Physics Test-4

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Physics Test-4

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Question 1
4 / -1

Two magnets, each of magnetic moment \(M\), are placed to form a cross at right angles to each other. The magnetic moment of the system will be:

A
\(2 M\)

B
\(M \sqrt{2}\)

C
\(\frac{M}{2}\)

D
\(\frac{M}{\sqrt{2}}\)

Solution

Given,

Two magnets, each of magnetic moment \(=M\)

The magnetic dipole moment is a vector quantity. Here, both dipoles have an angle of \(90^{\circ}\),

So, the net magnetic moment is:

\(\mathrm{M}_{\mathrm{net}}=\sqrt{\mathrm{M}^{2}+\mathrm{M}^{2}+2 \mathrm{M}^{2} \cos 90^{\circ}}\)

\(\Rightarrow M_{\mathrm{eq}}=\sqrt{2 \mathrm{M}^{2}}\)

\(\Rightarrow M_{\mathrm{eq}}=M \sqrt{2}\)
Question 2
4 / -1

The energy flux of sunlight reaching the surface of the earth is \(1.388 \times 10^{3} \mathrm{~W} \mathrm{m}^{-2}\). The number of photons (nearly) per square meter are incident on the earth per second is (Assume that the photons in the sunlight have an average wavelength of \(550 \mathrm{~nm}\) ).

A
\(3.838 \times 10^{21}\) photon \(\mathrm{m}^{-2} \mathrm{~s}^{-1}\)

B
\(3.838 \times 10^{23}\) photon \(\mathrm{m}^{-2} \mathrm{~s}^{-1}\)

C
\(5.838 \times 10^{21}\) photon \(\mathrm{m}^{-2} \mathrm{~s}^{-1}\)

D
\(5.838 \times 10^{23}\) photon \(\mathrm{m}^{-2} \mathrm{~s}^{-1}\)

Solution

Given,
Energy per unit area per second, \(p=1.388 \times 10^{3} \mathrm{~W} \mathrm{~m}\)
Let \(n\) be the number of photons incident on the earth per square metre.
The wavelength of each photon \(=550 \times 10^{-9} \mathrm{~m}\)
The energy of each photon,
\(E=\frac{h c}{\lambda}\)
\(=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{550 \times 10^{-9}}\)
\(=3.616 \times 10^{-19} \mathrm{~J}\)
Number of photons incident on the earth's surface,
\(n=\frac{P}{E}\)
\(=\frac{1.388 \times 10^{3}}{3.616 \times 10^{-19}}\)
\(=3.838 \times 10^{21}\)
\(=3.838 \times 10^{21}\) photon \(\mathrm{m}^{-2} \mathrm{~s}^{-1}\)
Question 3
4 / -1

The bob of a pendulum is released from a horizontal position \(A\) as shown in the figure. If the length of the pendulum is \(1.5\) m, what is the speed with which the bob arrives at the lowermost point \(B\), given that it dissipated \(5 \%\) of its initial energy against air resistance?

A
\(5.28\) m/s

B
\(3.28\) m/s

C
\(4.28\) m/s

D
\(5.21\) m/s

Solution

Length of pendulum \(l=1.5\) m
The potential energy of the bob at position \(A=m g h\)
As bob moves from position \(A\) towards position \(B\) its potential energy converted into kinetic energy. \(5 \%\) of its potential energy is dissipated against air resistance.
\(\therefore\) \(\mathrm{KE}\) at position \(\mathrm{B}=95 \%\) of its \(\mathrm{PE}\) at position \(\mathrm{A}\)
or \(\frac{1}{2} m{v}^{2}=\frac{95}{100} \times m g h\)
\(v=\sqrt{\frac{2 \times 95 \times g h}{100}}=\sqrt{\frac{19}{10} \times 9 .8 \times 1.5}\)
\(=5.28 \) m/s
Question 4
4 / -1

In the circuit shown in diagram, \(R=\sqrt{\frac{L}{C}}\). Switch \(S\) is closed at time \(t=0\). The current through \(C\) and \(L\) would be equal after a time \(t\), equal to:

A
\(C R\)

B
\(C R \ln (2)\)

C
\(\frac{1}{R \ln (2)}\)

D
\(L R\)

Solution

Given,
\(R=\sqrt{\frac{L}{C}}\)
\(R^{2}=\frac{L}{C}\)
\(C R=\frac{L}{R}\)
Thus, time constant of both the circuits are equal.
\(\tau_{C}=\tau_{L}=\tau\) (say)
\(i_{L}=\frac{v}{R}\left(1-e^{\frac{-t}{\tau}}\right)=i_{C}=\frac{V}{R} e^{\frac{-t}{\tau}}\)
\(i_{L}=i_{C}\)
And \(i_{C}=\frac{V}{R} e^{\frac{-t}{\tau}}\)
\(i_{L}=i_{C}\)
\(\therefore 1-\mathrm{e}^{\frac{-t}{\tau}}=e^{\frac{-t}{\tau}}\)
\(\Rightarrow 2 e^{\frac{-t}{\tau}}=1\)
\(\Rightarrow \mathrm{e}^{\frac{-t}{\tau}}=\frac{1}{2}\)
\(\Rightarrow \frac{t}{\tau}=\ln (2)\)
\(\Rightarrow t=\tau \ln (2)=C R \ln (2)\)
Question 5
4 / -1

Two masses A and B of \(10\) kg and \(5\) kg respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table as in figure. The coefficient of friction of A with the table is \(0.2\). The minimum mass of C that may be placed on A to prevent it from moving is equal to:

A
\(15\) kg

B
\(5\) kg

C
\(10\) kg

D
Zero

Solution

The free body diagram of the system.
The tension in the string is equal to,
\(T=m_{C} g\)
\(\Rightarrow T=5 \times 10\)
\(\Rightarrow T=50 N\)
The force that the combination of block A and C is \(50 \mathrm{N}\).
\(T=\mu\left(m_{A}+m_{C}\right) g\)
\(\Rightarrow T=0.2 \times\left(10+m_{C}\right) \times 10\)
\(\Rightarrow T=2 \times\left(10+m_{C}\right)\)
\(\Rightarrow T=20+2 m_{C}\)
\(\Rightarrow 50=20+2 m_{C}\)
\(\Rightarrow 2 m_{C}=30\)
\(\Rightarrow m_{C}=15\) kg
The minimum mass of the block B is equal to \(15\) kg.
Question 6
4 / -1

An electric dipole of length \(2\) cm is placed with its axis making an angle of \(30°\)to a uniform electric field \(10^5\) N/C. If it experiences a torque of \(10\sqrt{3}\) Nm, then potential energy of dipole -

A
\(-10\) J

B
\(-20\) J

C
\(-30\) J

D
\(-16\) J

Solution

We know torque on a dipole in an electric field making some angle, \(\theta\) with it is given by the formula,
\(\tau=p E \sin \theta\)
Dipole moment, \(\mathrm{p}=q \times 2 r\), here \(\mathrm{q}\) is the magnitude of each charge and \(r\) is separation between them.
Given the value of \(r\) is \(2 c m=0.02 m\)
Given the value of the filed and the angle that the dipole makes with it, substituting these values we get,
\(10 \sqrt{3}=q \times 2 r \times 10^{5} \sin 30^{\circ}\)
\(\Rightarrow 10 \sqrt{3}=q \times 0.02 \times 10^{5}\)
\(\Rightarrow q=8.66 \times 10^{-3} C\)
\(\Rightarrow q=8.66 \times 10^{-3} C\)
Thus, the charges are \(\pm 8.66 \times 10^{-3} C\) and they are separated by \(2 \mathrm{~cm}\) distance.
Now we need to find the value of potential energy, it is given by the formula,
\(U=-p E \cos \theta\)
\(\Rightarrow U=-q \times 2 r \times E \times \cos 30^{\circ}\)
\(\Rightarrow U=-8.66 \times 10^{-3} \times 10^{5} \times 2 \times 0.02 \times\frac{\sqrt{3}}{2}\)
\(\Rightarrow U=-30\) J
Question 7
4 / -1

A lens has focal length \(+20 \mathrm{cm}\). Its power will be:

A
\(\frac{1}{20}\) dioptre

B
\(\frac{1}{500}\) dioptre

C
\(\frac{1}{5}\) dioptre

D
\(5\) dioptre

Solution

Given:
\(\mathrm{f}=20 \mathrm{~cm}\)
The power of the lens is written as,
\(\Rightarrow P=\frac{100}{f(\mathrm{~cm})}\)
\(\Rightarrow P=\frac{100}{20}\)
\(\Rightarrow P=5 D\)
Question 8
4 / -1

In the circuit shown in diagram switch \(S\) is closed at time \(t = 0\). The charge which passes through the battery in one time constant is:

A
\(\frac{e R^{2} E}{ L}\)

B
\(E(\frac{L}{R})\)

C
\(\frac{EL}{eR^2}\)

D
\(\frac{eL}{ER}\)

Solution

The current at time \(t\) is given by \(i= {i}_{0}\left(1-\mathrm{e}^{-\frac{t}{\tau}}\right)\)
Where, \(i=\) current and \(i_{0}=\) initial current , \(\tau=\) torque
Here, \(i_{0}= \frac{E}{R}\) and \(\tau=\frac{{L}}{{R}}\)
\(\therefore {q}=\int_{0}^{\tau} {idt}=\int_{0}^{\tau} {i}_{0}\left(1-{e}^{-\frac{t} {\tau}}\right) {d} t\)
\(=\frac{i_{0} \tau}{e}=\frac{\left(\frac{E}{R}\right)\left(\frac{L}{R}\right)}{e}\)
\(=\frac{E L}{e R^{2}}\)
Question 9
4 / -1

A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is \(2.5 \mathrm{~cm} .\) What must be the least period of these oscillations so that the object is not detached from the platform.

A
\(\pi \sec\)

B
\(\frac{\pi}{ 5} \sec\)

C
\(\frac{\pi}{ 10} \sec\)

D
\(\frac{\pi}{ 15} \sec\)

Solution

Equating the equation for forces, we get,
\(\mathrm{m} \omega^{2} \mathrm{a}=\mathrm{mg}\), where \(m=\) mass, \(\omega=\) angular frequency and \(a=\) amplitude.
\(\omega=\sqrt{\frac{\mathrm{g}}{\mathrm{a}}} \)
Now, formula for time period of simple pendulum is,
\(\mathrm{T}=\frac{2\pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{a}}{\mathrm{g}}}=2 \pi \sqrt{\frac{2.5}{1000}}=\frac{\pi}{10} \mathrm{sec}\)
Question 10
4 / -1

Calculate the ratio of \(rms\) speeds of oxygen gas molecules to that of hydrogen gas molecules kept at the same temperature.

A
\(1:4\)

B
\(1:8\)

C
\(1:2\)

D
\(1:6\)

Solution

Root mean square velocity is given by,
\(v=\sqrt{\frac{3 R T}{M}}\) where, \(M=\) Molecular mass
As given in question temperature is constant and \(R\) is also constant.
So \(\frac{\left(v_{\text {rms }}\right)_{O_2}}{\left(v_{\text {rms }}\right)_{ {H_2}}}\) \(=\sqrt{\frac{M_{ {H_2}}}{M_{O_2}}}\)
\(=\sqrt{\frac{2}{32}}\) (Molecular mass of \(H_2\) \(=2\) and Molecular mass of \(O_2\) \(=32\))
\(=\frac{1}{4}\) or \(1:4\)
Question 11
4 / -1

A common emitter amplifier has a voltagegain of 50, an input impedance of \(100 \Omega\) and an output impedance of \(200 \Omega\). The power gain of the amplifier is:

A
\(500\)

B
\(1000\)

C
\(1250\)

D
\(100\)

Solution

Given,
\(A_{V}=50\)
\(R_{0}=200 \Omega\)
\(R_{i}=100 \Omega\)
\(A C\) power gain \(=\frac{\text { Change in output power }}{\text { Change in input power }}\)
\(=\frac{\Delta V_{c} \times \Delta i_{c}}{\Delta V_{i} \times \Delta i_{b}}\)
\(=\left(\frac{\Delta V_{c}}{\Delta V_{i}}\right) \times\left(\frac{\Delta i_{c}}{\Delta i_{b}}\right)\)
\(=A_{V} \times \beta_{\mathrm{AC}}\)
Where \(A_{V}\) is voltage gain and \((\beta)_{\mathrm{AC}}\) is \(\mathrm{AC}\) current gain.
Also,
\(A_{V}=\beta_{\mathrm{AC}} \times\) resistance gain
\(\Rightarrow A_{V}=\beta_{\mathrm{AC}} \frac{R_{0}}{R_{i}}\)
\(\Rightarrow 50=\beta_{\mathrm{AC}} \times \frac{200}{100}\)
\(\therefore \beta_{\mathrm{AC}}=25\)
Now,\(\mathrm{AC}\) power gain \(=A_{V} \times \beta_{\mathrm{AC}}\)
\(=50 \times 25\)
\(=1250\)
Question 12
4 / -1

A copper wire \(2 \mathrm{~m}\) long is stretched by \(1 \mathrm{~mm}\). If the energy stored in the stretched wire is converted to heat, calculate the rise in temperature of the wire. (Given, \(Y=12 \times 10^{11}\) dyne \(\mathrm{cm}^{-2}\), density of copper \(=9~\mathrm{gcm}^{-3}\) and specific heat of copper \(=0.1\) \(cal\) \(\left.\mathrm{g}^{-1 \circ} \mathrm{C}^{-1}\right)\)

A
\(252^{\circ} \mathrm{C}\)

B
\({\frac{1}{252}}^{\circ} \mathrm{C}\)

C
\(1000^{\circ} \mathrm{C}\)

D
\(2000^{\circ} \mathrm{C}\)

Solution

Formula for Energy in terms of Young's modulus \(Y\) is,
\(E=\frac{1}{2} \frac{Y A \Delta l^{2}}{l}\)
Where, \(\frac{\Delta l^{2}}{l}\) is strain multiply by change in length.
But \(m=A l d\) where, \(A=\) Area, \(l=\) length, \(d=\) density
or \(A=\frac{m}{l d}\)
\(\therefore E=\frac{Y m \Delta l^{2}}{2 l^{2} d}\)
Where,\(E\) is in calorie.
\(=\frac{Y m \Delta l^{2}}{2 l^{2} d}\)
For conversion of energy into joule we divide by \(4.2\) because \(1\) \(cal\) \(= 4.2 J\).
Now, \(m S \theta=\frac{Y m \Delta l^{2}}{2 l^{2} d J}\) \((S=\) specific heat of copper, \(\theta=\) rise in temperature)
or \(\theta=\frac{Y \Delta l^{2}}{2 l^{2} d J S}\)
or \(\theta=\frac{12 \times 10^{11} \times 10^{-1} \times 10^{-3} \times 10^{-3}}{2 \times 2 \times 2 \times 9 \times 10^{3} \times 4.2 \times 0.1 \times 10^{3}}\)
\(=\frac{12 \times 10^{5}}{72 \times 42 \times 10^{5}}\)
\(={\frac{1}{252}}^{\circ} \mathrm{C}\)
Question 13
4 / -1

If the shear modulus of a wire material is \(5.9 \times 10^{11}\) dyne \(\mathrm{cm}^{-2}\) then the potential energy of a wire of \(4 \times 10^{3} \mathrm{~cm}\) in diameter and \(5 \mathrm{~cm}\) long twisted through an angle of \(10^{\circ}\), is:

A
\(1.253 \times 10^{-12} \mathrm{~J}\)

B
\(2.00 \times 10^{-12} \mathrm{~J}\)

C
\(1.00 \times 10^{-12} \mathrm{~J}\)

D
\(0.8 \times 10^{-12} \mathrm{~J}\)

Solution

To twist the wire through the angle \(d \theta,\) it is necessary to do the work.
\(d W=\tau d \theta\) where, \(W=\) Work done, \(\tau=\) torque.
And \(\theta=10^{\circ}=\frac{10^{\circ}}{60} \times \frac{\pi}{180^{\circ}}=\frac{\pi}{1080} \mathrm{rad}\) \(W=\int_{0}^{\theta} \tau d \theta=\int_{0}^{\theta} \frac{\eta \pi r^{4} \theta d \theta}{2 l}=\frac{\eta \pi r^{4} \theta^{2}}{4 l}\)\(\begin{array}{l}W=\frac{5.9 \times 10^{11} \times 10^{-5} \times \pi\left(2 \times 10^{-5}\right)^{4} \pi^{2}}{10^{-4} \times 4 \times 5 \times 10^{-2} \times(1080)^{2}} \\W=1.253 \times 10^{-12} \mathrm{J}\end{array}\)
Question 14
4 / -1

For the given uniform square lamina \(\mathrm{ABCD}\), whose centre is \(\mathrm{O}\).

A
\(\mathrm{I}_{\mathrm{AC}}=\sqrt{2} \mathrm{I}_{\mathrm{EF}}\)

B
\(\sqrt{2} \mathrm{I}_{\mathrm{AC}}=\mathrm{I}_{\mathrm{EF}}\)

C
\(\mathrm{I}_{\mathrm{AD}}=3 \mathrm{I}_{\mathrm{EF}}\)

D
\(\mathrm{I}_{\mathrm{AC}}=\mathrm{I}_{\mathrm{EF}}\)

Solution

By perpendicular axes theorem,
\(\mathrm{I}_{\mathrm{EF}}=M \frac{(a^{2}+b^{2})}{12}\)
\(=\frac{M\left(a^{2}+a^{2}\right)}{12}\)
\(=M \frac{2 a^{2}}{12}\)
\(\mathrm{I_{z}}=\frac{M\left(2 a^{2}\right)}{12}+\frac{M\left(2 a^{2}\right)}{12}\)
\(=\frac{M a^{2}}{3}\)
Again by perpendicular axes theorem,
\(\mathrm{I_{A C}+\mathrm{I_{B D}}=\mathrm{I_{z}}}\)
\( \Rightarrow \mathrm{I_{A C}}=\frac{\mathrm{I_{z}}}{2}\)
\(=\frac{M a^{2}}{6}\)
By the same theorem
\(\mathrm{I_{E F}}=\frac{\mathrm{I_{z}}}{2}\)
\(=\frac{M a^{2}}{6}\)
\(\therefore \mathrm{I_{A C}}=\mathrm{I_{E F}}\)
Question 15
4 / -1

Two vectors both are equal in magnitude and have their resultant equal in magnitude of the either. Find the angle between the two vectors.

A
\(\frac{4 \pi}{3}\)

B
\(\frac{6 \pi}{3}\)

C
\(\frac{2 \pi}{3}\)

D
\(\frac{8 \pi}{3}\)

Solution

\(\vec{A}\) and \(\vec{B}\) have equal magnitude so \(|\vec{A}|=|\vec{B}|\).
The magnitude of resultant of \(\vec{A}\) and \(\vec{B}\) is equal to the magnitude of either of them i.e., \(|\vec{A}+\vec{B}|=|\vec{A}|=|\vec{B}|\).
The magnitude of resultant of \(\vec{A}\) and \(\vec{B}\) is given by
\(|\vec{A}+\vec{B}|=\sqrt{|\vec{A}|^{2}+|\vec{B}|^{2}+2|\vec{A}||\vec{B}| \cos \theta}\)
Where \(\theta=\) Angle between \(\vec{A}\) and \(\vec{B}\)
This magnitude is equal to the magnitude of the either of them i.e.,\(|\vec{A}|=|\vec{A}+\vec{B}|\)
So, \(|\vec{A}|=\sqrt{|\vec{A}|^{2}+|\vec{B}|^{2}+2|\vec{A}||\vec{B}| \cos \theta}\)
Taking square in both the sides,
\(|\vec{A}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+2|\vec{A}||\vec{B}| \cos \theta\)
Put \(|\vec{A}|=|\vec{B}|\)
And \(|\vec{A}|^{2}=|\vec{B}|^{2}\)
So,\(|\vec{A}|^{2}=|\vec{A}|^{2}+|\vec{A}|^{2}+2|\vec{A}||\vec{A}| \cos \theta\)
\(\Rightarrow 2|\vec{A}|^{2} \cos \theta+|\vec{A}|^{2}=0\)
\(\Rightarrow 2 \cos \theta=-1\)
\(\Rightarrow \cos \theta=\frac{-1}{2}\)
\(\Rightarrow \theta=\cos ^{-1}\left(\frac{-1}{2}\right)\)
\(\Rightarrow \theta=\frac{2 \pi}{3}\)
So, the required angle between \(\vec{A}\) and \(\vec{B}\) is \(\frac{2 \pi}{3}\).