Physics Test - 9

\(\frac{q_{H}}{q_\alpha}=\frac{1}{2}\)

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

For same momenta, \(r \propto \frac{1}{q}\)

\(\frac{r_{H}}{r_{a}}=\frac{q_{a}}{q_{H}}=\frac{2}{1}\)

Bernoulli's principle: For a streamlined flow of an ideal liquid in a varying cross-section tube the total energy per unit volume remains constant throughout the fluid.

• From above it is clear thatBernoulli's equationstates that thesummation of pressure head,kinetic head,anddatum/potential headisconstant for steady,incompressible,rotational,andnon-viscous flow.
• In other words, anincrease in the speed of the fluidoccurs simultaneously with adecrease in pressure or a decrease in the fluid's potential energyi.e. thetotal energy of a flowing system remains constantuntil anexternal force is applied.
• SoBernoulli’s equationrefers to theconservation of energy.
• All of the above are themeasuring deviceslikeVenturimeter,Orifice meter,andPitot tube meterworks on theBernoulli’s theorem.

Meter Bridge:It is an electrical instruments based on the principle ofWheatstone bridgeand is used tomeasurethe resistance of a resistor.

The metre bridge, also known as the slide wire bridge consists of a one metre long wire of uniform cross sectional area, fixed on a wooden block. A scale is attached to the block. Two gaps are formed on it by using thick metal strips in order to make the Wheat stones bridge.

The formula meter bridge is given below:

ρ = Lπr2X

Where, L be the length of the wire and r be its radius.

We know,

\(\frac{E}{3 K}=1-2 \mu\)

\(\mathrm{K}=\) Bulk modulus.

Also, Compressibility \(\beta=\frac{1}{\mathrm{~K}}\)

\(\therefore \frac{E \beta}{3}-1-2 \mu\)

\(\therefore \beta=\frac{3}{\mathrm{E}}(1-2 \mu)\)

For a nucleus

Volume:

\( \mathrm{V}=\frac{4}{3} \pi \mathrm{R}^3 \)

\( \mathrm{R}=\mathrm{R}_0(\mathrm{~A})^{1 / 3} \)

\(\mathrm{~V}=\frac{4}{3} \pi \mathrm{R}_0^3 \mathrm{~A} \)

\( \Rightarrow \frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{\mathrm{A}_2}{\mathrm{~A}_1}=4\)

In the given formula,

\(X=3 Y Z^{2}\)

\([Y]=\left[\frac{X}{Z^{2}}\right]\)

\(=\left[\frac{\text { Capacitance }}{(\text { Magnetic induction })^{2}}\right]\)

\(=\left[\frac{\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{Q}^{2} \mathrm{~T}^{2}}{\mathrm{M}^{2} \mathrm{Q}^{-2} \mathrm{~T}^{-2}}\right]\)

\(=\left[\mathrm{M}^{-3} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{Q}^{4}\right]\)

According to Hookes law stress is directly proportional to strain. But when a material is subjected to strains alternatively there is a loss of strength of the material called Elastic Fatigue. Due to elastic fatigue the behavior of the body shows that of a less elastic body. When fatigue is removed from an elastic body it regains the original degree of elasticity when allowed to rest for a while.

Equivalent resistance,

\(R_{\text {eq }}=\frac{4 R \times 4 R}{4 R+4 R}+R+\frac{6 R \times 12 R}{6 R+12 R}+R \)

\(=2 R+R+4 R+R \)

\( =8 R\)

Using, \(P=\frac{V^2}{R_{\text {eq }}} \Rightarrow 4=\frac{16^2}{8 R}\)

\(\therefore \mathrm{R}=\frac{16^2}{4 \times 8}=8 \Omega\)

The photoelectric cut-off voltage in a certain experiment is \(1.5 \mathrm{~V}\).

The kinetic energy of photoelectrons is given as

 K.E. \(=\mathrm{e}v_0\)

Where, the stopping potential is \(v_O\) and the charge on the electron is \(e\).

By substituting the values in above equation, we get 

K.E. \(=1.6 \times 10^{-19} \times 1.5=2.4 \times 10^{-19} \mathrm{~J}\)