1
MHT CET 2024 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A metal sphere of radius R, density $\rho_1$ moves with terminal velocity $\mathrm{V}_1$ through a liquid of density $\sigma$. Another sphere of same radius but density $\rho_2$ moves through same liquid. Its terminal velocity is $\mathrm{V}_2$. The ratio $\mathrm{V}_1: \mathrm{V}_2$ is

A
$\left(\rho_2+\sigma\right):\left(\rho_1-\sigma\right)$
B
$\left(\rho_1+\sigma\right):\left(\rho_2-\sigma\right)$
C
$\left(\rho_2-\sigma\right):\left(\rho_1-\sigma\right)$
D
$\left(\rho_1-\sigma\right):\left(\rho_2-\sigma\right)$
2
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Three liquids of densities $\rho_1, \rho_2$ and $\rho_3$ (with $\rho_1>\rho_2>\rho_3$ ) having same value of surface tension T , rise to the same height in three identical capillaries. Angle of contact $\theta_1, \theta_2$ and $\theta_3$ respectively obey

A
$\frac{\pi}{2}>\theta_1>\theta_2>\theta_3>0$
B
$0 \leqslant \theta_1<\theta_2<\theta_3<\frac{\pi}{2}$
C
$\frac{\pi}{2}<\theta_1<\theta_2<\theta_3<\pi$
D
$\pi>\theta_1>\theta_2>\frac{\pi}{2}$
3
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Let ' $n$ ' is the number of liquid drops, each with surface energy ' $E$ '. These drops join to form single drop. In this process

A
some energy will be absorbed
B
energy absorbed is $\left[E\left(n-n^{2 / 3}\right)\right]$
C
energy released will be $\left[\mathrm{E}\left(\mathrm{n}-\mathrm{n}^{2 / 3}\right)\right]$
D
energy released will be $\left[\mathrm{E}\left(2^{2 / 3}-1\right)\right]$
4
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The work done in splitting a water drop of radius R into 64 droplets is ( $\mathrm{T}=$ Surface tension of water)

A
$6 \pi \mathrm{TR}^2$
B
$12 \pi \mathrm{TR}^2$
C
$8 \pi \mathrm{TR}^2$
D
$24 \pi \mathrm{TR}^2$
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