When a body of mass $m$ just begins to slide as shown, match List-I with List-II:

List-I (a) Normal reaction (b) Frictional force ($f_s$) (c) Weight ($mg$) (d) $mg \sin \theta$

List-II (i) $P$ (ii) $Q$ (iii) $R$ (iv) $S$

A block A of mass $m_1$ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of the table and from its other end, another block B of mass $m_2$ is suspended. The coefficient of kinetic friction between block A and the table is $\mu_k$. When block A is sliding on the table, the tension in the string is:

A block of mass 4 kg is suspended through two light spring balances A and B connected in series. Then A and B will read respectively:

A lift of mass $1000 \text{ kg}$ is moving with an acceleration of $1 \text{ m/s}^2$ in the upward direction. Tension developed in the string, which is connected to the lift, is:

The distance covered by a body of mass $5 \text{ g}$ having linear momentum $0.3 \text{ kg m/s}$ in $5 \text{ s}$ is:

A particle of mass $m$ is projected with velocity $v$ making an angle of $45^\circ$ with the horizontal. When the particle lands on the level ground, the magnitude of the change in its momentum will be:

It is easier to draw up a wooden block along a smooth inclined plane than to haul it vertically, principally because:

When two surfaces are coated with a lubricant, then they:

A block of mass $m$ is in contact with the cart $C$ as shown in the figure. The coefficient of static friction between the block and the cart is $\mu$. The acceleration $a$ of the cart that will prevent the block from falling satisfies: