Physics Test - 7

The potential difference between the balls should be \(\mathrm{E}\).

Therefore,

\(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q_{1}}{r_{1}}-\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q_{2}}{r_{2}}=E\)

Where \(\mathrm{Q}_{1}\) and \(\mathrm{Q}_{2}\) are charges on the balls. According to law of conservation of charge

\(\mathrm{Q}_{1}+\mathrm{Q}_{2}=0\)

So, \(\mathrm{Q}_{1}=-\mathrm{Q}_{2}\)

\(\frac{1}{4 \pi \epsilon_{0}}\left[\frac{Q_{1}}{r_{1}}-\frac{Q_{2}}{r_{2}}\right]=E\)

\(\frac{1}{4 \pi \epsilon_{0}}\left[\frac{Q_{1}}{r_{1}}+\frac{Q_{1}}{r_{2}}\right]=E\)

\(\frac{Q_{1}}{4 \pi \epsilon_{0}}\left[\frac{r_{1}+r_{2}}{r_{1} r_{2}}\right]=E\)

\(\mathrm{Q}_{1}=\frac{4 \pi \epsilon_{0} E r_{1} r_{2}}{r_{1}+r_{2}}\)

So, from Coulomb's law

\(\mathrm{F}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q_{1} Q_{2}}{R^{2}}\)

\(=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{16 \pi^{2} \epsilon_{0}^{2} E^{2} r_{1}^{2} r_{2}^{2}}{R^{2}\left(r_{1}+r_{2}\right)}\)

\(F=\frac{4 \pi \epsilon_{0} E^{2} r_{1}^{2} r_{2}^{2}}{R^{2}\left(r_{1}+r_{2}\right)^{2}}\)

As we know,

Coefficient of viscosity, \(\eta=\frac{\mathrm{F}}{\frac{\mathrm{Adv}}{\mathrm{dx}}}\)

Then,

\([\mathrm{F}]=[\) Force \(]=\mathrm{MLT}^{-2}\)

\([\mathrm{~A}]=[\) Area \(]=\mathrm{L}^{2}\)

\(\left[\frac{\mathrm{dv}}{\mathrm{dx}}\right]=[\) Velocity gradient \(]=\frac{\mathrm{LT}^{-1}}{\mathrm{~L}}\)

\(=\mathrm{T}^{-1}\)

\(\therefore[\eta]=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^{2} \mathrm{~T}^{-1}}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] \)

Hence, the correct option is (D).

The tyres of the cars are made circular because it makes them roll on the road rather than slide. And Rolling friction is less than sliding friction.

Also, the circular shape maintains a uniform and smooth contact with the surface.

The capacity to do work or power consumed in a given time is called energy.

Power is the rate of doing work or the rate at which energy is spent.Kilowatt-hour is the unit of energy.

Moving from \(A\) to \(B\),

\(V A-V B=[1 \times 5-15+5 \times 10-3(-103)]\)

\(V=-15{V}\) 

or,

\({VB}-{VA}=15 {~V}\)

\([\frac{\mathrm{di}}{\mathrm{dt}}\) is negative as \(I\) is decreasing\(]\)

Given:

Moment of inertia \((I)=3 \mathrm{~kg}-\mathrm{m}^{2}\)

Angular Velocity \((\omega)=2 \mathrm{rad} / \mathrm{s}\)

Mass \((\mathrm{m})=12 \mathrm{~kg}\)

To Find velocity (v)

The velocity is given as,

\(\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{mv}^{2}\)

\(\frac{1}{2} \times 3(2)^{2}=\frac{1}{2} \times 12 \times \mathrm{v}^{2}\)

\(\mathrm{v}=1 \mathrm{~m} / \mathrm{s}\)

Thus, the speed of the body is \(1 \mathrm{~m} / \mathrm{s}\).

Given,

The lift is moving up with acceleration 'a'.

The normal reaction by the lift floor on the mass \(=\mathrm{N}_{1}\)

Mass of object on floor \(=\mathrm{m}\)

Weight due to the mass \(\mathrm{m}=\mathrm{F}_{\mathrm{g}}=\mathrm{mg}\) (downwards)

The force due to acceleration of lift = ma (upwards)

The net force \(=\mathrm{N}_{1}-\mathrm{F}_{\mathrm{g}}=\mathrm{ma}\)

\(\therefore\) Normal reaction by the lift floor on the mass,

\(N_{1}=m a+F_{g}\)

\(=m a+m g\)

\(=m(a+g)\)

From Newton's \(3^{\text {rd }}\) law, the force by mass on the floor = force by the floor on the mass \(=\mathrm{m}({g}+{a})\)

If a body is rolling down from a hill at certain height, the body will haveKinetic energy, Rotational energy and potential energy.

Kinetic energy:Since the body is moving in a linear direction, so it will have kinetic energy due to its velocity.

Rotational energy:Along with linear velocity, the body is also revolving i.e. rotating, so it will have rotation energy.

Potential energy:Since the body is at a certain height, it will have potential energy.

Given that:

$y=2\sin \frac{\pi }{4}\left(4t+\frac{x}{2}\right)=2\sin \left(\pi t+\frac{\pi x}{8}\right)$

$A=2;w=\pi rad/s$

$\omega =2\pi f$

$f=0.5Hz$

Hookes law states that within the elastic limit stress is directly proportional to strain"

The stress of an elastic material is the restoring force acting per unit area of an object. The strain is the ratio of change in dimension to the original dimension.