Question 7 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at March 22, 2023 by Teachoo
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:
Question 7
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 𝐬𝐢𝐧𝜽
cos〖 𝜃=(𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝑺𝒊𝒅𝒆)/𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 " " 〗
cos〖 𝜃=𝑂𝐵/𝐴𝐵 " " 〗
cos〖 𝜃=𝑦/500 " " 〗
𝒚=𝟓𝟎𝟎 𝒄𝒐𝒔𝜽
Differentiating 𝒙 w.r.t. 𝒕
𝑥=500 sin𝜃
𝒅𝒙/𝒅𝒕 = 500 × (𝑑(sin𝜃))/𝑑𝑡
𝑑𝑥/𝑑𝑡 = 500 × (𝒅(𝒔𝒊𝒏𝜽))/𝒅𝜽 × 𝑑𝜃/𝑑𝑡
𝒅𝒙/𝒅𝒕 = 500 × 𝑐𝑜𝑠𝜃 × 𝑑𝜃/𝑑𝑡
Putting 𝒅𝒙/𝒅𝒕 = -10 cm/sec
−10 = 500 cos𝜃 × 𝑑𝜃/𝑑𝑡
(−10)/(500 𝑐𝑜𝑠𝜃 )=𝑑𝜃/𝑑𝑡
𝑑𝜃/𝑑𝑡= (−10)/(500 cos𝜃 )
𝑑𝜃/𝑑𝑡= (−1)/(50 cos𝜃 )
Putting 𝐜𝐨𝐬𝜽=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)
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.
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