Question 1
According to the NCERT chapter on Electromagnetic Waves, which of the following radiations is commonly used in water purifiers to kill bacteria and germs?
✅ Correct Answer: Ultraviolet rays
Explanation: Ultraviolet (UV) radiation has sufficient energy to destroy the DNA of bacteria and viruses, making it highly effective for disinfecting water. This is a direct application mentioned in the NCERT EM spectrum uses.
Question 2
As per NCERT, what is the de Broglie wavelength associated with an electron accelerated through a potential difference of $100\text{ V}$?
✅ Correct Answer: $0.1227\text{ nm}$
Explanation: The de Broglie wavelength of an electron accelerated by a potential $V$ is given by the standard NCERT formula $\lambda = \frac{1.227}{\sqrt{V}}\text{ nm}$. Substituting $V=100\text{ V}$, we get $\lambda = \frac{1.227}{10} = 0.1227\text{ nm}$.
Question 3
According to the thermal properties of matter in NCERT, the density of water is maximum at:
✅ Correct Answer: $4^\circ\text{C}$
Explanation: Water exhibits anomalous expansion. Unlike most liquids that expand continuously upon heating, water contracts as it is heated from $0^\circ\text{C}$ to $4^\circ\text{C}$, reaching its minimum volume and maximum density exactly at $4^\circ\text{C}$.
Question 4
The ratio of the nuclear density of an Oxygen nucleus ($^{16}_8\text{O}$) to that of a Lead nucleus ($^{208}_{82}\text{Pb}$) is approximately:
✅ Correct Answer: $1: 1$
Explanation: Nuclear density is constant and independent of the mass number ($A$) for all nuclei. It is approximately $2.3 \times 10^{17}\text{ kg/m}^3$. Therefore, the ratio of densities of any two nuclei is always $1:1$.
Question 5
An electric dipole with dipole moment $p$ is placed entirely inside a closed hollow spherical surface. The total electric flux passing through the spherical surface is:
✅ Correct Answer: Zero
Explanation: According to Gauss's Law, the total electric flux through a closed surface depends only on the net charge enclosed ($\Phi = q_{\text{net}}/\varepsilon_0$). Since an electric dipole consists of equal and opposite charges ($+q$ and $-q$), the net charge enclosed is zero, making the net flux zero.
Question 6
A positively charged particle is projected into a uniform magnetic field such that its velocity vector is exactly parallel to the magnetic field lines. The trajectory of the particle will be:
✅ Correct Answer: A straight line
Explanation: The magnetic force on a moving charge is $\vec{F} = q(\vec{v} \times \vec{B})$. Since velocity and magnetic field are parallel, the angle between them is $0^\circ$. Therefore, $F = qvB\sin(0^\circ) = 0$. With no deflecting force, the particle continues moving in a straight line at a constant speed.
Question 7
A symmetric convex lens of focal length $f$ and power $P$ is cut exactly in half along its principal axis (horizontally). What is the focal length and power of one of the resulting halves?
✅ Correct Answer: Focal length $f$, Power $P$
Explanation: Cutting a lens along its principal axis does not change its radii of curvature ($R_1, R_2$) or its refractive index. By the Lens Maker's formula, the focal length $f$ remains unchanged. Since power $P=1/f$, the power also remains the same. Only the intensity of the image is halved because the aperture area is reduced.
Question 8
Which of the following logic gates are known as "Universal Gates" because they can be used to construct any other logic gate (AND, OR, NOT)?
✅ Correct Answer: NAND and NOR
Explanation: NAND and NOR gates are termed universal gates. By combining multiple NAND gates or multiple NOR gates in specific configurations, you can replicate the Boolean logic functions of all basic gates (NOT, AND, OR) without needing any other gate type.
Question 9
Assertion: The escape velocity of a body from the surface of the Earth is independent of the angle of projection.
Reason: The escape velocity depends on the mass of the body being projected.
✅ Correct Answer: Assertion is correct but Reason is incorrect.
Explanation: Escape velocity $v_e = \sqrt{2GM/R}$ is derived from the conservation of energy (a scalar approach). It strictly depends on the mass of the planet ($M$) and its radius ($R$), and is completely independent of the angle of projection. It is also independent of the mass of the object being launched, making the Reason false.
Question 10
Assertion: A step-up transformer increases the alternating voltage while simultaneously decreasing the alternating current.
Reason: In an ideal transformer, the input power is exactly equal to the output power.
✅ Correct Answer: Both Assertion and Reason are correct and Reason is the correct explanation.
Explanation: An ideal transformer operates on the principle of conservation of energy (Power In = Power Out). Since Power $P=VI$, if a step-up transformer increases the voltage ($V_{out} > V_{in}$), it must proportionally decrease the current ($I_{out} < I_{in}$) to maintain constant power.
Question 11
Consider the following two statements regarding thermodynamics:
Statement I: The molar specific heat of an ideal gas at constant pressure ($C_p$) is always greater than its molar specific heat at constant volume ($C_v$).
Statement II: In an adiabatic expansion, the temperature of an ideal gas always decreases.
✅ Correct Answer: Both Statement I and Statement II are correct.
Explanation: Statement I is true; $C_p > C_v$ because at constant pressure, heat supplied must do external work in expanding the gas as well as increase internal energy (Mayer's relation $C_p - C_v = R$). Statement II is also true; in adiabatic expansion, the gas does work at the expense of its own internal energy, causing a drop in temperature.
Question 12
Consider the following two statements:
Statement I: Torque and Work have the exact same dimensional formula.
Statement II: Because they have the same dimensions, Torque and Work are both vector quantities.
✅ Correct Answer: Statement I is correct but Statement II is incorrect.
Explanation: Both Torque and Work have the dimensions $[M L^2 T^{-2}]$ (Force $\times$ Distance). Thus, Statement I is correct. However, Statement II is incorrect because Work is a scalar quantity (dot product of Force and Displacement), while Torque is a vector quantity (cross product of position vector and Force).
Question 13
Two identical cells, each of EMF $E$ and internal resistance $r$, are connected in parallel. This combination is then connected across an external variable resistance $R$. What should be the value of $R$ to dissipate maximum power in the external circuit?
✅ Correct Answer: $R=r/2$
Explanation: According to the Maximum Power Transfer Theorem, maximum power is delivered to the external load when the external resistance $R$ equals the equivalent internal resistance of the power source. For two identical cells in parallel, the equivalent internal resistance is $r_{eq} = (r \times r)/(r+r) = r/2$. Thus, maximum power occurs when $R=r/2$.
Question 14
A man walking on a horizontal road at $3\text{ km/h}$ finds that the rain appears to fall vertically. When he increases his speed to $6\text{ km/h}$, the rain appears to hit him at an angle of $45^\circ$ to the vertical. What is the actual speed of the rain with respect to the ground?
✅ Correct Answer: $3\sqrt{2}\text{ km/h}$
Explanation: Let actual rain velocity $v_r = a\hat{i} - b\hat{j}$.
Case 1: Man's velocity $v_{m1} = 3\hat{i}$. Rain wrt man $v_{rm1} = (a-3)\hat{i} - b\hat{j}$. Since it appears vertical, horizontal component is zero $\implies a-3=0 \implies a=3$.
Case 2: Man's velocity $v_{m2} = 6\hat{i}$. $v_{rm2} = (3-6)\hat{i} - b\hat{j} = -3\hat{i} - b\hat{j}$. Since it appears at $45^\circ$, the magnitudes of horizontal and vertical components are equal $\implies |-3| = |-b| \implies b=3$.
Actual speed of rain $= \sqrt{a^2 + b^2} = \sqrt{3^2 + 3^2} = 3\sqrt{2}\text{ km/h}$.
Question 15
In a Young's Double Slit Experiment (YDSE), the fringe width obtained in air is $\beta$. If the entire apparatus is immersed in a transparent liquid of refractive index $\mu=4/3$ without disturbing the setup, what will be the new fringe width?
✅ Correct Answer: $3\beta/4$
Explanation: Fringe width is given by $\beta = \frac{\lambda D}{d}$. When immersed in a liquid, the wavelength of light decreases to $\lambda' = \frac{\lambda}{\mu}$. Since $D$ and $d$ are unchanged, the new fringe width becomes $\beta' = \frac{\lambda' D}{d} = \frac{\beta}{\mu}$. Substituting $\mu = 4/3$, we get $\beta' = \frac{3\beta}{4}$.
Question 16
From a uniform circular disc of radius $R$ and mass $M$, a smaller circular hole of radius $R/2$ is cut out. The center of the hole lies at a distance of $R/2$ from the center of the original disc. What is the moment of inertia of the remaining portion about an axis passing through the center of the original disc and perpendicular to its plane?
✅ Correct Answer: $\frac{13}{32} MR^2$
Explanation: Initial MOI of full disc $I_{total} = \frac{1}{2}MR^2$.
Mass of cut portion $m' = M(A_{cut}/A_{total}) = M(R/2)^2 / R^2 = M/4$.
MOI of cut portion about its own center $I_c = \frac{1}{2}m'(R/2)^2 = \frac{1}{2}(M/4)(R^2/4) = \frac{MR^2}{32}$.
Using parallel axis theorem, MOI of cut portion about the original center $I_{cut} = I_c + m'd^2 = \frac{MR^2}{32} + (M/4)(R/2)^2 = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{3MR^2}{32}$.
MOI of remaining $I = I_{total} - I_{cut} = \frac{16MR^2}{32} - \frac{3MR^2}{32} = \frac{13MR^2}{32}$.
Question 17
In studying the photoelectric effect, a graph is plotted showing maximum kinetic energy ($K_{\max}$) on the y-axis against the frequency of incident light ($\nu$) on the x-axis for two different metals, A and B. What is the ratio of the slopes of the graphs for metal A to metal B?
✅ Correct Answer: $1: 1$
Explanation: From Einstein's photoelectric equation: $K_{\max} = h\nu - \Phi$. Comparing this to $y=mx+c$, the slope $m$ of the graph is Planck's constant ($h$). Since $h$ is a universal constant, the slope is exactly the same for all metals. Only the x-intercepts (threshold frequencies) will differ. Thus, the ratio of slopes is $1:1$.
Question 18
A capacitor of capacitance $C$ is fully charged to a potential difference $V$ using a battery. The battery is then disconnected, and this charged capacitor is connected in parallel to an identical but uncharged capacitor. What percentage of the initial electrostatic energy is dissipated as heat during the charge sharing process?
✅ Correct Answer: $50\%$
Explanation: Initial energy $U_i = \frac{1}{2}CV^2$. When connected to an identical uncharged capacitor, the total charge $CV$ is shared equally, so each gets $CV/2$. The new common potential is $V' = V/2$. Final equivalent capacitance $C_{eq} = 2C$. Final total energy $U_f = \frac{1}{2}(2C)(V/2)^2 = \frac{1}{4}CV^2$. Energy lost $\Delta U = U_i - U_f = \frac{1}{2}CV^2 - \frac{1}{4}CV^2 = \frac{1}{4}CV^2$, which is exactly half ($50\%$) of the initial energy.
Question 19
A heavy, uniform metal rod of length $L$, uniform cross-sectional area $A$, material density $\rho$, and Young's modulus $Y$ is suspended vertically from a rigid support. What is the total elongation of the rod solely due to its own weight?
✅ Correct Answer: $\frac{\rho g L^2}{2Y}$
Explanation: For a rod under its own weight, the tension is not uniform; it is maximum at the top and zero at the bottom. We can treat the total weight as acting at the center of mass (distance $L/2$ from the top). Using the formula $\Delta L = \frac{F \times \text{Length}}{A \times Y}$, substitute $F = Mg = (\rho A L)g$ and effective length $= L/2$. Thus, $\Delta L = \frac{(\rho A L g) (L/2)}{AY} = \frac{\rho g L^2}{2Y}$.
Question 20
TRAP QUESTION: A uniform conducting wire of resistance $R$ is bent to form a perfect circle. A battery is connected across two points on the circumference of this circle such that the circumference is divided into two unequal arcs. What is the magnitude of the magnetic field at the center of the circular loop?
✅ Correct Answer: Zero
Explanation: This is a classic trap. The current divides into the two arcs inversely proportional to their resistances (and therefore inversely proportional to their arc lengths). The longer arc carries less current, but has more length to contribute to the magnetic field. The shorter arc carries more current, but has less length. The magnetic fields produced by the two arc segments at the center will always be exactly equal in magnitude and opposite in direction, completely canceling each other out.
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