Physics Test

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Physics Test - 31

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Weekly Quiz Competition

Question 1
4 / -1

Two SHMs are respectively represented by y₁ = a sin (ωt - kx), and y₂ = b cos (ωt - kx). The phase difference between the two is:

A
π/2

B
π/4

C
π/6

D
3π/4

Solution

As
y1 = a sin (ωt - kx), and y2 = b cos (ωt - kx)
y2 = b sin (ωt - kx + (π/2))
Hence phase difference between them is π/2.
Question 2
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 3
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 4
4 / -1

A uniform metre scale of length 1 m is balanced on a fixed semi-circular cylinder of radius 30 cm as shown in the figure. One end of the scale is slightly depressed and released. The time period (in seconds) of the resulting simple harmonic motion is

(Take g = 10 ms^-2)

A
π

B
π/2

C
π/3

D
π/4

Solution

Refer to Fig. The magnitude of the restoring torque \(=\) force \(\times\) perpendicular distance

\[
\begin{array}{l}
=m g \times A B \\
=m g \times R \sin \theta
\end{array}
\]

since \(\theta\) is small, \(\sin \theta=\theta\). Here \(\theta\) is expressed in radian. The equation of motion of the scale is
\[
I \frac{d^{2} \theta}{d t^{2}}=-m g R \theta
\]
or \(\frac{d^{2} \theta}{d t^{2}}=\left(-\frac{m g R}{I}\right) \theta\)
\(\therefore\)
\(\omega=\sqrt{\frac{m g R}{I}}\) or \(\frac{2 \pi}{T}=\sqrt{\frac{m g R}{I}}\)
or \(T=2 \pi \sqrt{\frac{I}{m g R}} .\) Now \(I=\frac{m L^{2}}{12} .\) Hence
\(T=\frac{\pi L}{\sqrt{3 g R}}\)
Using the values \(L=1 \mathrm{m}, g=10 \mathrm{ms}^{-2}\) and \(R=0.3 \mathrm{m}\), we get \(T=\pi / 3\) second. Hence the correct choice is
(3).
Question 5
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 6
4 / -1

Three particles, each of mass 'm' g, are situated at the vertices of an equilateral triangle ABC of side l cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to AB in the plane of ABC in gram-cm² unit will be

A
(3/4) ml²

B
2 ml²

C
(5/4) ml²

D
(3/2) ml²

Solution

The moment of inertia of the

system \(=m_{A} r_{A}^{2}+m_{B} r_{B}^{2}+m_{C} r_{C}^{2}\)
\(=m_{A}(0)^{2}+m(l)^{2}+m\left(l \sin 30^{\circ}\right)^{2}\)
\(=m l^{2}+m l^{2} \times(1 / 4)=(5 / 4) m l^{2}\)
Question 7
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 8
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 9
4 / -1

A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time, a viscous fluid of mass 'm' is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period-

A
Decreases continuously

B
Decreases initially and increases again

C
Remains unaltered

D
Increases continuously

Solution

According to law of conservation of momentum,
l_ω = constant
When viscous fluid of mass m is dropped and start spreading out then its moment of inertia increases and angular velocity decreases. But when it start falling then its moment of inertia again starts decreasing and angular velocity increases.
Question 10
4 / -1

A coil takes a current of 2 A and 200 W power from an alternating current source of 220 V, 50 Hz. What is the inductance of the coil?

A
0.312 H

B
0.142 H

C
0.285 H

D
1.07 H

Solution

\(R=\frac{P}{I^{2}}\)
\(=\frac{200}{4}=50 \Omega\)
\(Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{(50)^{2+X_{L}^{2}}}\)
\(Z=\mathrm{V} / \mathrm{I}=220 / 2=110 \mathrm{ohm}\)
\(X_{L}^{2}=12100-2500=9600\)
\(X_{L}=98 \mathrm{ohm}\)
\(L=\frac{X_{L}}{\omega}=\frac{X_{L}}{2 \pi f}=\frac{98}{2 \times 3.14 \times 50}=0.312 \mathrm{H}\)
Question 11
4 / -1

A needle made of bismuth is suspended freely in a magnetic field. The angle which the needle makes with the magnetic field is

A
0°

B
45°

C
90°

D
180°

Solution

Bismuth is a diamagnetic substance, so when placed in an external magnetic field it rotates such that its axis becomes perpendicular to the magnetic field.
Question 12
4 / -1

Three identical spheres, each of mass 1 kg, are kept as shown in the figure below, touching each other with their centres on a straight line. If their centres are marked respectively as P, Q and R, then the distance of the centre of mass of the system from P is

A
\(\frac{P Q+P R+Q R}{3}\)

B
PQ

C
\(\frac{P Q+Q R}{3}\)

D
\(\frac{P R+O R}{2}\)

Solution

As the spheres are uniform, the mass is equally distributed about Q along x-axis; centre of mass is \(\mathrm{Q}\); distance of the centre of mass is \(\mathrm{PQ}\). \(\mathrm{CM}\) can be calculated as:
First sphere is centred at origin \((x=0)\), second sphere is centred at \(x=P Q\), third sphere is centred at \(\mathrm{x}=\mathrm{PR}\). \(\mathrm{PR}=2 \mathrm{PQ}\)
\(x_{c m}=\frac{(m \times 0)+(m \times P Q)+(m \times 2 P Q)}{3 m}=\frac{3 m \times P Q}{3 m}=P Q\)
Question 13
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 14
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 15
4 / -1

For the given uniform square lamina ABCD whose centre is O,

A
I_AC = I_EF

B
I_AC =√2I_EF

C
√2 I_AC = I_EF

D
I_AD= 3I_EF

Solution

Using perpendicular axis theorem:
2I_AC= I₀, where I₀ is the moment of inertia about an axis perpendicular to the plane of the lamina and passing through O.
Both diagonals are perpendicular to each other.
Further, 2I_EF = Io
Two medians (lines dividing the sides of the square equally) are also perpendicular to each other, also perpendicular to the axis passing through the centre and perpendicular to the plane of lamina.

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Physics Test - 31

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Physics Test - 31

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Question 1

π/2

π/4

π/6

3π/4

Solution

Question 2

πωBR2

ωBR2

πωBR2/2

ωBR2/2

Solution

Question 3

1 cm

√2 cm

2 cm

2.5 cm

Solution

Question 4

π

π/2

π/3

π/4

Solution

Question 5

Q/2

Q/√2

Q/√3

Q/3

Solution

Question 6

(3/4) ml2

2 ml2

(5/4) ml2

(3/2) ml2

Solution

Question 7

5

4

3

2

Solution

Question 8

50 V

70 V

130 V

10 V

Solution

Question 9

Decreases continuously

Decreases initially and increases again

Remains unaltered

Increases continuously

Solution

Question 10

0.312 H

0.142 H

0.285 H

1.07 H

Solution

Question 11

0°

45°

90°

180°

Solution

Question 12

\(\frac{P Q+P R+Q R}{3}\)

PQ

\(\frac{P Q+Q R}{3}\)

\(\frac{P R+O R}{2}\)

Solution

Question 13

πωBR²

ωBR²

πωBR²/2

ωBR²/2

(3/4) ml²

2 ml²

(5/4) ml²

(3/2) ml²

M¹L¹A^-2T⁰

M⁰L^-1A^-1F¹

M¹L⁰A^-1 T⁰

M⁰L ⁰A^-1T⁰

I_AC = I_EF

I_AC =√2I_EF

√2 I_AC = I_EF

I_AD= 3I_EF