MHT CET 2021 23rd September Evening Shift | MHT CET Year Wise Previous Years Questions

An ice cube of edge $$1 \mathrm{~cm}$$ melts in a gravity free container. The approximate surface area of water formed is (water is in the form of spherical drop)

$$(36 \pi)^{1 / 3} \mathrm{~cm}^2$$

$$(24 \pi)^{1 / 3} \mathrm{~cm}^2$$

$$(28 \pi)^{1 / 3} \mathrm{~cm}^2$$

$$(12 \pi)^{1 / 3} \mathrm{~cm}^2$$

MHT CET 2021 23rd September Evening Shift

MCQ (Single Correct Answer)

-0

A plano-convex lens of refractive index ($$\mu_1^{\prime}$$ fits exactly into a plano-concave lens of refractive index $$\mu_2$$. Their plane surface are parallel to each other. 'R' is the radius of curvature of the curved surface of the lenses. The focal length of the combination is

$$\frac{\mathrm{R}}{\mu_1-\mu_2}$$

$$\frac{R}{2\left(\mu_1+\mu_2\right)}$$

$$\frac{2 R}{\mu_1-\mu_2}$$

$$\frac{\mathrm{R}}{2\left(\mu_1-\mu_2\right)}$$

MHT CET 2021 23rd September Evening Shift

MCQ (Single Correct Answer)

-0

A cylindrical rod has temperatures '$$T_1$$' and '$$T_2$$' at its ends. The rate of flow of heat is '$$Q_1$$' cal $$\mathrm{s}^{-1}$$. If length and radius of the rod are doubled keeping temperature constant, then the rate of flow of heat '$$\mathrm{Q}_2$$' will be

$$\mathrm{Q}_2=\frac{\mathrm{Q}_1}{2}$$

$$\mathrm{Q}_2=\frac{\mathrm{Q}_1}{4}$$

$$\mathrm{Q_2=4 Q_1}$$

$$\mathrm{Q}_2=2 \mathrm{Q}_1$$

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