Physics Test-33

The Carbon Nitrogen Cycle (CNO) (sometimes called Bethe – Weizsacker cycle) is one of the two known sets of fusion reactions by which stars convert hydrogen to helium, the other being the proton–proton chain reaction (p–p cycle), which is more efficient at the Sun's core temperature. The CNO cycle is hypothesized to be dominant in stars that are more than 1.3 times as massive as the Sun. The required equations are:

\({ }_6^{12} C +{ }_1^1 H \rightarrow{ }_7^{13} N + Q _1 \)

\({ }_7^{13} N \rightarrow{ }_6^{13} C+\left({ }_1^0 {e}\right) + {Q}_2\)

\({ }_6^{13} C +{ }_1^1 H \rightarrow{ }_7^{14} N + Q _3 \)

\({ }_7^{14} N +1 H \rightarrow{ }_8^{15} O + Q _4 \)

\({ }_8^{15} O \rightarrow{ }_7^{15} N +\left({ }_1^0 e \right)+ Q _5 \)

\({ }_7^{15} N +1 H \rightarrow{ }_6^{12} C +{ }_2^4 He + Q _6\)

Overall reaction:

\(4_1^1 H \rightarrow{ }_2^4 He +2\left({ }_1^0 e \right)+ Q \quad\) where \(Q = Q _1+ Q _2+ Q _3+ Q _4+ Q _5+ Q _6\) and theoretically it is net energy released in Carbon - Nitrogen cycle.

It can be seen that carbon along with hydrogen and in final reaction carbon is the product. Thus, carbon act as a catalyst in the overall reaction.

A coin, feather, and brick are dropped simultaneously in a vacuum, then feather, coin, and brick all will reach the bottom together.

It is first further to be assumed that both are dropped from the same height and both hit the same ground. As this is being carried out in a vacuum, no force other than that due to gravity is acting on them. Therefore they will hit the ground at the same time. The mass has no role here.

The distance (height) covered by them will be equal:

\(h=u t+\frac{1}{2} g t^{2}\)

Given that the masses will be 3 kg and 9 kg after the explosion.

Initially mass is at rest so P_initial = 0

Now the explosion occurs due to internal forces, there is no external force involved, so momentum will be conserved.

According to the law of conservation of momentum, the initial momentum of the system is zero (as there is no motion initial). So the final momentum of the system will also be zero.

This implies that the momentum of the divided parts of the bomb must move in opposition to each other.

\(P_{\text {initial }}=P_{\text {final }}\)

\(\therefore 0=P_{b i g}+P_{\text {small }}\)

\(\Rightarrow 0=9 \times v+3 \times 18\)

\(\Rightarrow v=-6 m / s\)

Here negative sign means 9 kg mass will move opposite to the 3 kg mass.

Torque is the rotational equivalent of force and causes angular acceleration. The relationship between torque and angular acceleration is direct and can be expressed as τ = Iα, where τ is torque, I is the moment of inertia, and α is the angular acceleration.

When electrodes are immersed in water and electricity passed, the bubbles formed on the negative terminal is actually hydrogen gas.

When an electric current is passed through acidulated water, it will produce a reaction called electrolysis. If an electric current is passed through acidified water, then bubbles of oxygen gas(O₂) and hydrogen(H₂) gas are produced at the two electrodes immersed in it.

Power produced:

\(\mathrm{P}=10^{7} \mathrm{~kW}=10^{10} \mathrm{~J} / \mathrm{s}\)

Time, \(\mathrm{t}=1\) day \(=24 \times 60 \times 60 \mathrm{~s}\)

Energy produced per day,

\(E=P \times t=10^{10} \mathrm{~J} / \mathrm{s} \times 864 \times 10^{2} \mathrm{~s}\)

Mass equivalence relation:

\(E=m c^{2}\),

Therefore,

Mass \(=m=\frac{E}{c^{2}}\)

\(=\frac{\left\{10^{10}\times 864 \times 10^{2 }J\right\}}{\left\{3 \times 10^{8} \right\}^{2}}\)

\(=9.6 \times 10^{-3}\mathrm{~kg}\)

\(=9.6 \mathrm{~gm}\)

The sensitivity of the potentiometer can be increased by increasing the length of potentiometer wire.

A potentiometer is considered to be sensitive if the potential gradient (dV/dl) is low. Such a potentiometer can measure very small changes in potential difference. Increasing the length of the potentiometer wire decreases the potential gradient. Its sensitivity increases. Increasing potential gradient decreases the sensitivity. increasing the emf of the primary cell and by decreasing the length, potential gradient increases.

Given, 

m = 10 Kg

g = 10 m/s²

First draw the free body diagram of the given figure,

Now resolve all forces into a horizontal and a vertical component.

​Horizontal Component \(= 100 \cos45^\circ =100\times \frac{1}{\sqrt{2}}\)

\(= 50\sqrt{2} N\)

Vertical Component \(= 100\sin45^\circ=100\times \frac{1}{\sqrt{2}}\)

\(= 50\sqrt{2} N\)

For the system to be in equilibrium \(T_{1}\) must be equal to the horizontal component but the direction must be opposite.

Here the direction of \(T_{1}\) is reversed for balancing the horizontal component.

\( T _{2}=50 \sqrt{2 } N\)

\( T _{1}=-50 \sqrt{2} N\)

So, the value of \(T_{1}\) is \(-50 \sqrt{2} N\).(Negative sign only show the direction of \(T_{1}\).)

The ability of an optical instruments to show the images of two adjacent point objects as separate is called resolving power.

By definition, resolving power of an optical instrument is its ability to show two closely adjacent point (closely spaced) as distinct as possible.

We should select a material which have low heat transfer capacity. So we should select a material which thermal conductivity is low. So that heat will pass through material is minimum. We should use low thermal conductivity material both side because after all we need to decrease the rate of heat transfer.