An iron sphere of mass 20 times 10^{-3} kg f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The terminal velocity $v$ of a sphere falling through a viscous liquid is given by $v = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$.Since the ma

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The terminal velocity $v$ of a sphere falling through a viscous liquid is given by $v = \frac{2}{9} \frac{r^2(ho - \sigma)g}{\eta}$.Since the material of the sphere is the same, the density $ho$ is constant, so $v \propto r^2$.Given the mass $M = \frac{4}{3} \pi r^3 ho$, we have $r \propto M^{1/3}$.Substituting this into the proportionality, we get $v \propto (M^{1/3})^2 = M^{2/3}$.Therefore, $\frac{v_1}{v_2} = \left(\frac{M_1}{M_2}ight)^{2/3}$.Given $M_1 = 20 	imes 10^{-3} \ kg$, $v_1 = 0.5 \ ms^{-1}$, and $M_2 = 54 	imes 10^{-2} \ kg = 540 	imes 10^{-3} \ kg$.$\frac{0.5}{v_2} = \left(\frac{20 	imes 10^{-3}}{540 	imes 10^{-3}}ight)^{2/3} = \left(\frac{1}{27}ight)^{2/3} = \left(\left(\frac{1}{3}ight)^3ight)^{2/3} = \left(\frac{1}{3}ight)^2 = \frac{1}{9}$.Thus, $v_2 = 0.5 	imes 9 = 4.5 \ ms^{-1}$.

An iron sphere of mass $20 \times 10^{-3} \ kg$ falls through a viscous liquid with terminal velocity $0.5 \ ms^{-1}$. The terminal velocity (in $ms^{-1}$) of another iron sphere of mass $54 \times 10^{-2} \ kg$ is (in $.5$)

Similar Questions

If the coefficients of the $(2r+1)$-th term and the $(r+2)$-th term in the expansion of $(1+x)^{43}$ are equal,then $r$ is equal to:

Which of the following is a tricarboxylic acid?

Which of the following complexes has the highest stability constant at $298 \ K$?

If the harmonic mean of the roots of the equation $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$ is $4$,then the value of $b$ is:

$\int_{-2}^2 |[x]| \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module