Correct Answer: B. 2.00

Explanation

1/U + 1/V = 1/f f = r/2 = 10cm; 1/15 + 1/v = 1/10 v = 30 = m = v/u = 2

Explanation

(a)(i)

(ii) Advantages of using a convex mirror as driving mirror.

– It has wide field of view

– Image is always erect.

Disadvantages;

– Image is always smaller than the object

– It gives a false impression of the distance as image seems further away.

(iii) The objective lens forms the image of an object placed beyond its focal length within the focal length of eye piece. This image is real inverted and magnified. This image serves as object to the eye piece which functions as a simple microscope to form a final virtual and magnified image.

(c) \(\lambda = \frac{v}{f} = \frac{350}{512} = 0.68m\)

1st resonance position = \(\frac{\lambda}{f} = \frac{0.68}{4} = 0.17m\)

2nd resonance position = \frac{3\lambda}{4} \frac{3 \times 0.68}{4} = 0.51m\)

3rd resonance position = \(\frac{5\lambda}{4} = \frac{5 \times 0.85}{4} = 0.85m\)

Since 0.85m is greater than the length of the jar (0.6m), therefore only the first and second resonance positions are possible.

Correct Answer: C.2.3 x \(10^{-7}\)m

Explanation

\(\theta = E - E_f = h_f = \frac{v}{\lambda} \Rightarrow = \lambda = \frac{h_v}{\theta}

\frac{6.6\times 10^{-34} \times 3 \times 10^{8}}{8.6 \times 10^{-19}} =

Correct Answer: D. Neon

Correct Answer: A. 100N

Explanation

W=\(\frac{mg}{4}= \frac{400}{4}=100N\)

Correct Answer: D. its frequency remains constant as it travels between different media

Explanation

ai. Parabolic mirrors, due to their ability to efficiently collect and focus incoming parallel light rays, are suitable for use in vehicle headlamps, ensuring a concentrated and directed beam of light for improved visibility and reduced glare.

ii.

bi. – Sonar Systems: Sonar (Sound Navigation and Ranging) systems use echoes to determine the depth and location of objects underwater. A sound signal is emitted, and the time it takes for the signal to bounce off an object and return as an echo is used to calculate the distance to the object. Sonar is commonly used in marine navigation, fishing, and underwater mapping.

- Medical Imaging: In medical ultrasound imaging, echoes are used to create images of the interior of the body. High-frequency sound waves are directed into the body, and the echoes that bounce back from different tissues are used to generate detailed images of organs, blood vessels, and developing fetuses during pregnancy.

- Architectural Design: Architects and engineers use echoes to evaluate the acoustics of buildings and design spaces with optimal sound qualities. The controlled use of echoes can enhance the sound in concert halls, auditoriums, and theaters, providing a more enjoyable listening experience for audiences.

- Geological Surveys: Seismic surveys use echoes generated by controlled explosions or mechanical sources to map subsurface geological features. By analyzing the time it takes for seismic waves to reflect off different rock layers, geologists can infer the composition and structure of the Earth's subsurface.

- Surveying and Distance Measurement: Echoes are used in surveying and distance measurement equipment like total stations and laser rangefinders. A laser or light signal is emitted, and the time it takes for the light to bounce off a target and return is used to measure distances accurately.

- Echo Sounding in Navigation: In addition to sonar, echo sounding is used in navigation to determine water depth, especially in shallow or coastal areas. A sound signal is sent to the seabed, and the time it takes for the echo to return helps sailors and navigators avoid underwater hazards.

- Musical Effects: Musicians and audio engineers often use echoes and reverb effects in music production. These effects add depth and dimension to music by simulating the reflection of sound in different environments. Echoes and reverb are commonly used in recording studios, live concerts, and digital audio effects.

ii. V = \(\frac{2d}{t}\)

Case1: V = \(\frac{2d}{1.2}\)

Case2: V = \(\frac{2(40 + d)}{1.45}\)

Equating case 1 and 2

\(\frac{2d}{1.2}\) = \(\frac{2(40 + d)}{1.45}\)

2.9d = 96 + 2.4d

2.9d - 2.4d = 96

0.5d = 96

d = \(\frac{96}{0.5}\)

= 192m

∴ Distance between P and H = 192m

put d = 192m into case(1)

V = \(\frac{2d}{1.2}\)

V = \(\frac{2\times192}{1.2}\) = 320m/s

ci. The absolute refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in the medium. It is a measure of how much a medium slows down light.

ii. Displacement = real depth - \(\frac{real depth}{refractive index }\) ( n = \(\frac{real depth}{apparentdepth }\))

= real depth( 1 - \(\frac{1}{n}\))

For liquid A, displacement,d = 8 (1 - \(\frac{1}{1.3}\)) = 9.23cm

For liquid B, displacement,d = 40( 1 - \(\frac{1}{1.4}\)) = 2.29cm

∴ Total displacement = 9.23 + 2.29 = 11.52cm

∴ the apparent position of the coin is 11.52cm from the bottom.

Correct Answer: D. The collision between the gases is perfectly inelastic

Explanation

The collision between the gases is perfectly elastic

Correct Answer: A) moon is between the sun and the earth

Correct Answer: B. 0.44m

Explanation

\(l_2 - l_1 = \alpha l_1 (\theta_2 - \theta_1) = 4.0 \times 10^{-3} \times 2(80 - 25)\)

= 0.44m