NTA JEE Main 4th September 2020 Shift 2

A quantity $$x$$ is given by $$(1Fv^2/WL^4)$$ in terms of moment of inertia $$I$$, force $$F$$, velocity $$v$$, work $$W$$ and length $$L$$. The dimensional formula for $$x$$ is same as that of:

A particle of charge $$q$$ and mass $$m$$ is subjected to an electric field $$E = E_0(1 - ax^2)$$ in the $$x$$-direction, where $$a$$ and $$E_0$$ are constants. Initially the particle was at rest at $$x = 0$$. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:

A small ball of mass $$m$$ is thrown upward with velocity $$u$$ from the ground. The ball experiences a resistive force $$mkv^2$$ where $$v$$ is its speed. The maximum height attained by the ball is:

A person pushes a box on a rough horizontal platform surface. He applies a force of 200 N over a distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100 N. The total distance through which the box has been moved is 30 m. What is the work done by the person during the total movement of the box?

Consider two uniform discs of the same thickness and different radii $$R_1 = R$$ and $$R_2 = \alpha R$$ made of the same material. If the ratio of their moments of inertia $$I_1$$ and $$I_2$$, respectively, about their axes is $$I_1 : I_2 = 1 : 16$$ then the value of $$\alpha$$ is:

For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is:

A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:

A cube of metal is subjected to a hydrostatic pressure 4 GPa. The percentage change in the length of the side of the cube is close to: (Given bulk modulus of metal, $$B = 8 \times 10^{10}$$ Pa)

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density $$d$$. The area of the base of both vessels is $$S$$ but the height of liquid in one vessel is $$x_1$$ and in the other $$x_2$$. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:

Match the thermodynamics processes taking place in a system with the correct conditions. In the table: $$\Delta Q$$ is the heat supplied, $$\Delta W$$ is the work done and $$\Delta U$$ is change in internal energy of the system.
Process    Condition
(I) Adiabatic    (A) $$\Delta W = 0$$
(II) Isothermal    (B) $$\Delta Q = 0$$
(III) Isochoric    (C) $$\Delta U \neq 0, \Delta W \neq 0, \Delta Q \neq 0$$
(IV) Isobaric    (D) $$\Delta U = 0$$