Question
A steel ball of diameter $3 \mathrm{~mm}$ falls through glycerine and covers a distance of $25 \mathrm{~cm}$ in $10 \mathrm{~S} .$ The specific gravity of steel and glycerine are $7.8$ and $1.26$ respectively. The viscosity of glycerine is about........ pascal-sec(A) $1.3$(B) $1.5$(C) $0.8$(D) $1.0$
Step 1
The diameter is given as $3 \mathrm{~mm}$, so the radius $r$ is half of the diameter, which is $1.5 \mathrm{~mm}$. We convert this to meters by multiplying by $10^{-3}$, so $r = 1.5 \times 10^{-3} \mathrm{~m}$. Show more…
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A wide jar is filled with glycerine having specific gravity $1.26$, in this jar, a steel ball of radius $0.25 \mathrm{~cm}$ has been dropped. After some time it has been observed that ball is taking equal interval of time $(1.8 \mathrm{~s})$ to cover equal successive distances, of $20 \mathrm{~cm} .$ [Take, $\left.\rho_{\text {steel }}=7.8 \times 10^{3} \mathrm{~kg}-\mathrm{m}^{3}, g=9.81 \mathrm{~ms}^{-2}\right]$. The viscosity of glycerine is [in $\mathrm{N}-\mathrm{sm}^{-2}$ ] (a) $0.802$ (b) $1.67$ (c) $0.76$ (d) $0.963$
Experimental Physics
Round 1
A steel ball of diameter $3.2 \mathrm{~mm}$ falls under gravity through an oil of density $920 \mathrm{~kg} \mathrm{~m}^{-3}$ and viscosity $1.64 \mathrm{Ns} \mathrm{m}^{-2}$. The density of steel. may be taken as $7820 \mathrm{~kg} \mathrm{~m}^{-3}$. What is the terminal velocity of the ball ? (A) $3.45 \mathrm{cms}^{-1}$ (B) $4.25 \mathrm{cms}^{-1}$ (C) $1.85 \mathrm{~cm}^{-1}$ (D) $2.39 \mathrm{cms}^{-1}$
If a ball of steel (density p, 7.8 g/cm) attains a terminal velocity of 10 cm/s when falling in a tank of water (coefficient of viscosity =8.5 x 10
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