A rectangle is inscribed in a semi-circle of radius r with one

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 Multiple Choice QuestionsShort Answer Type

321.

The combined resistance R of two resistors R1 and R2 (R1 , R2 > 0) is given by 1 over straight R space equals space 1 over straight R subscript 1 plus 1 over straight R subscript 2.
If R1 + R2 = C (a constant), show that the maximum resistance R is obtained by choosing R1 = R2.

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 Multiple Choice QuestionsLong Answer Type

322.

Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius is a square of side straight a square root of 2.

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323.

Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.

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 Multiple Choice QuestionsShort Answer Type

324.

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic
centimetres, find the dimensions of the can which has the minimum surface
area?
Or
Of all the closed cylindrical tin cans (right circular) which enclose a given volume of 100 cubic cm., which has the minimum surface area?

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325.

Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

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326.

Show that a cylinder of given volume open at the top has minimum total surface area provided its height is equal to the radius of its base.

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327.

Show that the height of the cylinder, open at the top, of given surface area and greatest volume is equal to the radius of its base.

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 Multiple Choice QuestionsLong Answer Type

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328.

A rectangle is inscribed in a semi-circle of radius r with one of its sides on the diameter of the semi-circle. Find the dimensions of the rectangle so that its area is maximum Find also this area.


Let PQRS be the rectangle inscribed in the semi-circle of radius r so that OR = r, where O in centre of circle.
Let PO = OQ =  x and QR = y so that sides of rectangle are of lengths 2x and y respectively.
   Let  Syntax error from line 1 column 49 to line 1 column 152. Unexpected '<mlongdiv '.


           In increment OQR comma

                 straight x over straight r space equals cos space straight theta 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 rightwards double arrow space space straight x space equals straight r space cos space straight theta

and        straight y over straight r space equals space sin space straight theta space space space space space space space space space space space rightwards double arrow space space space straight y space equals space straight r space sin space straight theta
Let A be area of rectangle PQRS.
 therefore space space space space space space straight A space equals space left parenthesis PQ right parenthesis thin space left parenthesis QR right parenthesis space equals space left parenthesis 2 space straight x right parenthesis thin space left parenthesis straight y right parenthesis space equals space left parenthesis 2 straight r space cos space straight theta right parenthesis thin space left parenthesis straight r space sin space straight theta right parenthesis
space space space space space space space space space space space space space space space equals straight r squared space left parenthesis 2 space sin space straight theta space cos space straight theta right parenthesis space equals space straight r squared space sin space 2 straight theta
therefore space space space space space dA over dθ space equals space 2 straight r squared space cos space 2 straight theta
For A to be maximum or minimum
                 dA over dθ space equals space 0 space space space space space space space rightwards double arrow space space 2 straight r squared space cos space 2 space straight theta space equals space 0 space space space space space space rightwards double arrow space space space cos space 2 straight theta space equals space 0

rightwards double arrow space space space 2 straight theta space equals space straight pi over 2 space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight theta space equals space straight pi over 4
Also comma space space space fraction numerator straight d squared straight A over denominator dθ squared end fraction space equals space minus 4 straight r squared space sin space 2 straight theta When space straight theta space equals space straight pi over 4 comma space space fraction numerator straight d squared straight A over denominator dθ squared end fraction space equals space minus 4 straight r squared sin straight pi over 2 space equals space minus 4 straight r squared cross times 1 space equals space minus 4 straight r squared less than 0
therefore space space space space straight A space is space maximum space when space straight theta space equals space straight pi over 4
therefore space space space straight x space equals space straight r space cos space straight pi over 4 space equals fraction numerator straight r over denominator square root of 2 end fraction comma space space space straight y space equals space straight r space sin straight pi over 4 space equals space fraction numerator straight r over denominator square root of 2 end fraction
therefore space space space sides space are space square root of 2 space straight r comma space space space space fraction numerator straight r over denominator square root of 2 end fraction space and space area space space equals space left parenthesis square root of 2 straight r right parenthesis space open parentheses fraction numerator straight r over denominator square root of 2 end fraction close parentheses space equals straight r squared space sq. space units. space

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329.

Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.

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 Multiple Choice QuestionsShort Answer Type

330. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water, show that the cost of the material will be least when the depth of the tank is half of its width.
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