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MathsMediumMCQ2019 · 09 Jan Shift 1

Q81.The maximum volume in 𝑐𝑢. 𝑚 of the right circular cone having slant height 3 𝑚 is: JEE Main 2019 (09 Jan Shift 1) JEE Main Previous Year Paper (1) 2√3 𝜋 (2) 3√3 𝜋 4 (3) 6 𝜋 (4) 3𝜋

What This Question Tests

This question requires finding the maximum volume of a cone given its slant height by expressing volume as a function of one variable and using differentiation for optimization.

Concepts Tested

Volume of a coneRelation between slant height, radius and heightDifferentiation for optimization

Formulas Used

V = (1/3)πr²h

l² = r² + h²

dV/dh = 0 for extremum

📚 NCERT Sections This Tests

9.1A Small Candle, 2.5 Cm In Size Is Placed At 27 Cm In Front Of A Concave

Physics Class 12 · Chapter 9

70% match

9.1 A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

9.5A Small Bulb Is Placed At The Bottom Of A Tank Containing Water To A

Physics Class 12 · Chapter 9

70% match

9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

9.17(A) Sin I¢C = 1.44/1.68 Which Gives I¢C = 59°. Total Internal Reflection

Physics Class 12 · Chapter 9

69% match

9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.