Physics Test

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Physics Test - 28

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Weekly Quiz Competition

Question 1
3 / -1

The binding energy per nucleon of ${}_{1}H^{2}$ is $1.1 M e V$ and the binding energy per nucleon of ${}_{2}H e^{4}$ is $7 M e V$ . The energy released when two ${}_{1}H^{2}$ combines to form ${}_{2}H e^{4}$ is:

A
$6.8 M e V$

B
$30.2 M e V$

C
$23.6 M e V$

D
None of these

Solution

As given, ${}_{1}H^{2} + {}_{2}H^{4} +$ energy

The binding energy per nucleon of a deuteron $({}_{1}H^{2}) = 1.1 M e V$

$∴$ Total binding energy $= 2 \times 1.1 = 2.2 M e V$

The binding energy per nucleon of helium $({}_{2}H e^{4}) = 7 M e V$

$∴$ Total binding energy $= 4 \times 7 = 28 M e V$

Hence, the energy released in the above process $= 28 - 2 \times 2.2 = 28 - 4.4 = 23.6 M e V$
Question 2
3 / -1

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

A
$10 H z$

B
$360 H z$

C
$216 H z$

D
$60 H z$

Solution

The number of holes in the disc determines the number of waves produced on each rotation.

$1 r p m$ is the number of rotations around a fixed axis in one minute or it's a measure of the frequency of rotation.

$1 r p m = \frac{1}{60} H z$

The total number of waves emitted per second is the frequency,

$f = 60 \times \frac{360}{60} = 360 H z$
Question 3
3 / -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 4
3 / -1

A convex lens made up of a material of refractive index $μ_{1}$ is immersed in a medium of refractive index $μ_{2}$ as shown in the figure. The relation between $μ_{1}$ and $μ_{2}$ is:

A
$μ_{1} < μ_{2}$

B
$μ_{1} > μ_{2}$

C
$μ_{1} = μ_{2}$

D
$μ_{1} = \sqrt{μ_{2}}$

Solution

In a medium of higher refractive index, a convex lens diverges a parallel beam of light, thus behaving as a diverging lens. Hence, $μ_{1} < μ_{2}$ .
Question 5
3 / -1

In the photoelectric effect, we assume that photon energy is proportional to its frequency and is completely absorbed by the electrons in a 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 another. The number of charges released will depend only on the number of electrons falling on the metal plates, given by intensity. Frequency increases only the kinetic energy of electrons.
Question 6
3 / -1

A toy of mass $M_{1}$ is pulled along a horizontal frictionless surface by the rope of mass $M_{2}$ . A force $F$ is applied to the free end of the rope. the force exerted on the toy is:

A
$F$

B
$\frac{M_{1} F}{(M_{1} + M_{2})}$

C
$\frac{M_{1} F}{(M_{1} - M_{2})}$

D
$\frac{M_{2} F}{(M_{1} + M_{2})}$

Solution

Given,

A toy of mass $= M_{1}$

Rope of mass $= M_{2}$

Force applied at the free end of rope $= F$

Consider this whole as a system. Let suppose it is moving with acceleration $a$ . Then we can write,

$F = m a$

Here $a$ is acceleration,

$F = (M_{1} + M_{2}) a$

$a = \frac{F}{(M_{1} + M_{2})}$

Now force exerted by the rope on the block is,

$F' = M a$

$F' = \frac{M_{1} F}{M_{1} + M_{2}}$
Question 7
3 / -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
Question 8
3 / -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.,
Question 9
3 / -1

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

The number of photoelectrons emitted per second is directly proportional to the intensity of light.

The intensity of light source is

$I \propto \frac{1}{d^{2}}$

When the distance is double, intensity becomes one-fourth.

A number of photoelectrons $\propto$ intensity, so the number of photoelectrons is a quarter of the initial number.
Question 10
3 / -1

The 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 a lower wavelength, thus higher frequency.

We know,

$h_{v} = ϕ + E$ where $v =$ Frequency of radiation, $ϕ =$ Work function, $E =$ Maximum energy of released electrons.

Stopping potential $= \frac{E}{e}$ , ( $e =$ charge of an electron)

As frequency increases, $E$ increases, and thus, stopping potential increases.

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Physics Test - 28

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Physics Test - 28

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Question 1

6.8MeV

30.2MeV

23.6MeV

None of these

Solution

Question 2

10Hz

360Hz

216Hz

60Hz

Solution

Question 3

0°

45°

90°

180°

Solution

Question 4

μ1<μ2

μ1>μ2

μ1=μ2

μ1=μ2

Solution

Question 5

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 6

F

M1F(M1+M2)

M1F(M1-M2)

M2F(M1+M2)

Solution

Question 7

zero

-100J

200J

100J

Solution

Question 8

9:4

2:3

3:2

4:9

Solution

Question 9

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 10

equal in both cases

more with ultraviolet light

more with visible light

varies randomly

Solution

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$6.8 M e V$

$30.2 M e V$

$23.6 M e V$

$10 H z$

$360 H z$

$216 H z$

$60 H z$

$0 °$

$45 °$

$90 °$

$180 °$

$μ_{1} < μ_{2}$

$μ_{1} > μ_{2}$

$μ_{1} = μ_{2}$

$μ_{1} = \sqrt{μ_{2}}$

$F$

$\frac{M_{1} F}{(M_{1} + M_{2})}$

$\frac{M_{1} F}{(M_{1} - M_{2})}$

$\frac{M_{2} F}{(M_{1} + M_{2})}$

$- 100 J$

$200 J$

$100 J$

$9 : 4$

$2 : 3$

$3 : 2$

$4 : 9$