Differential Equation

Study Material (PDF)
Self Assesment
Study Material (PDF)
Self Assesment

Click on each question to see answer & solution.

Question No: #181

The rate of increment of surface area of a spherical bubble is `2\ cm^2`/`s`. Find the rate of increment of its volume when radius of the bubble is 6 cm.

Question No: #182

The rate of change of radius of a sphere is `\frac{1}{2\pi}` with respect to time. Find the rate of change of area of its outer surface when radius is 5 cm.

Question No: #183

A balloon which always remains spherical on inflation, is being inflated by pumping in 40 c.c. per minute. Find the rate of change of surface area when radius is 8 c.m. Find also the increase in radius in the next `1/2` minute.

Question No: #184

A spherical snowball melts at a rate proportional to its volume at that time. If half the ice melts in 30 minutes, then prove that the volume of the remaining ice is `1/8` part of original volume after 90 minutes of start of melting ice.

Question No: #185

The rate of change of volume of a cube is constant. Prove that the rate of change of total surface area of this cube is inversely proportional to its side.

Question No: #186

The volume of a cube changes in such a way that it remains a cube after that change. Show for for a cube (whose volume is 1 cube unit), rate of change of volume = `3/2 \times` (rate of change of anyone surface area)

Question No: #187

If `\gamma` be the increase in volume per degree centigrade of a cube of unit volume and `\beta` be the increase in area per degree centigrade of each surface of the cube, then show that `2\gamma=3\beta`

Question No: #188

A right-circular conical water tank, with its vertex down and semi-vertical angle being `\pi/6`, loses water out of a circular hole at its bottom at a rate of `\pi\ cm^3//sec`. Find the rate of change of radius of water level when it is 3 cm deep.

Question No: #189

A right-circular inverted conical water tank has height 18 inch and diameter of base 10 inch. Water is poured in this tank at a rate of 4 cubic inches per minute. Find the rate of increment of surface (alt, rate of increment of height) of water when depth of water is 12 inch?

Question No: #190

Solve: `\tanx\frac{dy}{dx}=1+y^2`, given `y=1` when `x=\frac{\pi}{2}`
Ans: `\tan^{-1}y-log|sinx|=\frac{\pi}{4}`
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