MHT CET 2024 2nd May Morning Shift | Differential Equations Question 48 | Mathematics

A spherical metal ball at 80$^\circ$C cools in 5 minutes to 60$^\circ$C, in surrounding temperature of 20$^\circ$C, then the temperature of the ball after 20 minutes is approximately

(8.15)$^\circ$C

(11.85)$^\circ$C

(28.15)$^\circ$C

(31.85)$^\circ$C

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

-0

If the slope of the tangent of the curve at any point is equal to $$-y+\mathrm{e}^{-x}$$, then the equation of the curve passing through origin is

$$y+x \mathrm{e}^x=0$$

$$y \mathrm{e}^x+x=0$$

$$y \mathrm{e}^x-x=0$$

$$y-x \mathrm{e}^x=0$$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

-0

If a body cools from $$80^{\circ} \mathrm{C}$$ to $$50^{\circ} \mathrm{C}$$ in the room temperature of $$25^{\circ} \mathrm{C}$$ in 30 minutes, then the temperature of the body after 1 hour is

$$31.36^{\circ} \mathrm{C}$$

$$32.25^{\circ} \mathrm{C}$$

$$36.36^{\circ} \mathrm{C}$$

$$33.25^{\circ} \mathrm{C}$$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

-0

The differential equation representing the family of curves $$y^2=2 \mathrm{c}(x+\sqrt{\mathrm{c}})$$, where $$\mathrm{c}$$ is a positive parameter, is of

order 1, degree 4

order 2, degree 3

order 2, degree 4

order 1, degree 3

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