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JEE Main Practice Test- 8 - JEE MCQ


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30 Questions MCQ Test - JEE Main Practice Test- 8

JEE Main Practice Test- 8 for JEE 2024 is part of JEE preparation. The JEE Main Practice Test- 8 questions and answers have been prepared according to the JEE exam syllabus.The JEE Main Practice Test- 8 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main Practice Test- 8 below.
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JEE Main Practice Test- 8 - Question 1

In the figure shown the acceleration of A is, , then the acceleration of B is : (A remains in contact with B)

Detailed Solution for JEE Main Practice Test- 8 - Question 1

From wedge constraint


JEE Main Practice Test- 8 - Question 2

The speed of sound in hydrogen gas at N.T.P. is 1,328 ms-1. What will be its value in air at N.T.P., if density of hydrogen is 1/16th that of air?

Detailed Solution for JEE Main Practice Test- 8 - Question 2


Let vair be the velocity of sound in air at N.T.P



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JEE Main Practice Test- 8 - Question 3

A balloon that is initially flat, is inflated by filling it from a tank of compressed air. The final volume of the balloon is 5m3. The barometer reads 95 kPa. The work done in this process is

Detailed Solution for JEE Main Practice Test- 8 - Question 3

W = P Δ V⁡ = 95 × 103 × (5-0)
= 4.75 × 105 J

JEE Main Practice Test- 8 - Question 4

Two bodies of same mass tied with an inelastic string of length   together on a horizontal surface. One a horizontal surface of them is projected vertically upwards with velocity  ⁡Find the maximum height up to which the centre of mass of system of the two masses rises.

Detailed Solution for JEE Main Practice Test- 8 - Question 4


Velocity of B when string is just taut

After string in taut both will move up with same speed, 
By this time centre of mass in at 
Further rise in 

Hence maximum height 

JEE Main Practice Test- 8 - Question 5

The electric potential at a point (x, y, z) is given by V = -x2y - xz3 + 4 The electric field  at that point is

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

The electric potential at a point,

JEE Main Practice Test- 8 - Question 6

Binding energy per nucleon versus mass number curve for nuclei is shown in figure. W, X, Y and Z are four nuclei indicated on the curve. The process that would release energy is

Detailed Solution for JEE Main Practice Test- 8 - Question 6

Energy is released in a process when total binding energy of the nucleus (= binding energy per nucleon x number of nucleons) is increased or we can say, when total binding energy of products is more than the reactants.
Binding energy of reactants = 120 x 7. 5 = 900 MeV
and binding energy of products = 2 (60 x 8. 5)
= 1020 MeV > 900 MeV.

JEE Main Practice Test- 8 - Question 7

A wire of length 1.0 m and radius 10-3 m is carrying a heavy current and is assumed to radiate as a black body. At equilibrium, its temperature is 900 K while that of the surroundings is 300 K. The resistivity of the material of the wire at 300 K is π2 × 10-8 ohm-m and its temperature coefficient of resistance is 7.8× 10−3 Co. Find the current in the wire. [Given Stefan's constant = 5.68 × 10-8 W/m2K4]

Detailed Solution for JEE Main Practice Test- 8 - Question 7

According to Stefan's law power radiated by a black body is given by.


Now if I is the current in the resistance R,

According to the given problem Equations (i) and (ii) represent the same. 

JEE Main Practice Test- 8 - Question 8

50 V battery is supplying a steady current of 10 amp when connected to an external resistor. If the efficiency of the battery at this current is 25%, then internal resistance of battery is:

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


JEE Main Practice Test- 8 - Question 9

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. Now, the potential difference across the capacitor is.

Detailed Solution for JEE Main Practice Test- 8 - Question 9

Initially

After charge Q is given

Charge distribution is

JEE Main Practice Test- 8 - Question 10

Energy from the sun falls on the earth at a rate of 1353 W/m2, which is known as solar constant, i.e., the power incident per unit area per second at the top of atmosphere. Find the r.m.s values of the electric and magnetic fields in the sunlight reaching the top of the atmosphere.

Detailed Solution for JEE Main Practice Test- 8 - Question 10

For an electromagnetic wave of sinusoidal form



The mean value of energy flux is intensity

[The mean value of cos2θ over on cycle is 1/2] 

Hence   

JEE Main Practice Test- 8 - Question 11

The size of the image of an object, which is at infinity, is formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image

Detailed Solution for JEE Main Practice Test- 8 - Question 11


Image formed by convex lens at ^ will act as a virtual object for concave lens. For concave lens

Magnification for concave lens 

As height of the image at h is 2 cm.
Therefore, height of image at I2 will be 2 x 1.25 = 2.5 cm

JEE Main Practice Test- 8 - Question 12

A network of Four capacitors of capacity equal to C1 = C, C2 = 2C, C3 = 3C and C4 = 4C, are connected to a battery as shown in the figure. The ratio of the charges on C2 and C4 is

Detailed Solution for JEE Main Practice Test- 8 - Question 12



All the capacitors in upper branch are in series so the charge on each capacitor is 
Also charge on capacitor C4 is Q = 4CV

JEE Main Practice Test- 8 - Question 13

A rod AD, consisting of three segments AB, BC and CD joined together, is hanging vertically from a fixed support at A. The lengths of the segments are respectively 0.1 m, 0.2 m and 0.15 m. The cross - section of the rod is uniform 10-4 m2. A weight of 10 kg is hung from D. Calculate the displacements of point D if YAB = 2.5 x 1010 N/m2, YBC = 4 x 1010 N/m2 and YCD = 1 x 1010 N/m2. (Neglect the weight of the rod.)

Detailed Solution for JEE Main Practice Test- 8 - Question 13

By defination of Young's modulus,


JEE Main Practice Test- 8 - Question 14

A block of mass m & charge q is released on a long smooth inclined plane. Magnetic field B is constant, uniform, horizontal and out of the plane of paper as shown. Find the time from start when block loses contact with the surface.

Detailed Solution for JEE Main Practice Test- 8 - Question 14

Block will loose contact with surface when force due to magnetic field will become equal to mg cos θ

 (along the inclined plane)

JEE Main Practice Test- 8 - Question 15

A battery of internal resistance 4Ω is connected to the network of resistances as shown in figure. In order that the maximum power can be delivered to the network, the value of R in $$ should be

Detailed Solution for JEE Main Practice Test- 8 - Question 15

The given circuit is a balanced Wheatstone's bridge.

Thus, no current will flow across 6R of the side CD. The given circuit will now be equivalent to

 

For maximum power, net external resistance
= Total internal resistance
or 2R = 4
or R = 2Ω. 

JEE Main Practice Test- 8 - Question 16

In the circuit shown in diagram, the equivalent resistance between point A and B is

Detailed Solution for JEE Main Practice Test- 8 - Question 16

The circuit shown in diagram (1) can be redrawn as shown in diagram (2).

∴  RAB = 5R/8

JEE Main Practice Test- 8 - Question 17

Two trains, which are moving along different tracks in opposite directions, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when they are 300 m apart. Graphs given below show their velocities as a function of time as they slow down. The separation between the trains, when both have stopped, is:

Detailed Solution for JEE Main Practice Test- 8 - Question 17

Initial distance between trains is 300 m. Displacement of 1st train is calculated by area under V-t.


Displacement of train Which means it moves towards left.
∴ Distance between the two is = 300 - 280 = 20 m

JEE Main Practice Test- 8 - Question 18

A tuning fork sends out waves of wavelength 68.75 cm and 3 m in air and hydrogen gas respectively. If the velocity of sound in air is 330 ms-1, find the velocity of sound in hydrogen. Also, find the frequency of the tuning fork

Detailed Solution for JEE Main Practice Test- 8 - Question 18

In air : λa = 68.75 cm = 0.6875 m ; va = 330 ms-1 Let v be the frequency of the tuning fork. Then,

If vH is velocity of sound in hydrogen, then

JEE Main Practice Test- 8 - Question 19

A particle executes S.H.M. given by x = 0 · 24 cos (400 t - 0.5) in SI units. Find amplitude

Detailed Solution for JEE Main Practice Test- 8 - Question 19

Here, x = 0.24 cos (4001 - 0.5) ...(i)
The standard equation for S.H.M. is
 ... (ii)
Comparing the equations (i) and (ii),we have
r = 0.24 m

JEE Main Practice Test- 8 - Question 20

A lens (μ = 1.5) is coated with a thin film of refractive index 1.2 in order to reduce the reflection from its surface at λ = 4800 Å. Find the minimum thickness of the film which will minimise the intensity of the reflected light. (Assume near normal incidence)

Detailed Solution for JEE Main Practice Test- 8 - Question 20


*Answer can only contain numeric values
JEE Main Practice Test- 8 - Question 21

A ray of light from a liquid (μ = √3) is incident on a system of two right angled prism of refractive indices √3 and √2 as shown in the figure. The light suffers zero net deviation when it emerges into air from surface CD. If the angle of incidence (in degrees) is 5n. Find n ?


Detailed Solution for JEE Main Practice Test- 8 - Question 21


Hence 5*9 = 45

*Answer can only contain numeric values
JEE Main Practice Test- 8 - Question 22

A drop water of mass = 4.0 g is placed between two clean glass plates, the distance between the plates is 0.01cm. Find the force (103N) required to pull the plates away. Surface tension of water = 0.08 N/m and density of water is 1000 kg/m3


Detailed Solution for JEE Main Practice Test- 8 - Question 22

Let R be the radius of the circular layer of water. then πR2.d x r = m(l)
The pressure inside the water.

*Answer can only contain numeric values
JEE Main Practice Test- 8 - Question 23

A car is accelerating horizontally with constant acceleration a=10m/s2. One end of a light string is attached to the roof of the ceiling and there is a small bob at other end of string. The bob is given an initial velocity such that it continues to move in uniform circular motion with respect to an observer inside the car. The bob moves such the maximum vertical separation between two points of its path is h=1m. The length of the string is  and acceleration due to gravity g=10m/s2. If the angular speed of the bob in rad/s is √x .find the value of x.


Detailed Solution for JEE Main Practice Test- 8 - Question 23

First we will drive a result for conical pendulum in stationary car. see figure

Now in acceleration car, with respect to an inside observer, there will be pseudo force acting on the particle opposite
to the acceleration of the car, and hence perpendicular to the weight. since a=g, the effective gravity will be
  inclined at an angle of 450 with the verticle axis.

*Answer can only contain numeric values
JEE Main Practice Test- 8 - Question 24

Find the current (in A) through the battery after the switch S is closed if L/R = RC = 1 ms.


Detailed Solution for JEE Main Practice Test- 8 - Question 24

Since the battery is across the two branches in parallel the current through the RL branch is unaffected by the current of the RC branch

i.e there will be no transient current through the battery in this case

*Answer can only contain numeric values
JEE Main Practice Test- 8 - Question 25

Assuming that the law of gravitation is of the form  and attractive. A body of mass m revolves in a circular path of radius r around a fixed body of mass M. Find on what power of r will the square of time period depend.


Detailed Solution for JEE Main Practice Test- 8 - Question 25

The  centrepetal  acceleration is due  to the gravitational force

JEE Main Practice Test- 8 - Question 26

Place the following alcohols in decreasing order of rate of dehydration with concentrated H2SO4 :
1. CH3CH2CH(OH)CH2CH2CH3
2. (CH3)2C(OH)CH2CH2CH3
3. (CH3)2C(OH)CH(CH3)2
4. CH3CH2CH(OH)CH(CH3)2
​5. CH3CH2CH2CH2CH2CH2OH

Detailed Solution for JEE Main Practice Test- 8 - Question 26

The alcohols (3) and (2) are both 3o, but alcohol (3) gives a more substituted alkene. Alcohol (4) and (1) are both 2o, but alcohol (4) can give a more substituted alkene and alcohol (5) is 1o. Rate of dehydration of alcohols with concentrated H2SO4 follows the order 3o > 2o > 1o.

JEE Main Practice Test- 8 - Question 27

1.2 g of a salt with its empirical formula KxHy(C2O4)z was dissolved in 50 mL of water and its 10 mL portion required 11 mL of a 0.1 M HCl solution to reach the equivalence point. In a separate titration, 15 mL of the stock solution required 20 mL 0.2475 M KOH to reach the equivalence point. Identify the correct option.

Detailed Solution for JEE Main Practice Test- 8 - Question 27





∴ y = 3x.

JEE Main Practice Test- 8 - Question 28

The structure of H2O2 is

Detailed Solution for JEE Main Practice Test- 8 - Question 28

In H2O2, the O - H groups are not in the same plane. So it has non - planar structure. If has a half - opened book structure in which the two O - H groups lie on the two pages of the book. The angle between two pages of the book is 94o and H - O - O bond angle is 97o.

JEE Main Practice Test- 8 - Question 29

In nitroprusside ion the iron and NO exist as Fe II and NOrather than Fe III and NO. These forms can be differentiated by

Detailed Solution for JEE Main Practice Test- 8 - Question 29

Fe II and Fe III will have different values of magnetic moment due to different number of unpaired electrons in their d-orbitals.

JEE Main Practice Test- 8 - Question 30

Among the following pair of oxides, which pair cannot be reduced by carbon to give the respective metals ?

Detailed Solution for JEE Main Practice Test- 8 - Question 30

Potassium and calcium are strong reductant, hence their oxides cannot be reduced by carbon.

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