JEE Advanced Mix Test 22

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JEE Advanced Mix Test 22

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following is not possible ?

A

B

C

D

Solution

The pressure at any point can never have different values. Hence (A) & (D) are not possible. (Calculate the pressures at points A & D from both their left and right)

In case of insufficient length of capillary tube the shape of meniscus is as below :
Question 2
4 / -1

Two particle P and Q are in motion under gravity. Then

A
their relative acceleration is constant but not zero

B
their relative velocity is constant

C
their centre of mass has constant velocity

D
their centre of mass has constant acceleration.

Solution
Question 3
4 / -1

Statment–1 : Steady flow can be maintained in compressible liquids.

Statment–2 : Steady flow means all physical parameters related to the flow being constant with time though they may vary with position.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C
Statement-1 is True, Statement-2 is False

D
Statement-1 is False, Statement-2 is True.

Solution

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
Question 4
4 / -1

Statment–1 : In a circular motion, the force must be directed perpendicular to the velocity all the time.

Statment–2 : A centripetal force is required to provide the centripetal acceleration in a circular motion.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C
Statement-1 is True, Statement-2 is False

D
Statement-1 is False, Statement-2 is True.

Solution

Only in uniform circular motion force must be directed perpendicular to the velocity all the time. However in non-uniform circular motion, force will always not be perpendicular to velocity. Hence statement-1 is false.
Question 5
4 / -1

Statment–1 : The centre of gravity and centre of mass of a body may be at different positions.

Statment–2 : If the mass is uniformly distributed then only the centre of mass and centre of gravity of a body will be at the same place.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C
Statement-1 is True, Statement-2 is False

D
Statement-1 is False, Statement-2 is True

Solution

Centre of mass and centre of gravity of a uniformly distributed mass shall coincide if acceleration due to gravity at all point of mass distribution is same. Hence statement-2 is false.
Question 6
4 / -1

Statment-1 : Consider an object that floats in water but sinks in oil. When the object floats in water, half of it is submerged. If we slowly pour oil on top of water till it completely covers the object, the object moves up.

Statment-2 : As the oil is poured in the situation of statement-1, pressure inside the water will increase everywhere resulting in an increase in upward force on the object.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C
Statement-1 is True, Statement-2 is False

D
Statement-1 is False, Statement-2 is True

Solution

As the oil is poured till it covers the object completely, pressure in water at all points keeps on increasing. As a result upward force on object exerted by water increases and the object moves up for the given duration. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
Question 7
4 / -1

Directions For Questions

Consider a star and two planet system. The star has mass M. The planets A and B have the same mass m, radius a and they revolve around the star in circular orbits of radius r and 4r respectively (M > > m, r > > a). Planet A has intelligent life and the people of this planet have achieved a very high degree of technological advance. They wish to shift a geostationary satellite of their own planet to a geostationary orbit of planet B. They achieve this through a series of high precision maneuvers in which the satellite never has to apply brakes and not a single joule of energy is wasted. S₁ is a geostationary satellite of planet A and S₂ is a geostationary satellite of planet B. Neglect interaction between A and B, S₁ and S₂, S₁and B & S₂ and A.

...view full instructions

If the time period of the satellite in geostationary orbit of planet A is T, then its time period in geostationary orbit of planet B is:

A
T

B
4T

C
8T

D
Data insufficient

Solution

The time in which the planet rotates about its axis is not given for either planet.
Question 8
4 / -1

Directions For Questions

Consider a star and two planet system. The star has mass M. The planets A and B have the same mass m, radius a and they revolve around the star in circular orbits of radius r and 4r respectively (M > > m, r > > a). Planet A has intelligent life and the people of this planet have achieved a very high degree of technological advance. They wish to shift a geostationary satellite of their own planet to a geostationary orbit of planet B. They achieve this through a series of high precision maneuvers in which the satellite never has to apply brakes and not a single joule of energy is wasted. S₁ is a geostationary satellite of planet A and S₂is a geostationary satellite of planet B. Neglect interaction between A and B, S₁and S₂, S₁ and B & S₂and A.

...view full instructions

If the radius of the geostationary orbit in planet A is given by then the time in which the geostationary satellite will complete 1 revolution is

I. 1 planet year = time in which planet revolves around the star

II. 1 planet day = time in which planet revolves about its axis.

A
I

B
II

C
both I and II

D
neither I nor II

Solution
Question 9
4 / -1

Directions For Questions

Consider a star and two planet system. The star has mass M. The planets A and B have the same mass m, radius a and they revolve around the star in circular orbits of radius r and 4r respectively (M > > m, r > > a). Planet A has intelligent life and the people of this planet have achieved a very high degree of technological advance. They wish to shift a geostationary satellite of their own planet to a geostationary orbit of planet B. They achieve this through a series of high precision maneuvers in which the satellite never has to apply brakes and not a single joule of energy is wasted. S₁ is a geostationary satellite of planet A and S₂ is a geostationary satellite of planet B. Neglect interaction between A and B, S₁ and S₂, S₁ and B & S₂and A.

...view full instructions

If planet A and B, both complete one revolution about their own axis in the same time, then the energy needed to transfer the satellite of mass m0 from planet A to planet B is -

A

B

C

D
Zero

Solution

The energy of any geostationary satellite is the sum of kinetic energy of satellite, interaction energy of satellite and its own planet and interaction energy of satellite and star. Both planets have same mass and same length of day.

Geostationary satellite - planet system will have same interaction energy in either planet. Also kinetic energy of both satellites will be same. But the satellite-sun system will account for the energy difference.
Question 10
4 / -1

In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. its x-coordinate. The particle moves under the effect of this conservative force along the x-axis.

If the particle is released at the origin then :

A
it will move towards positive x-axis.

B
it will move towards negative x-axis.

C
it will remain stationary at the origin.

D
its subsequent motion cannot be decided due to lack of information.

Solution

If the particle is released at the origin, it will try to go in the direction of force. Here is positive and hence force is negative, as a result it will move towards – ve x-axis.