Physics Test - 8

The temperature at which the vapour pressure of a liquid becomes equals to the external (atmospheric) pressure is boiling point.

Boiling: As the temperature of a liquid is increased, the rate of evaporation also increases.

• A stage is reached when bubbles of vapour start forming in the body of the liquid which rise to the surface and escape.
• A liquid boils at a temperature at which the Saturated Vapour Pressure (S.V.P.) is equal to the external pressure.
• At boiling point, vapour pressure becomes equal to the external pressure.

Relation between \(E_{0}\) and \(B_{0}\)

\(E_{0}=c . B_{0}\)

where,

\(E_{0}=\) Electric field amplitude

\(B_{0}=\) Magnetic field amplitude

\(\mathrm{c}=\) speed of light

\(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s} \)

\(B_{0}=100 \times 10^{-6} \mathrm{~T}\)

so we have

\(E_{0}=B_{0} \times c\)

\(=100 \times 10^{-6} \times 3 \times 10^{8}\)

\(E_{0}=3 \times 10^{4} N / C\)

The convex lens forms a virtual image. This happens when the object is placed between the focal point and optic centre of the lens i.e. the image is formed on the same side as the object.

Therefore, \(v\) is negative and \(u\) is negative. \(\quad \cdots(1)\)

Size of the image \(=n\) times the size of object

Magnification, \(m=n \quad\cdots(2)\)

Substituting (1) and (2) in \(m=\frac{v}{u}\)

\(n=\frac{-v}{-u}\)

\(\Rightarrow v=nu\)

Using lens formula, \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

\(\frac{1}{f}=\frac{1}{-n u}-\frac{1}{-u} \quad\) (Using sign convention, \(\mathrm{v}=-n \mathrm{u}\) and \(\mathrm{u}=-\mathrm{u}\) )

\(\frac{1}{f}=\frac{1}{u}\left(1-\frac{1}{n}\right)\)

\(\therefore u=\left(\frac{n-1}{n}\right) f\)

Generally, nuclei tend towards those where a number of proton and neutron are equal. If in a given nucleus, the number of proton and neutron much higher, then binding energy per nucleon decreases and nuclei unstable in those cases.

Americium \(\left({ }^{243} Am_{95}\right)\) is a synthetic radioactive chemical element.

The isotope \({ }^{242} Am\) has the largest cross-sections for absorption of thermal neutrons, which result in a small critical mass for a sustained nuclear chain reaction. The liquid drop model explains the fusion \({ }^{235} U,{ }^{140} Ce\) and \({ }^{239} Pe\) but it fails to explain the fission in an element like \({ }^{243} Am\).

So, from the above discussion, we can say that, liquid drop model can not explain the fission in \({ }^{243} Am\).

According to Newton’s first law of motion, to change in velocity, (either a change in magnitude or direction), there must be a cause which is a net external force.

Example: On the frictionless surface, we would observe that if an object had a velocity, it would continue moving with that velocity until there comes some force to cause a change in the motion.

Similarly, if an object were at rest on the surface, it would remain at rest until there was a force to cause it to change to move.

So the formula of the force can be given on the basis of Newton’s second law of motion and the definition of the force is given on the basis of the first law of motion.

Given,

If two force \(F_{1}\) and \(F_{2}\) are equal and opposite to each then \(\mathrm{F}_{1}=\mathbf{F}_{2}\)

Angle between \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}, \theta=180^{\circ}\)

Thus, by applying the law of Parallelogram of forces we can conclude that:

\(R=\sqrt{F^{2}{ }_{1}+F^{2}{ }_{2}+2 F_{1} F_{2} \cos \theta}\)

\(R=\sqrt{F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos 180^{\circ}}\)

\(F_{1}=F_{2}\)

\(\therefore R=\sqrt{2 F_{1}^{2}+2 F_{1}^{2} \times(-1)} \quad\left(\because \cos 180^{\circ}=-1\right)\)

\(\Rightarrow R=0 N\)

Given:

mA = 600 ml, mB = 250 ml,

TA = 60°C, and TB = 26°C

According to the principle of calorimetry,

Heat lost by liquid A = Heat gained by liquid B

⇒ mA SW ΔTA = mB SW ΔTB

⇒ 600 (60 –T) = 250 (T – 26)

⇒ 720 - 12T = 5T - 130

⇒ 850 = 17T

⇒ T = 50° C

Given:

\(\mathrm{I}_{\mathrm{B}}=1.95 \mathrm{~mA}, \mathrm{I}_{\mathrm{C}}=0.5 \mathrm{~mA}\)

As we know,

The electrical relationship between the three transistor currents is

\(\mathrm{I}_{\mathrm{E}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}\)

Put the given values in above formula:

\(\mathrm{I}_{\mathrm{E}}=1.95+0.5 \)

\(\Rightarrow \mathrm{I}_{\mathrm{E}}=2.45 \mathrm{~mA}\)

Therefore, emitter current will be \(2.45 \mathrm{~mA}\).

Distance between their surfaces

\(=12 R-R-2 R=9 R\)

Since,

\(P \propto\) \(mass\)

\(a \propto\) \(mass\)

We know that,

\(Distance\) \(\propto\) \(acceleration\)

So, we can write

\(\frac{a_{1}}{a_{2}}=\frac{m}{5 m}=\frac{s_{1}}{s_{2}}\)

\(\frac{s_{1}}{s}=51\)

\(5 s~s_{1}=s_{2}\)\(\ldots \ldots(i)\)

\(s_{1}+s_{2}=9 R\)\(\ldots \ldots(ii)\)

On solving these equations;

\(s_{1}=1.5 R\)

\(s_{2}=7.5 R\)

Since smaller ball have more acceleration in same time interval, smaller ball will cover more distance.