Physics Test - 8

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.

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

Displacement Current:

• The idea of displacement current was firstly developed by famous physicist James Maxwell.
• The displacement current produces due to the change in electric flux (number of electric field lines through a cross-sectional area of a closed-loop) with respect to time.
• The SI unit of displacement current is Ampere.
• The magnitude of displacement current is zero in the case of steady electric fields in conducting wire.
• The idea of displacement current was introduced to the current for making ampere circuital law consistent.

Therefore, out of all the given statements, Only the statement of option (C) is incorrect.

Given,

Diameter of the sphere \(=2.4\)

\(\therefore\) Radius of sphere, \(r=\frac{2.4}{2}=1.2 {~m}\)

Surface charge density of conducting sphere, \(\sigma=80 \times 10^{-6} {C} / {m}^{2}\)

Therefore,

Charge on sphere will be:

\(q=\sigma A=\sigma 4 \pi r^{2}\)

\(q=80 \times 10^{-6} \times 4 \times 3.14 \times(1.2)^{2}\)

\(q=1.45 \times 10^{-3} {C}\)

Then, the total electric flux leaving the surface of the sphere will be calculated usingthe gauss formula, i.e.,

\(\phi=\frac{q}{\varepsilon_{0}}\)

\(\phi=\frac{1.45 \times 10^{-3}}{8.854 \times 10^{-12}} \quad\)\( (\because \epsilon_{0}=8.854 \times 10^{-12}) \)

\(\phi=1.6 \times 10^{8} {Nm}^{2} / {C}\)

Given,

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

Magnetic field, \(B= 800\) G

Torque, \(\tau = 0.016\) Nm

As we know,

Magnetic moment of the magnet

\(\tau=m B \sin \theta\)

\(0.016=m \times\left(800 \times 10^{-4} \mathrm{~T}\right) \times(\sin 30)\)

\(0.016=m \times\left(800 \times 10^{-4} \mathrm{~T}\right) \times( \frac 1 2)\)

\(m=160 \times \frac {2} { 800}=0.40 \mathrm{~A} \mathrm{~m}^{2}\)

Given:

Capacitance, (C) \(=1 \mu {F}\),

Rate of change in voltage, \(\frac{d V}{d t}=10^{6} {~V} / {s}\)

The expression for displacement current \({i}_{{d},}\) in the case of capacitance is given as:

\( i_{d}=\frac{d q}{d t}\)\(\quad\).....(i)

As we also know that, the charge on the capacitor is:

\( {q}={CV}\)

Where q = charge on the capacitors

On substituting the value of q = CV in equation (i), we get,

\( i_{d}=C \frac{d V}{d t}\)\(\quad\).....(ii)

On substituting the given values in equation (ii), we get,

\( i_{d}=\left(10^{-6} \times 10^{6}\right) A\)

\( {i}_{{d}}=1 {~A}\)

Given:

Two photons of same frequency are produced due to the annhiliation of a proton and antiproton.

We know that:

\(c=3 \times 10^{8}\)

Mass of a proton = \(1.67 \times 10^{-27}\)

\(E=m c^{2}\)

Put the values in above formula.

\(E=\left(2 \times 1.67 \times 10^{-27}\right) \times\left(3 \times 10^{8}\right)^{2} \mathrm{~J}\)

\(=3.006 \times 10^{-10} \mathrm{~J}\)

Also We know that:

\(2 h \nu=E\) or \(2 h \frac{c}{\lambda}=E\)

\(\therefore \lambda=\frac{2 h c}{E}\)

Put the values in above formula.

\(=\frac{2 \times 6.62 \times 10^{-34} \times 3 \times 10^{8}}{3.006 \times 10^{-10}} \mathrm{~m}\)

\(=1.323 \times 10^{-15} \mathrm{~m}\)

Given:

\(n_{1}=4\) and \(n_{2}=5\)

The wavelength of the radiations emitted from the hydrogen atom is given by:

\(\frac{1}{\lambda_{1}}=R\left[\frac{1}{(4)^{2}}-\frac{1}{(5)^{2}}\right]\)

\(= R\left[\frac{1}{16}-\frac{1}{25}\right]=\frac{9 R}{400}\).....(1)

When electrons jump from \(4^{\text {th }}\) to \(1^{\text {st }}\) orbit, then the wavelength of the radiations emitted from the hydrogen atom is given by:

\(\frac{1}{\lambda_{2}}=R\left[\frac{1}{(1)^{2}}-\frac{1}{(4)^{2}}\right]\)

\(= R\left[\frac{1}{1}-\frac{1}{16}\right]=\frac{15 R}{16}\).....(2)

Divide equation (1) and (2), we get,

\(\frac{\lambda_{1}}{\lambda_{2}}=\frac{400 \times 15 R}{9 R \times 16}\)

\(=\frac{125}{3}\)

\(=125 : 3\)

Semiconductors have negative temperature co-efficient. The reason for this is, when the temperature is increased, a large number of charge carriers are produced due to the breaking of covalent bonds and hence these electrons move freely and gives rise to conductivity.

Given:

\(N=200, a=0.16\) m^\(2\)

Change in magnetic flux density \((dB)=0.5-0.1=0.4\) Wb/m^\(2\) 

\(t=0.02\) sec

Induced emf \(e=\frac{-N d \phi}{d t}\)

\(d \phi=(dB) \cdot a\)

\(=0.4 \times 0.16\)

\(=0.064\) Wb

\(e=\frac{-200 \times(0.064)}{0.02}\)

\(=-640\) V

Given:

Net focal length \(f =10\) cm \(=0.1\) m

The focal length of one of the lens \(f_{1}=-20\) cm \(=-0.2\) m (Lens is converging so, negative sign is used)

Focal length of another lens \(f_{2}\)

Power of another lens \(=\frac{1}{f_{2}}\)

So, in order to get combined focal length:

\(\frac{1}{f} =\frac{1}{f_{1}}+\frac{1}{f _{2}}\)

\(\Rightarrow \frac{1}{f_{2}}=\frac{1}{f} \frac{-1}{f _{1}}\)

\(\Rightarrow \frac{1}{f_{2}}= \frac{1}{0.1}-\left(\frac{-1}{0.2}\right)\)

\(=\frac{1}{f_{2}}=15\)

So, power of lens is \(15\) D.