A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm^{3} /min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is

  • Option 1)

    \frac{1}{18\pi }cm/min

  • Option 2)

    \frac{1}{36\pi }cm/min

  • Option 3)

    \frac{5}{6\pi }cm/min

  • Option 4)

    \frac{1}{54\pi }cm/min

 

Answers (1)

As we learnt in 

Rate Measurement -

Rate of any of variable with respect to time is rate of measurement. Means according to small change in time how much other factors change is rate measurement:

\Rightarrow \frac{dx}{dt},\:\frac{dy}{dt},\:\frac{dR}{dt},(linear),\:\frac{da}{dt}


\Rightarrow \frac{dS}{dt},\:\frac{dA}{dt}(Area)


\Rightarrow \frac{dV}{dt}(Volume)


\Rightarrow \frac{dV}{V}\times 100(percentage\:change\:in\:volume)

- wherein

Where dR / dt  means Rate of change of radius.

 

 V=\frac{4}{3}\pi \left [ (R+x)^{3} -R^{3} \right ]

\frac{dv}{dt}=\frac{4}{3}\pi \left [ 3(R+x)^{2} \times \frac{dx}{dt}\right ]

50=\frac{4}{3}\pi \left [ 3\times (15)^{2} \right ]\times \frac{dx}{dt}

\frac{50\times 3}{4\pi \times 3\times 225}=\frac{dx}{dt}

=\frac{1}{18\pi }\ cm/min


Option 1)

\frac{1}{18\pi }cm/min

This option is correct

Option 2)

\frac{1}{36\pi }cm/min

This option is incorrect

Option 3)

\frac{5}{6\pi }cm/min

This option is incorrect

Option 4)

\frac{1}{54\pi }cm/min

This option is incorrect

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