Physics Test - 20

Given,

\(m _{ A }=9 \) kg

\(m _{ B }=25\) kg

\(KE _{ A }= KE _{ B }\)

As we know,

The relation between the kinetic energy and the momentum is given as,

\( P ^{2}=2 mKE\)

Where \(\mathrm{K.E} = \text{Kinetic Energy}\)

Momentum for body A, \((P_{A})^{2}=2 m_{A} \times K E_{A}^{2}\)

\(\therefore (P_{A})^{2}=2 \times 9 \times K E\)

\(\Rightarrow (P_{A})^{2}=18 \times K E\)...(1)

Momentum for body B \((P_{B})^{2}=2 m_{B} \times K E_{B}^{2}\)

\(\therefore (P_{B})^{2}=2 \times 25 \times K E\)

\(\Rightarrow (P_{B})^{2}=50 \times K E\)...(2)

On dividing equation (1) and (2), we get

\(\frac{(P_{A})^{2}}{(P_{B})^{2}}=\frac{18 \times K E}{50 \times K E}\)

\(\Rightarrow \frac{(P_{A})^{2}}{(P_{B})^{2}}=\frac{9}{25}\)

\(\Rightarrow \frac{P_{A}}{P_{B}}=\sqrt{\frac{9}{25}}\)

\(\Rightarrow \frac{P_{A}}{P_{B}}=\frac{3}{5}\)

\(\Rightarrow P_{A}: P_{B}=3: 5\)

So, the ratio of linear momentum is \(3:5\).

Given - n2 = 2 and n1 = 1

According to the Bohr atomic model, the time period of a revolution is given by:

⇒ $T_n=\frac{4{ϵ}_o^2{n}^3{h}^3}{m{Z}^2{e}^4}$

Where n = number of orbit, h = plank's constant, m = mass of electron, Z = atomic number and e = charge on electron

According to the questions, we have to find the time period of the revolution of an electron of a hydrogen atom. So, ϵ_o, h, m, Z, and e are constant.

∴ T α n³

$\Rightarrow \frac{T_2}{T_1}=\frac{n_2^3}{n_1^3}$

$\Rightarrow \frac{T_2}{T_1}=\frac{2^3}{1^3}=\frac{8}{1}$

⇒ T₂ = 8 T₁

Given that a clock keeps correct time at \(25^{\circ} C\). So initial temperature \(\theta_{1}=25^{\circ} C\).

The pendulum is made of metal. We need to find the number of seconds gained per day by the clock when the temperature falls to \(0^{\circ} \mathrm{C}\).

So final temperature \(\theta_{2}=0^{\circ} \mathrm{C}\)

So, \(\theta_{1}-\theta_{2}=25-0=25^{\circ} \mathrm{C}\)

The relation connecting gain or loss in time and the change in temperature is given as:

\(\Delta t=\frac{1}{2} \times \alpha \times \Delta \theta \times t\).....(1)

Where \(\Delta t\) is the gain or loss in time.

\(\alpha\) is the coefficient of linear expansion.

\(\Delta \theta\) is the change in temperature and \(\mathrm{t}\) is the time.

\(\alpha\) is given as \(1.9 \times 10^{-5}\) per \({ }^{\circ} \mathrm{C}\)

We need to calculate the gain of time per day.

Thus \(t=24 \times 60 \times 60 \mathrm{~s}\)

Change in temperature is \(\Delta \theta=25^{\circ} \mathrm{C}\)

Now substitute all the given values in the equation (1).

\(\Delta t=\frac{1}{2} \times 1 \cdot 9 \times 10^{-5} \times 25 \times 24 \times 60 \times 60 \)

\(\therefore \Delta t=20.520\) seconds / day

Current gain \(=\frac{{I}_{C}}{{I}_{B}}\)

\({I}_{{C}}=\) collector current

\(I_{B}=\) base current

So, \(I_{B}=\frac{I_{C}}{\text { current gain }}\)

\(=\frac{7.2}{0.96}\) mA

\(=7.5\) mA

As \(I_{E}=I_{B}+I_{C}\)

So, \(I_{B}=I_{E}-I_{C}\)

\(=(7.5-7.2)\) mA

\(=0.3\) mA

\(\approx 0.29\) mA

Given,

Electric field, \(E=9 \times 10^{4} N / C\)

Distance, \(r=2 \times 10^{-2} m\)

Using the formula of electric field for uniformly charged wire

\(E=\frac{\lambda}{2 \pi r \varepsilon_{0}} \)

\(\therefore \lambda=E \times 2 \pi r \varepsilon_{0}\)

where, \(\lambda\) is a linear charge density,

and, \(\epsilon_{0}=8.854 \times 10^{-12}\)

Then,

\(\lambda=9 \times 10^{4} \times 2 \pi \times 2 \times 10^{-2} \times 8.854 \times 10^{-12} \)

\(\lambda= 18 \times 10^{4} \times 2 \times 3.14 \times 10^{-2} \times 8.854 \times 10^{-12} \)

\(\lambda= 10^{3}\times 10^{-10} \)

\(\Rightarrow\lambda = 10^{-7}\)

Therefore,

Linear charge density, \(\lambda= 10^{-7}\ C / m\)

According to Einstein's quantum theory light propagates in the form of packets i.e. quanta of energy, which is called a photon.

The rest mass of photons is being zero. It can be shown, according to the relativity theory of light.

According to the relativistic theory equation, the mass of the photon is computed as:

\({m}=\frac{{m}_{0}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}\)

\(\Rightarrow {m}_{0}={m} \sqrt{1-\frac{{v}^{2}}{{c}^{2}}}\)

when \({v}=0\)

So, \(m_{0}=0\)

Where \(m_{0}\) is the rest mass of the Photon.

In the interface of Brewster's angle is written as;

μ = tan ib ----(1)

here, μ is the refractive index of the given material.

where μ lies between 1 to ∞.

1 < μ < ∞

putting eqaution (1) in the above equation we have;

1 < tan ib < ∞ 

where, ib is the polarization angle.

 tan^-1 (1) < ib < tan^-1 (∞)

 ⇒ 45° < ib < 90°

We know,

\(V_{r m s}=\sqrt\frac{3 k T}{ m}\) ,The symbols have their usual meaning 

\(V_{escape}  =2 g R\) where \(R\) is the radius of the moon.

Given \(V_{r m s}=V_{escape} \)

\(\therefore \sqrt\frac{3 k T}{ m}\)=\(\sqrt{2 g R} \)

\(T=\frac{2 m g R}{3 k}\)

​Positive charge: A body having a deficiency of electrons and the number of protons becomes more than the number of electrons.

​Negative charge: A body having an excess of electrons and the number of electrons becomes more than the number of protons.

Neutron is not responsible for the charge on the element.

The number of neutrons becomes more than the number of electrons in the element, by this statement, we can't predict whether the number of protons is more than, less than, or equal to the number of electrons in the element.

So we can't predict the type of charge on the element. Hence, option 4 is correct.

It is given that,

Water flows at a rate of \(3.0\) litre \(/ min =3 \times 10^{-3} m ^{3} / min\)

Density of water, \(\rho=10^{3} kg / m ^{3}\).

Clearly, mass of water flowing per minute \(=3 \times 10^{-3} \times 10^{3} kg / min =3 kg / min\)

The geyser heats the water, raising the temperature from \(27^{\circ} C\) to \(77^{\circ} C\).

Initial temperature, \(T_{1}=27^{\circ} C\)

Final temperature, \(T_{2}=77^{\circ} C\)

Thus, rise in temperature,

\(\Delta T=T_{2}-T_{1}\)

\(\Delta T=77^{\circ} C -27^{\circ} C \)

\(\Delta T=50^{\circ} C\)

Now, heat of combustion \(=4 \times 10^{4} J / g =4 \times 10^{7} J / kg\)

Specific heat of water \(=4.2 J / g^{\circ} C\)

It is known that total heat used, \(\Delta Q=m c \Delta T\)

\(\Delta Q=3 \times 4.2 \times 10^{3} \times 50 \)

\(\Delta Q=6.3 \times 10^{5} J / min\)

Now, consider \(m kg\) of fuel to be used per minute.

Thus, the heat produced \(=m \times 4 \times 10^{7} J / min\)

However, the heat energy taken by water \(=\) heat produced by fuel

Thus, equating both the sides,

\(\Rightarrow 6.3 \times 10^{5}=m \times 4 \times 10^{7} \)

\(\Rightarrow m=\frac{6.3 \times 10^{5}}{4 \times 10^{4}} \)

\(\Rightarrow m=15.75 g / min\)

Clearly, the rate of consumption of the fuel when its heat of combustion is \(4.0 \times 10^{4} J / g\) supposing the geyser operates on a gas burner is \(15.75 g / min\).