The combined resistance R of two resistors R1 and R2 (R1 , R

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 Multiple Choice QuestionsShort Answer Type

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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.


Here,   1 over straight R space equals space 1 over straight R subscript 1 plus 1 over straight R subscript 2 space space space space space space space space space space rightwards double arrow space space space space 1 over straight R space equals space fraction numerator straight R subscript 1 plus straight R subscript 2 over denominator straight R subscript 1 space straight R subscript 2 end fraction

  rightwards double arrow space space space space 1 over straight R space equals space fraction numerator straight C over denominator straight R subscript 1 left parenthesis straight C minus straight R subscript 1 right parenthesis end fraction                                                     open square brackets because space straight R subscript 1 plus straight R subscript 2 space equals space straight C close square brackets

rightwards double arrow space space space space space straight R space equals space fraction numerator straight R subscript 1 space left parenthesis straight C minus straight R subscript 1 right parenthesis over denominator straight C end fraction space space space space rightwards double arrow space space space straight R space equals space 1 over straight C left square bracket CR subscript 1 minus space straight R subscript 1 squared right square bracket space space rightwards double arrow space space dR over dR subscript 1 space equals space 1 over straight C left square bracket straight C minus 2 straight R subscript 1 right square bracket
space space space dR over dR subscript 1 space equals 0 space space space space space space space space space space space space space rightwards double arrow space space space space 1 over straight C left square bracket straight C minus 2 straight R subscript 1 right square bracket space equals space 0 space space space space space space space rightwards double arrow space space space straight R subscript 1 space space equals straight C over 2
space space space space space space space space fraction numerator straight d squared straight R over denominator dR subscript 1 squared end fraction equals space 1 over straight C left square bracket negative 2 right square bracket space equals space minus 2 over straight C

At   straight R subscript 1 space equals space straight C over 2 fraction numerator straight d squared straight R over denominator dR subscript 1 squared end fraction space equals space minus 2 over straight C less than 0 space space space space space space rightwards double arrow space space space straight R space has space straight a space local space maximum space at space straight R subscript 1 space equals space straight C over 2

But straight R subscript 1 equals space straight C over 2 space is space the space only space extreme space point. space
therefore space space space straight R space is space maximum space when space straight R subscript 1 space equals space straight C over 2 comma space space straight R subscript 2 space equals space straight C minus straight C over 2 space equals straight C over 2 space straight i. straight e. space when space straight R subscript 1 space equals space straight R subscript 2

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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.

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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

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.

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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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