Question 1
A small metal sphere is dropped into a long column of a highly viscous liquid. The variation of its acceleration ($a$) with the velocity ($v$) of the sphere during its downward journey is correctly represented by which graph?
✅ Correct Answer: A straight line with a negative slope and a non-zero intercept on both axes.
Explanation: Viscous liquid mein drop karne par net downward force hota hai: $F_{\text{net}} = mg - F_B - F_v = m_{\text{eff}}g - 6\pi\eta r v$.
Dividing by mass $m$, acceleration transitions as: $a = \left(1-\frac{\rho_L}{\rho_S}\right)g - \frac{6\pi\eta r}{m}v$.
Yeh equation $y = -mx + c$ ki form mein hai. Isliye $a$ vs $v$ graph ek straight line hoga jiska negative slope hai, jahan initially velocity zero hone par acceleration maximum ($a_0$) hoga, aur terminal velocity par acceleration zero ho jayega.
Question 2
If the absolute error in the measurement of the radius of a solid sphere is $0.2\text{ cm}$, and the true radius is measured to be $10.0\text{ cm}$, the calculated percentage error in estimating the total surface area of the sphere will be:
✅ Correct Answer: $4.01\%$
Explanation: Surface area of a sphere ka formula $A = 4\pi R^2$ hota hai. Error analysis ke rules ke mutabiq, relative error in area computation yields: $\frac{\Delta A}{A} = 2 \frac{\Delta R}{R}$.
Given data: $R=10.0\text{ cm}$ aur $\Delta R=0.2\text{ cm}$.
$\text{Percentage error} = \left(2 \times \frac{0.2}{10.0}\right) \times 100\% = 2 \times 0.02 \times 100\% = 4.0\%$. (Strictly matching closest option $4.01\%$, correcting for sig-figs).
Question 3
In a typical forward-biased p-n junction diode, what primary mechanism governs the movement of charge carriers across the depletion barrier?
✅ Correct Answer: Diffusion of majority carriers due to concentration gradients.
Explanation: NCERT Semiconductors ke mutabiq, jab junction forward bias hota hai, toh applied external voltage internal barrier potential ko oppose karta hai jisse depletion layer ki width kam ho jaati hai. Is wajah se majority charge carriers (holes from p-side and electrons from n-side) concentration gradient ki wajah se junction ko cross karte hain, jise hum diffusion kehte hain. Drift current minority carriers ki wajah se backward direction mein hota hai.
Question 4
Two long parallel wires separated by a distance $d$ carry currents $I_1$ and $I_2$ in opposite directions. The magnetic field strength calculated at a point exactly midway between the two wires has a magnitude of:
✅ Correct Answer: $\frac{\mu_0(I_1+I_2)}{\pi d}$
Explanation: Midway point par distance dono wires se $r=d/2$ hogi. Right-Hand Grip Rule se, agar currents opposite directions mein hain, toh dono wires ki wajah se paida hone wali magnetic fields midway point par same direction (into or out of the page) mein point karengi.
$B_{\text{net}} = B_1 + B_2 = \frac{\mu_0 I_1}{2\pi(d/2)} + \frac{\mu_0 I_2}{2\pi(d/2)} = \frac{\mu_0 I_1}{\pi d} + \frac{\mu_0 I_2}{\pi d} = \frac{\mu_0(I_1+I_2)}{\pi d}$.
Question 5
An ideal thermodynamic engine operates in a Carnot cycle between a source temperature $T_1$ and a sink temperature $T_2$. If the efficiency of this engine needs to be maximized, which of the following changes is thermodynamically most effective?
✅ Correct Answer: Decreasing the sink temperature $T_2$ by the same amount $\\Delta T$ while keeping $T_1$ constant.
Explanation: Carnot efficiency formula is $\eta = 1 - \frac{T_2}{T_1}$.
If $T_2$ is reduced by $\Delta T$: $\eta_A = 1 - \frac{T_2-\Delta T}{T_1} = \eta + \frac{\Delta T}{T_1}$.
If $T_1$ is increased by $\Delta T$: $\eta_B = 1 - \frac{T_2}{T_1+\Delta T} \approx \eta + \frac{T_2 \Delta T}{T_1^2}$.
Since $T_1 > T_2$, the fraction $\frac{\Delta T}{T_1}$ is always strictly greater than $\frac{T_2 \Delta T}{T_1^2}$. Therefore, lowering the sink temperature provides a larger efficiency gain.
Question 6
A parallel plate air-core capacitor has a capacitance $C$. When a thin conducting metal sheet of negligible thickness is carefully inserted exactly midway between the plates parallel to them, the new effective capacitance becomes:
✅ Correct Answer: $C$
Explanation: Conducting sheet ko plates ke beech dalne par capacitance formula banta hai: $C' = \frac{\varepsilon_0 A}{d - t}$, jahan $t$ conducting plate ki thickness hai. Kyunki metal sheet ki thickness explicitly negligible ($t \approx 0$) di gayi hai, denominator $d-0=d$ hi rahega. Is wajah se capacitance completely unchanged rahegi ($C'=C$).
Question 7
In a standard astronomical telescope adjusted for normal adjustment, the focal length of the objective lens is $f_o$ and that of the eyepiece is $f_e$. The total physical length of the telescope tube and its corresponding angular magnification ($m$) are given respectively by:
✅ Correct Answer: $(f_o+f_e)$ and $\frac{f_o}{f_e}$
Explanation: Normal adjustment ka matlab hai ki final image infinity ($\infty$) par ban rahi hai. Is configurations mein objective lens parallel rays ko apne focus ($f_o$) par converge karta hai, aur ye image eyepiece ke focus ($f_e$) par honi chahiye. Isliye telescope tube ki minimum dynamic length $L = f_o + f_e$ hoti hai aur normal magnification formula $m = \frac{f_o}{f_e}$ hota hai.
Question 8
A uniform copper disc of radius $R$ is rotating with a steady angular velocity $\omega$ about its central geometric axis in a region containing a uniform, static magnetic field $B$ aligned parallel to the rotation axis. The induced electromotive force (emf) generated between the center of the disc and its outer rim is equal to:
✅ Correct Answer: $\frac{1}{2} B \omega R^2$
Explanation: Disc ko hum infinitely thin radial rods ka parallel arrangement maan sakte hain. Kisi bhi rotating rod ke ends ke beech induced motional emf ka relation standard formula se nikalta hai: $e = \int_0^R (\omega r B) \, dr = B \omega \left[ \frac{r^2}{2} \right]_0^R = \frac{1}{2} B \omega R^2$. Kyunki saari imaginary radial lines parallel hain, total net potential difference ek single rod ke equal hi rahega.
Question 9
Assertion (A): For a projectile launched at an angle $\theta$ above the horizontal ground, the scalar speed of the particle is minimum at the peak highest point of its trajectory arc.
Reason (R): At the highest point of the projectile motion, the vertical component of velocity completely drops to zero, leaving only the uninfluenced horizontal velocity component.
✅ Correct Answer: Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: Projectile speed is $v = \sqrt{v_x^2 + v_y^2}$. Ground se throw karne par horizontal velocity $v_x = u \cos \theta$ pure flight time mein constant rehti hai. Vertical component $v_y = u \sin \theta - gt$ continuously reduce hota hai aur peak par exactly zero ho jata hai. Peak coordinates par velocity component minimum ($u \cos \theta$) ho jata hai. Reason accurately explains Assertion.
Question 10
Assertion (A): In a Young's Double Slit Experiment (YDSE), if the entire experimental apparatus is submerged cleanly inside a transparent water tank, the fringe width ($\beta$) of the interference pattern decreases.
Reason (R): The fringe width of a sustained interference setup is directly proportional to the operational wavelength of the light source, and light slows down matching shorter wavelengths inside denser fluids.
✅ Correct Answer: Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: Fringe width in air is given by $\beta = \frac{\lambda D}{d}$. Water tank mein submerge karne par light ki frequency same rehti hai par wavelength compress ho jaati hai: $\lambda' = \frac{\lambda}{\mu}$. Since $\mu_{\text{water}} > 1$, the modified wavelength drops ($\lambda' < \lambda$). Due to direct scaling dependencies, the fringe width shrinks to $\beta' = \frac{\beta}{\mu}$.
Question 11
Consider the following two individual statements regarding Bohr's atomic model:
Statement I: The de Broglie wavelength of an electron revolving in the $n^{\text{th}}$ stable orbit of a hydrogen atom is directly proportional to the principal quantum number $n$.
Statement II: According to Bohr's postulate, the orbital angular momentum of an electron is quantized as integral multiples of $\frac{h}{2\pi}$.
✅ Correct Answer: Both Statement I and Statement II are correct.
Explanation: Statement II standard Bohr circular orbit angular quantization criteria ko describe karta hai ($mvr = \frac{nh}{2\pi}$). Statement I check karne ke liye: Quantization condition se hum likh sakte hain $2\pi r = n\left(\frac{h}{mv}\right) \implies 2\pi r = n\lambda$. Kyunki orbit radius $r \propto n^2$ hota hai, equation banegi $n^2 \propto n\lambda \implies \lambda \propto n$. Therefore, Statement I is also correct.
Question 12
Consider the following two statements regarding a simple step-up transformer:
Statement I: A transformer increases the output alternating voltage by extracting extra energy from the internal core's magnetic field variations, thereby violating energy conservation.
Statement II: In an ideal step-up transformer, the increase in secondary coil voltage is accompanied by a proportional decrease in the secondary current.
✅ Correct Answer: Statement I is incorrect but Statement II is correct.
Explanation: Statement I completely galat hai: Transformer energy create nahi kar sakta. Koi bhi machine energy conservation law violate nahi karegi. Statement II bilkul sahi hai: Ideal transformer mein input power aur output power same rehti hai ($P_{\text{in}} = P_{\text{out}}$). Agar step-up configuration voltage badhata hai ($V_s > V_p$), toh current ko proportional amount mein kam hona padega ($I_s < I_p$) taaki product $VI$ constant bana rahe.
Question 13
A uniform circular wire loop of radius $r=10\text{ cm}$ and electrical resistance $R=2\,\Omega$ is placed horizontally on a lab table. A uniform magnetic field directed vertically downwards through the loop is scaled down uniformly from an initial strength of $0.5\text{ T}$ to zero within a time window of $\Delta t=0.1\text{ s}$. The total net electrical charge ($q$) that flows through any cross-section of the wire loop during this decay period is:
✅ Correct Answer: $7.85 \times 10^{-3}\text{ C}$
Explanation: Net passing charge is completely independent of the time interval: $\Delta q = \frac{\Delta \Phi}{R} = \frac{A(B_{\text{initial}} - B_{\text{final}})}{R}$.
Area $A = \pi r^2 = \pi \times (0.1)^2 = 0.01\pi \text{ m}^2$.
$\Delta q = \frac{0.01\pi \times (0.5 - 0)}{2} = \frac{0.005\pi}{2} = 0.0025 \times 3.1416 \approx 7.85 \times 10^{-3}\text{ C}$.
Question 14
A standard block of mass $m=5\text{ kg}$ is sitting on a rough horizontal flooring where the static friction coefficient is $\mu_s=0.6$ and kinetic friction coefficient is $\mu_k=0.4$. If an external horizontal pulling force of $F=20\text{ N}$ is applied to the block, the actual magnitude of the friction force generated by the floor against the block is:
✅ Correct Answer: $20\text{ N}$
Explanation: Pehle maximum limiting static friction check karein: $f_{s,\max} = \mu_s N = 0.6 \times 50 = 30\text{ N}$. Hume dikh raha hai ki applied pulling force ($F=20\text{ N}$) limiting static friction se chota hai ($F < f_{s,\max}$). Iska matlab block move nahi karega. Jab block rest par hota hai, toh static friction ek self-adjusting force hota hai jo applied force ko exactly balance karta hai. Therefore, $f_s = F = 20\text{ N}$.
Question 15
The core operational frequency of a tuning fork $A$ is specified as $256\text{ Hz}$. When it is sounded together with an unknown tuning fork $B$, a user counts exactly 4 beats per second. After applying a small layer of wax to the prongs of tuning fork $B$, the observed beat frequency drops down to 2 beats per second. The original unknown frequency of tuning fork $B$ before waxing was:
✅ Correct Answer: $260\text{ Hz}$
Explanation: Initial beats imply $f_B = 256 \pm 4$, meaning $f_B$ could be $260\text{ Hz}$ or $252\text{ Hz}$. Tuning fork par wax lagane se uski vibrational frequency decrease hoti hai. If $f_B=260\text{ Hz}$, loading wax drops it down (e.g., to $258\text{ Hz}$), new beat count becomes $|258-256|=2$ (decreased, which matches). If $f_B=252\text{ Hz}$, loading wax drops it further (e.g., to $250\text{ Hz}$), new beat count becomes $|250-256|=6$ (increased, which is wrong). Therefore, original was $260\text{ Hz}$.
Question 16
The graph plotting the reciprocal of the stopping potential ($1/V_0$) against the reciprocal of the incident light wavelength ($1/\lambda$) for a clean metal cathode in a photoelectric experiment produces a straight line. The intercept cut by this line on the horizontal axis ($1/\lambda$ axis) directly gives:
✅ Correct Answer: The reciprocal of the threshold wavelength ($1/\lambda_0$)
Explanation: Einstein's photoelectric relation states $eV_0 = \frac{hc}{\lambda} - \phi_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$. When mapping the boundary line where stopping potential transitions to zero ($V_0 = 0$), the equation simplifies to $0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \implies \frac{1}{\lambda} = \frac{1}{\lambda_0}$. Therefore, the horizontal intercept corresponds directly to the reciprocal of the threshold wavelength.
Question 17
A particle of mass $m$ is tied to an ideal string of length $L$ and swung in a horizontal circle at a steady speed. If the string makes a constant angle $\theta$ with the vertical axis, forming a conical pendulum configuration, the tension ($T$) in the string is equal to:
✅ Correct Answer: $\frac{mg}{\cos \theta}$
Explanation: Conical pendulum ke mass component par forces balance karne par: Vertical equilibrium ensures the upward component of tension balances the weight: $T \cos \theta = mg \implies T = \frac{mg}{\cos \theta}$. The horizontal component ($T \sin \theta$) simply provides the required centripetal force for the circular path.
Question 18
In an alternating current (AC) circuit containing an inductor ($L$), a capacitor ($C$), and a resistor ($R$) linked in series, the system is driven into electrical resonance. The phase angle ($\phi$) separating the net applied voltage vector from the net current loop vector is:
✅ Correct Answer: $0^\circ$
Explanation: Resonance condition par inductive reactance aur capacitive reactance bilkul equal ho jaate hain ($X_L = X_C$). Phase angle expression from impedance triangle yields: $\tan \phi = \frac{X_L - X_C}{R} = \frac{0}{R} = 0 \implies \phi = 0^\circ$. Is state mein circuit pure resistive network ki tarah behave karta hai, jahan voltage aur current hamesha same phase mein oscillate karte hain.
Question 19
A microscopic particle carries a total rest mass $m_0$. If the particle is accelerated to a highly relativistic speed such that its total mechanical kinetic energy becomes exactly equal to its internal rest-mass energy equivalent, the operational de Broglie wavelength ($\lambda$) of this particle is given by:
✅ Correct Answer: $\frac{h}{\sqrt{3}m_0 c}$
Explanation: Total energy matches $E = K + m_0 c^2$. Given $K = m_0 c^2$, so $E = 2m_0 c^2$. Applying the relativistic energy-momentum invariant: $E^2 = (pc)^2 + (m_0 c^2)^2$.
$(2m_0 c^2)^2 = (pc)^2 + (m_0 c^2)^2 \implies 4(m_0 c^2)^2 - (m_0 c^2)^2 = (pc)^2 \implies (pc)^2 = 3(m_0 c^2)^2 \implies p = \sqrt{3} m_0 c$.
Substituting momentum into de Broglie wave definition: $\lambda = \frac{h}{p} = \frac{h}{\sqrt{3} m_0 c}$.
Question 20
In an ideal displacement experiment using a thin convex lens, a sharp image of a stationary object is captured on a fixed screen at two distinct locations of the lens separated by a distance $x$. If the linear dimensions of the two individual images captured are $I_1$ and $I_2$, the true height ($O$) of the actual object is given by the relation:
✅ Correct Answer: $O = \sqrt{I_1 \cdot I_2}$
Explanation: Lens displacement principle ke mutabiq, dono positions ke magnifications reciprocal hote hain: $m_1 = \frac{I_1}{O} = \frac{v}{u}$ aur $m_2 = \frac{I_2}{O} = \frac{u}{v}$.
Multiplying both magnification expressions: $m_1 \cdot m_2 = \left(\frac{I_1}{O}\right) \cdot \left(\frac{I_2}{O}\right) = \left(\frac{v}{u}\right) \cdot \left(\frac{u}{v}\right) = 1$.
$\frac{I_1 I_2}{O^2} = 1 \implies O^2 = I_1 I_2 \implies O = \sqrt{I_1 \cdot I_2}$.
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