and the distance between slit and screen is ‘D’. In diffraction pattern, if slit width is equal to the width of the central maximum, then 'D’ is equal to

Detailed Solution for UGEE SUPR Mock Test-2 - Question 3

The diffraction pattern due to a single slit is shown below

2y = 2Dλ/a gives the width of central maximumDλ/a. Where, λ = wavelength of incident light.

Here, a = 2y, then from Eq.(i), we get a = 2Dλ/a ⇒ D = a2/2λ.

UGEE SUPR Mock Test-2 - Question 4

A stretched string fixed at both ends has 'm' nodes, then the length of the string will be

Detailed Solution for UGEE SUPR Mock Test-2 - Question 4
For p number of loops in a stretched string, the length is given by l = pλ/2 …(i)

As, number of harmonics = number of loops = number of anti-nodes = p …(ii)

Also, number of nodes = number of anti-nodes + 1 Here, number of nodes = m

Number of anti-nodes = m - 1 from Eq. (ii)

p = m - 1

Putting this value of p in Eq. (i), we get

UGEE SUPR Mock Test-2 - Question 5

Three identical rods each of mass ‘M’ and length ‘L’ are joined to form a symbol ‘H’. The moment of inertia of the system about one of the sides of 'H’ is

Detailed Solution for UGEE SUPR Mock Test-2 - Question 5
The given situation can be shown as

Let us take the moment of inertia of the system about rod R1 then the total moment of inertia is lT = l1 + l2 + l3 …(i)

For rod R1, l1 = 0

For rod R2, using perpendicular axis theorem, l2 = ML2/3.

For rod R3, using parallel axis theorem, l3 = lcm + l(at L) = 0 + ML2 = ML2

Now, putting the values of l1, l2 and l3 in Eq. (i), we get

UGEE SUPR Mock Test-2 - Question 6

A block of mass ‘m’ moving on a frictionless surface at speed V collides elastically with a block of the same mass, initially at rest. Now the first block moves at an angle 'θ’ with its Initial direction and has speed ‘v1’. The speed of the second block after the collision is

Detailed Solution for UGEE SUPR Mock Test-2 - Question 6

Applying the law of conservation of kinetic energy, KE (before collision) = KE (after collision)

Thus, the velocity of the second block after the collision is

UGEE SUPR Mock Test-2 - Question 7

Two pendulums begin to swing simultaneously. The first pendulum makes nine full oscillations when the other makes seven. The ratio of the lengths of toe two pendulums is

Detailed Solution for UGEE SUPR Mock Test-2 - Question 7
As two pendulums begin to swing simultaneously, then n1T1 = n2T2 where, n1 and n2 are the number of oscillations of first and second pendulum respectively and T1 and T2 be their respective time periods.

The time period of a simple pendulum is given by

Where, I = length of the pendulum and g = acceleration due to gravity ⇒ T2 ∝ l …(ii)

So. from Eqs. (i) and (i), we get

Here, n1 = 9, n2 = 7

Hence, the ratio of pendulum lengths l1:l2 = 49.81.

UGEE SUPR Mock Test-2 - Question 8

Which one of the following statements is correct?

Detailed Solution for UGEE SUPR Mock Test-2 - Question 8
Surface tension is the force applied per unit length or work done (or energy) per unit area of a liquid surface. While surface energy is the amount of work done per unit area by force.
UGEE SUPR Mock Test-2 - Question 9

A wire of length ‘L’ and area of cross section ‘A' is made of material of Young’s modulus ‘Y’. It is stretched by an amount 'x'. The work done in stretching the wire is

Detailed Solution for UGEE SUPR Mock Test-2 - Question 9
If a force F is applied along the length L of wire for stretching by an amount x, then Young's modulus is given by

where, A = area of cross - sectional ⇒ F = (YA/L)x

The work done in stretching the wire is given by

UGEE SUPR Mock Test-2 - Question 10

An aircraft is moving with uniform velocity 150 m/s in space. If all the forces acting on it are balanced, then it will

Detailed Solution for UGEE SUPR Mock Test-2 - Question 10
As all the forces acting on the aircraft are balanced, so the net force on it will be zero, i.e., no external force act on it. Thus, the aircraft will keep moving with the same velocity of 150 m/s in the space.
UGEE SUPR Mock Test-2 - Question 11

A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between then is µ = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g = 10 ms–2 is

UGEE SUPR Mock Test-2 - Question 12

The force ‘F’ acting on a body of density ‘d’ are related b y the relation F = y/√d. The dimensions of 'y' are

Detailed Solution for UGEE SUPR Mock Test-2 - Question 12
The dimensions of force(F) = [MLT-2] and density [d] = (ML-3T0]

From the given relation, F = y/√d

⇒ y = F√d

Substituting the above dimensions, we get [y] = [F][d]½ = [MLT-2][ML-3T0]½=[M3/2L-1/2T-2]

UGEE SUPR Mock Test-2 - Question 13

The dimensions of self or mutual inductance/ are given as

Detailed Solution for UGEE SUPR Mock Test-2 - Question 13
The self or mutual inductance of a coil is the flux change due to change in current, i.e. L or M = ∅/l

So. dimensions of Mutual Inductance or Self-Inductance can be given by

⇒ [L] = [0]/[l]

UGEE SUPR Mock Test-2 - Question 14

Magnetic susceptibility of a paramagnetic substance is

Detailed Solution for UGEE SUPR Mock Test-2 - Question 14

Key Idea For paramagnetic substances, the magnetic susceptibility is small and positive, because they get feebly magnetised when placed in a magnetic field.

The magnetic susceptibility of a substance shows how easily a substance am be magnetised and given by xm = I/H

where, I = intensity of magnetisation,

and H = magnetic intensity of the field

UGEE SUPR Mock Test-2 - Question 15

A simple harmonic progressive wave Is represented as y = 0.03 sin π(2t - 0.01x) m. At a given Instant of time, the phase difference between two particles 25 m apart Is

Detailed Solution for UGEE SUPR Mock Test-2 - Question 15
The given equation of SHM wave is y = 0.03 sin n(2t - 0.01x)m

= 0.03 sin(2πt - 0.01πx) m

Comparing it with general equation, we get y = asin(ωt-kx)

where, k = 2π/λ⇒λ = 200m

The phase deference between two particles is given by Δ∅ = kx = 2π/λ x x …(i)

Here, x = 25m

Substituting the values of x and λ in Eq. (i). we get Δ∅ = (2π/200) x 25 = (π/4)rad

UGEE SUPR Mock Test-2 - Question 16

The magnetic dipole moment of a short magnetic dipole at a distant point along the equator of magnet has a magnitude of ‘X’ in SI units, tf the distance between the point and the magnet is halved then the magnitude of dipole moment win be

Detailed Solution for UGEE SUPR Mock Test-2 - Question 16
The magnetic dipole moment is the product of either of pole strength and the magnetic length of dipole. Thus, it is independent of the distance of point at which it is measured. So. it remains unchanged, if the distance between point and the magnet is halved.
UGEE SUPR Mock Test-2 - Question 17

If ‘x’, ‘v’ and 'a' denote the displacement, velocity and acceleration of a particle respectively executing SHM of periodic time t, then which one of die following does not change with time?

Detailed Solution for UGEE SUPR Mock Test-2 - Question 17
Key Idea For checking the correctness of an equation dimensional analysis is done using the principle of homogeneity.

The dimensions of given variables of SHM are as Displacement, [x] = [M0LT0] Velocity, [v] = [M0LT-1] Acceleration, [a] = [M0LT-2] and time period, [T] = [M0L0T]

Now, checking each option for these values.

For option (a)

As it depends on time, so change with it.

For option (b), [a][T] + 2π[v] = [M0LT-2][M0L0T] + [M0L0T-1] = [M0LT-1]

It is also dependent on time and hence changes with it.

For option (c),

As it is a constant having no dimension, so it does not change with time.

For option (d), [a][T] + 4π2[d]2 = [M0LT-2][M0L0T] + [M0LT-1]2 = [LT-1]+[L2T-2]

As the term is dependent on time, so changes with it.

Also, it is dimensionally incorrect.

Hence, option (c) is correct.

UGEE SUPR Mock Test-2 - Question 18

The number of σ and π-bonds in 2-formylbenzoic acid are respectively

Detailed Solution for UGEE SUPR Mock Test-2 - Question 18

Structure of 2 -formyl benzoic acid Is

Thus, it has 17 σ and 5π bonds.

UGEE SUPR Mock Test-2 - Question 19

Veronal is used as a/an

Detailed Solution for UGEE SUPR Mock Test-2 - Question 19

Veronal is the trade name for barbiturate drugs. It is used as tmnquiizor, these are the drugs used for the treatment of anxiety, fear, tension and mental illness. It’s structure is

UGEE SUPR Mock Test-2 - Question 20

The oxidation number of sulphur in S8 molecule is

Detailed Solution for UGEE SUPR Mock Test-2 - Question 20
Oxidation number of sulphur is zero in S8 molecule as it is a monatomic molecule.
UGEE SUPR Mock Test-2 - Question 21

Which of the following acts as an oxidising agent in hydrogen-oxygen fuel ceil?

Detailed Solution for UGEE SUPR Mock Test-2 - Question 21
Hydrogen-oxygen fuel cell is an electrochemical cell that converts the chemical energy of hydrogen which is a fuel and oxygen which act as an oxidising agent into electricity through a pair of redox reactions. Thus, option (b) is correct.
UGEE SUPR Mock Test-2 - Question 22

According to Werner’s theory the geometry of the complex is determined by

Detailed Solution for UGEE SUPR Mock Test-2 - Question 22

Werner’s theory was used to describe the structure and formation of complex compounds or coordination compounds. According to this theory the primary valency gives the oxidation number and the secondary valency gives the coordination number. Also, the geometry of the complex is determined by the number and position of secondary valences in space as the ligand satisfying secondary valences are always Greeted towards the fixed position in space.

UGEE SUPR Mock Test-2 - Question 23

The correct representation of Nernst’s equation for half-cell reaction Cu2+(aq) + e- → Cu+(aq) is

Detailed Solution for UGEE SUPR Mock Test-2 - Question 23

Key Idea General Nernst equation for a reaction is given as

For the half-cell Cu2+(aq) + e- → Cu+(aq)

The correct Nernst's equation is

UGEE SUPR Mock Test-2 - Question 24

Identify the equation in which change in enthalpy is equal to change in internal energy

Detailed Solution for UGEE SUPR Mock Test-2 - Question 24

Key Idea Relationship between change in enthalpy and change in internal energy is given as:

ΔH = ΔU + ΔngRT

Is the given options,

(a) 2H2O2(l) → 2H2O(l) + O2(g) Δng = 1

∴ ΔH = ΔU + RT

(b) C(s) + O2(g) → CO2(g)

Δng = 1 - 1 = 0

∴ ΔH = ΔU

(c) PCl3(g) → PCl3(g) + Cl2(g)

Δng = 2 - 1 = 1

∴ ΔH = ΔU + RT

(d) N2(g) + 3H2(g) → 2NH3(g)

ΔngRT = 2 - 3 = -1

∴ ΔH = ΔU - RT

Equation given in option (b) has enthalpy change equal to internal change.

UGEE SUPR Mock Test-2 - Question 25

Which among the following does not form polyhalide ions?

Detailed Solution for UGEE SUPR Mock Test-2 - Question 25
When the halide ions combine with halogen molecules or interhalogen, univalent ions are obtained. These are known as polyhalide ions. Among the given options. F doesn't form polyhalide ion because it doesn't have d-orbitals and cannot show a higher oxidation state.
UGEE SUPR Mock Test-2 - Question 26

Which of following elements does not form amide when reacted with ammonia?

Detailed Solution for UGEE SUPR Mock Test-2 - Question 26
Lithium doesn't form amide when reacted with ammonia. It forms tetraammine lithium, Li(NH3)4.

The equation for the reaction is as follows:

UGEE SUPR Mock Test-2 - Question 27

α-chloro sodium acetate on boiling with aqueous sodium nitrite gives

Detailed Solution for UGEE SUPR Mock Test-2 - Question 27
α - Chlorthe sodium acetate on boiling with aqueous sodium nitrite gives nitromethane. The reaction can be written as

UGEE SUPR Mock Test-2 - Question 28

How the molecular formula C4H10O represents many metameric ethers?

Detailed Solution for UGEE SUPR Mock Test-2 - Question 28

There are 4-metameric ethers that are represented by the molecular formula C4H10. Which are as follows

UGEE SUPR Mock Test-2 - Question 29

Which of the following metals occurs in native state?

Detailed Solution for UGEE SUPR Mock Test-2 - Question 29
Among the given metals, platinum has the least reactivity thus, it occurs in the native state. While all other metals occur in their oxides, sulphides, and chloride forms.
UGEE SUPR Mock Test-2 - Question 30

The oxidation state of sulphur in H2S2O7 is

Detailed Solution for UGEE SUPR Mock Test-2 - Question 30
Let, the oxidation state of sulphur in H2S2O7 be x.

∴ 2(+1) + 2(x) + 7(-2) = 0

2 + 2x - 14 = 0

2x = 12 x = +6

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