The radius of a spherical soap bubble is increasing at the rate

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 Multiple Choice QuestionsShort Answer Type

21. The surface area of a spherical bubble is increasing at the rate of 2 cm2/sec. Find the rate at which the volume of the bubble is increasing at the instant its radius is 6 cm.
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22.

The volume of a spherical balloon is increasing at the rate of 25 cm3/sec. Find the rate of change of its surface area at the instant when its radius is 5 cm.

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 Multiple Choice QuestionsLong Answer Type

23. The radius of a spherical soap bubble is increasing at the rate of 0.3 cms–1. Find the rate of change of its (i) volume (ii) surface area when the radius is 8 cm. 
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24. The radius of a spherical soap bubble is increasing at the rate of 0.2 cms–1. Find the rate of change of its (i) volume (ii) surface area, when the radius is 4 cm.
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25. The radius of a spherical soap bubble is increasing at the rate of 0.4 cms–1 Find the rate of change of its (i) volume (ii) surface area, when the radius is 10 cm.


Let r be radius of spherical soap bubble.
therefore space space space space space dr over dt space equals space 0.4 space space cm space straight s to the power of negative 1 end exponent 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 1 right parenthesis
left parenthesis straight i right parenthesis space space space space space space space space space space space space space straight V space equals space fraction numerator 4 straight pi over denominator 3 end fraction straight r cubed space space where space straight V space is space volume.

therefore space space space space space space dV over dt space equals space fraction numerator 4 straight pi over denominator 3 end fraction cross times 3 straight r squared dr over dt space equals 4 πr squared space cross times space 0.4 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 1 right parenthesis close square brackets
space space space space space space space space space space space space space space space space space space space space equals space 1.6 space πr squared space
When space straight r space equals space 10 comma space space space dV over dt space equals space 1.6 space straight pi space cross times space 100 space equals space 160 space straight pi space cubic space cm divided by sec

(ii) Again,          straight S space equals space 4 πr squared comma space space space space where space straight S space is space surface space area.

therefore space space space space space space space space space space space space space dS over dt space equals space 8 πr dr over dt space equals space 8 space πr space cross times 0.4 space equals space 3.2 space πr space space space space space space space space space space space space space
when space straight r space equals 10 comma space space space space space dS over dt space equals space 3.2 space straight pi space cross times space 10 space equals space 32 space straight pi space sq. space cm divided by sec.
           


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 Multiple Choice QuestionsShort Answer Type

26.

A ballon which always remains spherical on inflation,  is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate of which the radius of the balloon is increasing when the radius is 15 cm.

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27.

The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 10 centimeters?

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28.

The volume of a cube is increasing at a rate of 7 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 12 centimeters?

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29. The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?
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30.

The length x of a rectangle is decreasing at the rate of 2 cm/s and the width y is increasing at the rate of 2 cm/s. When x = 12 cm and y = 5 cm, find the rate of change of (a) the perimeter and (b) the area of the rectangle.

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