Physics Test - 6

Given,

\(KE _{1}=50 J\)

\(KE _{2}=150 J\)

\(t =10 s\)

As we know,

The work-energy theorem states that the net work done by the forces on an object is equal to the change in its kinetic energy.

\( W =\Delta KE\)

Where \(W=\) work done and \(\Delta K E=\) change in kinetic energy

By the work-energy theorem,

\( W=\Delta KE \)

\(\therefore W=K E_{2}-K E_{1} \)

\(\Rightarrow W=150-50 \)

\(\Rightarrow W=100 J\)

So, the power of the body \( (P)=\frac{W}{t}\)

Where \(W=\) work and \(t=\) time

\(\therefore P=\frac{100}{10} \)

\(\Rightarrow P=10 W\)

Given:

Mass of a planet (Mp) = \(\frac{1}{4}\)th of the mass of the earth \((M_e)\)

The radius of a planet (rp) = \(\frac{1}{9}\)th the radius of the earth \((r_e)\)

Escape velocity from the earth (Ve) = 11 km/s

The escape velocity on the earth is given by:

\( V_{e}=\sqrt{\frac{2 G M_{e}}{r_{e}}}\).....(1)

The escape velocity on the planet is given by:

\( V_{p}=\sqrt{\frac{2 G M_{p}}{r_{p}}}\)

\(=\sqrt{\frac{2 G \frac{M_{e}}{4}}{\frac{r^{e}}{9}}} \)

\(=\sqrt{\frac{2\left(9 G M_{e}\right)}{4 e_{e}}}\)

\(=\frac{3}{2} \sqrt{\frac{2 G M_{e}}{r_{e}}}\).....(2)

On dividing equation (1) and (2), we get:

\( \frac{V_{e}}{V_{p}}=\frac{2}{3} \)

\(\Rightarrow V_{p}=\frac{3}{2} V_{e}\)

\(=\frac{33}{2} km / s\)

If N = number of turns in the coil, I = current in the loop, A = area enclosed by the loop, B = magnetic field intensity, and θ= angle between the normal to the plane of the coil and the direction of a uniform magnetic field.

Then the torque on the current-carrying rectangular loop is given as,

We know that the minimum and the maximum angle possible between the area vector and the magnetic field is 0^∘ and 90^∘.

We know that when the value of θ is increased from 0^∘ to 90^∘, the value of sinθ also increases.

Therefore, if the angle between the area vector and the magnetic field is increased then the value of sinθ increases and by equation (2) we can say that the torque on the loop will also increase.

As we know,

NAND Gate: The logic Gate is obtained after adding NOT Gate after AND Gate.

\(Y=\overline{A . B}\)

From above, we can see that NAND gate is one of the logic gates, which is opposite of AND and output of of different inputs (A & B) are as shown below:

The dimensional formula of the coefficient of thermal conductivity is

\({\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~K}^{-1}\right] }\)

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,

\(nBIA = C \theta\)

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

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

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

Thus, if the current flowing through the moving coil galvanometer is increased, then the deflection in the coil will also increase.

First, draw the free body diagram of the given figure and resolve it into a horizontal and vertical component.

As we know that the value of the horizontal component is always zero irrespective of the inclination of the body.

So, the Horizontal component will be,

Given:

Potential difference \(( V )=220 V _{1}\)

Power of the bulb \(( P )=100 W\)

Actual voltage \(\left( V ^{\prime}\right)=110 V\)

We know that:

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.

The resistance of the bulb can be calculated as,

\(R=\frac{V^{2}}{P}=\frac{(220)^{2}}{100}=484 \Omega\)

The power consumed by the bulb.

\(P=\frac{V^{'2}}{R}=\frac{(110)^{2}}{484}=25 W\)

Given,

Output power \(P=100 W\)

Voltage across primary \(V_{{p}}=220 V\)

Current in the primary \(I_{p}=0.5 A\)

Efficiency of a transformer

\(\eta=\frac{\text { output power }}{\text { input power }} \times 100\)

\(=\frac{P}{V_{p} I_{p}} \times 100\)

\(=\frac{100}{220 \times 0.5} \times 100=90 \%\)

The magnetic field \(B\) inside the toroid is given as,

\(B=\frac{\mu_{o} N I}{2 \pi r}\quad \ldots\) (1)

Where, \(N =\) number of turns, \(I=\) current, and \(r =\) average radius of the toroid

When the current in the toroid is doubled, then:

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

So, the magnetic field in the toroid when the current is doubled is given as,

\(B^{\prime}=\frac{\mu_{o} N I^{\prime}}{2 \pi r}\)

\(\Rightarrow B^{\prime}=\frac{2 \mu_{o} N I}{2 \pi r}\quad \ldots\) (3)

By equation (1) and equation (3),

\(B^{\prime}=2B\)

Let the current \(i\) in the large loop of radius \(\mathrm{R}=20 \mathrm{~cm}\) produces a magnetic field of magnitude \(B=\frac{\mu_{o} i}{2 R}\) at its centre.
Since the radius \(r=2\ \mathrm{cm}\) of the smaller loop is \(r<So, the mutual inductance of the loop is \(M=\frac{\Phi}{i} =\frac{\mu \pi r^{2}}{2 R} =\frac{4 \times 3.14 \times 10^{-7} \times 3.14 \times 2 \times 2}{2 \times 20}\)
\(M \approx 4.0 \times 10^{-7}\)