JEE Main 2011 — Mechanics

A mass $\mathrm{m}$ hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass $\mathrm{m}$ and radius $\mathrm{R}$. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass $m$, if the string does not slip on the pulley, is

A pulley of radius $2 \mathrm{~m}$ is rotated about its axis by a force $\mathrm{F}=\left(20 \mathrm{t}-5 \mathrm{t}^2\right)$ Newton (where $\mathrm{t}$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation made by the pulley before its direction of motion if reversed, is :

A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : $0 \mathrm{~mm}$ Circular scale reading : 52 divisions Given that $1 \mathrm{~mm}$ on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above date is :

A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc:

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is $\mathrm{v}$, the total area around the fountain that gets wet is :

An object, moving with a speed of $6.25 \mathrm{~m} / \mathrm{s}$, is decelerated at a rate given by : $$ \frac{\mathrm{dv}}{\mathrm{dt}}=-2.5 \sqrt{\mathrm{v}} $$ where $v$ is the instantaneous speed. The time taken by the object, to come to rest, would be:

If a wire is stretched to make it $0.1 \%$ longer, its resistance will :

Two bodies of masses $\mathrm{m}$ and $4 \mathrm{~m}$ are placed at a distance $\mathrm{r}$. The gravitational potential at a point on the line joining them where the gravitational field is zero is:

Water is flowing continuously from a tap having an internal diameter $8 \times 10^{-3} \mathrm{~m}$. The water velocity as it leaves the tap is $0.4 \mathrm{~ms}^{-1}$. The diameter of the water stream at a distance $2 \times 10^{-1} \mathrm{~m}$ below the lap is close to :

Work done in increasing the size of a soap bubble from a radius of $3 \mathrm{~cm}$ to $5 \mathrm{~cm}$ is nearly (Surface tension of soap solution $=0.03 \mathrm{Nm}^{-1}$ ):