Physics MCQs for NEET — Practice Questions with Answers

In metals, the motion of free electrons is indeed random in the absence of an electric field, which means that there is no net current. The assertion states that there is no current in metals without an electric field, and the reason given is that the motion of free electrons is random. Both statements are true, and the reason correctly explains the assertion.

Current in loop = 3 A

$V_A - V_B$ = 3 + 2 - (1 ohm) (3A) = 2 V

A potentiometer is a device used for measuring the potential difference accurately. A longer length potentiometer wire allows for a smaller potential gradient, which makes the measurements more accurate. Thus, the assertion that a potentiometer of longer length is used for accurate measurement is true, and the reason that a longer length results in a smaller potential gradient is also true and correctly explains the assertion.

A 200W bulb glows with more brightness than a 100W bulb because it consumes more power. The reason given is that a 100W bulb has more resistance than a 200W bulb, which is also true. According to the formula P = V^2 / R, for the same voltage, the bulb with higher power (200W) will have lower resistance than the bulb with lower power (100W). Therefore, both the assertion and the reason are true, and the reason correctly explains the assertion.

Both assertion and reason are true, and the reason correctly explains the assertion. In a series combination of cells, the total internal resistance is the sum of all the internal resistances, which is small compared to the external resistance. The given current equation $I = \frac{nE}{R+n}$ (where $n$ is the number of cells, $E$ is the emf of each cell, $R$ is the external resistance, and $n$ is the total internal resistance) shows that the current is primarily dependent on the external resistance when internal resistance is much smaller.

$ W = {1 \over 2} mv^2 = q V $ $ \therefore v = \sqrt {2qV/m} $ $ \therefore {V_A \over V_B} = \sqrt {q_A \over q_B} = \sqrt { q \over 4q} = 1/2 $ $ \therefore { V_A \over V_B } = 1:2 $

$ {1 \over 2 } mv^2_{max} = {hc \over \lambda } - \phi ....(1) $ $ \lambda' = \lambda - 0.25 \lambda = 0.75\lambda \,and \,v' = 2v $ $ \therefore {1 \over 2 } m ( 2v_{max})^2 = { hc \over 0.75 \lambda } -\phi.....(1)$ $4 ( { \lambda c \over \lambda } - \phi) = {4hc \over 3\lambda } - \phi $ $ \therefore {8hc \over 3\lambda} = 3 \phi $ $ \therefore \phi = { 8hc \over 9\lambda } $ $ { hc \over \lambda } = 2eV $ $ \phi = { 8 \over 9 } \times 2eV = 16/9 $ $ \phi = 1.8 eV$

$ p = { nhc \over \lambda } $ $ \therefore n = { p \lambda \over hc } = { 100 \times 540 \times 10^{-9} \over 6 \times 10^{-34} \times 3 \times 10^8 }$ $ = 3 \times 10^{20} $

$ {1 \over 2 } mv^2 = eV $ $ \therefore m^2 v^2 = p^2 =2meV$ $ \therefore p = {h \over \sqrt {2meV}}$ $ \therefore p \alpha { 1 \over \sqrt {me}} $ $ \therefore {\lambda_\infty \over \lambda _p } = \sqrt { m_p e_p \over m_\infty e_\infty} $ $ m_\infty = 4 mp ,e_\infty = 2 e_p $ $ \therefore { \lambda _\infty \over \lambda_p} = \sqrt{ m_p e _p \over 4m_p \times 2 e_p } = {1 \over \sqrt 8 } $ $ {\lambda_\infty \over \lambda _p } = { 1 \over 2\sqrt 2} $ $ \therefore {\lambda_\infty \over \lambda_0 } = { 1 \over 2\sqrt 2} $ $ \lambda _8 = {\lambda _0 \over 2\sqrt 2}$