Q9. A driver in a car, approaching a vertical wall notices that the frequency of his car horn has changed from 440 Hz to 480 Hz, when it gets reflected from the wall. If the speed of sound in air is 345 m s−1, then the speed of the car is: (1) 54 km/hr (2) 36 km/hr (3) 18 km/hr (4) 24 km/hr
What This Question Tests
This problem applies the Doppler effect twice: first for sound reaching the wall (as an observer) and then for the reflected sound reaching the driver (as an observer of the wall acting as a source).
Concepts Tested
Formulas Used
f_apparent = f_source * (v_sound ± v_observer) / (v_sound ± v_source)
📚 NCERT Sections This Tests
10.1 — Monochromatic Light Of Wavelength 589 Nm Is Incident From Air On A
Physics Class 12 · Chapter 10
10.1 Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.
9.27 — (A) M = ( Fo/Fe) = 28
Physics Class 12 · Chapter 9
9.27 (a) m = ( fO/fe) = 28 f O f O (b) m = 1 + = 33.6 f e 25 349 Reprint 2025-26 Physics 9.28 (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective h h = = f O 140 Equating the two, h = 4.7 cm. (c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. 9.29 The image formed by the larger (concave) mirror acts as virtual object for the smaller (convex) mirror. Parallel rays coming from the object at infinity will focus at a distance of 110 mm from the larger mirror. The distance of virtual object for the smaller mirror = (110 –20) = 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror formula, image is formed at 315 mm from the smaller mirror. 9.30 The reflected rays get deflected by twice the angle of rotation of the mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm. 9.31 n = 1.33 CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) l = 589 nm, n = 5.09 ´ 1014 Hz, c = 3.00 ´ 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) n = 5.09 ´ 1014Hz v = (c/n) = 2.26 × 108 m s–1, l = (v/n) = 444 nm 10.2 (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar). 10.3 (a) 2.0 × 108 m s–1 (b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. [When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour.] Now we know violet colour deviates more than red in a glass prism, i.e. nv > nr. Therefore, the violet component of white light travels slower than the red component. 1.2 10 – 2 0.28 10 – 3 10.4 m = 600 nm 4 14. 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 350 10.8 tan–1(1.5) ~ 56.3o Reprint 2025-26 Answers
14.6 — Reflection Of Waves
Physics Class 12 · Chapter 14
14.6 Reflection of waves some cork pieces on the disturbed surface, it is seen that
📋 Question Details
- Chapter
- Waves & Sound
- Topic
- Doppler Effect
- Year
- 2020
- Shift
- 05 Sep Shift 2
- Q Number
- Q9
- Type
- MCQ
- NCERT Ref
- Class 11 Physics Ch 15: Waves
More from this Chapter
Q38.Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube A and B is (1) 1 : 2 (2) 1 : 4 (3) 2 : 1 (4) 4 : 1
Q39.A tuning fork arrangement (pair) produces 4 beats / sec with one fork of frequency 288cps. A little wax is placed on the unknown fork and it then produces 2 beats /sec. The frequency of the unknown fork is (1) 286 cps (2) 292 cps (3) 294 cps (4) 288 cps
Q40.A wave y = a sin(ωt −kx) on a string meets with another wave producing a node at x = 0. Then the equation of the unknown wave is (1) y = a sin(ωt + kx) (2) y = −a sin(ωt + kx) (3) y = a sin(ωt −kx) (4) y = −a sin(ωt −kx)
Q41.When temperature increases, the frequency of a tuning fork (1) increases (2) decreases (3) remains same (4) increases or decreases depending on the material