Physics Test-13

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Physics Test-13

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Question 1
4 / -1

A body moves with a speed of \(10 {~m} / {s}\) in the curved path of \(25 {~m}\) radius of curvature. If the tangential acceleration is \(3 {~m} / {s}^{2}\), then total acceleration for the body will be:

A
\(3.3 {~m} / {s}^{2}\)

B
\(4 {~m} / {s}^{2}\)

C
\(5 {~m} / {s}^{2}\)

D
None of these

Solution

Given:
Speed, \(v=10 {~m} / {s}\)
Radius, \(r=25 {~m}\)
Tangential acceleration, \({a}_{{t}}=3 {~m} / {s}^{2}\)
Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration i.e.,
\(a=\sqrt{a_{c}^{2}+a_{t}^{2}}\)
Centripetal Acceleration\(\left(a_{c}\right):=\frac{v^{2}}{r}\)
\(\Rightarrow a_{c}=\frac{(10)^{2}}{25}=\frac{100}{25}=4 {~m} / {s}^{2}\)
Therefore, total acceleration,
\(\Rightarrow a=\sqrt{a_{t}^{2}+a_{c}^{2}}=\sqrt{4^{2}+3^{2}}=5 {~m} / {s}^{2}\)
Question 2
4 / -1

A body starts from rest and is uniformly accelerated for \(30\) sec. The distance travelled in the first \(10\)sec is \(X_{1}\), next \(10\)sec is \(X_{2}\)and the last \(10\) sec is \(X_{3}\). Then \(X_{1}: X_{2}: X_{3}\)is the same as:

A
\(1 : 2 : 4\)

B
\(1 : 2 : 5\)

C
\(1 : 3 : 5\)

D
\(1 : 3 : 9\)

Solution

Let the acceleration is \(a\).
For the displacement in first \(t\) sec,
\(S=u t+\frac{1}{2} a t^{2}\)
As initial speed,\(\mathrm{u}=0\)
⇒\(S=\frac{1}{2} a t^{2}\)
For \(X_{1}\), displacement in first \(10\) sec,
\(X_{1}=\frac{1}{2} a(10)^{2}\)
⇒ \(X_{1}=50 a\)
For \(\mathrm{X}_{2}\), displacement in next \(10\) sec,
\(X_{2}=\frac{1}{2} a(10+10)^{2}-\frac{1}{2} a(10)^{2}\)
⇒\(X_{2}=150 a\)
For \(\mathrm{X}_{3}\), displacement in next \(10\)sec,
\(X_{3}=\frac{1}{2} a(30)^{2}-\frac{1}{2} a(20)^{2}\)
⇒\(X_{3}=150a\)
Now, for ratio,
\(X_{1}: X_{2}: X_{3}=50 a: 150 a: 250 a\)
⇒\(X_{1}: X_{2}: X_{3}=1: 3: 5\)
Question 3
4 / -1

The amount of heat produced in an electric circuit depends upon the current \((I)\), resistance \((R)\) and time \((t)\). If the error made in the measurements of the above quantities are \(2 \%, 1 \%\) and \(1 \%\) respectively then the maximum possible error (in \(\%)\)in the total heat produced will be:

A
\(1\)

B
\(2\)

C
\(6\)

D
\(3\)

Solution

Given that,
\( \frac{\Delta I}{I} \times 100=2 \%\)
⇒ \(\frac{\Delta R}{R} \times 100=1 \%\)
⇒\(\frac{\Delta t}{t} \times 100=1 \%\)
Produced heat, \(\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}\)
Maximum possible error in produced heat is,
\(\left(\frac{\Delta H}{H} \times 100\right) \%=2\left(\frac{\Delta I}{I} \times 100\right) \%+\left(\frac{\Delta R}{R} \times 100\right) \%+\left(\frac{\Delta t}{t} \times 100\right) \%\)
⇒ \(\left(\frac{\Delta H}{H} \times 100\right) \%=(2 \times 2) \%+1 \%+1 \%\)
⇒ \(\left(\frac{\Delta H}{H} \times 100\right) \%=6 \%\)
Question 4
4 / -1

A stationary wave is given by, \(y=A \sin \frac{\pi x}{2} \cos \frac{\pi}{2} t\)where \(A,x,y\) are in \(cm\) and is in \(t\mathrm{~sec}\). Find the distance (in \(cm\)) between the adjacent nodes whose superposition can give rise to this vibration?

A
\(1\)

B
\(2\)

C
\(3\)

D
\(4\)

Solution

As given wave equation,
\(y=A \sin \frac{\pi x}{2} \cos \frac{\pi}{2} t\)
⇒ \(y=\frac{A}{2}\left[2 \sin \frac{\pi x}{2}+\frac{\pi}{2} t+\sin \left(\frac{\pi x}{2}-\frac{\pi}{2} t\right)\right]\)
⇒ \(y=\frac{A}{2} \sin \left(\frac{\pi x}{2}+\frac{\pi}{2} t\right)+\frac{A}{2} \sin \left(\frac{\pi x}{2}-\frac{\pi}{2} t\right)\)
By comparing, the amplitude is \(\frac{A}{2} c m\) and the velocity of wave is \(1 ~cm / \mathrm{sec}\).
As, \(\frac{2 \pi}{\lambda}=\frac{\pi}{2}\)
⇒ \(\lambda=4 c m\)
So, distance between adjacent nodes = \(\frac{4}{2}=2cm\)
Question 5
4 / -1

Let the refractive index of a denser medium with respect to a rarer medium be \(n≅12\) and its critical angle be \(θ_c\). At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is \(90°\). Angle \(A\) is given by:

A
\(\frac{1}{\cos ^{-1}\left(\sin \theta_{c}\right)}\)

B
\(\frac{1}{\tan ^{-1}\left(\sin \theta_{c}\right)}\)

C
\(\cos ^{-1}\left(\sin \theta_{c}\right)\)

D
\(\tan ^{-1}\left(\sin \theta_{c}\right)\)

Solution

When a light is an incident from denser medium to rarer medium, light ray move away from the normal.
According to Snell's law,
\(\frac{\sin i}{\sin r}=\frac{\mu_{R}}{\mu_{D}}\)
If the incident angle is critical then \(\mathrm{r}=90^{\circ}\)
⇒ \(\sin \theta_{c}=\frac{\mu_{R}}{\mu_{D}}\)
Given that the light is incident angle is \(i=A, r=90^{\circ}-A \).
Thus,
⇒ \(\frac{\sin A}{\sin \left(90^{\circ}-A\right)}=\frac{\mu_{R}}{\mu_{D}}\)
⇒ \(\frac{\sin A}{\cos A}=\frac{\mu_{R}}{\mu_{D}}\)
⇒ \(\frac{\sin A}{\cos A}=\sin \theta_{c}\)
⇒ \(\tan A=\sin \theta_{c}\)
⇒ \(A=\tan ^{-1}\left(\sin \theta_{c}\right)\)
Question 6
4 / -1

In a series \(LCR\) circuit \(\mathrm{R}=200 \Omega\) and the voltage and the frequency of the main supply is \(220 \mathrm{~V}\) and \(50 \mathrm{~Hz}\) respectively. On taking out the capacitance from the circuit the current lags behind the voltage by \(30^{\circ}\). On taking out the inductor from the circuit the current leads the voltage by \(30^{\circ} \). The power dissipated (in Watt) in the \(LCR\) circuit is

A
\(242\)

B
\(305\)

C
\(210\)

D
\(0\)

Solution

Given,
\(\mathrm{R}=200 \Omega, \mathrm{V_{rms}}=220 \mathrm{~V}, \mathrm{f}=50 \mathrm{~Hz}\)
When only thecapacitance is removed, the phase difference between the current and voltage is
\(\tan \phi=\frac{X_{L}}{R}\)
⇒ \(\tan 30^{\circ}=\frac{X_{L}}{R}\)
⇒ \(X_{L}=\frac{1}{\sqrt{3}} R\)
When only the inductance is removed, the phase difference between current and voltage is
\(\tan \phi_{i}=\frac{X_{c}}{R}\)
⇒ \(\tan 30^{\circ}=\frac{X_{c}}{R}\)
⇒ \(X_{c}=\frac{1}{\sqrt{3}} R\)
As \(X_{L}=X_{c},\) therefore the given series \(LCR\) is a resonance.
The impedance of the circuit is,
\(\mathrm{Z}=\mathrm{R}=200 \Omega\)
The power of the dissipated in the circuit is
\(P=V_{r m s} I_{r m s} \cos \phi\)
\(P=\frac{V_{\mathrm{rms}}^{2}}{2} \cos \phi\)
At resonance,
Power factor, \(\cos \phi=1\)
\(P=\frac{V_{\text {rms }}^{2}}{Z}\)
\(P=\frac{(220 \mathrm{~V})^{2}}{220 \Omega}\)
\(P=242 W\)
Question 7
4 / -1

Two-point charges \(+8q\) and \(-2q\) are located at \(x = 0\) and \(x = L\) respectively. The location of a point on the \(X\)-axis at which the net electric field due to these two point charges is zero is:

A
\(2L\)

B
\(\frac{L}{4}\)

C
\(8L\)

D
\(4L\)

Solution

Consider the region between \(x=-\infty\) and \(x=0\)
The electric field due to \(+8 \mathrm{q}\) is towards left and that due to \(-2 \mathrm{q}\) is towards right.
Electric field at a general point \(x=d(<0)\) is \(E=\frac{1}{4 \pi \epsilon_{0}}\left(-\frac{8 q}{d^{2}}+\frac{2 q}{(L-d)^{2}}\right)\)
If \(\mathrm{E}=0,\)
Then \(\frac{8 \mathrm{q}}{\mathrm{d}^{2}}=\frac{2 \mathrm{q}}{(\mathrm{L}-\mathrm{d})^{2}}\)
\(\Rightarrow \frac{\mathrm{d}^{2}}{(\mathrm{~L}-\mathrm{d})^{2}}=4\)
But, if \(\mathrm{d}<0,\) then \(|\mathrm{d}|<|\mathrm{L}-\mathrm{d}| \), no solution exists.
Consider the region between \(x=0\) and \(x=\mathrm{L}\)
The electric field due to \(+8 \mathrm{q}\) is towards right, and that due to \(-2 \mathrm{q}\) is towards right.
Hence, there is no point in \((0, \mathrm{~L})\) with zero electric field.
Consider the region between \(x=L\) and \(x=+\infty\)
The electric field due to \(+8 \mathrm{q}\) is towards right and \(-2 \mathrm{q}\) is towards left.
Electric field at a general point \(x=d(>L)\) is given by,
\(E=\frac{1}{4 \pi \epsilon_{0}}\left(+\frac{8 q}{d^{2}}-\frac{2 q}{(d-L)^{2}}\right)\)
If \(\mathrm{E}=0,\)
Then \(\frac{8 \mathrm{q}}{\mathrm{d}^{2}}=\frac{2 \mathrm{q}}{(\mathrm{d}-\mathrm{L})^{2}}\)
\( \Rightarrow \frac{\mathrm{d}}{\mathrm{d}-\mathrm{L}}=\pm 2\)
\(\Rightarrow \mathrm{d}=2 \mathrm{~L}\)
\(\Rightarrow \mathrm{d}=2\frac{\mathrm{~L}}{3}\)
But, \(\mathrm{d}>\mathrm{L}\)
\(\mathrm{d}=2 \mathrm{~L}\)
Therefore, at \(x=2 L\), the net electric field is zero.
Question 8
4 / -1

This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement – 1: For a mass \(M\) kept at the centre of a cube of side \(a\), the flux of gravitational field passing through its sides is \(4 πGM\).
Statement– 2: If the direction of a field due to a point source is radial and its dependence on the distance \(r\) from the source is given as \(r-2\), its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface.

A
Statement – 1 is True, Statement – 2 is True; Statement – 2 is not a correct explanation for Statement – 1

B
Statement – 1 is True, Statement – 2 is False

C
Statement – 1 is False, Statement – 2 is True

D
Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1

Solution

Let \(A\) be the Gaussian surface enclosing a spherical charge \(Q\).
⇒ \(\vec{E} \cdot 4 \pi \mathrm{r}^{2}=\frac{\mathrm{Q}}{\varepsilon_{0}}\)
⇒ \(\vec{E}=\frac{Q}{4 \pi r^{2} \varepsilon_{o}}\)
Flux passing through the surface is
\(\phi=\vec{E} \cdot 4 \pi \mathrm{r}^{2}=\frac{\mathrm{Q}}{\varepsilon_{0}}\)
Statement - 2
Every line passing through \(A\), has to pass through \(B\), whether \(B\) is a cube or any surface. It is only for the Gaussian surface, the lines of the field should be normal.
Assuming the mass is a point, gravitational field (\(\vec{g}\))\(=-\frac{G M}{r^{2}}\)
Flux,
\(\phi_{g}=\left|\vec{g} \cdot 4 \pi r^{2}\right|\)
⇒ \(\phi_{g}=\frac{4 \pi r^{2}}{r^{2}} \cdot G M\)
⇒ \(\phi_{g}=4 \pi \mathrm{GM}\)
Here, \(\mathrm{B}\) is a cube. So whatever be the shape, all the lines passing through \(A\) are passing through \(\mathrm{B}\), although all the lines are not normal.
Question 9
4 / -1

When a certain metal surface is illuminated with light of frequency \(\nu\), the stopping potential for photoelectric current is \(\mathrm{V}_{0}\). When the same surface is illuminated by light of frequency \(\frac{\mathrm{V}}{2}\) the stopping potential is \(\left(\frac{\mathrm{V}_{0}}{4}\right)\). The threshold frequency for photoelectric emission is

A

\(\frac{\nu}{6}\)

B
\(\frac{\nu}{3}\)

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

D
\(\frac{4 \nu}{3}\)

Solution

According to Einstein's photoelectric equation,
\(K_{\max }=\mathrm{h\nu}-\phi_{0}\)
Where \(\nu\) is the incident frequency and \(\phi_{0}\) is the work function of the metal.
As \(\mathrm{K}_{\max }=\mathrm{eV}_{0}\) where \(\mathrm{V}_{0}\) is the stopping potential
Therefore,
\(\Rightarrow\mathrm{eV}_{0}=\mathrm{h\nu}-\phi_{0} \ldots(\mathrm{i})\)
And \(e \frac{V_{0}}{4}=4 \frac{V}{2}-\phi_{0} \ldots(i i)\)
From (i) and (ii), we get
\(\Rightarrow\frac{h \nu}{4}-\frac{\phi_{0}}{4}=\frac{h \nu}{2}-\phi_{0}\)
or \(\Rightarrow\phi_{0}-\frac{\phi_{0}}{4}=\frac{h \nu}{2}-\frac{h \nu}{4}\)
\(\Rightarrow\frac{3}{4} \phi_{0}=\frac{h \nu}{4}\)
or \(\phi_{0}=\frac{h \nu}{3}\)
Therefore, threshold frequency
\(V_{0}=\frac{\phi_{0}}{h}=\frac{h \nu}{3} \times \frac{1}{h}=\frac{\nu}{3}\)
Question 10
4 / -1

The maximum number of 40 W tube lights connected in parallel which can safely be run from a 240 V supply with a 5 A fuse is:

A
5

B
15

C
20

D
30

Solution

Given, Power of each Bulb P = 40 W

Potential difference V = 240 V

Limit of fuse = 5 A

By Ohms Law,

V = I R

So, minimum resistance required so that a maximum of 5 A current will flow through the circuit is 48 Ω.

Now, we have to arrange the bulbs of 40 Watt, such that maximum resistance will be 48 Ω.

Resistance of each bulb,
Question 11
4 / -1

A particle of charge \(-16 \times 10^{-18}\) coulomb moving with velocity \(10 \mathrm{~ms}^{-1}\) along the \(x\) -axis enters a region where a magnetic field of induction \(B\) is along the \(\mathrm{y}\) -axis, and an electric field of magnitude \(10^{4} \mathrm{~V} / \mathrm{m}\) is along the negative \(z\)-axis. If the charged particle continues moving along the \(x\)-axis, the magnitude of \(B\) is:

A
\(10^{3} \mathrm{~Wb} / \mathrm{m}^{2}\)

B
\(10^{5} \mathrm{~Wb} / \mathrm{m}^{2}\)

C
\(10^{16} \mathrm{~Wb} / \mathrm{m}^{2}\)

D
\(10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\)

Solution

As we know that the force on a particle is
\( \mathrm{F}=\mathrm{q}(\mathrm{E}+\mathrm{v} \times \mathrm{B})\)
\(\mathrm{F}=\mathrm{F}_{\mathrm{e}}+\mathrm{F}_{\mathrm{m}}\)
\(\therefore F_{e}=q E=-16 \times 10^{-18} \times 10^{4}(-j)\)
⇒ \(\mathrm{F}_{\mathrm{e}}=16 \times 10^{-14} \mathrm{k}\)
\(F_{m}=-16 \times 10^{-18}(10 i \times B j)\)
⇒ \(\mathrm{F}_{\mathrm{m}}=-16 \times 10^{-17} \times \mathrm{B}(+\mathrm{k})\)
⇒ \(\mathrm{F}_{\mathrm{m}}=-16 \times 10^{-17} \mathrm{Bk}\)
Since, the particle will continue to move alone \(+x\) -axis, so resultant force is equal to zero.
\(\mathrm{F}_{\mathrm{e}}+\mathrm{F}_{\mathrm{m}}=0\)
⇒ \(16 \times 10^{-14}-16 \times 10^{-17} \mathrm{~B} = 0\)
⇒ \(16 \times 10^{-14}=16 \times 10^{-17} \mathrm{~B}\)
⇒ \(B=\frac{16 \times 10^{-14}}{16 \times 10^{-17}}=10^{3}\)
⇒ \(B=10^{3} W b / m^{2}\)
Question 12
4 / -1

\(1 \mathrm{gm}\) of an ideal gas expands isothermally. Heat flow will be:

A
From the gas to outside the atmosphere

B
From outside atmosphere to the gas

C
Zero

D
Both (A) and (B)

Solution

\(1 \mathrm{gm}\) of an ideal gas expands isothermally. Heat flow will be from outside atmosphere to the gas.
The isothermal process is a thermodynamic process where the temperature of the system remains constant.
The internal energy of a system is the measure of its temperature. So, in an isothermal process the change in internal energy \(\Delta U=0\).
Therefore, \(\Delta U=Q+W\)
\(Q=W\)
The work done is then equal to the heat energy absorbed.
So, heat is flown from the outside atmosphere to the inside.
Mechanical energy is the measure of work done, hence, the source of mechanical energy for gas to undergo isothermal expansion will be heat energy from the surrounding.
Question 13
4 / -1

The work done in carrying a charge q once around a circle of radius \(\mathrm{r}\) with a charge \(\mathrm{Q}\) placed at the center will be

A
\(Q q\left(4 \pi \varepsilon_{0} r^{2}\right)\)

B
\(\frac{Qq}{\left(4 \pi \varepsilon_{0} \mathrm{r}\right)}\)

C
\(0\)

D
\(\frac{Qq^{2}}{\left(4 \pi \varepsilon_{0} \mathrm{r}\right)}\)

Solution

As charge \(+Q\) is placed at the center of the circle, therefore the electric potential will be the same on its circumference as the distance is the same. Thus it is an equipotential surface.
As we know that the potential at each point on an equipotential surface is the same.
\(\therefore\) Potential difference is
\(\Delta V=V_{2}-V_{1}=V-V=0\)
Therefore the amount of work done in moving a charge \(q\) from one point to another around a circle of radius \(r\) is:
Work done \(=q \Delta V=q \times 0=0\)
Question 14
4 / -1

A force \(F=P y^{2}+Q y+R\) acts on a body in the \(y\) direction. The change in kinetic energy of the body during a displacement from \(y=-a\) to \(y=a\) is:

A
\(\frac{2 P a^{2}}{3}+2 R a\)

B
\(\frac{2 Q a^{2}}{3}+P a\)

C
\(\frac{2 P a^{2}}{3}+\frac{R a}{2}\)

D
\(\frac{2 P a^{2}}{3}+\frac{Q a}{2}\)

Solution

Work-energy theorem: The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
Work done, \(W=\Delta K E=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}\)
Where \(m\) is the mass of the object, \(v\) is the final velocity of the object and \(u\) is the initial velocity of the object.
Work-energy theorem for a variable force:
Kinetic energy, \(K=\frac{1}{2} m v^{2}\)
\(\Rightarrow \frac{d K}{d t}=\frac{d\left(1 / 2 m v^{2}\right)}{d t} \)
\(\Rightarrow \frac{d K}{d t}=m \frac{d v}{d t} v=m a v=F v \)
\(\Rightarrow \frac{d K}{d t}=F \frac{d x}{d t} \)
\(\Rightarrow d K=F d x\)
On integrating,
\(\Rightarrow \int_{K_{i}}^{K_{f}} d K=\int_{x_{i}}^{x_{f}} F d x \)
\(\Rightarrow \Delta K=\int_{x_{i}}^{x_{f}} F d x\)
From the work-energy theorem, a change in kinetic energy equals the work done.
\(\Rightarrow \Delta K=\int_{x_{i}}^{x_{f}} F d x \)
\(\Rightarrow \Delta K=\int_{-a}^{a}\left(P y^{2}+Q y+R\right) d y \)
\(\Rightarrow \Delta K=\left[\frac{P_{y}^{3}}{3}+\frac{Q y^{2}}{2}+R y\right]_{-a}^{a} \)
\(\Rightarrow \Delta K=\left[\left(\frac{P a^{3}}{3}+\frac{Q a^{2}}{2}+R a\right)-\left(\frac{-P a^{3}}{3}+\frac{Q a^{2}}{2}-R a\right)\right] \)
\(\Rightarrow \Delta K=\frac{2 P a^{3}}{3}+2 R a\)
Question 15
4 / -1

Which of the following is a correct relation for a transistor?

A
\(\alpha=\frac{\beta}{1-\beta}\)

B
\(\beta=\frac{\alpha}{1-\alpha}\)

C
\(\beta=\frac{\alpha}{1+\alpha}\)

D
\(\alpha=\frac{\beta}{1+\alpha}\)

Solution

A semiconductor device used to amplify or switch electronic signals and electrical power is called a transistor. It is made of semiconductor material with at least three terminals for connection to an external circuit.

Alpha gain (a): In the common-base configuration, the current gain that is defined as the ratio of change in collector current to change in emitter current is known as alpha gain.

\(\alpha=\frac{I_{c}}{I_{e}}\)

Beta gain \((\beta)\): It is the current gain in the commonEmitter configuration. the current gain that is defined as the change in collector current to base current is known as beta gain.

\(\beta=\frac{I_{c}}{I_{b}}\)

The relation between alpha gain and beta gain is given by:

\(\Rightarrow \alpha=\frac{\beta}{1+\beta}\)

The above equation can be written for \(\beta\) as,

\(\Rightarrow \beta=\frac{\alpha}{1-\alpha}\)