Physics Test

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Physics Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

In the circuit shown in the figure, the input voltage \(\mathrm{V_{i}}\) is \(20 \mathrm{~V}, \mathrm{~V}_{\mathrm{BE}}=0\) and \(\mathrm{V}_{\mathrm{CE}}=0 .\) The values of \(\mathrm{I_{B}, I_{C}}\) and \(\beta\) are given by:

A
\(\mathrm{I_{B}=20 \mu A, I_{C}=5 \mathrm{~mA}, \beta=250}\)

B
\(\mathrm{I_{B}=25 \mu A, I_{C}=5 \mathrm{~mA}, \beta=200}\)

C
\(\mathrm{I_{B}=40 \mu A, I_{C}=10 \mathrm{~mA}, \beta=250}\)

D
\(\mathrm{I_{B}=40 \mu A, I_{C}=5 \mathrm{~mA}, \beta=125}\)

Solution
Question 2
1 / -0

Mass spectroscopy is performed ____________ for a sample.

A
After detection of ions

B
Only after the sample is converted into gaseous state

C
By bombarding the sample with an electron beam

D
None of these

Solution

Mass spectroscopy is a process in which a compound is bombarded with electrons to generate ions. These ions are then separated by accelerating them through electric and magnetic fields. This is then separated according to their specific mass-charge ratio, where the radius of trajectory is proportional to the mass of a charged particle moving in uniform electric and magnetic fields. The sample must be in the gaseous state to be ionized. Therefore, the sample has to be converted into a gaseous state before the process can begin.
Question 3
1 / -0

A cricket bat is cut at the location of its center of mass as shown. Then,

A
The bottom piece will have a greater mass

B
The two pieces will have same mass

C
The handle piece will have a greater mass

D
The mass of the handle piece will be double the mass of the bottom piece

Solution

As there is more material near the bottom end of the bat, the centre of mass of the bat is nearer the bottom end of the bat. The bat will balance horizontally on a knife-edge placed at the centre of mass. Centre of the mass is the point about which the rigid body can be balanced. The centre of mass of the handle piece and the bottom piece are at distances d₁and d₂ respectively from the centre of mass of the entire bat.

For the rotational equilibrium, we have M₁d₁ = M₂d₂ where M1 and M₂are the masses of the handle piece and the bottom piece respectively. The distance d₁ is greater than d₂ since the handle piece has thinner regions compared to the bottom piece.

Therefore, M₂> M₁
Question 4
1 / -0

A pole is vertically submerged in swimming pool, such that it gives a length of shadow \(2.15 \mathrm{~m}\) within water when sunlight is incident at an angle of \(30^{\circ}\) with the surface of water. If swimming pool is filled to a height of \(1.5 \mathrm{~m}\), then the height of the pole above the water surface in centimeters is \(\left(\mathrm{n}_{\mathrm{w}}=4 / 3\right)\)_____________.

A
50

B
-50

C
45

D
-45

Solution
Question 5
1 / -0

In the relation \(p=\frac{\alpha}{\beta} \mathrm{e}^{-\frac{\alpha z}{k \theta}}, p\) is the pressure, \(z\) the distance, \(k\) is Boltzmann constant and \(\theta\) is the temperature, the dimensional formula of \(\beta\) will be:

A
\(\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\)

B
\(\left[\mathrm{ML}^{2} \mathrm{~T}\right]\)

C
\(\left[\mathrm{ML}^{0} \mathrm{~T}^{-1}\right]\)

D
\(\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]\)

Solution

The given relation,

\(p=\frac{\alpha}{\beta} \mathrm{e}^{-\frac{\alpha z}{k \theta}}\)

\(pe^0=\frac{\alpha}{\beta} \mathrm{e}^{-\frac{\alpha z}{k \theta}}\)

By comparing both sides, we get

\(\frac{\alpha z}{k \theta} = 0\)

\(\Rightarrow \alpha=\frac{k \theta}{z}\)

\(\Rightarrow [\alpha]=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~K}^{-1} \times \mathrm{K}\right]}{[\mathrm{L}]}\)

\(=\left[\mathrm{MLT}^{-2}\right]\)

And \( p=\frac{\alpha}{\beta}\)

\(\Rightarrow [\beta]=\left[\frac{\alpha}{p}\right]\)

\(=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}\)

\(=\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]\)
Question 6
1 / -0

The figure below shows three identical springs A, B, and C. When a \(4\; kg\) mass is hung on A, it descends by \(1 \;cm\). When a \(6\; kg\) mass is hung on C, it will descend by:

A
\(1.5\; cm\)

B
\(3\; cm\)

C
\(4.5\; cm\)

D
\(6\; cm\)

Solution
Question 7
1 / -0

Two batteries A and B connected in given circuit have equal emfs E and internal resistances r₁ and r₂, respectively (r₁ > r₂).

The switch S is closed at t = 0. After a long time, it was found that terminal potential difference across battery A is zero. Find the value of R.

A
\(\frac{3}{4}\left(r_{2}-r_{1}\right)\)

B
\(\frac{1}{2}\left(r_{2}-r_{1}\right)\)

C
\(\frac{4}{3}\left(r_{1}-r_{2}\right)\)

D
\(\frac{1}{4}\left(r_{2}-r_{1}\right)\)

Solution

After a long time, steady state is reached in which impendance due to inductor ( \(\omega L\) for \(d c\) ) is zero and that due to capcitance \(\left(\frac{1}{\omega C}\right)\) becomes infinite, so equivalent circuit is shows in Fig Net external resistance
\(R_{e x t}=\frac{\frac{R}{2}+R}{2}=\frac{3}{4} R\)
Net internal resistance \(R_{\text {int }}=r_{1}+r_{2}\)
\(\therefore\) Current in circuit \(I=\frac{2 E}{\frac{3}{4} R+r_{1}+r_{2}}\)

The potential difference across the terminals of cell \(A\) is zero, so
\(E-I r_{1}=0
\Rightarrow E-\frac{2 E r_{1}}{\frac{3}{4} R+r_{1}+r_{2}} \Rightarrow R=\frac{4}{3}\left(r_{1}-r_{2}\right)\)
Question 8
1 / -0

Two satellites of Earth, S1and S2, are moving in the same orbit. The mass of S1 is four times the mass of S2. Which of the following statements is true?

A
The time period of S1is four times that of S2.

B
The potential energy of Earth and satellite in both the cases is the same.

C
S1 and S2are moving with the same speed.

D
The kinetic energies of the two satellites are equal.

Solution
Question 9
1 / -0

A stainless steel chamber contains Ar gas at a temperature \(T\) and pressures \(P\). The total number of Ar atoms in the chamber is \(n\). Now Ar gas in the chamber is replaced by \(CO _{2}\) gas and the total number of \(CO _{2}\) molecules in the chamber is \(\frac{n}{2}\) at the same temperature \(T\). The pressure in the chamber now is \(P ^{\prime}\). Which one of the following relations holds true? (Both the gases behave as ideal gases.)

A
\(P^{\prime}=P\)

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

C
\(P^{\prime}=\frac{P}{2}\)

D
\(P^{\prime}=\frac{P}{4}\)

Solution

Both Ar and \(CO _{2}\) are ideal gases. So, we will apply the ideal gas equation.

For Ar: The temperature is \(T\), pressure is \(P\), and the number of molecules is \(n\), and volume \(( V )\) is the same for both the gases.

Use ideal gas equation:

\(PV = nRT \quad \ldots\) (1)

For \(CO _{2}\): The temperature is T, Pressure is \(P ^{\prime}\), number of molecules are \(\frac{n}{2}\) and Volume \(( V )\)

Apply Ideal gas equation:

\(P ^{\prime} V=\left(\frac{n}{2}\right) RT\quad \ldots\) (2)

Equation (2) divided by (1):

\(\frac{P ^{\prime}}{P} = \frac{n}{2}\)

\(P^{\prime}=\frac{P}{2}\)
Question 10
1 / -0

A liquid flows through pipes of different diameters. Velocity of liquid is \(2\)\(\mathrm{ms}^{-1}\).At the point where the diameter of the pipe is \(6\) cm.The velocity of liquid at a point where the diameter of the pipe is\(3\) cm will be

A
\(1\)\(\mathrm{ms}^{-1}\)

B
\(4\)\(\mathrm{ms}^{-1}\)

C
\(8\)\(\mathrm{ms}^{-1}\)

D
\(16\)\(\mathrm{ms}^{-1}\)

Solution

According to the principle of continuity,

\({A}_{1} V_{1}={A}_{2} {~V}_{2}\)

\(\Rightarrow{~V}_{2}\)=\(\frac {{A}_{1} {~V}_{1}} { {A}_{2}}\)

\(\therefore V_{2}=\left[\frac{\pi(3)^{2}} { \pi(1.5)^{2}}\right] \times 2=8\) \(\mathrm{ms}^{-1}\)