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JEE Main Practice Test- 5 - JEE MCQ


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30 Questions MCQ Test - JEE Main Practice Test- 5

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JEE Main Practice Test- 5 - Question 1

A metal wire PQ slides on parallel metallic rails having separation 0.25 m, each having negligible resistance. There is a 2Ω resistor and 10V battery as shown in figure. There is a uniform magnetic field directed into the plane of the paper of magnitude 0.5 T. A force of 0.5N to the left is required to keep the wire PQ moving with constant speed to the right. With what speed is the wire PQ moving? (Neglect self-inductance of the loop)

Detailed Solution for JEE Main Practice Test- 5 - Question 1



Solving V = 16 m/s.

JEE Main Practice Test- 5 - Question 2

In the given figure a ring of mass m is kept on a horizontal surface while a body of equal mass 'm' attached through a string, which is wounded on the ring. When the system is released the ring rolls without slipping. Consider the following statements and choose the correct option.

(i) acceleration of the centre of mass of ring is g/3
(ii) acceleration of the hanging particle is 2g/3
(iii) frictional force (on the ring) acts along forward direction
(iv) frictional force (on the ring) acts along backward direction

Detailed Solution for JEE Main Practice Test- 5 - Question 2

Free body diagram of ring and mass is 

Since the ring is in pure rolling so its above point acceleration will be 2a and similarly acceleration of mass will be 2a.
The equation of motion of the system is
mg – T = m(2a)      ......(1)
T = ma        .......(2)
Substitute the value of T in equation (1)
mg - ma = 2ma
3ma = mg
a = g/3
So the acceleration of the center of the ring is, g/3
And acceleration of the hanging mass is 2a = 2g/3

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JEE Main Practice Test- 5 - Question 3

A concave mirror gives a real image magnified 6 times when the object is moved 6 cm the magnification of the real image is 4 times. What will be the focal length of the mirror?

Detailed Solution for JEE Main Practice Test- 5 - Question 3

As we know that the magnification

And in the second case,

JEE Main Practice Test- 5 - Question 4

A spring mass system with unequal masses is placed at rest on a smooth horizontal surface. The spring is initially kept compressed with a thread. When the spring is cut, the mass m moves so as to enter in a vertical circular loop of radius r. The minimum compression x in the spring so that the mass m may negotiate the vertical loop is

Detailed Solution for JEE Main Practice Test- 5 - Question 4

Let x be the compression in the spring, v and V be the velocities of m and M respectively, when the spring released.
According to the law of conservation of momentum and energy
MV = mv ... (1)

From (2) and (1)

To negotiate a loop, minimum velocity required

JEE Main Practice Test- 5 - Question 5

A plane electromagnetic wave of angular frequency ω propagates in a poorly conducting medium of conductivity σ and relative permittivity ε. Find the ratio of conduction current density and displacement current density in the medium.

Detailed Solution for JEE Main Practice Test- 5 - Question 5

Jc = σE0 sin (ωt – kx)

= ∈ ∈0 × A E0ω cos(ωt–kx)
Jd = ∈ ∈0E0ω

JEE Main Practice Test- 5 - Question 6

A uniform electric field of magnitude 245 V/m is directed in the negative y direction as shown in figure. The coordinates of point P and Q are (- 0.5m, - 0.8m) and (0.3m, 0.7m) respectively. Calculate the potential difference VQ - VP along the path shown in the figure,

Detailed Solution for JEE Main Practice Test- 5 - Question 6

Potential difference between points P and Q is

JEE Main Practice Test- 5 - Question 7

A cylinder of radius r = 1 m and height 3 m filled with a liquid upto 2 m high and is rotated about its vertical axis as shown in the Figure.
The speed of rotation of the cylinder when the point at the centre of the base is just exposed is

Detailed Solution for JEE Main Practice Test- 5 - Question 7


v0 = 0: H = 3 m; r = 1 m: g = 10 ms− 2

JEE Main Practice Test- 5 - Question 8

A charge particle q of mass m0 is projected along the y-axis at t = 0 from origin with a velocity V0. If a uniform electric field E0 also exists along the x-axis, then the time at which the de-Broglie wavelength of the particle becomes half of the initial value is:

Detailed Solution for JEE Main Practice Test- 5 - Question 8


Force on charged particle is =qE0 in x direction.
At any time t velocity in y direction will be same as at t=0 as there is no force in that direction.
At time t velocity in x direction is given by
vx =axt
vx =(qE0/m)t
so speed of particle at time t is- 
debroglie wavelength is inversly proportional to speed of particle and given by- 
(Where R is a constant)
So for half the wavelength speed should be doubled.

JEE Main Practice Test- 5 - Question 9

The beat frequency produced when the following two waves x1 = 12 sin (484πt − 7πx) and x2 = 12 sin (480πt − 7πx) are sounded together is

Detailed Solution for JEE Main Practice Test- 5 - Question 9

x1 = 12 sin (484 πt − 7 π x)

∴ frequency n1 = 242 Hz
x2 = 12 sin (480 πt − 7 πx) 
∴ frequency n2 = 240 Hz
Beat frequency = n1 − n2 = 2

JEE Main Practice Test- 5 - Question 10

An 80 kg man standing on ice throws a 4 kg body horizontally at 6 ms−1. The frictional coefficient between the ice and his feet is 0.02. The distance through which the man slips is (g = 10 ms−2)

Detailed Solution for JEE Main Practice Test- 5 - Question 10

While throwing the body, velocity gained by the man is

JEE Main Practice Test- 5 - Question 11

Liquid cools from 50ºC to 45 ºC in 5 minutes and from 45ºC to 41.5ºC in the next 5 minutes. The temperature of the surrounding is :- [Assume newton's law of cooling is applicable]

Detailed Solution for JEE Main Practice Test- 5 - Question 11

From Newton's law of cooling

in second case

From (i) and (ii)
θ0 = 33.3oC

JEE Main Practice Test- 5 - Question 12

A long solenoid of self-inductance L and area of cross section A not carrying any current is placed in a uniform magnetic field of strength B with its axis parallel to the field direction. The total magnetic flux linked with the solenoid is φ. The energy of the magnetic field stored in the solenoid is

Detailed Solution for JEE Main Practice Test- 5 - Question 12

Flux linked with each turn of the solenoid = BA
Total number of turns = nℓwhereℓ is the length of the solenoid
Total flux linked 
Energy per unit volume 
∴total energy of the magnetic field stored in the solenoid

JEE Main Practice Test- 5 - Question 13

All wires have same resistance and equivalent resistance between A and B is R. Now keys are closed, then the equivalent resistance will become:-

Detailed Solution for JEE Main Practice Test- 5 - Question 13



As R0 = R/3
Req= 7R/9

JEE Main Practice Test- 5 - Question 14

Steam at 100°C is passed into 1.4 kg of water kept in a calorimeter of water equivalent 0.03 kg at 20 °C till the temperature of the calorimeter and contents reach 80°C. Latent heat of steam is 2.26 × 106 J kg − 1 and sp.heat capacity of water is 4200 J /kg/°C. The mass of steam condensed in kilogram is nearly equal to

Detailed Solution for JEE Main Practice Test- 5 - Question 14

(M + Mw) C(80-20) = mL + mC(100 − 80)
(1.4 + 0.03) 4200 (80 − 20)= m × 2.26 × 106 + m × 4200 × 20
1.43 × 4200 × 60= m (2.26 × 106 + 4200 × 20)

= 0.1537 kg ≈ 0.15 kg

JEE Main Practice Test- 5 - Question 15

A satellite orbiting around the earth of radius R is shifted to an orbit of radius 2R. The time taken for one revolution will increase nearly by

Detailed Solution for JEE Main Practice Test- 5 - Question 15

The time period of a satellite orbiting around the earth is

When the radius of the orbit is increased to 2R

T’=  2√2 T
T' = 2 × 1.414 T
T' = 2.828 T ≈ 2.8 T

JEE Main Practice Test- 5 - Question 16

Screw gauge shown in Figure has 50 divisions and in one complete rotation of circular scale the main scale moves 0.5 mm.

Detailed Solution for JEE Main Practice Test- 5 - Question 16

As shown in first figure the zero error of the screw gauge is:

ZE = 5×0.5 / 50 = 0.05mm

Least count of circular scale = 0.5 / 50 = 0.01mm

Circular scale reading = 25×0.01=0.25mm

Main scale reading = 2×0.5=1mm

Actual measurement is = MSR+CSR − ZE 

1+0.25−0.05=1.20mm

JEE Main Practice Test- 5 - Question 17

The given graph shows the extension (Δl) of a wire of length 1.0 m suspended from the top of a roof at one end and loaded at the other end. If the cross sectional area of the wire is 10− 6m2, the Young’s modulus Y of the material of the wire is

Detailed Solution for JEE Main Practice Test- 5 - Question 17

JEE Main Practice Test- 5 - Question 18

In Young’s double slit experiment, a monochromatic light of wavelength 5500 Å is used. The slits are 2 mm apart. The fringes are formed on a screen placed 20 cm away from the slits. It is found that the interference pattern shifts by 16 mm when a transparent plate of thickness 0.4 mm is introduced in the path of one of the slits. The refractive index of the transparent plate is

Detailed Solution for JEE Main Practice Test- 5 - Question 18

Δy = 16 mm = 16 × 10− 3 m
d = 2 mm = 2 × 10− 3 m λ = 5500 Å
D = 20 cm = 0.2 m
t = 0.4 mm = 4 × 10− 4 m

∴ μ = 1 + 0.4 = 1.4

JEE Main Practice Test- 5 - Question 19

The average degree of freedom per molecule of a gas is 6. The gas performs 25 J work, while expanding at constant pressure. The heat absorbed by the gas is:-

Detailed Solution for JEE Main Practice Test- 5 - Question 19

Change in the internal energy of gas is

For a constant pressure process. Work done (W) = nRΔT
So heat absorbed by the gas is

JEE Main Practice Test- 5 - Question 20

The radioactive sources A and B have half lives of 2hr and 4hr respectively, initially contain the same number of radioactive atoms. At the end of 2 hours, their rates of distintegration are in the ratio:-

Detailed Solution for JEE Main Practice Test- 5 - Question 20

Given that the half-life of radioactive source A is, t1/2 = 2 hr
half-life of radioactive source B is, t1/2 = 4 hr
Decay constant of radioactive source A is,

Decay constant of radioactive source B is,

The rate of disintegration (A) of source A at the end 2 hours is,

The rate of disintegration (A) of source B at the end 2 hours is,

The ratio of rate of distintegration of source A to B is

 

*Answer can only contain numeric values
JEE Main Practice Test- 5 - Question 21

A Young's double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and interference pattern is observed on the screen at a distance 1.33 m from plane of slits. The wavelength of light used in air is 6300A°. When one of the slit is covered by a glass sheet of thickness   and refractive index 1.53, the position of maxima and minima get interchanged. Find the maximum possible value of n.


Detailed Solution for JEE Main Practice Test- 5 - Question 21

*Answer can only contain numeric values
JEE Main Practice Test- 5 - Question 22

An oscillator of frequency 425 Hz drives two speakers. The speaker are fixed on a vertical pole at a distance 2.4m from each other. A person whose height (of ears) is same as that of lower speaker runs horizontally away from the two speakers. Find the maximum distance (in m) of person from the pole where he hears no sound, (velocity of sound in air 340 m/s)


Detailed Solution for JEE Main Practice Test- 5 - Question 22

Let x be maximum distance of person from speaker | when he hears no sound (i.e, minimal)

*Answer can only contain numeric values
JEE Main Practice Test- 5 - Question 23

A particle of mass m = 1 kg and having charge Q = 2C is projected on a rough horizontal xy plane from (4, 0, 0) m with velocity   and in the region there exist a uniform electric field  and a uniform magnetic field  The coefficient of friction between the particle and the horizontal surface is μ = 1/3 and the particle comes to rest by moving a distance 1000/n metres. Find n.


Detailed Solution for JEE Main Practice Test- 5 - Question 23

*Answer can only contain numeric values
JEE Main Practice Test- 5 - Question 24

A wedge of inclination 45° is moved towards right with a constant acceleration of 2√6 m/s2 as shown in figure. Find magnitude of acceleration of the ball placed between an inclined wall and the wedge.


Detailed Solution for JEE Main Practice Test- 5 - Question 24

*Answer can only contain numeric values
JEE Main Practice Test- 5 - Question 25

In the situation shown in figure, a particle having charge q = 2C and mass = 1 kg is projected from bottom of an inclined plane of inclination 45° at an angle 45° with horizontal, so that the particle strike the inclined plane normally due to a uniform horizontal electric field E. Find the value of E (in NC-1)


Detailed Solution for JEE Main Practice Test- 5 - Question 25

JEE Main Practice Test- 5 - Question 26

Ratio of Boyle temperature and critical temperature of gas is:

Detailed Solution for JEE Main Practice Test- 5 - Question 26

Boyle Temperature Tb= a/Rb
Critical Temperature Tc= 8a/27Rb
Dividing both
Tb/Tc =27/8
Hence option (b) is correct.

JEE Main Practice Test- 5 - Question 27

Which of the following is reducing sugar?

Detailed Solution for JEE Main Practice Test- 5 - Question 27

A reducing sugar is one which is capable of acting as reducing agent.
All monosaccharaides are reducing sugars because all monosaccharaides have an aldehyde group (if they are aldoses) or can tautomerize in solution to form an aldehyde group (if they are ketoses) and can themselves get oxidized hence can act as reducing agent.



Galactose
All of them are monosaccharides.
Hence option (d) is correct.

JEE Main Practice Test- 5 - Question 28

Which of the following complex square having planner complex can exhibit geometrical isomerism?

Detailed Solution for JEE Main Practice Test- 5 - Question 28


Trans-isomer

Cis –isomer
Square planar complexes having symmetric bidentate ligands carrying one or more substituents can form geometrical isomers. Because, such ligand show kind of asymmetry. Here pn (propylenediamine) shows this kind of nature hence leading to geometrical isomerism.
All the rest options are of symmetrical in nature around metal ion so no geometrical isomerism is possible in these.
Hence option (a) is correct.

JEE Main Practice Test- 5 - Question 29

An element has a bcc structure with a cell edge of 288 pm. The density of the element is 7.2 g cm–3. Number of atoms present in 208 g of the element ?

Detailed Solution for JEE Main Practice Test- 5 - Question 29

Volume of unit cell = (288 pm)3 = (288 × 10–12 m)3
= (288 × 10–10 cm)3
=2.39 × 10–23 cm3
For bcc, zeff = 2

∴ Aw = 7.2 × 3 × 1023 × 2.39 × 10–23 g
∴ 7.2 × 3 × 2.39 g of element contains = NA atoms = 6 × 1023 atoms.
∴ 208 g of the element contains 
= 24.17 × 1023 atoms
Alternatively
Volume of 208 g of the element 
=28.88 cm3
Number of unit cells in this volume

= 12.08 × 1023 unit cells
Since each bcc cubic unit contains 2 atoms, therefore the total number of atoms in 208 g
= 2 (atoms/unit cell) × 12.08 × 1023 unit cells
= 24.16 × 1023 atoms
Hence option (b) is correct.

JEE Main Practice Test- 5 - Question 30

The most acidic oxyacid of halogen among the given option is

Detailed Solution for JEE Main Practice Test- 5 - Question 30

The acidity of oxyacids of halogen can be compared by checking the oxidation state of halogen in the acid. Higher the oxidation state, higher the acidity. Here H5IO6 has an oxidation state of +7 and hence is the strongest acid.

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