Physics Test - 9

Fluids offer resistance to motion due to internal friction, this property is called viscosity.

Viscous force (F):

• When a layer of fluid slips or tends to slip on adjacent layers in contact, the two layers exert tangential force on each other which tries to destroy the relative motion between them.
• The property of a fluid due to which it opposes the relative motion between its different layers is called viscosity (or fluid friction or internal friction) and the force between the layers opposing the relative motion is called viscous force.
• The force acting between the different layers of a fluid is given by:

\(F=-\eta A \frac{d v}{d x}\)

Where η = coefficient of viscosity, A = area of the plane and dv/dx = velocity gradient

Any object which is submerged in a fluid is acted upon by an upward force whose magnitude is equal to the weight of the fluid displaced. The upward force applied on the object is called the buoyant force.

Specific gravity: It is also known as relative gravity, it is the ratio between the density of an object, and a reference substance.

Given,

The reversible heat engine \((\Delta H) 3000 K J\)

\(T_{1}=650 K \)

\(T_{2}=295 K\)

To find the availability of heat energy we have to use the formula:

\(\Delta G=\Delta H-T_{2} \Delta S \)

\(\Delta S=\frac{\Delta Q}{\Delta T} \text { Where } \Delta Q=3000 K J,\left(\Delta T=T_{1}+T_{2}\right)=650+295 \)

\(=945 K \)

\(\Delta S=\frac{3000}{945}=3.17 K J / K \)

\(\Delta G=\Delta H-T_{2} \Delta S\)

On putting the value from above we get

\(\Delta G=3000-295 \times 3.17 \)

\(=2064.85 K J\)

Given,

\(c=3 \times 10^{8} {~m} / {sec}\) and \(\mu=2\)

We know that the absolute refractive index of a medium for a wave of a given wavelength is defined as the ratio of the speed of the wave in a vacuum to the speed of the wave in that medium.

\(\mu=\frac{c}{v} \ldots(1)\)

Where \(\mu=\) absolute refractive index, \(c=\) speed of the wave in vacuum, and \(v=\) speed of the wave in the medium

By equation (1),

\(v=\frac{c}{\mu} \)

\(v=\frac{3 \times 10^{8}}{2} \)

\(v=1.5 \times 10^{8} {~m} / {sec}\)

The image formed is real, the object is placed behind the focal point.

So, object distance, \({u}={- ( f}+{p})\)

Using mirror formula, \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)

\(\frac{1}{-f}=\frac{1}{-v}+\frac{1}{-(f+p)} \quad\) (Using sign convention: \({f}, {V}\), and \({u}\) are negative)

\(\frac{1}{v}=\frac{1}{f}-\frac{1}{(f+p)}\)

\(v=\frac{f(f+p)}{p}\)

The ratio of the size of the real image to the size of the object is magnification.

Magnification, \(m=\frac{-v}{u}=-\frac{-\left[\frac{f(f+p)}{p}\right]}{-[f+p]}\)

\(=-\frac{f}{p}\)

Measuring a parallax angle less than 0.01 arcsec from Earth is very difficult due to the influence of the Earth's atmosphere. The parallax method is influenced by the path of light that travels from a distant planet to the points of observation on Earth. Light reaching an observer from a distant object passes through several layers of Earth's atmosphere. These layers of the atmosphere allow light to undergo multiple refractions and reduce the accuracy of the method. The parallax method is used to determine large distances, such as the distance from Earth to a planet or a star.

From de-Broglie equation:

\(\lambda=\frac{\mathbf{h}}{\mathbf{p}}\) ......(i)

Where \(\mathrm{h}\) and \(\mathrm{p}\) are the Planck's constant and momentum of the electron respectively.

Kinetic energy of the electron:

\(E=\frac{p^{2}}{2 m}\)

\(\Rightarrow p=\sqrt{2 \mathrm{mE}}\)

\(\therefore \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}\quad\quad\) [From eq (i)]

\(\Rightarrow \lambda \propto \frac{1}{\sqrt{\mathrm{E}}}\)

Thus, wavelength associated with the wave will decrease with the increase in kinetic energy of electron.

The alternating emf in the circuit is:

\(e=e_{o} \sin \omega t\)

Current in the capacitive circuit is:

\(I=I_{o} \sin \left(\omega t+\frac{\pi}{2}\right)\)

From above it is clear that electric current through the capacitor leads the applied voltage by \(\frac{\pi}{2}\).

\(\therefore\)The current through the capacitor is of the applied voltage is \(\frac{\pi}{2}\) ahead.

Given that the threshold frequency for metal surface corresponds to an energy \((\nu)=6.2 {eV}\)

From Einstein's photoelectric equation:

\(h v=h v_{0}+e V_{0} \)

\(=6.2+5=11.2 {eV} \)

\(\Rightarrow \frac{h c}{\lambda}=11.2 {eV} \)

\(\lambda=\frac{h c}{11.2 {eV}} \)

\(=\frac{6.6 \times 10^{-34} \times 3.0 \times 10^{8}}{11.2 \times 1.6 \times 10^{-19}} \)

\(=1.1049 \times 10^{-7} \)

\(=1104.9 \mathring{\mathrm{A}}\)

This incident radiation lies in ultraviolet region,

Infrared waves are called waves of heat energy.

• The infrared waves present in the light ray coming from the sun is responsible for the heat energy.
• Thus, infrared waves are called as waves of heat energy.
• The Radio waves, Ultraviolet waves and Microwaves don’t have the heat energy.

Given,

Mass (m) = 5 kg

Coefficient of friction (μ) = 0.3

g = 10 m/s2

Free body diagram of the given condition,

In vertical direction forces are balanced.

∴ \(N = mg\)

The friction force is calculated by,

\(F = μ N =  μ mg\)

∴ \(F = 0.3 × 5 × 10 = 15\) N