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UGEE SUPR Mock Test- 5 - JEE MCQ


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30 Questions MCQ Test - UGEE SUPR Mock Test- 5

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UGEE SUPR Mock Test- 5 - Question 1

A pipe open at both ends and a pipe closed at one end have the same length. The ratio of frequencies of their Pth overtone is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 1
Frequency of pth overtone of the open organ pipe

where L = length of an organ pipe and that of closed organ pipe is

So, the ratio of pth overtone of open to closed.

UGEE SUPR Mock Test- 5 - Question 2

Maximum kinetic energy gained by the charged particle in the cyclotron is independent of

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 2
Maximum kinetic energy gained by a charged particle in a cyclotron is given by

where q = charge of the cyclotron,

B = intensity of the magnetic field,

R = radius of orbit

And the m = mass of the particle,

Hence, EK is independent of frequency of revolution.

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UGEE SUPR Mock Test- 5 - Question 3

The magnifying power of a telescope is nine. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal length of the objective and eyepiece are respectively

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 3
For the final image at infinity, magnifying power of a telescope is given by

where, m = magnification,

fο = focal length of objective

and fe = focal length of eyepiece

⇒ fο = 9fe …(i)

Also, distance between objective and eyepiece

⇒ fο = fe = (given)

⇒ 9fe + fe = 20 ⇒ fe = 2 cm

⇒ fο = 9fe = 18 cm

UGEE SUPR Mock Test- 5 - Question 4

In a hydrogen atom, an electron of charge e revolves in a orbit of radius r with speed v. Then, magnetic moment associated with electrons is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 4
Magnetic moment of the revolving electron,

μe = i x A

where i = current and A = area.

where T = time period of revolution and r = radius orbit

where, v = velocity of revolving electron

⇒ μe = evr/2

Hence, the magnetic moment of a revolving electron is evr/2.

UGEE SUPR Mock Test- 5 - Question 5

The stopping potential of the photoelectrons from a photocell is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 5
The stopping potential of photoelectrons is the potential needed to stop the electrons from reaching the collector depends only on the frequency of incident radiation, ft is directly proportional to it.

This is because, more the frequency of incident light, more will be the energy of the photons incident photons. (E = hv)

Thus, more will be the energy (kinetic) acquired from the electrons.

The more the kinetic energy of emitted electrons, higher is the potential needed to stop them (stopping potential).

UGEE SUPR Mock Test- 5 - Question 6

Consider a particle of mass m suspended by a string at the equator. Let R and M denote the radius and mass of the earth. If ω is the angular velocity of rotation of the earth about its own axis, then the tension on the string will be (cos 0º = 1)

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 6
When a body suspended by the string situated at position P as shown in the figure, where latitude is λ, then the body is also rotated with angular frequency (ω) of earth. Hence tension on the string is given by

T = mg - mω2 cos λ

When the body is suspended at the equator, then λ = 0 and r = R

∴ From Eq. (i), we have

UGEE SUPR Mock Test- 5 - Question 7

The fundamental frequency of the sonometer wire Increases by 9 Hz, if its tension is increased by 69%, keeping the length constant. The frequency of the wire is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 7
We know that, frequency of vibration of a hed string is given by

where, T = tension in string

m = mass/length

and l = length of string.

New frequency,

⇒ 13v = 10v + 90

v = 30Hz

UGEE SUPR Mock Test- 5 - Question 8

A potentiometer wire has length L For a given cell of emf E, the balancing length is L/3 from the positive end of the wire. If the length of potentiometer wire is increased by 50% , then for the same cell, the balance point is obtained at length

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 8
For the cell of emf E balancing length

Increased length of wire

Since the cell is the same, therefore the new balancing length = 1/3 (new increased length)

= L/2

Therefore, for the same cell the balance point will be as length L/2 from positive end.

UGEE SUPR Mock Test- 5 - Question 9

A mass is whirled in a circular path with constant angular velocity and its linear velocity is v. If the string is now halved keeping the angular momentum the same, the linear velocity is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 9
Angular velocity = ω

linear velocity = v

Length of string = Radius = r

If string is halved, r’ = r/2

Angular momentum. L = r x p

where , r = radius vector

and p = linear momentum

= r x mv

= mvt

As L remain constant,

mvr = constant

⇒ r' → r/2 , v' = 2v such that

mv' r' = m 2v x (r/2)

= m v r

Hence, linear velocity will be 2 v.

UGEE SUPR Mock Test- 5 - Question 10

In Young’s double slit experiment, a fifth dark fringe is formed opposite to one of the slit. If D is the distance between the slits and the screen and d is the separation between toe slits, then toe wavelength of fight used is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 10
Position of nth dark fringes in Young's double slit experiment is given

Since, fifth dark fringe is formed opposite to one of the slit.

∴ 2x5 = d (distance between slits]

Put n = 5 in Eq. (i). we get

UGEE SUPR Mock Test- 5 - Question 11

A 220 V input is supplied to a transformer. The output circuit draws a current of 2.0 A at 440 V. If the ratio of output to input power is 0.8, then the current drawn by primary winding is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 11
Given, input voltage supplied to the transformer, V1 = 220 V

Current at output circuit of transformer, i2 = 2A

V2 = 440V

Hence, the current drawn by the primary windings is 5A.

UGEE SUPR Mock Test- 5 - Question 12

If the speed of an electron of hydrogen atom in the ground state is 2.2 x 106 m/s, then its speed in the third excited state will be

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 12

speed of electron of the hydrogen atom in the ground state,

v1 = 2.2 x 106 m/s

Since

[for third excited state. n2 = 4]

5.5 x 105 m/s

UGEE SUPR Mock Test- 5 - Question 13

The vectors (A + B) and (A - B) are at right angles to each other. This Is possible under the condition

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 13
Vectors (A + B) and (A - B) are at the right angle to each other, therefore

(A + B) · (A - B) = 0

A·A + B·A - A·B - B·B = 0

|A| + BAcosθ - AB cosθ - |B|2 = 0

|A|2 = |B|2

Hence. |A| = |B|

UGEE SUPR Mock Test- 5 - Question 14

A person measures a time period of a simple pendulum inside a stationary lift and finds it to be T. If the lit starts accelerating upwards with an acceleration (g/3) the time period of the pendulum will be

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 14

Time period of simple pendulum inside a stationary lift

..(i)

where, l = length of string and

g = gravitational acceleration

Acceleration of lift in upward direction,

a = g/3

∴ Time period of the pendulum

UGEE SUPR Mock Test- 5 - Question 15

In damped SHM, the Sl unit of damping constant is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 15
In damped SHM, damping force is proportional to the velocity of the oscillator

i.e., Fd ∝ v ⇒ Fd = bv

where, b is a damping constant.

∴ Damping constant. b = Fd/v

SI unit of damping constant

Hence, the SI unit of damping constant is kg/s.

UGEE SUPR Mock Test- 5 - Question 16

When a certain metal surface is illuminated with a light of wavelength λ, the stopping potential is V, When the same surface is illuminated by light of wavelength 2λ, the stopping potential is (V/3). The threshold wavelength for the surface is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 16
Given for a metal, the wavelength of light used = λ,

Stopping potential = V

If λ0 be the threshold wavelength, then the maximum kinetic energy of emitted electrons.

Again, the wavelength of used light

λ' = 2λ

Stopping potential V' = V/3

Then

From Eqs (i) and (ii), we have

⇒ λ0 = 4λ

So, the threshold wavelength is 4 times of wavelength of light.

UGEE SUPR Mock Test- 5 - Question 17

A hole is drilled half way to the centre of the earth. A body weighs 300 N on the surface of the earth. How much wil, it weigh at the bottom of the hole?

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 17
Given, the distance of the bottom of the hole from the surface of the earth (d) = half the radius of earth = Rο/2

If g be the value of gravitational acceleration on the surface of the earth, then weight of body mg = 300N

If g' be the gravitational acceleration at the bottom of the hole, then

∴ Weight of the body on the bottom erf hole,

mg' = mg/2 = 300/2 = 150N

UGEE SUPR Mock Test- 5 - Question 18

The monomers used in the preparation of dextron are

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 18

The monomers used in the preparation of dextron are lactic acid and glycolic acid. There action for the formation of dexron can be written as

UGEE SUPR Mock Test- 5 - Question 19

In which oxidation state, group 15 elements act as Lewis base?

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 19
Group 15 elements acts as Lewis base in - 3 oxidation state due to availability of lone pair of electrons on the central atom. This basic character of group 15 decreases down the group with an increase in the size of the central atom.
UGEE SUPR Mock Test- 5 - Question 20

Which of the following elements does not react with hot concentrated sulphuric acid?

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 20
Nitrogen doesn’t react with concentrated sulphuric acid as its bond dissociating energy is highest among the given elements. This is because of its small size.
UGEE SUPR Mock Test- 5 - Question 21

The tonic charges of m anganate and permanganate ion are respectively

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 21
Chemical formula for manganate and permanganate ion respectively are MnO2-4 and MnO-4. Thus, the ionic charge of manganate and permanganate ion are respectively -2 and -1.
UGEE SUPR Mock Test- 5 - Question 22

The enzyme which converts maltose to glucose is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 22

The enzyme which converts maltose to glucose is maltase. During digestion, starch is partially transformed into maltose by pancreatic or salivary enzymes called amylases; maltase secreted by the intestine men converts maltose into glucose. The glucose so produced is either utilised by the body or stored in the liver as glycogen.

UGEE SUPR Mock Test- 5 - Question 23

What is the atomicity of aluminium phosphate?

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 23
The chemical formula for aluminium phosphate is AIPO4.

Atomicity of heteronuclear molecule = total number of atoms present

∴ Atomicity of AlPO4 = 1 +1 + 4 = 6

UGEE SUPR Mock Test- 5 - Question 24

Standard hydrogen electrode (SHE) is a

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 24
Standard hydrogen electrode (SHE) is the primary reference electrode. SHE is represented by Pt(s)|B2(g)|H+(ag), is assigned a zero potential at all temperatures corresponding to reaction.

UGEE SUPR Mock Test- 5 - Question 25

Which of the following sets of solutions of urea (mol, mass 60 g mol-1) and sucrose (mol. mass 342 g mol-1) is isotonic?

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 25
Key idea: Isotonic solutions are those solutions which have the same osmotic pressure at a given temperature.

Formula for osmotic pressure, π = CRT

considering the set given in option (d), i.e. 30 gL-1 urea and 17.1 gL-1 sucrose.

Given, the molecular mass of urea 60 g mol-1 and molecular mass of sucrose 342 g mol-1.

For urea.

conc, C = 3/60 = 1/20

Osmotic pressure π1 = (1/20) x R x T

For sucrose conc, C = 17.1/342 = 1/20

∴ Osmotic pressure, π2 = (1/20) RT

Thus, the set of solutions of urea and sucrose given in option (d) is isotonic.

UGEE SUPR Mock Test- 5 - Question 26

Which of the following is a natural polymer?

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 26
Natural polymers are those polymers which are found in plants and animals. Among the given polymers liner is a natural polymer which is made from cellulose (polymer of glucose molecules).
UGEE SUPR Mock Test- 5 - Question 27

When a mixture of manganese dioxide, potassium hydroxide and potassium chlorate is fused, the product obtained is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 27
When a mixture of manganese dioxide, potassium hydroxide and potassium chlorate is fused then potassium chlorate first decomposes to give potassium chloride and oxygen gas. The formed O2 gas then reacts with MnO2 and KOH to give K2MnO4. The equations for the above reaction can be written as

2MnO2 + 4KOH + O2 → 2K2MnO2 + 2H2O

UGEE SUPR Mock Test- 5 - Question 28

Relationship between van't Hoff's factor (i) and degree of dissociation (α) is

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 28
Relationship between van't Hoff factor(i) and degree of dissociation (α) is given by

α = 1-i/n'-1

where, n is the number of ions formed alter dissociation.

The relationship can be obtained as follows;

For the reaction, A ⇌ n'B

Initially 1 mole 0

After dissociation (1 - α) mole n' α

Total number of moles present in the solution

= (1 - α) + n'α = 1 + (n'-1)α

van't Hoff factor, i = 1 + (n' - 1), α > 1 if n' ≥ 2

∴ α = i-1/n'-1

UGEE SUPR Mock Test- 5 - Question 29

In the reaction iodide ton acts as

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 29
For there action,

Iodide ion acts a homogenous catalyst as the phase of I- and reactant, H2O2 is aqueous (same).

UGEE SUPR Mock Test- 5 - Question 30

How many gram of sodium (atomic mass 23 u) is required to prepare one mole of ethane from methyl chloride by Wurtz reaction?

Detailed Solution for UGEE SUPR Mock Test- 5 - Question 30
The Wortz reaction between methyl chloride and sodium metal in presence of dry ether can be written as

⇒ 2 moles of sodium metal reacts to give 1 mole of ethane.

Also,

1 mole of Na metal = 23 g

∴ 2 motesol Na metal = 23x 2 = 46g

Thus, 46 g of sodium is required to prepare one mole of ethane from methyl chloride by Wurtz reaction.

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