Physics Test

Result

Physics Test - 8

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

A particle in SHM is described by the displacement equation x(t) = A cos (ωt + θ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is π cm/s, then what is its amplitude? (The angular frequency of the particle is πs^-1.)

A
1 cm

B
√2 cm

C
2 cm

D
2.5 cm

Solution

Rate of change of displacement gives velocity. Given, \(\quad x=A \cos (\omega t+\theta)\)
Velocity \(v=\frac{d x}{d t}=A \frac{d}{d t} \cos (\omega t+\theta)\)
\(v=-A \omega \sin (\omega t+\theta)\)
Using \(\sin ^{2} \theta+\cos ^{2} \theta=1,\) we have
\(v=-A \omega \sqrt{1-\cos ^{2}(\omega t+\theta)}\)
\(\therefore\) From equation \(x=A \cos (\omega t+\theta),\) we have \(v=-A \omega \sqrt{1-x^{2} / A^{2}}\)
\(\Rightarrow \quad v=-\omega \sqrt{A^{2}-x^{2}}\)
Given, \(v=\pi \mathrm{cm} / \mathrm{s}, x=1 \mathrm{cm}, \omega=\pi \mathrm{s}^{-1}\)
\(\therefore \quad \pi=-\pi \sqrt{A^{2}-1}\)
\(\Rightarrow \quad 1=A^{2}-1\)
\(\Rightarrow A=\sqrt{2} \mathrm{cm}\)
Question 2
4 / -1

Three weights w, 2w and 3w are connected to an identical spring suspended from a rigid horizontal rod. The assembly of the rod and the weights fall freely. The position of the weight from the rod are such that

A
3w will be farthest

B
w will be farthest

C
all will be at the same distance

D
2w will be farthest

Solution

For \(w, 2 w, 3 w\) apparent weight will be zero because the system is falling freely. So, the distances of the weights from the rod will be same.
Question 3
4 / -1

In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor, when energy is stored equally between the electric and magnetic fields, is

A
Q/2

B
Q/√2

C
Q/√3

D
Q/3

Solution

If energy is stored equally in electric and magnetic fields, then energy in electric field will be half of the maximum value, which occurs when the charge on the capacitor is maximum.

Energy in capacitor \(=\frac{1}{2}\left(\frac{Q^{2}}{2 c}\right)\) \(\frac{Q^{\prime 2}}{2 c}=\frac{Q^{2}}{4 c}\)
\(Q^{\prime}=\frac{Q}{\sqrt{2}}\)
Question 4
4 / -1

A bar magnet of magnetic moment 2.0 JT^-1 lies aligned with the direction of a uniform magnetic field of 0.25 T. What is the work done to turn the magnet so as to align its magnetic moment opposite to the field direction?

A
0.25 J

B
0.5 J

C
0.75 J

D
1.0 J

Solution

Potential energy of the dipole \(=M B\left(\cos \Theta_{1}-\cos \Theta_{2}\right)=\)
\(M B(\cos 0-\cos 180)\)
\(=2 M B\)
\(=2 \times 2.0 \times 0.25=1.0 \mathrm{J}\)
Question 5
4 / -1

Directions: The following question has four choices out of which ONLY ONE is correct.

An open knife edge of mass M is dropped from a height 'h' on a wooden floor. If the blade penetrates 's' into the wood, the average resistance offered by the wood to the blade is

A
Mg

B
\(\operatorname{Mg}\left(1+\frac{h}{s}\right)\)

C
\(M g\left(1-\frac{h}{s}\right)\)

D
\(\operatorname{Mg}\left(1+\frac{h}{s}\right)^{2}\)

Solution

Let the average resistance be R.
\[
\begin{array}{l}
\therefore \quad M g(h+s)=R \times S \\
\Rightarrow R=\frac{M g(h+s)}{s} \\
=M g\left(1+\frac{h}{s}\right)
\end{array}
\]
Question 6
4 / -1

We wish to see inside an atom. Assume the atom to have a diameter of 100 pm. This means that one must be able to resolve a width of say 10 pm. If an electron microscope is used, the energy required should be

A
1.5 keV

B
15 keV

C
150 keV

D
1500 keV

Solution

As the de-Broglie wavelength is given by
\(\lambda=\frac{h}{p} \Rightarrow \frac{h}{\sqrt{2 m E}}, E\) is the kinetic energy
\(E=\frac{h^{2}}{2 m \lambda^{2}}\)
\(E=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times 9.1 \times 10^{-31} \times\left(10 \times 10^{-12}\right)^{2}}=2.39 \times 10^{-15} \mathrm{J}\)
or \(E=14958.7 \mathrm{eV} \approx 15 \mathrm{keV}\)
Question 7
4 / -1

The binding energy per nucleon of ₁H² is 1.1 MeV and binding energy per nucleon of ₂He⁴ is 7 MeV. The energy released when two ₁H² combine to form ₂He⁴ is

A
6.8 MeV

B
30.2 MeV

C
23.6 MeV

D
None of these

Solution

Energy Absorbed to break apart 2₁H² atoms = 1.1 × 4 = 4.4 MeV

Energy Released in formation of 1 ₂He⁴ atom = 7 × 4 = 28 MeV

Hence, energy released = (28 - 4.4) MeV = 23.6 MeV
Question 8
4 / -1

An observer is moving towards a stationary source with 1/10 the speed of sound. The ratio of apparent frequency to real frequency is

A
11/10

B
10/9

C
(11/10)²

D
(9/10)²

Solution

Frequency as observed by the observer is

\(f=\left[\frac{v+v_{0}}{v}\right] f_{0}\) where vo is the speed of the observer, v is the speed of the sound
\(\frac{f}{f_{0}}=\left[\frac{v+v_{0}}{v}\right]=\frac{v+\frac{v}{10}}{v}\)
\(\frac{f}{f_{0}}=\frac{11}{10}\)
Question 9
4 / -1

What is the dimension of L/RCV,where L, R, C and V have their usual meanings?

A
M¹L¹A^-2T⁰

B
M⁰L^-1A^-1F¹

C
M¹L⁰A^-1 T⁰

D
M⁰L ⁰A^-1T⁰

Solution

\(\mathrm{emf}=\mathrm{V}=\mathrm{L} \mathrm{d} \mathrm{I} / \mathrm{dt}\)
\(\mathrm{L}=\mathrm{V} \mathrm{dt} / \mathrm{d}\)
\(\mathrm{RC}\) is time constant. Its dimension is equal to time. So, dimension of \(\frac{L}{R C V}=V d t / d l\) \(\mathrm{RCV}=1 / \mathrm{A}=\mathrm{M}^{0} \mathrm{L}^{0} \mathrm{A}^{-1} \mathrm{T}^{0}\)
Question 10
4 / -1

Li nucleus has three protons and four neutrons. Mass of lithium nucleus is 7.016005 amu. Mass of proton is 1.007277 amu and mass of neutron is 1.008665 amu. Mass defect for lithium nucleus in amu is

A
0.04048 amu

B
0.04059 amu

C
0.04062 amu

D
0.04055 amu

Solution

Mass defect = mass of nucleons - mass of nucleus \(=(3 \times 1.007277+4 \times 1.008665)-7.016005\)
\(=0.040486\) amu
\(\approx 0.04050 \mathrm{amu}\)

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Physics Test - 8

Result

Physics Test - 8

Score

-

Rank

-

SHARING IS CARING

Question 1

1 cm

√2 cm

2 cm

2.5 cm

Solution

Question 2

3w will be farthest

w will be farthest

all will be at the same distance

2w will be farthest

Solution

Question 3

Q/2

Q/√2

Q/√3

Q/3

Solution

Question 4

0.25 J

0.5 J

0.75 J

1.0 J

Solution

Question 5

Mg

\(\operatorname{Mg}\left(1+\frac{h}{s}\right)\)

\(M g\left(1-\frac{h}{s}\right)\)

\(\operatorname{Mg}\left(1+\frac{h}{s}\right)^{2}\)

Solution

Question 6

1.5 keV

15 keV

150 keV

1500 keV

Solution

Question 7

6.8 MeV

30.2 MeV

23.6 MeV

None of these

Solution

Question 8

11/10

10/9

(11/10)2

(9/10)2

Solution

Question 9

M1L1A-2T0

M0L-1A -1F1

M1L0A -1 T0

M0L 0A-1T0

Solution

Question 10

0.04048 amu

0.04059 amu

0.04062 amu

0.04055 amu

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

(11/10)²

(9/10)²

M¹L¹A^-2T⁰

M⁰L^-1A^-1F¹

M¹L⁰A^-1 T⁰

M⁰L ⁰A^-1T⁰