Physics Test - 26

Given:

The value of charge on the first sphere is \(2 \times 10^{-7} \mathrm{C}\).

The value of charge on the second sphere is \(3 \times 10^{-7} \mathrm{C}\).

The distance between spheres is \(30 \mathrm{~cm}=30 \mathrm{~cm} \times\left(\frac{\mathrm{m}}{100 \mathrm{~cm}}\right)=0.3 \mathrm{~m}\)

We are required to calculate the value of force between the given two spheres due to their charge.

Let us write the expression for force of attraction or repulsion between the given spheres.

\(F=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r^{2}} \quad\quad\ldots \ldots .(1)\)

Here \(q_{1}\) is the charge on the first sphere, \(q_{2}\) is the charge on the second sphere, \(r\) is the distance between spheres and \(\varepsilon_{0}\) is free space permittivity.

We know the value of free space permittivity is:

\(\varepsilon_{0}=8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-2}\)

Substitute \(2 \times 10^{-7} \mathrm{C}\) for \(q_{1}, 3 \times 10^{-7} \mathrm{C}\) for \(q_{2}, 0.3 \mathrm{~m}\) for \(r\) and \(8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\) for \(\varepsilon_{0}\) in equation (1).

\(F=\frac{\left(2 \times 10^{-7} \mathrm{C}\right)\left(3 \times 10^{-7} \mathrm{C}\right)}{4 \pi\left(8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)(0.3 \mathrm{~m})^{2}}\)

\(=\frac{\left(2 \times 10^{-7} \mathrm{C}\right)\left(3 \times 10^{-7} \mathrm{C}\right)}{4 \times 3.14\left(8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)(0.3 \mathrm{~m})^{2}}\)

\(=5.994 \times 10^{-3} \mathrm{~N}\)

Given that:

Input Voltage,\(V_{P}=220 \) Volt

Output power = 44 Watt

We know that:

In ideal transformer:

Input power = Output power

\(\Rightarrow V_{P} I_{p}=V_{S} I_{s}=\) Given power

\(\Rightarrow 220 \times I_{P}=44\)

\(\Rightarrow I_{P}=0.2 \mathrm{~A}\)

Incident Energy is increased by \(20 \%\).

Initial kinetic energy \((KE_{1})=0.5 \mathrm{eV}\) 

Final kinetic energy \((K E_{2})=0.8 \mathrm{eV}\) 

\(E_{1}=E=\) initial incident energy

According to the question.

\(E_{2}=E_{1}+E_{1} \times 20 \%=E_{1}+0.2 E_{1}=1.2 \mathrm{E}_{1}=1.2 \mathrm{E}\)

Let \(\phi\) is the work function of metal. 

According to Einstein equation:

\( E=\phi+K E \)

\(E_{1} =\phi+K E_{1}=E \)

Put the value of \(K E_{1}\) in above equation,

\(E_{1} =\phi+0.5=E \cdots(i)\)

\(E_{2} =\phi+K E_{2}=1.2 E \)

Put the value of \(K E_{2}\) in above equation,

\( E_{2}=\phi+0.8=1.2 E \cdots(ii)\)

Divide equation (i) by equation (ii),

\(\frac{E_{1}}{E_{2}}= \frac{\phi+0.5}{\phi+0.8}=\frac{E}{1.2E}=\frac{1}{1.2}\)

\(\Rightarrow 1.2 \phi +0.6=\phi+0.8 \)

\(\Rightarrow 0.2 \phi =0.2 \Rightarrow \phi =\frac{0.2}{0.2}=1 \mathrm{eV}\)

Work function of metal is \(\phi=1 \mathrm{ev}\).

Given:

distance \((r)=0.2\) m

charge \((q)=1.6 \times 10^{-7}\) C

We know that:

Electric field at a point \(0.2\) m from the center of the ball,

\(E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}\)

Where, \(\varepsilon_{0}=\) the permittivity of free space,

\(E=\frac{9 \times 10^{9} \times 1.6 \times 10^{-7}}{(0.2)^{2}}\)

\(=\frac{14.4 \times 10^{2}}{0.04}\)

\(=360 \times 10^{2}\)

\(=3.6 \times 10^{4} N / C\)

\(\therefore\) The electric field at a point \(0.2\) m from the center of the iron ball \(=3.6 \times 10^{4} NC ^{-1}\)

From figure, \(\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{d}}{\mathrm{a}} \Rightarrow \mathrm{y}=\frac{\mathrm{d}}{\mathrm{a}} \mathrm{x}\)

\(\mathrm{dy}  =\frac{\mathrm{d}}{\mathrm{a}}(\mathrm{dx}) \Rightarrow \frac{1}{\mathrm{dc}}=\frac{\mathrm{y}}{\mathrm{K} \varepsilon_0 \mathrm{adx}}+\frac{(\mathrm{d}-\mathrm{y})}{\varepsilon_0 \mathrm{adx}} \)

\(\frac{1}{\mathrm{dc}}  =\frac{\mathrm{y}}{\varepsilon_0 \mathrm{abx}}\left(\frac{\mathrm{y}}{\mathrm{k}}+\mathrm{d}-\mathrm{y}\right) \)

\(\int \mathrm{dc}  =\int \frac{\varepsilon_0 \mathrm{adx}}{\frac{\mathrm{y}}{\mathrm{k}}+\mathrm{d}-\mathrm{y}}\)

or, \( c=\varepsilon_0 a \cdot \frac{a}{d} \int_0^d \frac{d y}{d+y\left(\frac{1}{k}-1\right)}\)

\( =\frac{\varepsilon_0 a^2}{\left(\frac{1}{k}-1\right) d}\left[l n\left(d+y\left(\frac{1}{k}-1\right)\right)\right]_0^d \)

\( =\frac{k \in_0 a^2}{(1-k) d} ln \left(\frac{d+d\left(\frac{1}{k}-1\right)}{d}\right) \)

\( =\frac{k \in_0 a^2}{(1-k) d} l n\left(\frac{1}{k}\right)=\frac{k \in_0 a^2 \ell n k}{(k-1) d}\)

The intensity of magnetic induction field

\(\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}\)

\(=\frac{4 \pi \times 10^{-7} \times 0.9}{2 \times 5 \times 10^{-2}}\)

\(\mathrm{B}=36 \pi \times 10^{-7} \mathrm{~T}\)

Given,

Resistance of coil \(=10 \Omega\)

\(\phi=8 t^{2}-4 t+1\)

We know that,

E.m.f. induced, \(E=-\frac{d \phi}{d t}\)

\(\Rightarrow E=-\frac{d\left(8 t^{2}-4 t+1\right)}{d t}\)

By differentiating, we get

\(E=-16 t+4\)

Here, \(t=0.1 \mathrm{sec}\),

\(E=-1.6+4=2.4 \mathrm{~V}\)

Current induced, \(I=\frac{E}{R}\)

\(\Rightarrow I=\frac{2.4}{10}\)

\(\Rightarrow I=0.24 \mathrm{~A}\)

Given:

Depth of the fish from the surface of the water \(=4 \mathrm{~m}\)

Height of the flame \(\mathrm{BC}=2 \mathrm{~m}\)

Refractive index of the water, \(\mu=\frac{4}{3}\)

We have to find the the apparent height of the flame from the eyes of the fish. The ray diagram is shown below:

When viewed from the rarer medium, the apparent height is

Apparent height \(=\frac{\text { Real height }}{\text { Refractive index of the denser medium }}\)

But, when viewed from the denser medium, the apparent height is

Apparent height \(=\) Real height \(\times\) Refractive index of the densermedium

\(A C=B C \times \mu \)

Put the given valuesin above formula:

\(A C=2 \times \frac{4}{3}=\frac{8}{3} \)

\(\Rightarrow A C=\frac{8}{3}\)

Thus, the total apparent height of the flame from the eyes of the fish is

\(A D=A C+C D \)

\(\Rightarrow A D=\frac{8}{3}+4 \)

\(=\frac{20}{3} m\)

Given:

Refractive index of air, \(\mu_{1}=1\)

Refractive index of glass, \(\mu_{2}=1.5\)

Radius of curvature, \(\mathrm{R}=20 \mathrm{~cm}\)

Object distance, \(\mathrm{u}=-\mathrm{1 0 0} \mathrm{cm}\)

We have to find the distance of the image from the spherical surface.

We know that,

\(\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R} \)

\(\Rightarrow \frac{1.5}{v}-\frac{1}{(-100)}=\frac{1.5-1}{20} \)

\(\Rightarrow \frac{1.5}{v}=\frac{0.5}{20}-\frac{1}{100} \)

\(\Rightarrow \frac{1.5}{v}=\frac{2.5-1}{100} \)

\(\Rightarrow v=\frac{1.5 \times 100}{1.5} \)

\(=100 \mathrm{~cm}\)

The correct expression for the amplification factor is\(\left(\frac{\Delta I_{E}}{\Delta I_{B}}\right)_{V_{C E}}\).

Amplification factor can be defined as theratio of change in emitter current (ΔI_E) to the change in base current (ΔI_B) is known as Current Amplification factor in common collector (CC) configuration. It is denoted by γ. The current gain in CC configuration is same as in CE configuration. The voltage gain in CC configuration is always less than 1.

The melting point of ice is 0ºC. At this temperature the ice in equilibrium with its liquid state, water.

When only ice used to make ice cream, at 0ºC ice starts melting by absorbing the energy from its environment in the form of heat.

Addition of salt to ice while making ice cream lowers the melting point of ice.

The temperature at which the equilibrium between ice and its liquid state will be achieved is lowered.

Therefore, salt is added to ice creams to decrease the melting point of ice cream.

Suppose that the distance between two protons \({p}_{1}\) and \({p}_{2}\) be \({r}_{12}\) and electron \(\left({e}^{-}\right)\) is placed at \({r}_{13}\) and distance from protons \({p}_{1}\) and \({p}_{2}\) respectively.

Given, \(r_{12}=1.5 \mathring A=1.5 \times 10^{-10} {~m}\); \(r_{13}=r_{23}=1 \mathring A=10^{-10} {~m}\); \(q_{1}=q_{2}=1.6 \times 10^{-19} {C}\) (protons); \({q}_{3}=-1.6 \times 10^{-19} {C}\) (electrons)
Now, the potential energy of the system \(W=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{q_{1} q_{2}}{r_{12}}+\frac{q_{1} q_{3}}{r_{13}}+\frac{q_{2} q_{3}}{r_{23}}\right)\) \(=9 \times 10^{9}\left[\frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{1.5 \times 10^{-10}}+\frac{1.6 \times 10^{-19} \times\left(-1.6 \times 10^{-19}\right)}{10^{-10}}+\frac{1.6 \times 10^{-19} \times\left(1.6 \times 10^{-19}\right)}{10^{-10}}\right]\)
\(W=\frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19} \times 9 \times 10^{9}}{10^{-10}}\left[\frac{1}{1.5}-\frac{1}{1}-\frac{1}{1}\right]\) \(=1.6 \times 1.6 \times 9 \times 10^{-19-19+9+10}\left[\frac{2}{3}-2\right]\) \(=-2.56 \times 9 \times 10^{-19} \times \frac{4}{3}=-2.56 \times 12 \times 10^{-19} \mathrm{~J}\)
\(W=\frac{-2.56 \times 12 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}\)\(=-1.6 \times 12 \mathrm{eV}\) \(=-19.2 \mathrm{e} \mathrm{V}\)

Given that:

Inductance, L = 5.0 H

Capacitance, C = 80 \(\mu \mathrm{F}\)\(=80 \times 10^{-6}\) F

We know that:

The resonance frequency of LCR series circuit is given as:

\(\omega_{0}=\frac{1}{\sqrt{L C}}\)

\(=\frac{1}{\sqrt{5 \times 80 \times 10^{-6}}}\)

\(=50 \mathrm{rad} / \mathrm{s}\)

Now half power frequencies are given as:

\(\omega=\omega_{0} \pm \frac{R}{2 L}\)

i.e.,

\(\omega_{L}=50-\frac{40}{2 \times 5}\)

\(=46 \mathrm{rad} / \mathrm{s}\)

\(\omega_{H}=50+\frac{40}{2 \times 5}=54 \mathrm{rad} / \mathrm{s}\)

Energy of photons \(=\mathrm{h\nu}\), Work function of surface \(=\mathrm{E_0}\), Maximum kinetic energy \(=\mathrm{K}\)

Now, frequency is doubled and the energy of photon is used in liberating the electron from metal surface and in imparting the kinetic energy to emitted photoelectron.

According to Einstein's photoelectric effect energy of photon = KE photoelectron + Work function of metal i.e.,

\(\mathrm{h\nu=\frac{1}{2} m v^{2}+E_{0}}\) 

\(\mathrm{h \nu=K_{\max }+E_{0}}\)

\(\mathrm{ K_{\max }= h \nu-E_{0}}  \cdots(i)\)

Now, we have given,

\(\nu'=2 \nu\)

Therefore, \(\mathrm{K_{\max }=h(2\nu)-E_{0}}\)

\(\mathrm{K_{\max }^{\prime}=2 h \nu-E_{0}}\)

From Equation (i) and (ii), we have

\(\mathrm{K'_{\max }=2\left(K_{\max }+E_{0}\right)-E_{0}}\)

\(=2 \mathrm{K_{\max }+E_{0}}\)

\(=\mathrm{K_{\max }+\left(K_{\max }+E_{0}\right)}\)

\(=\mathrm{K_{\max }+h \nu}\quad\) [From Equation (i)]

\(\mathrm{K_{\max }=K}\)

\(\mathrm{\therefore K'_{\max }=K+h\nu}\)