Water is running into a conical vessel, 15 cm deep and 5 cm in r

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 Multiple Choice QuestionsLong Answer Type

41. A point source of light along a straight road is at a height of ‘a’ metres. A boy ‘b’ metres in height is walking along the road. How fast is his shadow increasing if he is walking away from the light at the rate of c metres per minute?
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42.

Water is dripping out from a conical funnel, at the uniform rate of 2 cc/sec through a tiny hole at the vertex of the funnel. When the slant height of water is 5 cm, find the rate of decrease of the slant height of the water.

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43. Water is running into a conical vessel, 15 cm deep and 5 cm in radius, at the rate of 0.1 cm2/sec. When the water is 6 cm deep, find at what rate is
(i) the water level rising?
(ii) the water surface area increasing?
(iii) the wetted surface of the vessel increasing?


Let V be the volume of the water in the cone i.e. the volume of the water cone CA 'B' at any time t.
Let CO' = h, O' A' = r and CA' = I.
Let α be the semi-vertical angle of the cone. CAB where CO = 15 cm, OA = 5 cm
CO = 15 cm
OA = 5 cm
Then,      tan space straight alpha space equals space OA over CO space equals space 5 over 15 space equals space 1 third space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

 Also,         tan space straight alpha space equals space fraction numerator straight O apostrophe straight A apostrophe over denominator CO apostrophe end fraction space equals space straight r over straight h                              ...(2)
From (1) and (2), we get,
       therefore space space space space space space space space space space space space space space space space space space space space space space space space 1 third space equals space straight r over straight h

rightwards double arrow space space space space space space space space space 3 straight r space equals space straight h                                                        ...(3)
Now,      straight V space equals space 1 third πr squared straight h space equals space space 1 third straight pi open parentheses straight h over 3 close parentheses squared straight h space equals space straight pi over 27 straight h cubed             
therefore space space space space space space space dV over dt space equals space fraction numerator 3 straight pi over denominator 27 end fraction straight h squared space dh over dt
rightwards double arrow space space space space space space 0.1 space equals space fraction numerator 3 straight pi over denominator 27 end fraction straight h squared dh over dt space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space dV over dt space equals space 0.1 space cm cubed divided by sec space left parenthesis given right parenthesis close square brackets
rightwards double arrow space space space space space space space dh over dt space equals space fraction numerator 2.7 over denominator 3 πh squared end fraction
rightwards double arrow space space space space space space space space space open parentheses dh over dt close parentheses subscript straight h equals 6 end subscript space equals space fraction numerator 2.7 over denominator 3 space straight pi space left parenthesis 36 right parenthesis end fraction space equals space fraction numerator 1 over denominator 40 space straight pi end fraction
therefore space space space space space the space water space level space is space rising space at space the space rate space of space fraction numerator 1 over denominator 40 space straight pi end fraction cm divided by sec.
(ii) Let A be the water surface area at any time t. Then, A = πr squared 
therefore space space space space space space straight A space equals space straight pi straight h squared over 9 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space of space left parenthesis 3 right parenthesis close square brackets
rightwards double arrow space space space space space dA over dt space equals space fraction numerator 2 πh over denominator 9 end fraction space dh over dt
When space straight h space equals space 6 comma space space space space dh over dt space equals space fraction numerator 1 over denominator 40 space straight pi end fraction comma space space we space have
space space space space space space space space dA over dt space equals space fraction numerator 2 straight pi cross times 6 over denominator 9 end fraction cross times space fraction numerator 1 over denominator 40 space straight pi end fraction space equals space 1 over 30 cm squared divided by sec
therefore space space space space the space water space surface space area space is space increasing space at space the space rate space of space 1 over 30 cm squared divided by sec.
(iii) Let S be the wetted surface area of the vessel at any time t. Then. S = πrl.
Now,       straight l squared space equals space CA to the power of apostrophe squared end exponent space equals space CO to the power of apostrophe squared end exponent space plus space straight O apostrophe straight A to the power of apostrophe 2 end exponent
space rightwards double arrow space space space space straight l squared space equals space straight h squared plus straight r squared
space rightwards double arrow space space space space straight l squared space equals space straight h squared plus straight h squared over 9 space equals space fraction numerator 10 space straight h squared over denominator 9 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 3 right parenthesis close square brackets
space rightwards double arrow space space space space space straight l space equals space fraction numerator square root of 10 space straight h over denominator 3 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 4 right parenthesis
therefore space space space space space space straight S space equals space πrl
rightwards double arrow space space space space space straight S space equals space straight pi open parentheses straight h over 3 close parentheses space open parentheses fraction numerator square root of 10 space straight h over denominator 3 end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 4 right parenthesis close square brackets
rightwards double arrow space space space space space space straight S space equals space straight pi over 9 square root of 10 space straight h squared
rightwards double arrow space space space space space space dS over dt space equals space fraction numerator 2 straight pi square root of 10 straight h over denominator 9 end fraction dh over dt
Since space straight h space equals 6 comma space space space dh over dt space equals space fraction numerator 1 over denominator 40 space straight pi end fraction
therefore space space space space space space space dS over dt space equals space fraction numerator 2 space straight pi space square root of 10 over denominator 9 end fraction cross times 6 cross times fraction numerator 1 over denominator 40 space straight pi end fraction space equals space fraction numerator square root of 10 over denominator 30 end fraction space cm squared divided by sec.
therefore space space space the space wetted space surface space area space of space the space vessel space is space increasing space at space the space rate space of space fraction numerator square root of 10 over denominator 30 end fraction cm squared divided by sec.

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44.

An inverted cone has a depth of 10 cm and a base of radius 5 cm. Water is poured into it at the rate of 3 over 2 straight c. straight c. space per space minute. Find the rate at which the level of water in the cone is rising when the depth is 4 cm.

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45. Water is running out of a conical funnel at the rate of 5 cm3/sec. If the radius of the base of the funnel is 10 cm and the altitude is 20 cm, find the rate at which the water level is dropping when it is 5 cm from the top.
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 Multiple Choice QuestionsMultiple Choice Questions

46. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
  • 10 straight pi
  • 12 straight pi
  • 8 straight pi
  • 8 straight pi
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47.

The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x 2 + 36x + 5.

  • 116
  • 96

  • 90

  • 90

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48.

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of 

  • m3/h

  • 0 · 1 m3/h
  • 1 · 1 m3/h
  • 1 · 1 m3/h
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 Multiple Choice QuestionsShort Answer Type

49. Prove that the tangents to the curve Y = x2 – 5x + 6 at the points (2, 0) and (3, 0) are at right angles.
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50.

Show that the tangent to the curve y = 7x3 + 11 at the points where x = 2 and x = – 2 are parallel.

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