Physics Test

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Physics Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

A current of i amp flows in the loop having circular arc r metre, subtending an angle as shown in the figure. The magnetic field at centre O of circle is :

A

\(\frac{\mu_{0} i}{4 \pi r} \theta\)

B

\(\frac{2 \mu_{0} i}{4 \pi r^{2}} \sin \theta\)

C

\(\frac{2 \mu_{0} i}{2 r} \sin \theta\)

D

\(\frac{\mu_{0}^{2}}{2 r} \sin \theta\)

Solution

Magnetic field due to the current carrying arc at the centre of curvature.

\(B=\frac{\mu_{0} i}{4 \pi r} \theta\)

Hence option A is the correct answer.
Question 2
1 / -0

A coin is placed on a horizontal platform, which undergoes vertical simple harmonic motion of angular frequency ω. The amplitude of oscillation is gradually increased. The coil will leave contact with the platform for the first time.

A
at the mean position of the platform

B
for an amplitude g/ω²

C
for an amplitude g²/ω²

D
at the highest position of the platform

Solution

As the amplitude is increased, the maximum acceleration of the platform (along with coin as long as they do not get separated) increases. We draw the FBD for coin at one of the extreme positions.

At the extreme position, a = ω²A

Then, from Newton's second law,

mg − N = mω²A

For loosing contact with the platform,

N = 0

So, A = g/ω²

Hence option B is the correct answer.
Question 3
1 / -0

Time taken by sunlight to pass through a window of thickness 4 mm with refractive index 3/2, is :

A
2 × 10^-4 s

B
2 × 10⁴ s

C
2 × 10^-11 s

D
2 × 10¹¹ s

Solution

\(\begin{aligned} \text { Time taken } &=\frac{\mu t}{c} \\ &=\frac{\frac{3}{2} \times 4 \times 10^{-3}}{3 \times 10^{8}} \\ &=2 \times 10^{-11} \mathrm{s} \end{aligned}\)

Hence option C is the correct answer.
Question 4
1 / -0

A horizontal force \(F\) is applied such that the block remains stationary because \(N\) will produce torque. The torque produced by friction force is equal and opposite the torque produce due to normal reaction (N).

A
If both assertion and reason are true and reason is the correct explanation of assertion.

B
If both assertion and reason are reason are true but reason is not the correct explanation of assertion.

C
If assertion is true but reason is false

D
If both assertion and reason are false.

Solution
Question 5
1 / -0

The logic gate represented in following figure is:

A
NOR gate

B
OR gate

C
AND gate

D
XOR gate

Solution
Question 6
1 / -0

In the following common emitter circuit, if β = 100, V_CE = 7V, V_BE = Negligible and R_C = 2 kΩ, then IB is equal to :

A
0.01 mA

B
0.04 mA

C
0.02 mA

D
0.03 mA

Solution

Considering the output circuit, we have
\(V_{C C}=i_{C} R_{C}+V_{C E}\)
\(i _{ C } R _{ C }=15-7=8 V\)
Hence, \(i_{C}=\frac{8}{R_{C}}=\frac{8}{2 \times 10^{3}}=4 mA\)
\(i_{B}=\frac{I_{C}}{\beta}=\frac{4 \times 10^{-3}}{100}=0.04 mA\)

Hence option B is the correct answer.
Question 7
1 / -0

A circular disc of radius R is removed from a bigger circular disc of radius 2R as shown in the figure. The centre of mass of the new disc is αR from the centre of the bigger disc. What is the value of α?

A
1/3

B
1/2

C
1/6

D
1/4

Solution

\(\mathrm{X}_{\mathrm{CM}}=\frac{\sigma \pi(2 \mathrm{R})^{2} \times 0-\sigma\left(\pi \mathrm{R}^{2}\right) \mathrm{R}}{\sigma \pi(2 \mathrm{R})^{2}-\sigma(\pi \mathrm{R})^{2}}\)

\(\frac{\mathrm{R}}{3} \Rightarrow \alpha=1 / 3\)

Hence option A is the correct answer.
Question 8
1 / -0

A copper bar of length 1 m and cross-sectional area 10 × 10^-2 m² has its one end maintained at 100^oC by means of 0.4 kW electric heater. What is the temperature of the other end in steady state, given that the thermal conductivity of copper is 400 W/mK?

A
60^oC

B
70^oC

C
80^oC

D
90^oC

Solution

\(\mathrm{As}, \frac{d Q}{d t}=\frac{K A}{d}\left[T_{H}-T_{C}\right]\)

\(\left[\frac{d Q}{d t}\right] \frac{d}{K A}=\left[T_{H}-T_{C}\right] \Rightarrow T_{C}=T_{H}-\left[\frac{d Q}{d t}\right] \frac{d}{K A}\)

\(T_{C}=100-400 \times \frac{1}{400 \times 10 \times 10^{-2}}=90^{\circ} \mathrm{C}\)

Hence option D is the correct answer.
Question 9
1 / -0

Number of particles passing in unit time through unit area perpendicular to z-axis is given by N= D (N₂ - N₁)/(Z₂ - Z₁) , where N₂ and N₁ are number of particles in unit volume at Z₂ and Z₁. The dimensional formula of D is

A
M⁰L^-2T²

B
M⁰L^-1T^-1

C
M⁰L²T^-1

D
M⁰L^-2T^-2

Solution

Dimensional formula for number of particles per unit area per unit time is \(\left[M^{0} L^{-2} T^{-1}\right]\)
\(N _{1}\) and \(N _{2}\) are the number of the particles in unit volume. Dimensional formula for \(N _{1}\) is \(\left[M^{0} L^{-3} T^{0}\right]\)
Dimensional formula for \(z_{1}\) and \(z_{2}\) is \(\left[M^{0} L^{1} T^{-0}\right]\)
Therefore, dimensional formula for D
\([D]=\frac{[N]\left[Z_{1}\right]}{\left[N_{1}\right]}\)
\([D]=\frac{\left[M^{0} L^{-2} T^{-1}\right][L]}{\left|M^{0} L^{-3} T^{0}\right|}=\left[M^{0} L^{2} T^{-1}\right]\)

Hence option C is the correct answer.
Question 10
1 / -0

The equation of a wave is \(y=5 \sin \left(\frac{t}{0.04}-\frac{x}{4}\right),\) where \(x\) is in \(cm\) and \(t\) in seconds. The maximum velocity of the particle will be :

A
1 ms^-1

B
2 ms^-1

C
1.5 ms^-1

D
1.25 ms^-1

Solution

Equation of wave

\(y=5 \sin \left(\frac{t}{0.04}-\frac{x}{4}\right)\)

The standard equation of a wave in the given form is

\(y=a \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)\)

Comparing the given equation with the standard equation, we get

\(a=5\) and \(\omega=\frac{1}{0.04}=25\)

Therefore, maximum velocity of particles of the medium,

\(\begin{aligned} v_{\max } &=a \omega \\ &=5 \times 25 \\ &=125 \mathrm{cm} \mathrm{s}^{-1}=1.25 \mathrm{ms}^{-1} \end{aligned}\)

Hence option D is the correct answer.

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Physics Test - 8

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Physics Test - 8

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Question 1

\(\frac{\mu_{0} i}{4 \pi r} \theta\)

\(\frac{2 \mu_{0} i}{4 \pi r^{2}} \sin \theta\)

\(\frac{2 \mu_{0} i}{2 r} \sin \theta\)

\(\frac{\mu_{0}^{2}}{2 r} \sin \theta\)

Solution

Question 2

at the mean position of the platform

for an amplitude g/ω2

for an amplitude g2/ω2

at the highest position of the platform

Solution

Question 3

2 × 10-4 s

2 × 104 s

2 × 10-11 s

2 × 1011 s

Solution

Question 4

If both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are reason are true but reason is not the correct explanation of assertion.

If assertion is true but reason is false

If both assertion and reason are false.

Solution

Question 5

NOR gate

OR gate

AND gate

XOR gate

Solution

Question 6

0.01 mA

0.04 mA

0.02 mA

0.03 mA

Solution

Question 7

1/3

1/2

1/6

1/4

Solution

Question 8

60oC

70oC

80oC

90oC

Solution

Question 9

M0L-2T2

M0L-1T-1

M0L2T-1

M0L-2T-2

Solution

Question 10

1 ms-1

2 ms-1

1.5 ms-1

1.25 ms-1

Solution

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for an amplitude g/ω²

for an amplitude g²/ω²

2 × 10^-4 s

2 × 10⁴ s

2 × 10^-11 s

2 × 10¹¹ s

60^oC

70^oC

80^oC

90^oC

M⁰L^-2T²

M⁰L^-1T^-1

M⁰L²T^-1

M⁰L^-2T^-2

1 ms^-1

2 ms^-1

1.5 ms^-1

1.25 ms^-1