Physics Test - 21

Given, EMF of battery, \(\varepsilon=10 \mathrm{~V}\)

Internal resistance of battery, \(\mathrm{r}=3 \Omega\)

Current flowing in the circuit, I \(=0.5 \mathrm{~A}\)

Using the formula \(\mathrm{I}=\frac{\varepsilon}{\mathrm{R}+\mathrm{r}}\)

or \(\mathrm{R}=\frac{\varepsilon}{\mathrm{I}}-\mathrm{r}\)

where, \(R\) is the external resistance.

\(\therefore \quad \mathrm{R}=\frac{10}{0.5}-3=17 \Omega\) is the required resistance.

Terminal voltage, \(\mathrm{V}\) \(=\mathrm{IR} \)

\(=0.5 \times 17 \)

\(=8.5 \mathrm{~V}\)

In the above experiment when the \(N\)-pole of the magnet is moved towards the coil. Thus work has to be done against the force of repulsion in bringing the magnets close. When the \(N\) pole of the magnet is moved away, the south pole develops on the right face of the coil. Therefore work has to be done in this case. This mechanical work in moving the magnet changes into electrical energy producing induced current.

Hence energy transformation takes place. We can say that the law is a consequence of the conservation of energy.

The thermionic emissions and photoelectric emissions are not the same. During thermionic emissions, the electrons are emitted from the metal surface by providing heat energy, whereas, during photoelectric emission light energy is emitted when electrons are emitted from the surface of the metal. So, it is the opposite in operation.

The relation between \(E _{0}\) and \(B _{0}\) is given by 

\(\frac{ E _{0}}{ B _{0}}= c \ldots\) (i)

Here, \(c=\) Speed of the electromagnetic wave,

The relation between \(\omega\) (the angular frequency) and \(k\) (wave number), 

\(\frac{\omega}{ k }= c \ldots\) (ii)

Therefore, from (i) and (ii), we get \(\frac{ E _{0}}{ B _{0}}=\frac{\omega}{ k }= c \)

\(E _{0} k = B _{0} \omega\)

Given:

\(v \propto g^{p} h^{q}\)

We know that:

\(v=\left[\mathrm{L} \mathrm{T}^{-1}\right]\)

\(g=\left[\mathrm{L} \mathrm{T}^{-2}\right]\)

\(h=[\mathrm{L}]\)

By substituting the dimension of each quantity and comparing the powers in both sides we get:

\({\left[\mathrm{L} \mathrm{T}^{-1}\right]=\left[\mathrm{L} \mathrm{T}^{-2}\right]^{\mathrm{p}}[\mathrm{L}]^{\mathrm{q}}} \)

\(\Rightarrow {\left[\mathrm{L} \mathrm{T}^{-1}\right]=\left[\mathrm{L}^{\mathrm{p}+\mathrm{q}} \mathrm{T}^{-2 \mathrm{p}}\right]}\)

On comparing the powers

\( p+q=1,-2 p=-1 \)

\(\therefore p=\frac{1}{2}, q=\frac{1}{2}\)

Given that,

a∝−x

⇒a=−kx

Now using \(a=\frac{v d v}{d x}\)

\(\Rightarrow v d v=-k x d x\)

\(\Rightarrow \quad \int_{v_1}^{v_2} v d v=-k \int_0^x x d x\)

\(\Rightarrow \quad \frac{v_2^2-v_1^2}{2}=\frac{-k \cdot x^2}{2}\)

or, \(\left(v_1^2-v_2^2\right)=k x^2\)

Now, loss in kinetic energy is given by

\(\Delta K \cdot E=\frac{1}{2} m v_1^2-\frac{1}{2} m v_2^2\)

\(=\frac{1}{2} m\left(v_1^2-v_2^2\right)=\frac{k m x^2}{2}\)

Thus, the loss in kinetic energy is proportional to \(x^2\).

Given \(L =1 H\) and \(C =25 \mu F\)

The current in the LC circuit becomes maximum when resonance occurs. So,

\(\Rightarrow \nu=\frac{1}{2 \pi \sqrt{L C}}\)

Angular frequency,

\(\Rightarrow \omega=2 \pi v\)

From equation 1 and equation 2 ,

\( \Rightarrow \omega=\frac{1}{\sqrt{L C}} \)

\(\Rightarrow \omega=\frac{1}{\sqrt{1 \times 25 \times 10^{-6}}} \)

\( \Rightarrow \omega=200 rad / sec\)

Given, 

m = 10 Kg

g = 10 m/s²

First draw the free body diagram of the given figure,

Now resolve all forces into a horizontal and a vertical component.

​Horizontal Component \(= 100 \cos45^\circ =100\times \frac{1}{\sqrt{2}}\)

\(= 50\sqrt{2} N\)

Vertical Component \(= 100\sin45^\circ=100\times \frac{1}{\sqrt{2}}\)

\(= 50\sqrt{2} N\)

For the system to be in equilibrium \(T_{1}\) must be equal to the horizontal component but the direction must be opposite.

Here the direction of \(T_{1}\) is reversed for balancing the horizontal component.

\( T _{2}=50 \sqrt{2 } N\)

\( T _{1}=-50 \sqrt{2} N\)

So, the value of \(T_{1}\) is \(-50 \sqrt{2} N\).(Negative sign only show the direction of \(T_{1}\).)

Kepler's first law explains that planets orbit the sun in a path described as an ellipse. It is not circular.

That is why the distance between the earth and the sun is not constant. It is changing.

Let initial velocity of the bullet = u

After penetrating \(3 \mathrm{~cm}\) its velocity becomes \(\frac{\mathrm{u}}{2}\) 

From \(v^2=u^2-2 a s\)

\(\left(\frac{u}{2}\right)^2=u^2-2 a(3)\)

\(\Rightarrow 6 a=\frac{3 u^2}{4} \Rightarrow a=\frac{u^2}{8}\)

Let further it will penetrate through distance \(x\) and stops at point \(C\).

For distance \(B C, v=0, u=\frac{u}{2}, s=x, a=\frac{u^2}{8}\)

From \(v^2=u^2-2 a s \Rightarrow 0=\left(\frac{u}{2}\right)^2-2\left(\frac{u^2}{8}\right) \cdot x \Rightarrow x=1 \mathrm{~cm}\).

Given,

Mass of the ice, \(m=3 \mathrm{~kg}\)

Specific heat capacity of ice, \(S_{i c e}=2100 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\)

Specific heat capacity of water, \(S_{\text {water }}=4186 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\)

Latent heat of fusion of ice, \(L_{\text {ice }}=3.35 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}\)

Latent heat of steam, \(L_{\text {steam }}=2.256 \times 10^{6} \mathrm{~J} \mathrm{~kg}^{-1}\)

\(Q_{1}=\) Heat required to convert ice at \(-12^{\circ} \mathrm{C}\) to ice at \(0^{\circ} \mathrm{C}\)

\(=m \mathrm{~S}_{\mathrm{ice}} \Delta T_{1}\)

\(=(3 \mathrm{~kg})\left(2100 \mathrm{~J} \mathrm{~kg}^{-1}\right. \mathrm{K}^{-1}\) ) \([\mathrm{0}-(-12)]^{\circ} \mathrm{C}\)

\(=75600 \mathrm{~J}\)

\(Q_{2}=\) Heat required to melt ice at \(0^{\circ} \mathrm{C}\) to water at \(0^{\circ} \mathrm{C}\)

\(=m L_{\text {ice }}\)

\(=(3 \mathrm{~kg})\left(3.35 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}\right)\)

\(=1005000 \mathrm{~J}\)

\(Q_{3}=\) Heat required to convert water at \(0^{\circ} \mathrm{C}\) to water at \(100^{\circ} \mathrm{C}\)

\(=m S_{water} \Delta T_{2}\)

\(=(3 \mathrm{~kg})\left(4186 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right) \left(100^{\circ} \mathrm{C}\right)\)

\(=1255800 \mathrm{~J}\)

\(Q_{4}=\) Heat required to convert water at \(100^{\circ} \mathrm{C}\) to steam at \(100^{\circ} \mathrm{C}\)

\(=m L_{\text {steam }}\)

\(=(3 \mathrm{~kg})\left(2.256 \times 10^{6}\right. \mathrm{J} \mathrm{kg}^{-1}\) )

\(=6768000 \mathrm{~J}\)

Heat required to convert \(3 \mathrm{~kg}\) of ice at \(-12^{\circ} \mathrm{C}\) to steam at \(100^{\circ} \mathrm{C}\),

\(Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\)

\(=75600 \mathrm{~J}+1005000 \mathrm{~J}+1255800 \mathrm{~J}+6768000 \mathrm{~J}\)

\(=9.1 \times 10^{6} \mathrm{~J}\)

As the electron is at rest initially, only the electric field will exert force on it. There will be no magnetic force on the electron in the stating. So, the electron will start moving along the electric field.

Given,

A silicon diode has a threshold voltage of \(0.7 \mathrm{~V}\).

Input voltage =\(2 \sin (\pi \mathrm{t})\)

Peak value of rectified output voltage \(=\) Peak value of input voltage \(-\) Input voltage

\(=2-0.7\)

\(=1.3 \mathrm{~V}\)

Given,

\(|\vec{A}|=10\) units

\(|\vec{B}|=20\) units

Angle between \(\vec{A}\) and \(\vec{B}\),

\(\theta=180^{\circ}-30^{\circ}-30^{\circ}\)

\(=120^{\circ}\)

Magnitude of the resultant,

\(R=\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

\(=\sqrt{10^{2}+20^{2}+2 \times 10 \times 20 \cos 120^{\circ}}\)

\(=\sqrt{100+400-200}\)

\(=\sqrt{300}\)

\(=10 \sqrt{3}\) units