1
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $x \frac{\mathrm{~d} y}{\mathrm{~d} x}=y(\log y-\log x+1)$, then general solution of this equation is

A
$\log \left(\frac{x}{y}\right)=\mathrm{cy}$, where c is a constant of integration.
B
$\log \left(\frac{x}{y}\right)=\mathrm{c} x$, where c is a constant of integration.
C
$\log \left(\frac{y}{x}\right)=\mathrm{cy}$, where c is a constant of integration.
D
$\log \left(\frac{y}{x}\right)=c x$, where $c$ is a constant of integration.
2
MHT CET 2024 2nd May Morning Shift
MCQ (Single Correct Answer)
+2
-0

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

A
(8.15)$^\circ$C
B
(11.85)$^\circ$C
C
(28.15)$^\circ$C
D
(31.85)$^\circ$C
3
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-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

A
$$y+x \mathrm{e}^x=0$$
B
$$y \mathrm{e}^x+x=0$$
C
$$y \mathrm{e}^x-x=0$$
D
$$y-x \mathrm{e}^x=0$$
4
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-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

A
$$31.36^{\circ} \mathrm{C}$$
B
$$32.25^{\circ} \mathrm{C}$$
C
$$36.36^{\circ} \mathrm{C}$$
D
$$33.25^{\circ} \mathrm{C}$$
MHT CET Subjects
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