CBSE Class 12 Sample Paper for 2025 Boards
Question 2
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Question 4
Question 5 Important
Question 6
Question 7
Question 8 Important
Question 9 Important
Question 10 Important
Question 11
Question 12 Important
Question 13 Important
Question 14
Question 15 Important
Question 16 Important
Question 17
Question 18
Question 19 [Assertion Reasoning] Important
Question 20 [Assertion Reasoning] Important
Question 21 Important
Question 22
Question 23 (A)
Question 23 (B)
Question 24 (A)
Question 24 (B) Important
Question 25 Important
Question 26 Important You are here
Question 27 Important
Question 28 (A)
Question 28 (B)
Question 29 (A) Important
Question 29 (B)
Question 30 Important
Question 31 (A) Important
Question 31 (B)
Question 32 Important
Question 33 Important
Question 34 (A)
Question 34 (B)
Question 35 (A)
Question 35 (B) Important
Question 36 (i) [Case Based]
Question 36 (ii)
Question 36 (iii) (A) Important
Question 36 (iii) (B) Important
Question 37 (i) [Case Based]
Question 37 (ii) Important
Question 37 (iii) (A) Important
Question 37 (iii) (B)
Question 38 (i) [Case Based] Important
Question 38 (ii) Important
CBSE Class 12 Sample Paper for 2025 Boards
Last updated at Dec. 13, 2024 by Teachoo
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Question 26 A kite is flying at a height of 3 metres and 5 metres of string is out. If the kite is moving away horizontally at the rate of 200" " cm/s, find the rate at which the string is being released.Let AB be the string of kite & OA be the height of kite & OA be the ground. Given Height of kite is 5 m OA = 3 cm Since kite is moving horizontally, Height is fixed, And string length and horizontal distance is variable Let OB = 𝑥 cm & OB = 𝑦 cm Let OB = 𝑥 cm & AB = 𝑦 cm Given that kite is moving away horizontally at the rate of 200" " cm/s, i.e. 𝒅𝒙/𝒅𝒕 = 200 cm/sec We need to calculate rate at which the string is being released when 5 metres of string is out i.e. We need to calculate 𝒅𝒚/𝒅𝒕 when 𝒚 = 5 cm. Since ∆AOB is a right angled triangle Using Pythagoras theorem (AB)2 = (OB)2 + (OA)2 𝑦2= 𝑥2+3^2 𝒚𝟐= 𝒙𝟐+𝟗 Now, Differentiating w.r.t time (𝑑(𝑦2))/𝑑𝑡 =(𝑑(𝑥2))/𝑑𝑡+(𝑑(9))/𝑑𝑡 (𝑑 (𝑦2))/𝑑𝑡 = (𝑑(𝑥2))/𝑑𝑡+0 (𝑑(𝑦2))/𝑑𝑡 × 𝑑𝑦/𝑑𝑦 = (𝑑(𝑥2))/𝑑𝑡 × 𝑑𝑥/𝑑𝑥 (𝑑(𝑦2))/𝑑𝑦 × 𝑑𝑦/𝑑𝑡 = (𝑑(𝑥2))/𝑑𝑥 × 𝑑𝑥/𝑑𝑡 2𝑦 × 𝒅𝒚/𝒅𝒕 = 2𝑥 × 𝑑𝑥/𝑑𝑡 𝒚 × 𝒅𝒚/𝒅𝒕 = 𝒙 × 𝒅𝒙/𝒅𝒕 Now, 𝒅𝒙/𝒅𝒕 = 200, y = 5, but we do not know x Finding x when y = 5 From (1) 𝑦2= 𝑥2+9 52= 𝑥2+9 25 = 𝑥2+9 25 – 9 = 𝑥2 16 = 𝑥2 𝑥2=16 𝑥2=4^2 𝒙=𝟒 Putting x = 4, 𝒅𝒙/𝒅𝒕 = 200, y = 5 in equation (2) 𝑦 × 𝑑𝑦/𝑑𝑡 = 𝑥 × 𝑑𝑥/𝑑𝑡 5 × 𝒅𝒚/𝒅𝒕 = 4 × 200 𝑑𝑦/𝑑𝑡 = 4 × 200 × 1/5 𝑑𝑦/𝑑𝑡 = 4 × 40 𝒅𝒚/𝒅𝒕 = 160 Thus, rate at which the string is being released is 160 cm/s