VT=92 gηR2[ρB−ρL]⇒VT=92×9.8×10−610×(10−4)2[105−103]⇒VT=224.5 when ball fall from height (h) $\begin{aligned}& \mathrm{V}=\sqrt{2 \mathrm{gh}} \
& \mathrm{h}=\left(\frac{\mathrm{V}^2}{2 \mathrm{g}}\right)=2518 \mathrm{m}\end{aligned}$
A spherical ball of radius 1×10−4 m and density 105 kg/m3 falls freely under gravity through a distance h before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of h is approximately: (The coefficient of viscosity of water is 9.8×10−6 N s/m2 )
Held on 9 Apr 2024 · Verified 6 Jul 2026.
2296 m
2518 m
2249 m
2396 m
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