SATHEE: Application of Derivatives 2 Question 1

Application of Derivatives 2 Question 1

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

(2019 Main, 10 April II)

(a) $\frac{1}{9 π}$

(b) $\frac{1}{18 π}$

(d) $\frac{5}{6 π}$

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Answer:

(b)

Solution:

Let the thickness of layer of ice is $x cm$ , the volume of spherical ball (only ice layer) is

$V = \frac{4}{3} π [(10 + x)^{3} - 10^{3}]$

On differentiating Eq. (i) w.r.t. ’ $t$ ‘, we get

$\frac{d V}{d t} = \frac{4}{3} π (3 (10 + x)^{2}) \frac{d x}{d t} = - 50$

[-ve sign indicate that volume is decreasing as time passes].

$\Rightarrow 4 π (10 + x)^{2} \frac{d x}{d t} = - 50$

At $x = 5 cm$

$\frac{d x}{d t} [4 π (10 + 5)^{2}] = - 50$

$\Rightarrow \frac{d x}{d t} = - \frac{50}{225 (4 π)} = - \frac{1}{9 (2 π)} = - \frac{1}{18 π} cm / \min$

So, the thickness of the ice decreases at the rate of $\frac{1}{18 π} cm / \min$ .

Application of Derivatives 2 Question 1

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Application of Derivatives 2 Question 1

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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