Physics Test-35

The restriction on application of Bernoulli theorem is that the fluids must be incompressible.

Bernoulli's principle states that the sum of pressure energy, kinetic energy, and potential energy per unit volume of an incompressible, non- viscous fluid in a streamlined irrotational flow remains constant along a streamline. This means that in steady flow the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline.

\(P+\frac{1}{2} \rho V^{2}+\rho g h=A\) constant

The electric field at a point isdiscontinuous if there is a charge at that point.

The electric field due to any charge will be continuous, if there is no other charge in the medium. It will be discontinuous if there is a charge at the point under consideration.

Intrinsic semiconductor conductivity is due to breaking off the covalent bond.

• In an intrinsic semiconductor, the number of electrons is equal to the number of holes. Hence at room temperature, no free electron is available for conduction
• If some energy is applied to the atoms of intrinsic semiconductor the covalent bonds will break and electrons will be free
• Each electron will leave behind a hole and the process will continue, and charge flows through the intrinsic semiconductor.

Given,

\(m=20 g =20 \times 10^{-3} kg\)

Initial speed \(m=1 ms ^{-1}\)

Thickness, \(s=20 cm =20 \times 10^{-2} m\)

Resistance offered by the wall, \(F=-2.5 \times 10^{-2} N\)

So, deacceleration of bullet,

\(F=m a\)

\(a=\frac{F}{m}\)

\(=\frac{-2.5 \times 10^{-2}}{20 \times 10^{-3}}\)

\(=-\frac{5}{4} ms ^{-2}\)

Now, using the equation of motion,

\(v^{2}=u^{2}+2 a s\)

\(v^{2}=1+2\left(-\frac{5}{4}\right)\left(20 \times 10^{-2}\right)\)

\(v^{2}=\frac{1}{2}\)

\(v=\frac{1}{\sqrt{2}}=0.7 ms ^{-1}\)

Ratio of buoyancy to viscous forces is known asGrashof number.

Grashof number in natural convection is analogous to the Reynolds number is forced convection. Grashof number indicates the ratio of the buoyancy force to the viscous force. Higher Gr number means increased natural convection flow.

The bending of a beam of light around the corners of the obstacles is called diffraction.

The silver lining which we witness in the sky is caused due to diffraction of light. When the sunlight passes through or encounters the cloud, a silver lining is seen in the sky.

Given - The dimension of the power is \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\)

The equation is given

\(\Rightarrow \mathrm{P}=\mathrm{At}^{2}+\mathrm{Ct}\)

The dimension of A will be

\(\mathrm{At}^{2}=\) Dimension of power

\(\Rightarrow \mathrm{At}^{2}=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]\)

The dimension of A will be

\(\Rightarrow\) Dimesion of \(A=\frac{[M L^{2} T^{-3}]}{[T^{2}]}=[M L^{2} T^{-5}]\)

Similarly, the dimension of \(\mathrm{C}\) will be

\(\Rightarrow\) Dimension of \(C=\frac{\left[M L^{2} T^{-3}\right]}{[T]}=\left[M L^{2} T^{-4}\right]\)

Let the charge distribution be as shown in the figure:

\(\therefore V_{A}-V_{B}=\frac{q}{C_{1}}-E+\frac{q}{C_{2}}\)

or, \(\left(V_{A}-V_{B}\right)+E=q\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right)\)

\(\left(V_{A}-V_{B}\right)+E=\frac{q\left(C_{2}+C_{1}\right)}{C_{1} C_{2}}\)

\(\therefore q=\frac{\left[\left(V_{A}-V_{B}\right)+E\right] C_{1} C_{2}}{C_{1}+C_{2}}\)

Voltage across \(C_{1}\) is \(V_{1}=\frac{q}{C_{1}}\)

\(=\frac{\left[\left(V_{A}-V_{B}\right)+E\right] C_{2}}{C_{1}+C_{2}}\)

\(=\frac{(5+10) 2.0}{1.0+2.0}\)

\(=10\) Volt

Voltage across \(C_{2}\) is \(V_{2}=\frac{q}{C_{2}}\)

\(=\frac{\left[\left(V_{A}-V_{B}\right)+E\left[C_{1}\right.\right.}{C_{1}+C_{2}}\)

\(=\frac{(5+10) 1.0}{1.0 + 2.0}\)

\(=5\) Volt

In the above circuit we have the resistor \(10 \Omega\) and the \(10 H\) inductor in parallel combination and \(20 \Omega\) is in series with them. The potential difference by the battery is \(2 V\).

We have to calculate the currents in the circuit at \(t=0\) and at \(t \rightarrow \infty\)

If we consider the case of \(t=0\), we know that the inductor acts like an open circuit and does not allow the flux to change, therefore the current passes through the resistor in parallel and we can ignore the inductor.

The both resistors become in series with each other therefore the formula of resistors in series is applied. In this case the equivalent resistance of the complete circuit is

\(R_{e q}=R_1+R_2 \)

\(\Rightarrow R_{e q}=10+20 \)

\(\Rightarrow R_{e q}=30 \Omega\)

Now if we calculate the current through the battery it is given by Ohm's Law:

\(V=I R \)

\(\Rightarrow I=\frac{V}{R}\)

If we put the values in the above formula, we get current as:

\(I=\frac{2}{30} \)

\(\Rightarrow I=\frac{1}{15} A\)

So, the current through the battery at time \(t=0\) is \(\frac{1}{15} A\).

Now we will consider the case of \(t \rightarrow \infty\), the inductor acts like a short circuit and in this case, we have to assume that it lets all of the current to pass through it with no resistance. Once we have established this, we can conclude that there is only one resistancein the circuit and that is the \(20 \Omega\) resistor.

Now we have the equivalent resistance as \(R_{e q}=20 \Omega\), we can again use Ohm's Law to calculate the current through the battery.

\(I=\frac{V}{R} \)

\(\Rightarrow I=\frac{2}{20} \)

\(\therefore I=\frac{1}{10} A\)

So, the current through the battery at time \(t \rightarrow \infty\) is \(\frac{1}{10} A\).

The initial \((t=0)\) and final \((t \rightarrow \infty)\) currents through the battery are\(\frac{1}{15} A , \frac{1}{10} A\).