Differential Calculus jee mains questions | JEE Main Maths Quadratic Equations Previous Year Question

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Q 1 :

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm, the ice-cream melts at the rate of 81 ${cm}^{3}$ /min and the thickness of the ice-cream layer decreases at the rate of $\frac{1}{4 π}$ cm/min. The surface area (in ${cm}^{2}$ ) of the chocolate ball (without the ice-cream layer) is: [2025]

128 $π$
225 $π$
196 $π$
256 $π$

(4)

Volume of spherical chocolate,

$V = \frac{4}{3} π r^{3}$

Derivative w.r.t. t, we get

$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$

$\Rightarrow 81 = 4 π r^{2} \times \frac{1}{4 π}$

$\Rightarrow r^{2} = 81 \Rightarrow r = 9$

$∴$ Surface area of chocolate = $4 π {(r - 1)}^{2}$

= $4 π {(9 - 1)}^{2}$

= $256 π {cm}^{2}$

Q 2 :

The rate of change in area of a triangle having sides 10 cm and 12 cm when the variable angle between them is $θ = 60 °$ , is:

$30 {cm}^{2} / radian$
$120 {cm}^{2} / radian$
$30 \sqrt{3} {cm}^{2} / radian$
$60 \sqrt{3} {cm}^{2} / radian$

(1)

Ans. $30 {cm}^{2} / radian$

Explanation:

$a = 10 cm,$ $b = 12 cm$

$θ = 60 °$

$A = \frac{1}{2} a b \sin θ$

Differentiating with respect to $θ$ ,

$\frac{d A}{d θ} = \frac{1}{2} a b \cos θ$

$= \frac{1}{2} \times 10 \times 12 \times \cos 60 °$

$= 60 \times \frac{1}{2} = 30 {cm}^{2} / radian$

Topic Question Set

Q 2 :

The rate of change in area of a triangle having sides 10 cm and 12 cm when the variable angle between them is $θ = 60 °$ , is:

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Topic Question Set

Q 2 : The rate of change in area of a triangle having sides 10 cm and 12 cm when the variable angle between them is θ=60°, is:

Want Us to Email you About Special Offers & Updates?

Q 2 :

The rate of change in area of a triangle having sides 10 cm and 12 cm when the variable angle between them is $θ = 60 °$ , is: