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MathsMediumMCQ2022 · 25 Jul Shift 1

Q68.A tower 𝑃𝑄 stands on a horizontal ground with base 𝑄 on the ground. The point 𝑅 divides the tower in two parts such that 𝑄𝑅= 15m. If from a point 𝐴 on the ground the angle of elevation of 𝑅 is 60° and the part 𝑃𝑅 of the tower subtends an angle of 15° at 𝐴, then the height of the tower is (1) 52√3 + 3m (2) 5√3 + 3m (3) 10√3 + 1m (4) 102√3 + 1m

What This Question Tests

This problem involves solving a classic height and distance scenario using angles of elevation and trigonometric ratios, requiring careful setup of geometric relations and possibly using trigonometric identities.

Concepts Tested

Angle of elevationTrigonometric ratios (tan)Geometric properties of angles

Formulas Used

tan θ = opposite/adjacent

Compound angle formula for tan

📚 NCERT Sections This Tests

4.4A Horizontal Overhead Power Line Carries A Current Of 90 A In East To

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4.4 A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

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Physics Class 12 · Chapter 9

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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

9.21At What Angle Should A Ray Of Light Be Incident On The Face Of A Prism

Physics Class 12 · Chapter 9

70% match

9.21 At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

📋 Question Details

Chapter
Trigonometric Functions & Equations
Topic
Heights and Distances
Year
2022
Shift
25 Jul Shift 1
Q Number
Q68
Type
MCQ
NCERT Ref
Class 10 Mathematics Ch 9: Some Applications of Trigonometry

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