Practice free Physics NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
What is the unit of Polarization (P) as listed in the 'Physical quantity Symbol Dimensions Unit Remark' table?
The table explicitly states: 'Polarisation P [L$^{-2}$ AT] C m$^{-2}$ Dipole moment per unit volume'.
A capacitor has its plates separated by a dielectric. The product $\epsilon_0 K$ is referred to as:
The NCERT states: 'The product $\epsilon_0 K$ is called the permittivity of the medium and is denoted by $\epsilon = \epsilon_0 K$ (2.52).'
In the presence of an external electric field, a dielectric material develops surface charges ($\pm \sigma_p$) on its faces normal to the field. How do these induced surface charges affect the total electric field inside the dielectric?
The text states: 'The field produced by these surface charges opposes the external field. The total field in the dielectric is, thereby, reduced from the case when no dielectric is present.'
Why does a volume element ($\Delta v$) within a polarized dielectric slab, far from its surfaces, have no net charge, even though it possesses a net dipole moment?
The NCERT mentions: 'Anywhere inside the dielectric, the volume element $\Delta v$ has no net charge (though it has net dipole moment). This is, because, the positive charge of one dipole sits close to the negative charge of the adjacent dipole.'
Which type of molecule, when polarized by an external field, primarily exhibits an alignment effect of its existing permanent dipoles rather than a significant induced dipole moment effect?
The text states for polar molecules: 'there may be, in addition, the 'induced dipole moment' effect as for non-polar molecules, but generally the alignment effect is more important for polar molecules.'
Which of the following statements correctly describes the calculation of work done by a variable force in one dimension?
As per the NCERT text, 'if the displacements are allowed to approach zero, then the number of terms in the sum increases without limit, but the sum approaches a definite value equal to the area under the curve in Fig. 5.3(b). Then the work done is $W = \int_{x_i}^{x_f} F(x) dx$'. This integral represents the area under the force-displacement curve.
A woman pushes a trunk on a railway platform. She applies a force of 100 N over the first 10 m, and then her applied force reduces linearly to 50 N over the next 10 m. The frictional force opposing the motion is constant at 50 N. What is the total work done by the woman on the trunk over the entire 20 m displacement?
According to Example 5.5, the work done by the woman for the first 10 m (rectangle ABCD) is $100 \text{ N} \times 10 \text{ m} = 1000 \text{ J}$. For the next 10 m (trapezium CEID), the work done is $(1/2) \times (100 + 50) \text{ N} \times 10 \text{ m} = 750 \text{ J}$. The total work done by the woman is $1000 \text{ J} + 750 \text{ J} = 1750 \text{ J}$.
In the scenario described in the previous question (woman pushing a trunk), what is the work done by the frictional force over the total 20 m displacement?
As per Example 5.5, the frictional force $f$ is constant at 50 N and opposes motion, hence it is negative. The work done by friction is $W_f = (-50 \text{ N}) \times 20 \text{ m} = -1000 \text{ J}$. The area on the negative side of the force axis has a negative sign.
For a variable force $F(x)$, if the displacement $\Delta x$ is small, the work done $\Delta W$ can be approximated as:
The NCERT text states: 'If the displacement $\Delta x$ is small, we can take the force $F(x)$ as approximately constant and the work done is then $\Delta W = F(x) \Delta x$.'
The work-energy theorem for a variable force in one dimension can be derived from Newton’s Second Law. The intermediate step involves rewriting the time rate of change of kinetic energy as:
As shown in the NCERT derivation for the work-energy theorem for a variable force: $\frac{dK}{dt} = \frac{d}{dt} (\frac{1}{2}mv^2) = mv \frac{dv}{dt}$. Since $m \frac{dv}{dt} = F$ (from Newton's Second Law), then $\frac{dK}{dt} = Fv$.
Ready to ace NEET?
Free access · No credit card required
Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.
No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.
The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.