Examples

Example 1

Example 2

Example 3

Example 4 Important

Example 5

Example 6

Example 7

Example 8 Important

Example 9 Important

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14

Example 15

Example 16 Important

Example 17

Example 18 Important

Example 19

Example 20 Important

Example 21 Important

Example 22

Example 23 Important

Example 24

Example 25 Important

Example 26 Important

Example 27

Example 28 Important

Example 29 Important

Example 30 Important

Example 31 Important You are here

Example 32 Important

Example 33 Important

Example 34 Important

Example 35

Example 36 Important

Example 37

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Deleted for CBSE Board 2025 Exams

Question 4 Important Deleted for CBSE Board 2025 Exams

Question 5 Deleted for CBSE Board 2025 Exams

Question 6 Deleted for CBSE Board 2025 Exams

Question 7 Deleted for CBSE Board 2025 Exams

Question 8 Deleted for CBSE Board 2025 Exams

Question 9 Deleted for CBSE Board 2025 Exams

Question 10 Deleted for CBSE Board 2025 Exams

Question 11 Deleted for CBSE Board 2025 Exams

Question 12 Deleted for CBSE Board 2025 Exams

Question 13 Important Deleted for CBSE Board 2025 Exams

Question 14 Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

Example 31 A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is tanβ1 (0.5). Water is poured into it at a constant rate of 5 cubic meter per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 4 m.Water tank is in shape of cone Let r be the radius of cone, h be the height of cone, & πΆ be the semiβvertical angle Given Semi-vertical angle is tan^(β1)β‘(0.5) Ξ±=tan^(β1)β‘(0.5) tan Ξ± =(0.5) π/β = 0.5 π/β = 1/2 π = π/π Also, Water is poured at a constant rate of 5 cubic meter per hour π π½/π π=π π^π/ππ Where V is volume of cone Now, π=1/3 ππ^2 β π=1/3 π(π/π)^2 β π=1/3 π(γh/4γ^2 )β π½=(π π^π)/ππ Differentiate w.r.t. t ππ/ππ‘=π((πβ^3)/12)/ππ‘ ππ/ππ‘=π/12 . (πβ^3)/ππ‘ ππ/ππ‘=π/12 . (πβ^3)/πβ .πβ/ππ‘ ππ/ππ‘=π/12. 3β^2 . πβ/ππ‘ ππ/ππ‘=(πβ^2)/4 . πβ/ππ‘ Putting π π½/π π=π 5 =(πβ^2)/4 . πβ/ππ‘ π π/π π=ππ/(π π^π ) We need to find, Rate at which level of water is rising when depth is 4 m i.e. β π π/π πβ€|_(π = π π) Putting h = 4 m in (3) β πβ/ππ‘β€|_(β = 4π)=20/(π(β)^2 )=20/16 Γ 1/π=5/4 Γ 1/(22/7)=5/4 Γ 7/22=35/88 Hence, rate of change of water level is ππ/ππ m/hr.