Physics Test - 2

Candela is a unit of luminous intensity. In photometry, the wavelength-weighted power emanating from a light source in a particular direction, in a unit solid angle, is called luminous intensity. It is based on the optical formula, which is a standardized model of the sensitivity of the human eye.

The relationship between the energy of a photon (E) and the wavelength of the light (λ) is an inverse relationship and is described by the following by the equation:

\({E}=\frac{{hc}}{\lambda}={R}_{0}\left[\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right] \)

\({h} v={R}_{0}\left[\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right]\)

Put the values of \(n_{1}\) and \(n_{2}\) for each options and then we get lower frequency when \(n=6\) to \(n=2\).

Given,

Radius, \(r=1000 m\)

Banking angle, \(\theta=45^{\circ}\) 

Coefficient of friction, \(\mu=0.5\)

Acceleration due to gravity, \(g=9.8 m/s^2\)

As we know,

Maximum safe velocity is calculated by,

\(v=\sqrt{\frac{r g(\tan \theta+\mu)}{1-\mu \tan \theta}} \)

\(\therefore v=\sqrt{\frac{1000 \times 9.8\left(\tan 45^{\circ}+0.5\right)}{1-0.5 \times \tan 45^{\circ}}} \)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 (1+0.5)}{1-0.5 \times 1}}\)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 \times 1.5}{1-0.5}}\)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 \times 1.5}{0.5}}\)

\(\Rightarrow v=\sqrt{29400}\)

\(\therefore v =172 m / s\)

It can never decrease is correct for entropy of an isolated system,

Second law of thermodynamics:

The entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium.

In other words, The second law of thermodynamics states that the total entropy of an isolated system always increases over time, or remains constant in ideal cases where the system is in a steady-state or undergoing a reversible process. This increase in entropy accounts for the irreversibility of natural processes.

The second law gives the definition of entropy term.

\(\Delta S=\frac{\Delta Q}{T}\)

Where \(\Delta S\) is the entropy change, \(\Delta Q\) is the energy change and \(T\) is the temperature.

Isolated system: The system which doesn't interact with the surrounding is called isolated system. The universe is assumed as an isolated system.

According to the second law of thermodynamics, the entropy of an isolated system increases with time It can never decrease.

Since the universe is assumed as an isolated system and the entropy of an isolated system always increases with time.

As the entropy of an isolated system increases always, so the entropy change of an isolated system can never be negative.

Pressure at any point can be calculated as

P = ρgh

Where, ρ = density of the liquid, g = acceleration due to gravity, h = height at which the pressure is to be calculated.

The pressure is dependent on two things (as g is constant) i.e density of the liquid and height at which the pressure is to be calculated,

As mentioned in the Qs the liquid used is same in both the vessels, hence in this case pressure will be only dependent on the height.

Point P and Q both are at same height and hence the pressure at both the points will be same.

As we know,

Work done by a variable force is given by,

\(W=\int_{a}^{b} F(x) d x\)

Let \(x\) be the distance covered.

Given that \(F \propto \frac{1}{x}\)

Work done, \((W)=\int_{a}^{b} F(x) d x\)

\(\therefore W \propto \int_{a}^{b} \frac{1}{x} d x \)

\(W \propto[\ln (x)]_{a}^{b} \)

\(W \propto \ln (b)-\ln (a) \)

\(W \propto \ln \left(\frac{b}{a}\right)\)

The kinetic energy is given by \({KE}=\frac{{p}^{2}}{2 {~m}}\)

\(\Delta {KE}=\frac{2 {p} \Delta {p}}{2 {~m}}=\frac{{p} \Delta {p}}{{m}}\)

\(\frac{\Delta {KE}}{{KE}}=\frac{{p} \Delta {p}}{{m}} \times \frac{2 {~m}}{\mathrm{p}^{2}}=\frac{2 \Delta {p}}{{p}}\)

Since the momentum p increases by \(20 \%\), so the final momentum becomes \(1.2~ {p}\).

\({KE}_{\text {final }}=\frac{(1.2 {p})^{2}}{2 {~m}}=1.44 \frac{{p}^{2}}{2 {~m}}=1.44 {~KE}\)

So, \(\%\) change in \({KE}=44 \%\).

Let a small strip of mop has width dx and radius x, as shown below,

Torque applied to move this strip is

dτ = Force on strip × Perpendicular distance from the axis

dτ = Force per unit area × Area of strip × Perpendicular distance from the axis.

Given,

Radius of the mop \(=\mathrm{R}\)

Force applied on the mop \(=F\)

Friction coefficient between the mop and the floor \(=\mu\)

Here, Force per unit area \(=\frac{F}{A}=\frac{F}{\pi R^{2}}\)

Area of the strip \(=2 \pi r=2 \pi x d x\)

Perpendicular distance from the axis \(=\mathrm{x}\)

Applying all these in torque,

\(\Rightarrow d \tau=\frac{\mu F}{\pi R^{2}} 2 \pi x d x . x\)

\(\Rightarrow d \tau=\frac{2 \mu F x^{2}}{R^{2}} \cdot d x\)

So, the total torque to be applied on the mop is

\(\Rightarrow \tau=\int_{x=0}^{x=R} d \tau=\int_{0}^{R} \frac{2 \mu F x^{2}}{R^{2}} \cdot d x\)

\(\Rightarrow \tau=\frac{2 \mu F}{R^{2}} \times\left[\frac{x^{3}}{3}\right]_{0}^{R}\)

\(\Rightarrow \tau=\frac{2 \mu F}{R^{2}} \times \frac{R^{3}}{3}\)

\(\therefore \tau=\frac{2}{3} \mu F R(\mathrm{Nm})\)

Therefore, the torque applied by the machine on the mop is \(\frac{2}{3} \mu F R\) (Nm).

Given,

\(N=3000 \)

\(r=10 {~cm}=0.1 {~m} \)

\(n=\frac{N}{2 \pi {r}} \)

\(\mu_{r}=2000 \)

\(I=1 {~A}\)

For a solenoid magnetic field:

\(B=\mu n I=\mu_{r} n_{0} n I\)

\(B=2000 \times 4 \pi \times 10^{-7} \times \frac{\mathrm{N}}{2 \pi {I}} \times 1\)

\(=2000 \times 4 \times 10^{-7} \times \frac{3000}{2 \pi \times 0.1} \times 1\)

\(=12~ T\)

The distance \(s_1\) covered by the car during the time it is accelerated is given by \(2 a s_1=v^2\), which gives \(s_1=\frac{v^2}{2 a}\). The distance \(s_2\) covered during the time the car is decelerated is similarly given by \(s_2=\frac{\mathrm{v}^2}{2 \beta}\)

Therefore, the total distance covered is

\(s=s_1+s_2=\frac{v^2}{2}\left(\frac{1}{a}+\frac{1}{\beta}\right) \ldots \ldots \text { (i) }\)

If \(t_1\) is the time of acceleration and \(t_2\) that of deceleration, then \(v=a t_1=\beta t_2 \Rightarrow \mathrm{t}_1=\frac{\mathrm{v}}{a}\) and \(\mathrm{t}_2=\frac{\mathrm{v}}{\beta}\)

Therefore, the total time taken is

\(t=t_1+t_2=v\left(\frac{1}{a}+\frac{1}{\beta}\right) \ldots \ldots\)(ii)

From (i) and (ii), the average speed of the car is given by,

\(\frac{\text { Total distance }}{\text { Total time }} \frac{s}{t}=\frac{v}{2}\)