Question 1
According to NCERT, which of the following electromagnetic waves is widely used in LASIK eye surgery?
✅ Correct Answer: Ultraviolet rays
Explanation: Ultraviolet (UV) radiation is highly energetic and can be focused into very narrow beams. NCERT specifically mentions that UV rays are used in LASIK (Laser-assisted in situ keratomileusis) eye surgery to reshape the cornea.
Question 2
According to NCERT, materials suitable for making permanent magnets should ideally have:
✅ Correct Answer: High retentivity and high coercivity
Explanation: A permanent magnet must retain a strong magnetic field after the magnetizing field is removed (High Retentivity) and it must not easily lose its magnetization due to stray external magnetic fields, temperature changes, or rough handling (High Coercivity). Steel is a common example.
Question 3
As explicitly mentioned in the NCERT chapter on Atoms, which of the following phenomena is the Bohr model fundamentally unable to explain?
✅ Correct Answer: The relative intensities of the frequencies in the emission spectrum.
Explanation: The Bohr model successfully explained the stability of the atom and the frequencies of spectral lines for hydrogenic atoms. However, it could not explain why certain spectral lines are brighter (more intense) than others, which requires quantum mechanics (transition probabilities) to fully explain.
Question 4
According to NCERT, which of the following statements is true for elastomers (such as the elastic tissue of the aorta or natural rubber)?
✅ Correct Answer: They do not have a well-defined plastic region.
Explanation: Elastomers can be stretched to large values of strain (up to several hundred percent) but they do not obey Hooke's law (the stress-strain curve is not a straight line). Furthermore, they do not have a well-defined plastic region; they simply break if stretched too far beyond their elastic limit.
Question 5
Kepler's second law (the law of areas) for planetary motion is a direct consequence of the conservation of:
✅ Correct Answer: Angular momentum
Explanation: The areal velocity of a planet around the sun is constant ($dA/dt = L/2m$). Since the gravitational force is a central force, the torque acting on the planet is zero. Therefore, angular momentum ($L$) is conserved, making areal velocity constant.
Question 6
In which of the following thermodynamic processes is the heat supplied to an ideal gas completely converted into external work done by the gas?
✅ Correct Answer: Isothermal expansion
Explanation: According to the First Law of Thermodynamics, $Q = \Delta U + W$. For an ideal gas undergoing an isothermal process, the temperature remains constant, meaning the change in internal energy ($\Delta U$) is zero. Therefore, $Q = W$. Heat is fully converted to work.
Question 7
In a standing wave formed on a stretched string, the rate of transfer of energy across any node is:
✅ Correct Answer: Zero
Explanation: Standing waves (stationary waves) do not transfer energy or momentum through the medium. The energy is strictly confined between the nodes. Hence, the net energy transfer across any point (node or antinode) in a perfect standing wave is zero.
Question 8
A solid uncharged conducting sphere has a non-centric spherical cavity. A point charge $+q$ is placed inside the cavity, but not at its center. The electric field outside the conductor at a distance $r$ from the center of the sphere is:
✅ Correct Answer: Radially outward and spherically symmetric.
Explanation: Due to electrostatic induction, a charge $-q$ appears on the inner surface of the cavity, perfectly canceling the field of $+q$ in the conducting material. A charge $+q$ appears on the outer surface of the sphere. Because the outer surface is perfectly spherical and unaffected by the internal geometry, this outer charge distributes uniformly, creating a symmetric radial field outside.
Question 9
Assertion: The terminal voltage of a real battery is always less than its electromotive force (EMF).
Reason: Some potential drop always occurs across the internal resistance of the battery when it is connected in a circuit.
✅ Correct Answer: Assertion is incorrect but Reason is correct.
Explanation: The Assertion is false. While it is true that terminal voltage is less than EMF when the battery is discharging ($V = E - ir$), the terminal voltage is greater than the EMF when the battery is being charged by an external source ($V = E + ir$). The Reason is a fundamentally true statement regarding internal resistance drops.
Question 10
Assertion: The resolving power of a telescope increases when the wavelength of the light being observed is decreased.
Reason: The resolving power of a telescope is directly proportional to the diameter (aperture) of the objective lens and inversely proportional to the wavelength of light.
✅ Correct Answer: Both Assertion and Reason are correct and Reason is the correct explanation.
Explanation: The resolving power of a telescope is given by the formula $RP = D / (1.22\lambda)$, where $D$ is the diameter of the objective lens and $\lambda$ is the wavelength. Because $\lambda$ is in the denominator, decreasing the wavelength increases the resolving power. The reason correctly outlines this inverse proportionality.
Question 11
Consider the following two statements:
Statement I: A physical quantity can have a unit but no dimensions.
Statement II: A physical quantity can have dimensions but no unit.
✅ Correct Answer: Statement I is correct but Statement II is incorrect
Explanation: Statement I is correct (e.g., plane angle has the unit radian but is dimensionless $[M^0 L^0 T^0]$). Statement II is incorrect because if a quantity has physical dimensions (like mass, length, or time), it must inherently be measurable in some system of units corresponding to those specific dimensions.
Question 12
Consider the following two statements regarding nuclear decay:
Statement I: In $\beta^-$ decay, a neutron inside the nucleus transforms into a proton, an electron, and an antineutrino.
Statement II: The total rest mass energy of the products in a spontaneous $\beta^-$ decay must be greater than the rest mass energy of the parent nucleus.
✅ Correct Answer: Statement I is correct but Statement II is incorrect
Explanation: Statement I is the correct basic mechanism of beta-minus decay. Statement II is incorrect. For any spontaneous nuclear decay to occur, the $Q$-value must be positive. This requires the mass of the parent nucleus to be strictly greater than the total mass of the decay products, so that the missing mass is successfully converted into kinetic energy.
Question 13
A ball is dropped from rest from a height $H$ above the ground. It takes a total time of $T$ seconds to reach the ground. What is the height of the ball from the ground at time $T/2$?
✅ Correct Answer: $3H/4$
Explanation: Using the equation of motion $s = ut + \frac{1}{2}at^2$, the total distance fallen is $H = \frac{1}{2}gT^2$. The distance fallen in half the time $T/2$ is $y = \frac{1}{2}g(T/2)^2 = \frac{1}{4}(\frac{1}{2}gT^2) = H/4$. Since the ball has fallen a distance of $H/4$ from the top, its actual height from the ground is $H - H/4 = 3H/4$.
Question 14
An alternating voltage $V=200\sqrt{2}\sin(100t)$ is applied across a pure capacitor of $10\ \mu\text{F}$. The RMS value of current in the circuit is:
✅ Correct Answer: $200\text{ mA}$
Explanation: From the voltage equation, peak voltage $V_0 = 200\sqrt{2}\text{ V}$ and angular frequency $\omega = 100\text{ rad/s}$. The RMS voltage is $V_{rms} = V_0/\sqrt{2} = 200\text{ V}$. Capacitive reactance $X_C = 1/(\omega C) = 1/(100 \times 10 \times 10^{-6}) = 1000\ \Omega$. The RMS current is $I_{rms} = V_{rms}/X_C = 200/1000 = 0.2\text{ A} = 200\text{ mA}$.
Question 15
A uniform rod of mass $M$ and length $L$ is pivoted at its top end and hangs vertically at rest. A particle of mass $m$ moving horizontally with velocity $v$ strikes the free bottom end of the rod and sticks to it. The angular velocity of the system immediately after the collision is:
✅ Correct Answer: $\frac{3mv}{L(M+3m)}$
Explanation: Angular momentum is conserved about the pivot point during the collision. Initial angular momentum $L_i = mvr_{\perp} = mvL$. Final angular momentum $L_f = I_{\text{system}} \omega$. The system's inertia is $I_{\text{system}} = I_{\text{rod}} + I_{\text{particle}} = \frac{ML^2}{3} + mL^2 = L^2(\frac{M}{3} + m)$. Equating them: $mvL = L^2(\frac{M}{3} + m)\omega \implies \omega = \frac{mv}{L(M/3+m)} = \frac{3mv}{L(M+3m)}$.
Question 16
The ratio of the de Broglie wavelengths of a proton and an alpha particle, when both are accelerated from rest through the same potential difference $V$, is:
✅ Correct Answer: $2\sqrt{2}:1$
Explanation: De Broglie wavelength $\lambda = h/\sqrt{2mK}$. For charges accelerated by potential $V$, $K = qV$. Thus, $\lambda \propto 1/\sqrt{mq}$.
For a proton: mass $= m$, charge $= e$.
For an alpha particle: mass $= 4m$, charge $= 2e$.
Ratio $\lambda_p/ \lambda_\alpha = \sqrt{(m_\alpha q_\alpha) / (m_p q_p)} = \sqrt{(4m \cdot 2e)/(m \cdot e)} = \sqrt{8} = 2\sqrt{2}:1$.
Question 17
A parallel plate air capacitor has a capacitance $C$ and plate separation $d$. A dielectric slab of dielectric constant $K=5$ and thickness $d/2$ is introduced between the plates, perfectly parallel to them. The new capacitance of the system is:
✅ Correct Answer: $5C/3$
Explanation: Inserting the slab parallel to the plates divides the capacitor into two capacitors in series, each with thickness $d/2$.
Air part: $C_1 = \frac{\varepsilon_0 A}{d/2} = 2C$.
Dielectric part: $C_2 = \frac{K \varepsilon_0 A}{d/2} = 2KC = 10C$.
Series equivalent: $C_{eq} = \frac{C_1 C_2}{C_1+C_2} = \frac{(2C)(10C)}{2C+10C} = \frac{20C^2}{12C} = \frac{5C}{3}$.
Question 18
A conducting rod of length $L$ is rotated about one of its ends with a constant angular velocity $\omega$ in a uniform magnetic field $B$ directed perpendicular to the plane of rotation. If the magnetic field is doubled and the angular velocity is halved, the induced EMF across the ends of the rod will:
✅ Correct Answer: Remain unchanged
Explanation: The motional EMF induced across a rotating rod is given by $e = \frac{1}{2} B \omega L^2$. If $B \rightarrow 2B$ and $\omega \rightarrow \omega/2$, the new EMF is $e' = \frac{1}{2} (2B) (\omega/2) L^2 = \frac{1}{2} B \omega L^2 = e$. The EMF remains completely unchanged.
Question 19
An electron and a proton are injected into a uniform magnetic field perpendicular to the field lines, with the exact same kinetic energy. Which of the following statements is true regarding their paths?
✅ Correct Answer: The radius of the proton's path will be larger than that of the electron's path.
Explanation: The radius of a charged particle in a magnetic field is $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$. Since the kinetic energy $K$, charge magnitude $q$, and magnetic field $B$ are identical for both particles, $R \propto \sqrt{m}$. Since the proton is much heavier than the electron ($m_p \approx 1836 m_e$), the radius of the proton's circular path will be significantly larger.
Question 20
TRAP QUESTION: A block of mass $2\text{ kg}$ is kept on a rough horizontal surface with coefficient of static friction $\mu_s=0.5$ and kinetic friction $\mu_k=0.4$. A horizontal force varying with time as $F=5t$ (where $t$ is in seconds and $F$ is in Newtons) is applied to the block. What is the exact magnitude of the frictional force acting on the block at $t=1\text{ s}$? (Take $g=10\text{ m/s}^2$)
✅ Correct Answer: $5\text{ N}$
Explanation: Limiting static friction $f_{lim} = \mu_s N = \mu_s (mg) = 0.5 \times (2 \times 10) = 10\text{ N}$. At $t=1\text{ s}$, the applied force is $F = 5(1) = 5\text{ N}$. Since the applied force ($5\text{ N}$) is less than the limiting friction ($10\text{ N}$), the block does not move. Static friction is self-adjusting, so it will exactly match the applied force to keep the net force zero. Thus, frictional force = $5\text{ N}$.
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