Q.37. A spherical body of radius r and density falls freely through a viscous liquid having density

Q: What is the correct answer for JEE Main Physics Q37?

The correct answer is B. The terminal velocity $v_0$ for a spherical body falling through a viscous liquid is given by Stokes' law:$$v_0 = \frac{2r^2g(\sigma - \rho)}{9\eta}$$From this, we can express viscosity $\eta$ as:$$\eta = \frac{2r^2g(\sigma - \rho)}{9v_0}$$The fractional error in $\eta$ is given by:$$\frac{\Delta\eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta g}{g} + \frac{\Delta(\sigma - \rho)}{\sigma - \rho} +

Q: Q.37. A spherical body of radius r and density falls freely through a viscous liquid having density and viscosity and attains a terminal velocity . Estimated maximum error in the quantity is :…

The correct answer is B. The terminal velocity $v_0$ for a spherical body falling through a viscous liquid is given by Stokes' law:$$v_0 = \frac{2r^2g(\sigma - \rho)}{9\eta}$$From this, we can express viscosity $\eta$ as:$$\eta = \frac{2r^2g(\sigma - \rho)}{9v_0}$$The fractional error in $\eta$ is given by:$$\frac{\Delta\eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta g}{g} + \frac{\Delta(\sigma - \rho)}{\sigma - \rho} +

Q: What are the options for this JEE Main Physics question?

A. $$2\frac{\Delta r}{r} - \frac{\Delta v_0}{v_0}$$ · B. $$2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$$ · C. $$\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$$ · D. $$\frac{\Delta r}{r} - \frac{\Delta v_0}{v_0}$$

Question

Q.37. A spherical body of radius r and density  falls freely through a viscous liquid having density  and viscosity  and attains a terminal velocity . Estimated maximum error in the quantity  is : (Ignore errors associated with ,  and g, gravitational acceleration)

FirstInTest · Accepted Answer

The terminal velocity $v_0$ for a spherical body falling through a viscous liquid is given by Stokes' law:$$v_0 = \frac{2r^2g(\sigma - \rho)}{9\eta}$$From this, we can express viscosity $\eta$ as:$$\eta = \frac{2r^2g(\sigma - \rho)}{9v_0}$$The fractional error in $\eta$ is given by:$$\frac{\Delta\eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta g}{g} + \frac{\Delta(\sigma - \rho)}{\sigma - \rho} + \frac{\Delta v_0}{v_0}$$We are asked to ignore errors associated with $\sigma$, $\rho$, and g. So, $\Delta g = 0$ and $\Delta(\sigma - \rho) = 0$.Therefore, the maximum estimated error in $\eta$ is:$$\frac{\Delta\eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$$

Answer

$$2\frac{\Delta r}{r} - \frac{\Delta v_0}{v_0}$$

Answer

$$2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$$

Answer

$$\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$$

Answer

$$\frac{\Delta r}{r} - \frac{\Delta v_0}{v_0}$$

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