Physics Test - 7

For infinite long line charge,

\(E=\frac{\lambda}{2 \pi \epsilon_{0} r}=\frac{2 \lambda}{4 \pi \epsilon_{0} r}\)

\(\lambda=\) linear charge density

Distance, \(r=2 {~cm} \)

We have,

\({E}=9 \times {10}^{4} {~N} / {C}\)

Therefore,

\(9 \times 10^{4}=\frac{2 \times \lambda \times 9 \times 10^{9}}{2 \times 10^{-2}} \quad\) \((\because \frac {1}{4 \pi \epsilon_0} =9 \times 10^{9} \text { N m}^2 / \text{C}^2 ) \)

\(\lambda=\frac{10^{4}}{10^{11}}=10^{-7} {~cm}\)

When one joule of energy is consumed in one second, then the power used is said to be One watt.

The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.

\[P=\frac{W}{t}=V I=I^{2} R=\frac{V^{2}}{R}\]

Where, \(V=\) Potential difference, \(R=\) Resistance and \(I=\) current.

If,\(W=1\) Joule and \(t=1\) second, then \(P=1\) watt

Thus,The power of an electric circuit is said to be one watt orone ampere volt if one joule of energy is consumed in one second.

From above it is clear that when a current-carrying rectangular coil is placed in a magnetic field it experiences a torque.

\(\therefore\) The moment of the deflecting couple \((\tau)=n B i A\)

Where, \(n =\) number of turns, \(B =\) magnetic field, \(i =\) current and \(A =\) area of coil

When the coil deflects, the suspension wire is twisted. On account of elasticity, a restoring couple is set up in the wire. This couple is proportional to the twist.

If \(\theta=\) the angular twist, then,

The moment of the restoring couple \(=C \theta\)

Where, \(C=\) restoring couple per unit twist.

At equilibrium,

Deflecting couple \(=\) Restoring couple

So, we can write,

\(nBi A = C \theta\)

\(\Rightarrow\left(\frac{n B A}{C}\right) i=\theta\)

Here, \(n , B\), and \(A\) are constant

Therefore, the deflection produced in the moving coil galvanometer is directly proportional to the amount of current passing through it.

When the widths of the two slits are increased, the fringes become brighter. However, the width of each slit should be considerably smaller than the separation between the slits. When the slits become so wide that this condition is not satisfied, the interference pattern disappears i.e., The fringes become less distinct.

We know that the voltage sensitivity is given as,

\(V_{s}=\left(\frac{I_{s}}{R}\right) \quad \ldots\)(1)

Where, \(V _{ s }=\) voltage sensitivity, \(I _{ s }=\) current sensitivity, and \(R =\) resistance

When the resistance is increased by factor 2,

\(R^{\prime}=2 R\quad \ldots\)(2)

\(I_{s}^{\prime}=I_{s}+20 \% \text { of } I_{s}\)

\(\Rightarrow I_{s}^{\prime}=I_{s}+\frac{20}{100} I_{s}\)

\(\Rightarrow I_{s}^{\prime}=1.2 I_{s}\quad \ldots\)(3)

From equation (1) we get,

\( V_{s}^{\prime}=\frac{I_{s}^{\prime}}{R^{\prime}}\)

\(\Rightarrow V_{s}^{\prime}=\frac{1.2 I_{s}}{2 R}\quad \ldots\)(4)

By equation (1) and equation (4)

\(V_{s}^{\prime}=0.6 V_{s}\quad \ldots\)(5)

The change in voltage sensitivity is given as,

\(\% \Delta V=\frac{V_{s}^{\prime}-V_{s}}{V_{s}} \times 100\)

\(\Rightarrow \% \Delta V=\frac{0.6 V_{s}-V_{s}}{V_{s}} \times 100\)

\(\Rightarrow \% \Delta V=-40 \%\)

Volt-ampere (VA) is a measurement of electric power in a direct current ( DC ) electrical circuit. The VA specification is also used in alternating current ( AC ) circuits, but it is less precise in this application, because it represents apparent power, which often differs from true power .

In a DC circuit, 1 VA is the equivalent of one watt (1 W). The power, P (in watts) in a DC circuit is equal to the product of the voltage V (in volts) and the current I (in amperes):

P = VI

Polarization proves the wave nature of light.

• Polarization of light occurs when light is reflected, refracted, and scattered.
• A wave is an oscillation that carries energy from one place to another without transporting matter.

Given,

Radius, \(r=1000 m\)

Banking angle, \(\theta=45^{\circ}\)

Coefficient of friction, \(\mu=0.5\)

Acceleration due to gravity, \(g=9.8 m/s^2\)

As we know,

Maximum safe velocity is calculated by,

\(v=\sqrt{\frac{r g(\tan \theta+\mu)}{1-\mu \tan \theta}} \)

\(\therefore v=\sqrt{\frac{1000 \times 9.8\left(\tan 45^{\circ}+0.5\right)}{1-0.5 \times \tan 45^{\circ}}} \)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 (1+0.5)}{1-0.5 \times 1}}\)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 \times 1.5}{1-0.5}}\)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 \times 1.5}{0.5}}\)

\(\Rightarrow v=\sqrt{29400}\)

\(\therefore v =172 m / s\)

Given that:

Axial force, F = 18 kN = 18 × 10³ N

Diameter of the rod, d = 3 cm = 30 mm

Cross-sectional area, A \(=\frac{\pi}{4}\) d\(^{2}=\frac{\pi}{4}(30)^{2}=706.85\) mm2

We know that, 

Stress, \(\sigma=\frac{F}{A}=\frac{18 \times 10^{3}}{706.85}=25.47\) N/mm2

Since, the train is moving at a constant speed, so the horizontal velocity of the man and the coin will be the same even after tossing the coin because there is no acceleration in the horizontal direction.

So, during the whole flight of the coin, the relative velocity in the horizontal direction between the man the coin is zero because the horizontal velocity of the man and the coin is equal.

So, the coin will fall back into his hand.