Physics Test - 25

Given a mass of copper slab \(=2\) g contains \(2 \times 10^{22}\) atoms.

We have to find the fraction of electrons that must be removed from sphere to give it \(+2 \mu \mathrm{C}\) charge.

Charge on nucleus of each atom \(=29 \mathrm{e}\)

\(\therefore\) Net charge on \(2 \mathrm{gm}\) sphere \(=(29 \mathrm{e}) \times\left(2 \times 10^{22}\right)=5.8 \times 10^{23} \mathrm{e} \mathrm{c}\)

\(\therefore\) No of electrons on sphere \(=5.8 \times 10^{23}\)

\(\therefore\)Number of electrons removed to give\(2 \mu \mathrm{c}\) charge \(=\frac{\mathrm{q}}{\mathrm{e}}\)

\( =\frac{2 \times 10^{-6}}{1.6 \times 10^{-19}}\)

\(=1.25 \times 10^{13}\)

Fraction of electrons removed \(=\frac{1.25 \times 10^{13}}{\text { Total number of electrons in sphere }}\)

\(=\frac{1.25 \times 10^{13}}{29 \times 2 \times 10^{22}}\)

\(=2.16 \times 10^{-11}\)

Kepler's law tells us that the square of the Time period (T) of a satellite is proportional to the cube of its distance (R) from the orbiting planet.

∴T²∝R³

The time period of earth's orbit around the sun is 1 year, and assuming the mean distance between earth and sun as R,we can write the mean distance between mars and the sun as 1.5R.

Let, T be the time period of mars' orbit around the sun.

Thus from the above proportionality,

$\frac{T^2}{R^3}$is a constant.

$\frac{{(1year)}^2}{R^3}=\frac{T^2}{{(1.5R)}^3}$

T² = (1.5)³

$T={(1.5)}^{\frac{3}{2}}$

≈1.85 years

Nicol prism is used for producing a polarized beam of light from an un-polarized beam. It is based on the principle of action which involves refraction as it passes into the lower half of the prism. It leaves the prism as polarized light after undergoing another refraction as it exits the far right side of the prism. Thus its action is based on double refraction.

Double refraction is an optical property in which a single ray of unpolarized light entering an anisotropic medium is split into two rays, each traveling in a different direction.

As we know,

Work done by a variable force is given by,

\(W=\int_{x_{1}}^{x_{2}} F(x) d x\)

Work done, \((W)=\int_{0}^{t} F . d x=\int_{0}^{t} m a . d x\)

We know that, acceleration, \(a=\frac{d v}{d t}\)

\(\therefore W=\int_{0}^{t} m \frac{d v}{d t} \cdot d x \)

\(\Rightarrow W=\int_{0}^{t} m \frac{d(k \sqrt{x})}{d t} \cdot d x \)

\(\Rightarrow W=\int_{0}^{t} m \frac{k}{2 \sqrt{x}} \frac{d x}{d t} \cdot d x \)

\(\Rightarrow W=\int_{0}^{t} m \frac{k}{2 \sqrt{x}} v \cdot d x\)

\(\Rightarrow W=\int_{0}^{t} m \frac{k}{2 \sqrt{x}} k \sqrt{x} . d x \)

\(\Rightarrow W=\int_{0}^{t} m \frac{k^{2}}{2} \cdot d x \)

\(\Rightarrow W=m \frac{k^{2}}{2}[x]_{t=0}^{t=t}\) ...(1)

Given,

Velocity \(v=k\sqrt{x}\)

As we know,

Velocity, \(v=\frac{d x}{d t}\)

\(\therefore k \sqrt{x}=\frac{d x}{d t} \)

\(\Rightarrow k d t=\frac{d x}{\sqrt{x}}\)

On integrating, we get

\(kt =2 \sqrt{ x }\)

At \(t=0, x=0\)

\(\therefore x=\frac{k^{2} t^{2}}{4}\)...(2)

Substituting (2) in (1), we get

\(W=m \frac{k^{2}}{2}\left[\frac{k^{2} t^{2}}{4}\right]\)

\(\Rightarrow W=\frac{m k^{4} t^{2}}{8}\)

When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.

The net resistancelequivalent resistance(R) of resistances in parallel is given by:

\(\frac{1}{R_{e f f}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\)

Given:

\(R_{1}=10 \Omega\) and \(R_{2}=30 \Omega\)

Since, they are parallel connected \(\frac{1}{R_{\text {eff }}}=\frac{1}{10}+\frac{1}{30}\)

\(R_{e f f}=\frac{10 \times 30}{10+30}=\frac{300}{40}=7.5 \Omega\)

Here, 4 Ω across AB and AC are  in a series combination, the equivalent resistance is:

\(S = R _{1}+ R _{2}=4 \Omega+4 \Omega=8 \Omega\)

Now 4 Ω and 8 Ω are connected in parallel combination, the equivalent resistance is:

\(\frac{1}{R_{e q}}=\frac{1}{8}+\frac{1}{4}=\frac{3}{8}\)

\(\Rightarrow R _{ eq }=\frac{8}{3} \Omega\)

Given that:

Current sensitivity \(\left( S _{ i }\right)=5\) div/(mA) \(=5 \times 10^{3}\) div/A

Voltage sensitivity \(( S_{v} )=20\) div/V

We know that:

Voltage sensitivity \(\left( S _{ v }\right)=\frac{S _{i}}{R}\)

Resistance \(( R )= \frac{S_i}{S_{v}}=\frac{(5 \times 10^{3} \operatorname{div}/A)}{(20 \operatorname{div}/V)}=250 \Omega\)

Given: Radius of Coil \(A=a\)

Current in coil \(A=1\)

Field due to Coil A at center,

\(B_{A}=\frac{\mu_{0} I}{2a} \quad \ldots\) (i)

Radius of Coil \(B=2 a\)

Current in coil \(B=1\)

Field due to Coil B at center,

\(B_{B}=\frac{\mu_{0} I }{2(2 a )}=\frac{1}{2}\left(\frac{\mu_{0} I }{2 a }\right)\quad \ldots\) (ii)

Comparing (i) and (ii),

\(B_{B}=\frac{B_{A}}{2}\)

\(\Rightarrow B _{ A }: B _{ B }=2: 1\)

So, the required ratio is \(2: 1\).

Electromagnetic waves are the transverse waves.

Transverse Waves:

• Those waves whose direction of propagation and direction of disturbance is always perpendicular, are known as transverse waves.
• These waves produced in a medium that can sustain shearing strain.
• Example: Electromagnetic Waves, Ripples on the surface of water, Vibrations in a guitar string.

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 \(2.45 \mathrm{~mA}\).