Physics Test-15

The velocity at which the laminar flow stops is known as lower critical velocity.

Lower critical Velocity:

• It is the velocity at which the laminar flow stops.
• In this, the flow enters from laminar to transition period which exists between the laminar and turbulent flow.

Critical Velocity is the maximum velocity of a fluid, above which, the streamline flow get the change to turbulent flow.

Electric field lines form closed loops this statement about electric field lines associated with electric charges is false.

The imaginary lines which are used to represent the electric field are called electric field lines. The field lines emerge from a positive charge and terminate at a negative charge. They originate and end at right angles to the surface of the charge. Electric field lines do not make a loop. The magnitude of the electric field will be maximum where the number of field lines is maximum.

Given:

From the curve, \(I_{1} = 20\) mA, when \(V_{1} = 0.8 V\)

\(I_{2} = 10\) mA when \(V_{2} = 0.7\) V

Now, \(R = \frac{Δ V}{Δ I}\)

\(=\frac{V_{1}-V_{2}}{I_{1}-I_{2}}\).....(i)

Put all the given values in equation (i):

\(= \frac{0.8 -0.7}{20-10}\) mA

\(= \frac{0.1 V}{10}\) mA

\(= 10 Ω\)

Given:

A balloon is rising with a velocity of \(10\) m/s. So when a packet is released its initial velocity will be the same as the balloon i.e. \(15\) m/s.

So,

\(u=10\) m/s; \(h=-40\) m; \(a=\)\(-g=-10\) m/s^\(-2\)

\(s=u t+(\frac{1}{2}) a t^{2}\)

\(-40=10 \times t+(\frac{1}{2})(-10) t^{2}\)

\(-40=10t-5t^{2}\)

\(5t^{2}-10t-40=0\)

Solving equation \(t=-2\) (not possible) and \(4\)

\(t=4\) s.

The Grashof number (Gr) is a dimensionless number in fluid dynamics and heat transfer which approximates the ratio of the buoyancy to the viscous force acting on a fluid. It frequently arises in the study of situations involving natural convection and is analogous to the Reynolds number.

For a microscope, the limit of resolution is given by,

\(X =\frac{\lambda}{2 A }\)

where \(\lambda\) is the wavelength of light used, and \(A\) is the numerical aperture.

So, substituting the values, \(X =\frac{600}{2 \times 0.12}\),

which gives, \(X =2.5 \mu m\)

Angular displacement is measured in units of radians. Two pi radians equals 360 degrees. The angular displacement is not a length (not measured in meters or feet), so an angular displacement is different than a linear displacement.

Given that:

Charge, \(q_{1}=12 C\) and \(q _{2}=-6 C\)

Distance \(=r=3 m\)

Force \(= F =K \frac{q_{1} \times q_{2}}{r^{2}}=9 \times 10^{9} \frac{12 \times(-6)}{3^{2}}\)

Force \(=F=-72 \times 10^{9} N\)

The negative sign shows the attractive nature of the force.

Given that:

\(\mathrm{V}=200 \sin 300 \mathrm{t}\)

\(\mathrm{V}_{0}=200 \mathrm{~V},\)

\( \omega=300 \mathrm{rad} / \mathrm{s} ~(\because \mathrm{V}=\mathrm{V}_{0} \sin \omega \mathrm{t})\)

\(\mathrm{R}=10 \Omega \)

\(\mathrm{L}=800 \mathrm{mH}\)

\(=800 \times 10^{-3} \mathrm{H}\)

Inductive reactance, \(X_{L}=L \omega=800 \times 10^{-3} \times 300 \Omega\)

We know that,

\(Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)

Now,

\(Z=\sqrt{R^{2}+\left(X_{L}-0\right)^{2}}\)

\(\Rightarrow Z=\sqrt{10^{2}+\left(800 \times 10^{-3} \times 300\right)^{2}}\)

\(\Rightarrow Z=240.2 \Omega\)

From Ohm's law,

\(\mathrm{V}_{0}=\mathrm{I}_{0} \mathrm{R} \)

\(\Rightarrow \mathrm{I}_{0}=\frac{V_{0}}{R}\)

For an LCR circuit, the net resistance is the impedance Z.

\(\therefore I_{0}=\frac{V_{0}}{Z}\)

\(=\frac{200}{240.2}\)

\(=0.832 \mathrm{~A}\)