Physics Test

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Physics Test - 7

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Weekly Quiz Competition

Question 1
4 / -1

A radioactive nuclide can decay simultaneously by two different processes, which have individual decay constantsλ₁ anaλ₂ . The effective decay constant of the nuclide (λ) is given by

A
\(\lambda=\sqrt{\lambda_{1} \lambda_{2}}\)

B
\(\frac{1}{\lambda}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}\)

C
\(\lambda=\frac{1}{2}\left(\lambda_{1}+\lambda_{2}\right)\)

D
\(\lambda=\lambda_{1}+\lambda_{2}\)

Solution

The effective decay constant of the nuclide is given by λ₁ + λ₂
Question 2
4 / -1

A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at distances of 2 m and 3 m, respectively, from the source. The ratio of intensities of the waves at P and Q is

A
9 : 4

B
2 : 3

C
3 : 2

D
4 : 9

Solution

The average power per unit area that is incident perpendicular to the direction of propagation is called the intensity i.e.,
Or
\[
I=\frac{P}{4 \pi r^{2}} \quad \frac{I_{1}}{I_{2}}=\left(\frac{r_{2}}{r_{1}}\right)^{2} \quad I \propto \frac{1}{r^{2}}
\]
Here,
\[
r_{1}=2 \mathrm{m}, r_{2}=3 \mathrm{m}
\]
\(\therefore \quad \frac{I_{1}}{I_{2}}=\left(\frac{3}{2}\right)^{2}=\frac{9}{4}\)
As amplitude \(A \propto \sqrt{I},\) a spherical harmonic wave emanating from a point source can therefore, be written as:
\[
y(r, t)=\frac{A}{r} \sin (k r-\omega t)
\]
Question 3
4 / -1

Two weights 'w₁' and 'w₂' are suspended by two strings on a frictionless pulley. When the pulley is pulled up with an acceleration 'g', then the tension in the string is-

A
\(\frac{4 w_{1} w_{2}}{w_{1}+w_{2}}\)

B
\(\frac{w_{1} w_{2}}{w_{1}+w_{2}}\)

C
\(\frac{2 w_{1} w_{2}}{w_{1}+w_{2}}\)

D
\(\frac{w_{1}+w_{2}}{2}\)

Solution

Let m1 > m2.

As the lift is accelerating with acceleration (a = g), then the tension in the string is

\(T=\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}(g+a)\)
\(T=\frac{2 w_{1} w_{2}}{\left(w_{1}+w_{2}\right) \times g}(2 g)\)
As \(w_{1}=m_{1}g, w_{2}=m_{2} g\)
\(T=\frac{4 w_{1} w_{2}}{\left(w_{1}+w_{2}\right)}\)
Question 4
4 / -1

In photoelectric effect, we assume that photon energy is proportional to its frequency and is completely absorbed by the electrons in metal. Then, the photoelectric current

A
decreases when the frequency of the incident photon increases

B
increases when the frequency of the incident photon increases

C
does not depend upon the photon freqeuncy but only on the intensity of the incident beam

D
depends both on the intensity and frequency of the incident beam

Solution

Photocurrent depends upon the rate of transmission of charges from one end to other. The number of charges released will depend only on number of electrons falling on the metal plates, given by intensity. Frequency increases only the kinetic energy of electrons.
Question 5
4 / -1

In an LCR series circuit, the potential difference between the terminals of the inductance is 60 V, that between the terminals of the capacitor is 30 V and that across the resistance is 40 V. Then, the supply voltage will be

A
50 V

B
70 V

C
130 V

D
10 V

Solution

For series LCR circuit, effective voltage is \(\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}\)
\(V=\sqrt{40^{2}+(60-30)^{2}}=\sqrt{1600+900}\)
\(V=\sqrt{2500}-50 \mathrm{V}\)
Question 6
4 / -1

A toy of mass M₁ is pulled along a horizontal frictionless surface by rope of mass M₂. A force F is applied to the free end of the rope. the force exerted on the toy is:

A
F

B
M₁F/(M₁ + M₂)

C
M₁F/(M₁ - M₂)

D
M₂F/(M₁ + M₂)

Solution

a = F/(M₁ + M₂)

force on mass M₁ = M₁ × a = M₁F/(M₁+ M₂)
Question 7
4 / -1

The disc of a siren containing 60 holes rotates at a constant speed of 360 rpm. The emitted sound is in unison with a tuning fork of frequency

A
10 Hz

B
360 Hz

C
216 Hz

D
60 Hz

Solution

The number of holes in the disc determines the number of the waves produced on each rotation. The total number of waves emitted per second is the frequency,

\(f=60 \times \frac{360}{60}=360 \mathrm{Hz}\)
Question 8
4 / -1

The displacement y of a particle executing periodical motion is given by y = 4 cos² (t/2) sin (1000t). This expression may be considered to be a result of the superposition of how many independent harmonic motions?

A
5

B
4

C
3

D
2

Solution

\(\begin{aligned} y &=4 \cos ^{2}\left(\frac{t}{2}\right) \sin (1000 t) \\ y &=2 \times 2 \cos ^{2}\left(\frac{t}{2}\right) \sin (1000 t) \\ &=2 x(1+\cos t) \sin (1000 t) \\ &=2 \sin (1000 t)+2 \cos t \sin (1000 t) \\ &=2 \sin (1000 t)+\sin [(1000+1) t]+\sin [(1000-1) t] \\ &=2 \sin (1000 t)+\sin [(1001) t]+\sin [(999) t] \\ y &=y_{1}+y_{2}+y_{3} \end{aligned}\)
Hence above expression involves three different waves the displacement given is a result of superposition of three different waves.
Question 9
4 / -1

A circular metal plate of radius 'R' is rotating with a uniform angular velocity 'ω' with its plane perpendicular to a uniform magnetic field B. Then, emf developed between the centre and the rim of the plate is

A
πωBR²

B
ωBR²

C
πωBR²/2

D
ωBR²/2

Solution

The change will accumulate on the rim due to half effect until their fields are balanced
by the magnetic field's force at a radius \(r\).
Let the field be E.
\(\mathrm{dq} \mathrm{E}=\mathrm{Br} \omega \mathrm{dq}(\mathrm{q} \dot{\mathrm{V}} \times \mathrm{B})\)
\(E=B r^{\omega}\)
Now, total emf \(=\int E \cdot d r=\int_{0}^{R} B r \omega d r=\frac{-B R^{2} \omega}{2}\)
\(\mathrm{emf}=\frac{\mathrm{BR}^{2} \omega}{2}\)
Question 10
4 / -1

During the adiabatic expansion of 2 moles of a gas, the internal energy was found to have decreased by 100 J. The work done by the gas in this process is

A
zero

B
- 100 J

C
200 J

D
100 J

Solution

dU = 100 J

dQ = dU + dW

dQ = 0 (Adiabatic process)

dW = -dU = + 100 J

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Physics Test - 7

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Physics Test - 7

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Question 1

\(\lambda=\sqrt{\lambda_{1} \lambda_{2}}\)

\(\frac{1}{\lambda}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}\)

\(\lambda=\frac{1}{2}\left(\lambda_{1}+\lambda_{2}\right)\)

\(\lambda=\lambda_{1}+\lambda_{2}\)

Solution

Question 2

9 : 4

2 : 3

3 : 2

4 : 9

Solution

Question 3

\(\frac{4 w_{1} w_{2}}{w_{1}+w_{2}}\)

\(\frac{w_{1} w_{2}}{w_{1}+w_{2}}\)

\(\frac{2 w_{1} w_{2}}{w_{1}+w_{2}}\)

\(\frac{w_{1}+w_{2}}{2}\)

Solution

Question 4

decreases when the frequency of the incident photon increases

increases when the frequency of the incident photon increases

does not depend upon the photon freqeuncy but only on the intensity of the incident beam

depends both on the intensity and frequency of the incident beam

Solution

Question 5

50 V

70 V

130 V

10 V

Solution

Question 6

F

M1F/(M1 + M2)

M1F/(M1 - M2)

M2F/(M1 + M2)

Solution

Question 7

10 Hz

360 Hz

216 Hz

60 Hz

Solution

Question 8

5

4

3

2

Solution

Question 9

πωBR2

ωBR2

πωBR2/2

ωBR2/2

Solution

Question 10

zero

- 100 J

200 J

100 J

Solution

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M₁F/(M₁ + M₂)

M₁F/(M₁ - M₂)

M₂F/(M₁ + M₂)

πωBR²

ωBR²

πωBR²/2

ωBR²/2