Physics Test-22

The instrument having its deflection depends upon average value is rectifier type.

The deflection of rectifier type instrument depends upon average value because the deflection produce corresponds to the average value of voltage. Moving iron and hot wire type instruments responds to rms values.

Moment of inertia depends on the distribution of particles, mass and position of the axis of rotation.

Here, the temperature inside the refrigerator can be provided as,

\(T_{1}=9^{\circ} C=282 K\)

Room temperature is given as,

\(T_{2}=36^{\circ} C=309 K\)

Coefficient of performance can be given by the relation,

\(C O P=\frac{T_{1}}{\left(T_{2}-T_{1}\right)} \)

\(\Rightarrow C O P=\frac{282}{(309-282)} \)

\(\Rightarrow C O P=10.44\)

Clearly, the coefficient of performance of the mentioned refrigerator is \(10.44\).

Given,\(\text{Potential}~\text{Energy}=9800 {J} , \) \(mass \)\(=50 {~kg}, {~g}=9.8 {~m} / {s}^{2}\)

Potential energy \(={mgh}\),

\(9800=50 \times 9.8 \times {h}\)

\(9800=490 \times {h}\)

\(h=\frac{9800}{490}=20 {~m}\)

The specific weight of a body \(=w_{1}\), Specific weight of a liquic \(=w_{2}\)

The body floats on the liquid.

Let assume \(y \%\) of the part of a body is submerged into the liq

Therefore, by the principle of floatation,

Buoyancy force acted by liquid = weight of the body

\(y \times\left(\rho_{ f } \times g \times V \right)=\rho_{ b } \times g \times V\)

Therefore, \(y \times\left( w _{2}\right)= w _{1}\)

\(\Rightarrow y=\frac{w_{1}}{w_{2}} \ldots \ldots \ldots .(\) Part submerged into the liquid)

Now, the part of a body outside the liquid \(=1-\frac{w_{1}}{w_{2}}=\frac{w_{2}-w_{1}}{w_{2}}\)

Given,

Thickness of the dielectric slab, \(t=1 cm =10^{-2} m\)

Dielectric constant, \(\varepsilon_{r}=K=5\)

Area of the plates of the capacitor, \(A=0.01 m ^{2}=10^{-2} m ^{2}\)

Distance between parallel plates of the capacitor, \(d=2 cm =2 \times 10^{-2} m\)

We know that:

Capacity with air in between the plates,

\(C_{0}=\frac{\epsilon_{0} A}{d}\)

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

\(=\frac{8.85 \times 10^{-12} \times 10^{-2}}{2 \times 10^{-2}}\)

\(C_{0}=4.425 \times 10^{-12}\) Farad

Capacity with dielectric slab in between the plates,

\(C=\frac{\epsilon_{0} A}{d-t\left(1-\frac{1}{K}\right)}\)

\(=\frac{8.85 \times 10^{-12} \times 10^{-2}}{\left(2 \times 10^{-2}\right)-10^{-2}\left(1-\frac{1}{5}\right)}\)

\(C=7.375 \times 10^{-12}\) Farad

Increase in capacity on introduction of dielectric:

\(C-C_{0}=\left(7.375 \times 10^{-12}\right)-\left(4.425 \times 10^{-12}\right)\)

\(=2.95 \times 10^{-12}\) Farad

Electroplating is based on the chemical effect of electricity.

A direct current (DC) of electricity is passed through the solution, affecting the transfer of metal ions onto the cathode. So the transfer of ions from one electrode to another is because of the heating effect of the current and thus, electroplating is based on the chemical effect of electricity.

Standing waves are formed when two sinusoidal wave trains of the same frequency are moving in opposite directions in the same space and interfere with each other. EM waves can be standing waves when electromagnetic waves reflected from the end of a transmission line.

When the voltage is \(-5 \mathrm{~V}\), the diode is reverse baised and the voltoge across resistor will be zero as current will be zero, and when the voltoge is \(+5 \mathrm{~V}\) the diode in that case is forward biased,the voltoge across resistor will be same as input voltage.