Question 2 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives

Transcript

Question 2 A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is: (A) 1/10 radian/sec (B) 1/20 radian/sec (C) 20 radian/sec (D) 10 radian/sec Let AB be the ladder & OA be the wall & OB be the ground. Given Length of ladder is 5 m AB = 5 m AB = 500 cm Let OA = 𝑥 cm, OB = 𝑦 cm & ∠𝐴𝐵𝑂= 𝜃 Given that Top of the ladder slides downwards at the rate of 10 cm/sec i.e. 𝒅𝒙/𝒅𝒕 = −10 cm/sec We need to calculate at which the rate the angle between the floor and the ladder decreases when lower end of ladder is 2 metres from the wall. i.e. We need to calculate 𝒅𝜃" " /𝒅𝒕 when 𝒚 = 2 m or 𝒚=𝟐𝟎𝟎 𝒄𝒎. Now, We need to find 𝒅𝜽/𝒅𝒕 , so for that we will differentiate 𝑥 w.r.t. 𝑡, since value of 𝒅𝒙/𝒅𝒕 is given. sin⁡〖 𝜃=(𝑶𝒑𝒑𝒐𝒔𝒊𝒕𝒔𝒆 𝑺𝒊𝒅𝒆)/𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 " " 〗 sin⁡〖 𝜃=𝑂𝐴/𝐴𝐵 " " 〗 sin⁡〖 𝜃=𝑥/500 " " 〗 𝒙=𝟓𝟎𝟎 𝐬𝐢𝐧⁡𝜽 cos⁡〖 𝜃=(𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝑺𝒊𝒅𝒆)/𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 " " 〗 cos⁡〖 𝜃=𝑂𝐵/𝐴𝐵 " " 〗 cos⁡〖 𝜃=𝑦/500 " " 〗 𝒚=𝟓𝟎𝟎 𝒄𝒐𝒔⁡𝜽 Differentiating 𝒙 w.r.t. 𝒕 𝑥=500 sin⁡𝜃 𝒅𝒙/𝒅𝒕 = 500 × (𝑑(sin⁡𝜃))/𝑑𝑡 𝑑𝑥/𝑑𝑡 = 500 × (𝒅(𝒔𝒊𝒏⁡𝜽))/𝒅𝜽 × 𝑑𝜃/𝑑𝑡 𝒅𝒙/𝒅𝒕 = 500 × 𝑐𝑜𝑠⁡𝜃 × 𝑑𝜃/𝑑𝑡 Putting 𝒅𝒙/𝒅𝒕 = -10 cm/sec −10 = 500 cos⁡𝜃 × 𝑑𝜃/𝑑𝑡 (−10)/(500 𝑐𝑜𝑠⁡𝜃 )=𝑑𝜃/𝑑𝑡 𝑑𝜃/𝑑𝑡= (−10)/(500 cos⁡𝜃 ) 𝑑𝜃/𝑑𝑡= (−1)/(50 cos⁡𝜃 ) Putting 𝐜𝐨𝐬⁡𝜽=𝒚/𝟓𝟎𝟎 from equation (2) 𝒅𝜽/𝒅𝒕= (−1)/(50 × 𝒚/𝟓𝟎𝟎) 𝒅𝜽/𝒅𝒕 = (−1)/( 𝑦/10) 𝒅𝜽/𝒅𝒕= (−𝟏𝟎)/( 𝒚) Putting 𝒚=𝟐𝟎𝟎 ├ 𝒅𝜽/𝒅𝒕┤|_(𝒚 =𝟐𝟎𝟎) " "=(−10)/200 ├ 𝑑𝜃/𝑑𝑡┤|_(𝑦 =200) " "=(−𝟏)/𝟐𝟎 Hence, angle between ladder and the ground is decreasing at rate of 𝟏/𝟐𝟎 radian/sec. So, the correct answer is (B)