Physics Test-16

When the magnetic field through a coil changes, then then the number of magnetic fields passing through the coil changes, and hence emf is induced in the coil. Since the plane of the coil is parallel to the magnetic field so the angle between the area and the magnetic field is 90°.

We know that the magnetic flux associated with the coil is given as,

⇒ ϕ = B.A.cosθ​

⇒ ϕ = B.A.cos90

⇒ ϕ = 0 weber

So in this case the orientation of the coil is parallel to the magnetic field so here whatever be the value of the magnetic field, the magnetic flux associated with the coil will remain zero. Since in this case the magnetic flux associated with the coil is not changing so neither emf nor current will induce in the coil.

We know that when a particle enters the magnetic field in a direction perpendicular to the direction of the field, it undergoes circular motion.

The magnetic force is then equivalent to the centripetal force and is given as:

\(F=\frac{m v^{2}}{r}=q v B\)

\(\Rightarrow r=\frac{m v}{q B}\)

So, the radius of curvature \(r=\frac{m v}{q B}\)

Where, \(m=\) mass, \(q=\) magnitude of charge, \(v=\) speed of charge, \(B=\) magnetic field

It is given that the particles have the same charge (q) and initial velocity (v). Also, the magnetic field (B) is uniform.

Thus, r ∝ m

​Therefore, the radius of curvature is greater for the particle that has more mass.

In a steady flow reversible adiabatic process, work done is equal to change in enthalpy.

The first law of thermodynamics:

This law states that the heat and mechanical work are mutually convertible. According to this law, a definite amount of mechanical work is needed to produce a definite amount of heat and vice versa.

From first Law of Thermodynamics,

\(q=\Delta u+p d v\) for a closed system and,

\(q=\Delta h-v d p\) for an open system

And for a reversible adiabatic process \(q=0\), i.e. \(\Delta \mathrm{h}=\) vdp

From Steady Flow Energy Equation:

\(\left(h_{1}+\frac{v_{1}^{2}}{2}+g z_{1}\right)+q=\left(h_{2}+\frac{v_{2}^{2}}{2}+g z_{2}\right)+w\)

When potential and kinetic energy changes are zero or negligible,

\(h_{1}+q=h_{2}+w\)

Given reversible adiabatic process so \(q=0\),

\(w=h_{1}-h_{2}=\Delta h\)

As we know,

Energy of the \(\mathrm{nth}\) orbit by Bohr was given by:

\(E_{n}=R_{H} \times \frac{Z^{2}}{n^{2}} e V\)

where,

\(E=\) energy

\(R_{H}=\) Rydberg's Constant

\(Z=\) atomic number \(=3\) (for lithium)

\(n=\) number of orbit

Putting the values, in above equation, we get

Energy of the first shell \((n=1)\) in hydrogen atom:

\(z=1\)

\(-13.6 e V=R_{H} \times \frac{1^{2}}{1^{2}}\)

\(R_{H}=-13.6 \mathrm{eV}\)

To find energy value of electron in the excited state of \(L i^{2+}\) is:

\(L i: 1 s^{2} 2 s^{1} \)

\(L i^{2+}: 1 s^{1} \)

\(Z=3, n=1 \)

\(E_{n}=-13.6 \times \frac{3^{2}}{12} \mathrm{eV}\)

\(=-122.4 \mathrm{eV}\)

Given,

\(\mathrm{x}=\left[\frac{\mathrm{a}^{2} \mathrm{~b}^{2}}{\mathrm{c}}\right]\)

The percentage error in the measurement of physical quantities,

\(\frac{\Delta \mathrm{a}}{\mathrm{a}} \times 100= 2\)

\(\frac{\Delta \mathrm{b}}{\mathrm{b}} \times 100= 3\)

\(\frac{\Delta \mathrm{c}}{\mathrm{c}} \times 100= 4\)

The given quantity in terms of percentage error,

\( \frac{\Delta \mathrm{x}}{\mathrm{x}} \times 100=2 \frac{\Delta \mathrm{a}}{\mathrm{a}} \times 100+2 \frac{\Delta \mathrm{b}}{\mathrm{b}} \times 100+\frac{\Delta \mathrm{c}}{\mathrm{c}} \times 100 \)

\(\Rightarrow \frac{\Delta \mathrm{x}}{\mathrm{x}} \times 100=2 \times 2+2 \times 3+4\)

\(\Rightarrow \frac{\Delta \mathrm{x}}{\mathrm{x}} \times 100=14 \%\)

Thus the percentage error in the measurement of \(\mathrm{x}\) is 14%.

A girl swings on the cradle in a sitting position. If she stands then the time period of girl and cradle will decrease.

The time period, \(T=2 \pi \sqrt{\frac{l}{g}}\) where \({l}\) is the length of simple pendulum. On standing the effective length of the pendulum measured from the point of suspension decreases. So, the time period of the cradle decreases.

G is the universal gravitational constant which remains constant irrespective of the place and time. G is the force of attraction between two bodies of unit mass and unit distance apart.

Ground wave propagation is mode of wave propogation in which the ground has a strong influence on the propogation of signal waves from the transmitting antenna to receiving antenna. In this propogation, the signal wave glides over the surface of earth.

Skywave propagation is a mode of wave propogation in which the radiowaves emitted from the transmitter antenna reach the receiving antenna after reflection by the ionosphere.

Space wave propagation is mode of wave propogation in which the radio waves emitted from transmitter antenna reach the receiving antenna directly through space.These radiowaves are called space waves or tropospheric waves.

Surface Tension: It is the property of a fluid to resist any external force on its surface.

• Surface tension is due to the cohesive force between the fluid molecules at the surface.
• Surface tension is the force required to keep a unit length of a film of fluid in equilibrium.

Surface tension \((S)=\frac{F}{L}\)

where, \(S=\) surface tension, \(F=\) force, and \(L=\) length of fluid film.

Surface tension is dimensionally equal to the ratio of force and length. Hence, the SI unit of surface tension is a newton per meter \((N / m)\).

We know that,

Power \((P)=\) Voltage \(({V}) \times\) Current \(({I})\)

From ohms law, \(V=I R\)

So, \(P=\frac{V^2}R\)

Now,

\(P=500 {~W}\)

\(V=100\) volt

\(P=\frac{V^2}R\)

\(\Rightarrow R=\frac{(100 \times 100)}{500}=20\) ohm

Now, When \({V}=150 {~v}\)

Power \(=\frac{(150 \times 150)}{20}\)

\(=\frac{2250}2\)

\(=1125\) Watt