MHT CET 2024 3rd May Evening Shift | Differential Equations Question 40 | Mathematics

If a body cools from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in the room temperature of $30^{\circ} \mathrm{C}$ in 30 min , then the temperature of a body after one hour is

$42^{\circ} \mathrm{C}$

$24^{\circ} \mathrm{C}$

$48^{\circ} \mathrm{C}$

$56^{\circ} \mathrm{C}$

MHT CET 2024 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

The differential equation $\left[\frac{1+\left(\frac{d y}{d x}\right)^2}{\left(\frac{d^2 y}{d x^2}\right)}\right]^{\frac{3}{2}}=\mathrm{kx}$ is of

order $=2$, degree $=3$

order $=3$, degree $=2$

order $=2$, degree $=2$

order $=3$, degree $=3$

MHT CET 2024 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

The differential equation of family of circles, whose centres are on the X -axis and also touch the Y -axis is

$4\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=\left(x^2+y^2\right)^2$

$\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=\left(x^2+y^2\right)^2$

$2\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=\left(x^2+y^2\right)^2$

$\left(x+y \frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2 x^2=4\left(x^2+y^2\right)^2$

MHT CET 2024 2nd May Evening Shift

MCQ (Single Correct Answer)

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A body cools according to Newton's law of cooling from $100^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in 15 minutes. If the temperature of the surrounding is $20^{\circ} \mathrm{C}$, then the temperature of the body after cooling down for one hour is

$30^{\circ} \mathrm{C}$

$25^{\circ} \mathrm{C}$

$35^{\circ} \mathrm{C}$

$40^{\circ} \mathrm{C}$

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