Physics Test-17

Unit of magnetic flux is weber.

Magnetic flux is a physical quantity that measures the absolute magnitude of a magnetic field passing through a plane (such as a coil of a conducting wire). It is denoted by abbreviation . Its SI unit is Weber.

Since magnetic flux \((\varphi)=\mathrm{B} \mathrm{A}\)

Where, \(\mathrm{B}\) = megnatic field and \(\mathrm{A}\) = area

The \(\mathrm{SI}\) unit of magnetic flux \(=\mathrm{SI}\) unit of magnetic field × \( \mathrm{SI}\) unit of area \(=\) tesla meter\(^{2}=\mathrm{T} \mathrm{m}^{2}\)

Since, 1 Weber \(=1 \mathrm{~T} \mathrm{m}^{2}\)

Thus the SI unit of magnetic flux is \(\mathrm{T} \mathrm{m}^{2}\) and which is equal to weber \((\mathrm{Wb})\).

From \(\mathrm{E}=\mathrm{W}_{0}+\frac{1}{2} \mathrm{mv}_{\max }^{2}\)

\( \Rightarrow \mathrm{v}_{\max }=\sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}-\frac{2 \mathrm{~W}_{0}}{\mathrm{~m}}}\)

Where, \(\mathrm{E}=\frac{\mathrm{hc}}{\lambda}\)

If wavelength of incident light charges from \(\lambda\) to \(\frac{3 \lambda}{4}\) (Decreases).

Let, energy of incident light charges from \(\mathrm{E}\) and speed of fastest electron changes from \(\mathrm{v}\) to \(\mathrm{v}^{\prime}\) then,

\(\mathrm{v}^{\prime}=\sqrt{\frac{\mathrm{2} \mathrm{E}^{\prime}}{\mathrm{m}}-\frac{\mathrm{2} \mathrm{W}_{0}}{\mathrm{m}}}\)

As \(\mathrm{E} \propto \frac{1}{\lambda}\)

\( \Rightarrow \mathrm{E}^{\prime} = \frac{4}{3} \mathrm{E}\)

So, \(\mathrm{v}^{\prime}=\sqrt{\frac{2\left(\frac{4}{3}\right) \mathrm{E}}{\mathrm{m}}-\frac{2 \mathrm{W}_{0}}{\mathrm{~m}}}\)

\(\Rightarrow \mathrm{v}^{\prime}=(\frac4 3)^{\frac1 2} \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}-\frac{2 \mathrm{~W}_{0}}{\mathrm{~m}(\frac4 3)^{\frac1 2}}}\)

\(\therefore \mathrm{v}^{\prime}>\sqrt{\frac{4}{3}} \mathrm{v}\)

Work done by the steam engine per minute, \(W=5.4 \times 10^{8} J\)

Heat supplied from the boiler, \(H=3.6 \times 10^{9} J\)

Efficiency of the engine \(=\frac{\text { Output Energy }}{\text { Input Energy }}\)

\(\Rightarrow \eta=\frac{W}{H} \)

\(\Rightarrow \eta=\frac{5.4 \times 10^{8}}{3.6 \times 10^{9}} \)

\(\Rightarrow \eta=0.15\)

Thus, the percentage efficiency of the engine is \(15\) .

Amount of heat wasted \(=3.6 \times 10^{9}-5.4 \times 10^{8}\)

\(=30.6 \times 10^{8}\)

\(=3.06 \times 10^{9} J\)

Clearly, the amount of heat wasted per minute is \(3.06 \times 10^{9} J\).

A semiconductor in its purest form is known as Intrinsic semiconductor.

An intrinsic (pure) semiconductor, also called an undoped semiconductor or i-type semiconductor, is a pure semiconductor without any significant dopant species present. The number of charge carriers is therefore determined by the properties of the material itself instead of the amount of impurities.

Concept:

Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.

These equations are only valid when the acceleration of the body is constant and they move on a straight line.

There are three equations of motion:

\(V=u+a t\)

\(V^{2}=u^{2}+2 a S\)

\(S=u t+\frac{1}{2} a t^{2}\)

Where, \(V=\) final velocity, \(u=\) initial velocity, \(s=\) distance traveled by the body under motion, \(a=\) acceleration of body under motion, and \(t=\) time taken by the body under motion.

Given that:

Initial velocity \(( u )=0\)

Distance \(( S )=20~ m\)

Time \(( t )=4~ sec\)

Use \(S=u t+\frac{1}{2} a t^{2}\)

\(20=0+\frac{1}{2} \times a \times 4^{2}\)

acceleration \(= a =\frac{20}{8}=2.5~ m / s ^{2}\)

Given,

Input Voltage, \(\mathrm{V}_{\mathrm{i}}=2200 \mathrm{~V}\)

Output Volatge, \(\mathrm{V}_{\mathrm{o}}=220 \mathrm{~V}\)

\(P_{\text {output }}=880 \mathrm{~W}\)

Efficiency \((\eta)=88 \%\)

\(=\frac{88}{100}\)

\(=0.88\)

\(\eta=\frac{P_{\text {output }}}{P_{\text {input }}} \)

\(\Rightarrow 0.88=\frac{880}{P_{\text {input }}}\)

\(\Rightarrow \mathrm{P}_{\text {input }}=1000 \mathrm{~W}\)

Now,

Power, \(\mathrm{P}=\mathrm{VI}\)

\(\Rightarrow \mathrm{I}=\frac{\mathrm{P}}{\mathrm{V}}\)

\(\Rightarrow \mathrm{I}_{\text {input }}\)\(=\frac{\mathrm{P}_{\text {input }}}{V_{i}}\)

\(\Rightarrow \frac{1000}{2200}\)

\(\Rightarrow 0.45 \mathrm{~A}\)

2-Methyl-2-butene will be represented as:

\(\mathrm{^1CH}_{3}-{\overset{\mathrm{CH_3}}{\overset{|}{\mathrm{^2C}}}}=\mathrm{^3CH}-\mathrm{^4CH}_{3}\)

CH₃ is numbered first and in this, the Methyl group has been added so we will name Methyl.

Here,

Given that:

Main scale reading (MSR) = 0 mm

Circular scale reading (CSR) = 52 divisions

Pitch of the screw gauge, \(P=1 \mathrm{~mm}\)

Number of circular division, \(n=100\)

Thus,

Least count \(L C=\frac{P}{n}\)

\(=\frac{1}{100}\)

\(=0.01 \mathrm{~mm}\)

\(=0.001 \mathrm{~cm}\)

We know that:

Diameter of the wire = MSR + (CSR × LC)

\(=0+(52 \times 0.001)\) cm

\(=0.052 \mathrm{~cm}\)

Both Dyne and Newton are units of force in the CGS and S.I system.

Newton and dyne can also be written as,

\(\left(1 \mathrm{~N}=\mathrm{kg}\mathrm{m} / \mathrm{s}^{2}\right)\) and \(\left(1\right.\) dyne \(=\mathrm{gm}\mathrm{cm} / \mathrm{s}^{2}\) )

\(\Rightarrow 1\) dyne \(=1 \mathrm{gm}\mathrm{cm} / \mathrm{s}^{2}\) and 1 newton \(=1 \mathrm{~kg}\mathrm{m} / \mathrm{s}^{2}\)

\(\Rightarrow 1 \mathrm{~kg}=1000 \mathrm{gm}\)

\(\Rightarrow 1 \mathrm{~m}=100 \mathrm{~cm}\)

\(\Rightarrow 1 \mathrm{~kg}\mathrm{m} / \mathrm{s}^{2}=1000 \mathrm{gm} \times 100 \mathrm{~cm} / \mathrm{s}^{2}\)

\( \Rightarrow 10^{5} \mathrm{gm}\mathrm{cm} / \mathrm{s}^{2}=\) \(10^{5}\) dyne

\(\Rightarrow 1\) dyne \(=10^{-5} \mathrm{~kg}\mathrm{m} / \mathrm{s}^{2} \)

\(\Rightarrow 10^{-5} \mathrm{~N}\)