Physics Test

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Physics Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in

A
ultraviolet region

B
infrared region

C
visible region

D
X-ray region

Solution

From Einstein's photoelectric equation
or
$$
\begin{array}{l}
h v=h v_{0}+e V_{0} \\
-6.2+5=11.2 e V \\
\Rightarrow \frac{h c}{\lambda}=11.2 e V \\
\lambda=\frac{h c}{11.2 \cdot 0 V}
\end{array}
$$
$$
\begin{array}{l}
=\frac{6.6 \times 10^{-34} \times 3.0 \times 10^{8}}{11.2 \times 1.6 \times 10^{-19}} \\
=1.1049 \times 10^{-7} \\
=1104.9
\end{array}
$$
This incident radiation lies in ultra violet region,

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

A photoelectric cell is illuminated by a point source of light 1 m away. When the source is shifted to 2 m, then

A
each emitted electron carries half the initial energy

B
number of electrons emitted is a quarter of the initial number

C
each emitted electron carries one quarter of the initial energy

D
number of electrons emitted is half the initial number

Solution

Key Idea Number of photoelectrons einitted per second is directly proportional to intensity of light. Intensity of light source is
$$
I \propto \frac{1}{d^{2}}
$$
When distance is doubled, intensity becomes one-fourth. As number of photoelectrons $\propto$ intensity, so number of photoelectrons is quarter of the initial number.

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

The figure shows variation of photocurrent with anode potential for a photosensitive surface for three different radiations. Let I_a, I_b and I_c be the intensities and _a, _b and _c be the frequencies for the curves a, b and c, respectively. Then,

A
$\nu_{a}=\nu_{b}$ and $I_{a} \neq I_{b}$

B
$v_{a}=v_{c}$ and $I_{a}=I_{b}$

C
$v_{a}=v_{b}$ and $I_{a}=I_{b}$

D
$v_{b}=v_{c}$ and $I_{b} \neq I_{c}$

Solution

Saturation current is proportional to intensity while stopping potential increases with increase in frequency. Hence, $\mathbf{v}_{a}=v_{b}$ while (I_{a}
Hence , option ,A is correct answer
Question 4
1 / -0

Sodium surface is illuminated by ultraviolet and visible radiations, successively and the stopping potential is determined. This stopping potential is

A
equal in both cases

B
more with ultraviolet light

C
more with visible light

D
varies randomly

Solution

Ultraviolet light has lower wavelength, thus higher frequency.
We know, $h_{v}=\phi+E, v=$ Frequency of radiation, $\phi=$ Work function, $E=$ Maximum
energy of released electrons
Stopping potential= $E / e,$ (e $=$ charge of an electron) As frequency increases, $E$ increases, and thus, stopping potential increases.

Hence , option ,B is correct answer
Question 5
1 / -0

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

A
first increase and then decrease to the original value

B
first decrease and then increase to the original value

C
remain unchanged

D
increase towards a saturation value

Solution

When water begins to come out, the centre of mass begins to lower. There is an increase in the effective length and hence an increase in the time period : When water flows out completely, the centre of mass begins to shift back to the geometrical centre of the follow bob. So, there is a decrease in the effective length and hence a decrease in time period. When water flows out completely, the centre of mass is again at the geometrical centre. Now, the time period is the same as in the beginning.

Hence , option ,A is correct answer
Question 6
1 / -0

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 $\frac{d^{2} \theta}{d t^{2}}=\left(-\frac{m g R}{I}\right) \theta$
$\therefore \quad \omega=\sqrt{\frac{m g R}{I}}$ or $\frac{2 \pi}{T}=\sqrt{\frac{m g R}{I}}$
or $T=2 \pi \sqrt{\frac{I}{m g R}} \cdot$ 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).

Hence , option ,C is correct answer
Question 7
1 / -0

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.

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

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{equation}
\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 [1999)t] \\
y &=y_{1}+y_{2}+y_{3}
\end{aligned}
\end{equation}

Hence above expression involves three different waves the displacement given is a a resut of superpostion of three different waves

Hence , option ,C is correct answer
Question 9
1 / -0

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, $v=\pi \mathrm{cm} / \mathrm{s}, x=1 \mathrm{cm}, \omega=\pi \mathrm{s}^{-1}$
$\therefore \quad \pi=-\pi \sqrt{A^{2}-1}$
$\Rightarrow 1=A^{2}-1$
$\Rightarrow A=\sqrt{2} \mathrm{cm}$

Hence , option ,B is correct answer
Question 10
1 / -0

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
$\frac{P Q \cdot Q R}{3}$

D
$\frac{P R \cdot O R}{2}$

Solution

As the spheres are uniform, the mass is equally distributed about $\mathrm{Q}$ along $\mathrm{x}$ -axis:
centre of mass is $\mathrm{Q}$ : distance of the centre of mass is $\mathrm{PQ}$.
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 $x=P R$ $P R=2 P Q$
$I_{C m}=\frac{(m \times 0)+(m \times P Q)+(m \times 2 P Q)}{3 m}=\frac{3 m \times P Q}{3 m}=P Q$

Hence , option ,B is correct answer
Question 11
1 / -0

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

A
I_AC = I_EF

B
I_AC =√2 I_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.

Hence , option ,A is correct answer
Question 12
1 / -0

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

\begin{equation}
\begin{array}{l}
\text { system }=m_{A} r_{A}^{2}+m_{B} r_{B}^{2}+m_{\mathrm{C}} r_{\mathrm{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) \mathrm{m} l^{2}
\end{array}
\end{equation}

Hence , option ,C is correct answer
Question 13
1 / -0

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.

Hence , option ,C is correct answer
Question 14
1 / -0

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.

Hence , option ,B is correct answer
Question 15
1 / -0

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

$\mathrm{emf}=\mathrm{V}=\mathrm{L} \mathrm{d} 1 / \mathrm{dt}$
$\mathrm{L}=\mathrm{V} \mathrm{dt} / \mathrm{dl}$
$\mathrm{RC}$ is time constant. Its dimension is equal to time.
So, dimension of $\frac{L}{R C V}=V d t / d l$
$R C V=1 / A=M^{0} L^{O} A^{-1} T^{0}$

Hence , option ,D is correct answer

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Physics Test - 2

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Physics Test - 2

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SHARING IS CARING

Question 1

ultraviolet region

infrared region

visible region

X-ray region

Solution

Question 2

each emitted electron carries half the initial energy

number of electrons emitted is a quarter of the initial number

each emitted electron carries one quarter of the initial energy

number of electrons emitted is half the initial number

Solution

Question 3

\(\nu_{a}=\nu_{b}\) and \(I_{a} \neq I_{b}\)

\(v_{a}=v_{c}\) and \(I_{a}=I_{b}\)

\(v_{a}=v_{b}\) and \(I_{a}=I_{b}\)

\(v_{b}=v_{c}\) and \(I_{b} \neq I_{c}\)

Solution

Question 4

equal in both cases

more with ultraviolet light

more with visible light

varies randomly

Solution

Question 5

first increase and then decrease to the original value

first decrease and then increase to the original value

remain unchanged

increase towards a saturation value

Solution

Question 6

π

π/2

π/3

π/4

Solution

Question 7

π/2

π/4

π/6

3π/4

Solution

Question 8

5

4

3

2

Solution

Question 9

1 cm

√2 cm

2 cm

2.5 cm

Solution

Question 10

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

PQ

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

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

Solution

Question 11

IAC = IEF

IAC =√2 IEF

√2 IAC = IEF

IAD = 3IEF

Solution

Question 12

(3/4) ml2

2 ml2

(5/4) ml2

(3/2) ml2

Solution

I_AC = I_EF

I_AC =√2 I_EF

√2 I_AC = I_EF

I_AD= 3I_EF

(3/4) ml²

2 ml²

(5/4) ml²

(3/2) ml²

M¹L¹A^-2T⁰

M⁰L^-1A^-1F¹

M¹L⁰A^-1 T⁰

M⁰L ⁰A^-1T⁰