Physics MCQs for NEET — Practice Questions with Answers

The NCERT text explicitly states 'Av gives the volume flux or flow rate and remains constant throughout the pipe of flow.'

The text mentions, 'Steady flow is achieved at low flow speeds. Beyond a limiting value, called critical speed, this flow loses steadiness and becomes turbulent.'

From the equation of continuity, as the area decreases, the velocity increases. An increase in velocity over time implies acceleration. The text states, 'From (Fig 9.7b) it is clear that $A_R > A_Q$ or $v_R < v_Q$, the fluid is accelerated while passing from R to Q'.

The NCERT text specifies that 'For flow of incompressible fluids $\rho_P = \rho_R = \rho_Q$' and then states 'Equation (9.9) reduces to $A_P v_P = A_R v_R = A_Q v_Q$ (9.10) which is called the equation of continuity and it is a statement of conservation of mass in flow of incompressible fluids.'

The definition of an incompressible fluid implies that its density remains constant, regardless of changes in pressure or flow characteristics. The text states, 'For flow of incompressible fluids $\rho_P = \rho_R = \rho_Q$'.

The NCERT text shows the derivation and final formula: 'Total amount of work done in establishing the current I is $W = \int dW = \int_0^I L I' dI' = \frac{1}{2} L I^2$.' (Equation 6.17).

From Example 6.9 (a) and (b), the total magnetic energy is $U_B = \frac{1}{2} L I^2$. The magnetic energy per unit volume is $u_B = \frac{U_B}{V} = \frac{U_B}{Al}$ (where V is the volume of the solenoid, $Al$). So, $u_B = \frac{1}{2} L I^2 / (Al)$. This is explicitly shown in the derivation for $u_B$ to reach $\frac{B^2}{2\mu_0}$.

The NCERT text explains this clearly: 'L is analogous to m (i.e., L is electrical inertia and opposes growth and decay of current in the circuit).' This highlights L's role as a measure of electrical inertia.

The formula derived is $u_B = \frac{B^2}{2\mu_0}$. This implies that the magnetic energy per unit volume depends solely on the magnetic field B and the permeability of free space $\mu_0$ (assuming a vacuum core or air core solenoid). Although B itself depends on current and solenoid properties, the question asks what $u_B$ depends on directly from the given formula.

The NCERT text explicitly states: 'If we ignore the resistive losses and consider only inductive effect, then using Eq. (6.14), $\frac{dW}{dt} = I L \frac{dI}{dt}$.' The subsequent integration to find W leads to $W = \frac{1}{2} L I^2$. Therefore, this expression assumes negligible resistive losses.