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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

 

This question is similar to Ex 6.1, 10 - Chapter 6 Class 12 - Application of Derivatives

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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 = π‘₯ cm, OB = 𝑦 cm & βˆ π΄π΅π‘‚= πœƒ Given that Top of the ladder slides downwards at the rate of 10 cm/sec i.e. 𝒅𝒙/𝒅𝒕 = βˆ’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 π’…πœƒ" " /𝒅𝒕 when π’š = 2 m or π’š=𝟐𝟎𝟎 π’„π’Ž. Now, We need to find π’…πœ½/𝒅𝒕 , so for that we will differentiate π‘₯ w.r.t. 𝑑, since value of 𝒅𝒙/𝒅𝒕 is given. sin⁑〖 πœƒ=(π‘Άπ’‘π’‘π’π’”π’Šπ’•π’”π’† π‘Ίπ’Šπ’…π’†)/π‘―π’šπ’‘π’π’•π’†π’π’–π’”π’† " " γ€— sin⁑〖 πœƒ=𝑂𝐴/𝐴𝐡 " " γ€— sin⁑〖 πœƒ=π‘₯/500 " " γ€— 𝒙=πŸ“πŸŽπŸŽ 𝐬𝐒𝐧⁑𝜽 cos⁑〖 πœƒ=(𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 π‘Ίπ’Šπ’…π’†)/π‘―π’šπ’‘π’π’•π’†π’π’–π’”π’† " " γ€— cos⁑〖 πœƒ=𝑂𝐡/𝐴𝐡 " " γ€— cos⁑〖 πœƒ=𝑦/500 " " γ€— π’š=πŸ“πŸŽπŸŽ π’„π’π’”β‘πœ½ Differentiating 𝒙 w.r.t. 𝒕 π‘₯=500 sinβ‘πœƒ 𝒅𝒙/𝒅𝒕 = 500 Γ— (𝑑(sinβ‘πœƒ))/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = 500 Γ— (𝒅(π’”π’Šπ’β‘πœ½))/π’…πœ½ Γ— π‘‘πœƒ/𝑑𝑑 𝒅𝒙/𝒅𝒕 = 500 Γ— π‘π‘œπ‘ β‘πœƒ Γ— π‘‘πœƒ/𝑑𝑑 Putting 𝒅𝒙/𝒅𝒕 = -10 cm/sec βˆ’10 = 500 cosβ‘πœƒ Γ— π‘‘πœƒ/𝑑𝑑 (βˆ’10)/(500 π‘π‘œπ‘ β‘πœƒ )=π‘‘πœƒ/𝑑𝑑 π‘‘πœƒ/𝑑𝑑= (βˆ’10)/(500 cosβ‘πœƒ ) π‘‘πœƒ/𝑑𝑑= (βˆ’1)/(50 cosβ‘πœƒ ) Putting 𝐜𝐨𝐬⁑𝜽=π’š/πŸ“πŸŽπŸŽ from equation (2) π’…πœ½/𝒅𝒕= (βˆ’1)/(50 Γ— π’š/πŸ“πŸŽπŸŽ) π’…πœ½/𝒅𝒕 = (βˆ’1)/( 𝑦/10) π’…πœ½/𝒅𝒕= (βˆ’πŸπŸŽ)/( π’š) Putting π’š=𝟐𝟎𝟎 β”œ π’…πœ½/𝒅𝒕─|_(π’š =𝟐𝟎𝟎) " "=(βˆ’10)/200 β”œ π‘‘πœƒ/𝑑𝑑─|_(𝑦 =200) " "=(βˆ’πŸ)/𝟐𝟎 Hence, angle between ladder and the ground is decreasing at rate of 𝟏/𝟐𝟎 radian/sec. So, the correct answer is (B)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.