Physics Test-23

Since the particle carrying positive charge is revolving around another charge,

Electrostatic force \(=\) Centrifugal force

\(\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}=m r \omega^{2}\)

\(\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}=\frac{4 \pi^{2} m r}{T^{2}}\)

\(T^{2}=\frac{\left(4 \pi \epsilon_{0}\right) r^{2}\left(4 \pi^{2} m r\right)}{q_{1} q_{2}}\)

\(T=4 \pi r \sqrt{\frac{\pi \epsilon_{0} m r}{q_{1} q_{2}}}\)

EMF induced in an electrical circuit

\(\mathrm{e}=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}\) (numerically)

or \(\mathrm{L}=\mathrm{e} \frac{\mathrm{dt}}{\mathrm{dI}}\)

\(=\frac{\mathrm{W}}{\mathrm{Q}} \cdot \frac{\mathrm{dt}}{\mathrm{d} I}\left(\because \mathrm{e}=\mathrm{V}=\frac{\mathrm{W}}{\mathrm{Q}}\right)\)

\(=\frac{\mathrm{Wdt}}{\mathrm{It} \cdot \mathrm{dl}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right][\mathrm{T}]}{[\mathrm{A}][\mathrm{T}][\mathrm{A}]}\)

\(=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]\)

When a nucleus is formed, then the relation is given by,

\(( A - Z ) m _{ n }+ Z m _{ p }= M + Q\);

Where \(Q\) is the binding energy, \(Z\) is the atomic number, \(A\) is the mass number, \(M\) is mass of nucleus, \(m_n\) is mass of neutron, \(m_p\) is mass of proton.

When the nucleus gets broken into its constituents then the binding energy disappears.

So, the final obtained expression is:

\(M <( A - Z ) m _{ n }+Z m _{ p }\)

When the coil deflects, the suspension wire is twisted. On account of elasticity, a restoring couple is set up in the wire. This couple is proportional to the twist.

If \(\theta=\) the angular twist, then,

The moment of the restoring couple \(= C \theta\)

Where \(C=\) restoring couple per unit twist.

At equilibrium,

Deflecting couple \(=\) Restoring couple

So, we can write,

\(nBIA = C \theta\)

\(\Rightarrow \theta=\left(\frac{n B A}{C}\right) I\)

Here, \(n, B\), and \(A\) are constant.

Therefore, the deflection produced in the moving coil galvanometer is directly proportional to the amount of current 'I' passing through it.

Thus, if the current flowing through the moving coil galvanometer is increased, then the deflection in the coil will also increase.

The center of mass of a rigid body lies either inside or outside.

Archimedes' principle states that the upward force which is exerted by an object that is either partially or fully immersed in a fluid is equal to the weight of the liquid displaced by the object. This principle helps to prove the Law of floatation.

• The law of floatation states that the weight of a floating body is equal to the weight of the liquid displaced by it.
• For Example- A ship or submarine in the ocean does not sink until the weight of the water it displaces is equal to its own weight.

"Eddy currents converts useful energy into heat and waste it" is correct regarding eddy currents in the coil.

Eddy Current is the loops of electrical current induced within conductors by changing magnetic fields in the conductor are called eddy currents.Eddy currents transform useful energy, into heat, which isn’t generally useful.Eddy currents cause a loss of energy because they have the tendency to oppose.

Out of carbon, germanium, and silicon, carbon has the maximum energy bandgap whereas germanium has the least energy bandgap.

For all these elements the energy band gap can be related as:\(\left(E_{g}\right)_{C}>\left(E_{g}\right)_{S i}>\left(E_{g}\right)_{G e}\)

After division, the mass of the bomb divides into two parts ratio of masses is 1 : 3.

So after division, the masses will be 3 kg and 9 kg and the kinetic energy of the smaller part is 216 J.

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.

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

\(\Rightarrow 0=P_{\text {big }}-P_{\text {small }} \)

\(\Rightarrow P_{b}=P_{s}\)

The kinetic energy of the bigger part, \(K_{b}=\frac{P_{b}^{2}}{2 M_{b}}\)

The kinetic energy of the smaller part, \(K_{s}=\frac{P_{s}^{2}}{2 M_{s}}\)

Since, \(P _{ b }= P _{ s }\)

\(\therefore P _{ b }^{2}= P _{ s }^{2}\)

\(\Rightarrow P_{b}^{2}=K_{s}\left(2 M_{s}\right) \quad [\because P _{ s }^{2}=K_{s}(2 M_{s})]\)

\(\Rightarrow P_{b}^{2}=216(2 \times 3) \)

\(\Rightarrow P_{b}^{2}=216 \times 6 \)

\(\Rightarrow P_{b}=\sqrt{1296}\)

\(\Rightarrow P_{b}=36 \)kg-m/s

\(\therefore\) The momentum of the bigger part is \(36\)kg-m/s.

A light bulb and an open coil induction are connected in series to an ac source as shown in figure. Now an iron rod is inserted into the interior of the inductor. The glow of the light bulb will be decreases.

When an iron rod is inserted into the interior of the inductor then inductance \((L)\) of the coil increases. Here due to the iron rod, the inductance \((X_{L})\) of the coil increases.

Current \((I)\) = \(\frac{V}{X_L}\)

If the inductance increase then the current will decrease and the bulb will glow less brightly.