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  • Help me solve this - Limit , continuity and differentiability - JEE Main-13

A cone of having semivertical angle \frac{\pi}{}4 is such that its radius is increasing at the rate of  2cm/sec, then the rate  of increase of its volume when the radius is \frac{1}{\sqrt{\pi}} cm will be ?

  • Option 1)

    2 cm3 /sec

  • Option 2)

    4 cm3 /sec

  • Option 3)

    6 cm3 /sec

  • Option 4)

    8 cm3 /sec

 

Answers (1)

best_answer

As we have learned

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.

 

 

 

\tan 45\degree= r/h \Rightarrow r= h

\therefore volume = 1/3\pi r^{2}h= 1/3\pi r^{3}\Rightarrow \frac{dv}{dt}= 1/3\pi \times 3r^{2}\frac{dr}{dt}= \pi r^{2}\frac{dr}{dt}

\therefore dv/dt .. at.. r= 1/\sqrt\pi cm   will be dv/dt=\pi \times 1/\sqrt\pi\times 2=2

dv/dt=2cm^{3}/sec

 

 

 

 

 

 


Option 1)

2 cm3 /sec

Option 2)

4 cm3 /sec

Option 3)

6 cm3 /sec

Option 4)

8 cm3 /sec

Posted by

Himanshu

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