Physics Test

Result

Physics Test - 1

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
1 / -0

In the following figure, the height of the object is H₁ = + 2.5 cm. What is the height of the image of H₁?

A
- 5 cm

B
+ 5 cm

C
+ 7.5 cm

D
- 7.5 cm

Solution

Since the given mirror is a concave mirror, the image formed will be inverted and double the size of the object.
Height of the object = 2.5 cm
Height of the image = – 5 cm; -ve sign indicates that the image is inverted.
Hence, option , A is correct answer
Question 2
1 / -0

A convex lens made up of a material of refractive index μ₁ is immersed in a medium of refractive index μ₂ as shown in the figure. The relation between μ₁ and μ₂ is

A
μ₁ < μ₂

B
μ₁ > μ₂

C
μ₁ = μ₂

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 a diverging lens. Hence, $μ_{1} < μ_{2}$
Hence, option , A is correct answer
Question 3
1 / -0

A Carnot engine works as a refrigerator between 250 K and 300 K. If it receives 750 calories of heat from the reservoir at the lower temperature, then the amount of heat rejected at the higher temperature is

A
900 cal

B
625 cal

C
750 cal

D
1000 cal

Solution

$\begin{aligned} As we have, \frac{Q_{1}}{Q_{2}} = \frac{T_{1}}{T_{2}} \Rightarrow \frac{Q_{1}}{Q_{2}} - 1 = \frac{T_{1}}{T_{2}} - 1 \\ \frac{Q_{1} - Q_{2}}{Q_{2}} = \frac{T_{1} - T_{2}}{T_{2}} \\ \frac{Q_{1} - Q_{2}}{750} = \frac{300 - 250}{250} = \frac{50}{250} \\ Q_{1} = \frac{50}{250} \times 750 + 750 = 900 cal \end{aligned}$
Hence, option , A is correct answer
Question 4
1 / -0

F₁ and F₂ are focal lengths of the objective and the eyepiece of a telescope, respectively. The angular magnification for this telescope is equal to

A
F₁/F₂

B
F₂/F₁

C
$\frac{F_{F} F_{2}}{F_{1} + F_{2}}$

D
$\frac{F_{1} + F_{2}}{F_{1} F_{2}}$

Solution

Angular magnification is the ratio of the focal length of the objective to the focal length of the eyepiece.
Therefore, $M = \frac{F_{1}}{F_{2}}$
Hence, option , A is correct answer
Question 5
1 / -0

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
$f = [\frac{v + e_{0}}{v}] f_{0}$ where vo is the speed of the observer, v is the speed of the sound
$\frac{f}{f_{0}} = [\frac{v + v_{0}}{v}] = \frac{v + \frac{r}{10}}{v}$
$\frac{f}{f_{0}} = \frac{11}{10}$
Hence, option , A is correct answer
Question 6
1 / -0

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,
$f = 60 \times \frac{360}{60} = 360 Hz$
Hence, option , B is correct answer
Question 7
1 / -0

A police car horn emits sound at a frequency of 240 Hz when the car is at rest. If the speed of sound is 330 m/s, the frequency heard by an observer who is approaching the car at a speed of 11 m/s, is

A
248 Hz

B
244 Hz

C
240 Hz

D
230 Hz

Solution

Observer $\to \leftarrow_{V = 330 m / s}^{S o u n d}$
v_o=11m/s
+Source
v_s=0

$n' = \frac{330 + 11}{330} \times 240$
=248Hz
Hence , option ,A is correct answer
Question 8
1 / -0

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
$I = \frac{P}{4 π r^{2}} \frac{I_{1}}{I_{2}} = {(\frac{r_{2}}{r_{1}})}^{2} I \propto \frac{1}{r^{2}}$
$\begin{array}{ll} Here, & r_{1} = 2 m, r_{2} = 3 m \\ ∴ & \frac{I_{1}}{I_{2}} = {(\frac{3}{2})}^{2} = \frac{9}{4} \end{array}$
As amplitude $A \propto \sqrt{I},$ a spherical harmonic wave emanating from a point source can therefore, be written as :
$y (r, t) = \frac{A}{r} \sin (k r - ω t)$
Hence , option ,A is correct answer
Question 9
1 / -0

Directions: The following question contains 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct
In the experiment to determine the speed of sound using a resonance column,

A
prongs of the tuning fork are kept in a vertical plane

B
prongs of the tuning fork are kept in a horizontal plane

C
in one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound in air

D
in one of the two resonance observed, the length of the resonating air column is close to half of the wavelength of sound in air

Solution

Correct Answer : prongs of the tuning fork are kept in a vertical plane
Hence , option ,A is correct answer
Question 10
1 / -0

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.
Hence, option, C is correct answer

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Physics Test - 1

Result

Physics Test - 1

Score

-

Rank

-

SHARING IS CARING

Question 1

- 5 cm

+ 5 cm

+ 7.5 cm

- 7.5 cm

Solution

Question 2

μ1 < μ2

μ1 > μ2

μ1 = μ2

μ1=μ2

Solution

Question 3

900 cal

625 cal

750 cal

1000 cal

Solution

Question 4

F1/F2

F2/F1

FFF2F1+F2

F1+F2F1F2

Solution

Question 5

11/10

10/9

(11/10)2

(9/10)2

Solution

Question 6

10 Hz

360 Hz

216 Hz

60 Hz

Solution

Question 7

248 Hz

244 Hz

240 Hz

230 Hz

Solution

Question 8

9 : 4

2 : 3

3 : 2

4 : 9

Solution

Question 9

prongs of the tuning fork are kept in a vertical plane

prongs of the tuning fork are kept in a horizontal plane

in one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound in air

in one of the two resonance observed, the length of the resonating air column is close to half of the wavelength of sound in air

Solution

Question 10

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

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.

μ₁ < μ₂

μ₁ > μ₂

μ₁ = μ₂

$μ_{1} = \sqrt{μ_{2}}$

F₁/F₂

F₂/F₁

$\frac{F_{F} F_{2}}{F_{1} + F_{2}}$

$\frac{F_{1} + F_{2}}{F_{1} F_{2}}$

(11/10)²

(9/10)²