Question 1
In a certain system of absolute units, the velocity of light in vacuum ($c$), Newton's gravitational constant ($G$), and Planck's constant ($h$) are chosen as the fundamental base dimensions. The dimensional formula for mass in this new alternative system is given by:
✅ Correct Answer: $[c^{1/2}G^{-1/2}h^{1/2}]$
Explanation: Let mass $M \propto c^x G^y h^z$. Putting the standard dimensions: $[c]=[LT^{-1}]$, $[G]=[M^{-1}L^3T^{-2}]$, and $[h]=[ML^2T^{-1}]$.
$[M^1L^0T^0] = [LT^{-1}]^x [M^{-1}L^3T^{-2}]^y [ML^2T^{-1}]^z$
Comparing powers of $M$, $L$, and $T$, we solve the equations:
$-y+z=1$, $x+3y+2z=0$ and $-x-2y-z=0$
Solving these simultaneously gives $x=1/2$, $y=-1/2$, and $z=1/2$. Therefore, $M = [c^{1/2} G^{-1/2} h^{1/2}]$.
Question 2
A solid copper sphere of radius $R$ is given a total net positive charge $+Q$. The variation of the electric potential ($V$) as a function of the distance $r$ measured from the absolute center of the sphere is best represented by:
✅ Correct Answer: $V$ remains completely constant from $r=0$ to $r=R$, and decreases as $1/r$ for $r > R$.
Explanation: Copper ek metal (conductor) hai. NCERT Electrostatics ke mutabiq, kisi bhi conductor ke andar electric field zero hoti hai ($E_{\text{inside}}=0$). Kyunki $E=-\frac{dV}{dr}$, agar $E=0$ hai toh potential $V$ pure interior volume mein strictly constant rahega aur uski value surface potential ($\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$) ke barabar hogi. Outside the sphere, it behaves like a point charge ($V \propto 1/r$).
Question 3
The true value of the angle of dip at a certain geographic location is $\delta$. If the vertical plane of the magnetic compass needle is turned or misaligned by an angle $\theta$ away from the local magnetic meridian, the apparent angle of dip observed will be $\delta'$. The correct mathematical relationship linking these parameters is:
✅ Correct Answer: $\tan \delta' = \frac{\tan \delta}{\cos \theta}$
Explanation: True dip definition: $\tan \delta = \frac{B_V}{B_H}$. When the dip circle plane is rotated by an angle $\theta$ relative to the magnetic meridian, the vertical component $B_V$ remains exactly the same, but the horizontal component becomes $B_H' = B_H \cos \theta$.
Apparent dip layout: $\tan \delta' = \frac{B_V}{B_H'} = \frac{B_V}{B_H \cos \theta} = \frac{\tan \delta}{\cos \theta}$.
Question 4
In a classic thermodynamics cycle, an ideal monoatomic gas expands to exactly double its initial volume. This volume doubling can be achieved via three different individual processes: Isothermal, Isobaric, and Adiabatic. Let $W_{\text{iso}}$, $W_{\text{bar}}$, and $W_{\text{ad}}$ represent the work done by the gas in these respective expansions. The correct order of magnitude comparing these work values is:
✅ Correct Answer: $W_{\text{bar}} > W_{\text{iso}} > W_{\text{ad}}$
Explanation: On a standard $P-V$ diagram starting from identical initial coordinates ($P_1, V_1$), the isobaric expansion line is horizontal (highest pressure sustained), the isothermal curve drops gradually, and the adiabatic curve drops most steeply because its slope is $\gamma$ times greater. Work done corresponds to the total area bounded under the curve on the $V$-axis. Therefore, the area is maximum for isobaric and minimum for adiabatic expansions.
Question 5
An open water reservoir has a tiny structural pinhole leak in its side wall at a depth $h$ below the upper free surface of the water. If the top opening of the reservoir is tightly sealed with a heavy airtight piston maintaining an extra external gauge pressure $P$, the velocity of efflux ($v$) of the escaping water jet is given by:
✅ Correct Answer: $v=\sqrt{2gh+\frac{2P}{\rho}}$
Explanation: Applying Bernoulli's Theorem between the top surface (Point 1) and the exit pinhole (Point 2):
$P_{\text{top}} + \rho g h + \frac{1}{2}\rho v_1^2 = P_{\text{atm}} + 0 + \frac{1}{2}\rho v^2$
Since the reservoir area is much greater than the pinhole ($A_1 \gg A_2$), $v_1 \approx 0$.
Given $P_{\text{top}} = P_{\text{atm}} + P$:
$(P_{\text{atm}} + P) + \rho g h = P_{\text{atm}} + \frac{1}{2}\rho v^2 \implies P + \rho g h = \frac{1}{2}\rho v^2$
Solving for velocity: $v=\sqrt{2gh+\frac{2P}{\rho}}$.
Question 6
A parallel plate capacitor filled with air between its plates has a capacitance $C_0$. The space between the plates is now completely filled by inserting two identical dielectric slabs of dielectric constants $K_1$ and $K_2$ in a stacked horizontal arrangement, such that each slab occupies exactly half of the original separation distance ($d/2$). The new modified capacitance $C$ of the system is:
✅ Correct Answer: $C=\frac{2C_0 K_1 K_2}{K_1+K_2}$
Explanation: Jab slabs ko stacked horizontal layer profile ($d/2$ motai) mein insert karte hain, toh system do individual capacitors ka series combination ban jata hai.
$C_1 = \frac{K_1 \varepsilon_0 A}{d/2} = 2K_1 C_0$, \quad $C_2 = \frac{K_2 \varepsilon_0 A}{d/2} = 2K_2 C_0$
Using the series equivalent formula ($C = \frac{C_1 C_2}{C_1 + C_2}$):
$C = \frac{(2K_1 C_0)(2K_2 C_0)}{2K_1 C_0 + 2K_2 C_0} = \frac{4K_1 K_2 C_0^2}{2C_0(K_1+K_2)} = \frac{2C_0 K_1 K_2}{K_1+K_2}$.
Question 7
In a standard single-slit diffraction experiment, the width of the primary slit is chosen as $a$. If the slit is illuminated using a monochromatic light source of wavelength $\lambda$, the angular width of the central diffraction maximum is given by:
✅ Correct Answer: $\frac{2\lambda}{a}$
Explanation: Single-slit diffraction mein, first minimum key position $\sin \theta \approx \theta = \frac{\lambda}{a}$ standard conditional line par milti hai. Central maximum dono taraf ke first minima ke beech mein spread hota hai. Isliye total angular width central maxima ki $\omega_\theta = 2\theta = \frac{2\lambda}{a}$ hoti hai.
Question 8
A massive metal block is suspended from the ceiling via an ideal spring. The system oscillates vertically with a time period $T$. If this entire mechanical assembly is carefully transferred inside a frictionless liquid tank where the liquid density is exactly half the density of the metal block, the new time period of vertical oscillations will be:
✅ Correct Answer: $T$
Explanation: The time period of a spring-mass system is given strictly by $T = 2\pi\sqrt{\frac{m}{k}}$. Inside the liquid, the constant upward buoyant force alters the static equilibrium position of the block, shifting its mean center down. However, it does not alter the mass ($m$) or the spring constant ($k$). Since the restoring force gradient ($F=-kx$) remains unchanged, the time period stays exactly $T$.
Question 9
Assertion (A): In a radioactive decay process, the mean life ($\tau$) of a sample is always numerically greater than its corresponding half-life ($T_{1/2}$).
Reason (R): The decay constant ($\lambda$) is inversely related to both the mean life and the half-life parameters.
✅ Correct Answer: Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
Explanation: The mathematical relationships are $\tau = \frac{1}{\lambda}$ and $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}$. Since $1 > 0.693$, it is always true that $\tau > T_{1/2}$, so Assertion is true. Reason is also completely true because $\lambda = \frac{1}{\tau}$ and $\lambda = \frac{0.693}{T_{1/2}}$. However, the mere fact that both are inversely proportional to $\lambda$ does not explain why mean life is longer; that depends specifically on the fractional factor $\ln 2 < 1$.
Question 10
Assertion (A): When a metallic surface is illuminated with light of a frequency less than the threshold frequency, no photoelectrons are emitted, regardless of how intense the light is.
Reason (R): According to the photon picture of light, an increase in intensity merely increases the number of photons striking the surface per second, but does not increase the energy of individual photons.
✅ Correct Answer: Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: NCERT Dual Nature explicitly states that photoelectric emission is a one-to-one collision event between a single incoming photon and a bound electron. If the individual photon's energy ($E=h\nu$) is less than the work function ($\phi_0=h\nu_0$), no emission occurs. Increasing the intensity simply means more photons are delivered per second, but because each photon still lacks sufficient energy, the electron cannot be liberated.
Question 11
Consider the following two statements regarding a transistor switch circuit:
Statement I: In a common-emitter transistor configuration operating as a digital logic switch, the transistor is driven into its cutoff state to represent a logical 'LOW' output state.
Statement II: When the transistor is completely biased into its saturation region, the collector-emitter voltage ($V_{CE}$) drops close to zero, acting like a closed switch.
✅ Correct Answer: Statement I is incorrect but Statement II is correct.
Explanation: Statement I is incorrect: In an inverter/switch circuit, when the input is 'LOW', the transistor is in the cutoff region (no current flows). At this point, the output voltage taken across the collector is $V_{\text{out}} = V_{CC}$, which corresponds to a logical 'HIGH' state. Statement II is correct: When input is 'HIGH', it enters saturation, maximum current flows, and $V_{\text{out}} = V_{CE} \approx 0\text{V}$ ('LOW' output state, closed switch behavior).
Question 12
Consider the following two statements regarding mechanics and torque vectors:
Statement I: For a rigid body in structural equilibrium, the net vector sum of all external forces acting on it must be zero, but the net external torque can be non-zero.
Statement II: The numerical value of the torque acting on a body depends strictly on the choice of the origin or reference point about which the tracking coordinates are calculated.
✅ Correct Answer: Statement I is incorrect but Statement II is correct.
Explanation: Statement I is incorrect: For complete structural equilibrium, both the net external force $\sum \vec{F}_{\text{ext}} = 0$ (translational equilibrium) and the net external torque $\sum \vec{\tau}_{\text{ext}} = 0$ (rotational equilibrium) must be zero simultaneously. Statement II is correct: Since torque is defined by the cross product $\vec{\tau} = \vec{r} \times \vec{F}$, changing the origin shifts the position vector $\vec{r}$, which directly alters the torque value.
Question 13
A car of mass $m$ starts from rest at time $t=0$ and accelerates along a straight horizontal road such that the instantaneous power ($P$) delivered by the engine remains a constant value over time. The velocity ($v$) of the car as a function of time ($t$) scales proportional to:
✅ Correct Answer: $v \propto t^{1/2}$
Explanation: Power is defined as $P = F \cdot v = \left(m \frac{dv}{dt}\right) \cdot v$. Rearranging terms to integrate:
$v \, dv = \frac{P}{m} \, dt \implies \int_0^v v \, dv = \frac{P}{m} \int_0^t dt$
$\frac{v^2}{2} = \frac{P}{m} t \implies v = \sqrt{\frac{2P}{m}} \cdot t^{1/2}$
Since $P$ and $m$ are constants, $v \propto t^{1/2}$.
Question 14
In a moving coil galvanometer, the magnetic field is made radial by using concave pole pieces of a magnet. If the linear current sensitivity of the galvanometer is increased by exactly 20% by increasing the number of turns of the coil, while the internal resistance of the coil simultaneously increases by 50%, the voltage sensitivity of the galvanometer will change by:
✅ Correct Answer: Decreases by 20%
Explanation: Current sensitivity $I_s = \frac{\theta}{I} = \frac{NBA}{C}$. Voltage sensitivity $V_s = \frac{\theta}{V} = \frac{\theta}{IR} = \frac{I_s}{R}$.
New current sensitivity $I_s' = 1.2 I_s$. New resistance $R' = 1.5 R$.
The new modified voltage sensitivity is: $V_s' = \frac{I_s'}{R'} = \frac{1.2 I_s}{1.5 R} = 0.8 V_s$.
A drop to 0.8 times the initial value means it has decreased by 20% ($1 - 0.8 = 0.2$ or 20%).
Question 15
A satellite of mass $m$ is revolving around Earth in a stable circular orbit of radius $R$. If the satellite is to be shifted to a higher stable circular orbit of radius $2R$, the minimum amount of extra energy that must be supplied to the satellite from an external mechanism is:
✅ Correct Answer: $\frac{G M m}{4R}$
Explanation: Total energy of a satellite in orbit is $E = -\frac{GMm}{2r}$.
Initial Energy ($E_1$) at radius $R = -\frac{GMm}{2R}$.
Final Energy ($E_2$) at radius $2R = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$.
Energy to be supplied: $\Delta E = E_2 - E_1 = -\frac{GMm}{4R} - \left(-\frac{GMm}{2R}\right) = \frac{GMm}{4R}$.
Question 16
The temperature-resistance ($R$ vs $T$) graph for a clean semiconducting material (such as pure Germanium or Silicon) follows which general trend path?
✅ Correct Answer: A smooth curve sloping downwards non-linearly towards the temperature axis.
Explanation: Semiconductors have a negative temperature coefficient of resistance ($\alpha < 0$). As temperature increases, more covalent bonds rupture, releasing a large number of charge carriers (electrons and holes). This exponential rise in carrier concentration outweighs the increased lattice scattering, causing the net resistance and resistivity to drop non-linearly with temperature.
Question 17
A radioactive atomic nucleus at rest undergoes alpha decay according to the equation $X^A \rightarrow Y^{A-4} + \alpha^4$. If the total nuclear reaction energy release ($Q$-value) is $Q$, the kinetic energy ($K_\alpha$) carried away specifically by the emitted alpha particle is:
✅ Correct Answer: $\frac{A-4}{A} Q$
Explanation: By momentum conservation, the daughter nucleus $Y$ and the alpha particle must have equal and opposite momentum magnitudes ($p_Y = p_\alpha = p$). The total kinetic energy released is the $Q$-value: $Q = \frac{p^2}{2m_Y} + \frac{p^2}{2m_\alpha}$. Given $m_Y \propto (A-4)$ and $m_\alpha \propto 4$:
$Q = \frac{p^2}{2} \left[ \frac{1}{A-4} + \frac{1}{4} \right] = \frac{p^2}{2} \left[ \frac{A}{4(A-4)} \right]$
Since $K_\alpha = \frac{p^2}{2m_\alpha} = \frac{p^2}{2 \times 4}$, we substitute this back into the expression:
$Q = K_\alpha \cdot \frac{A}{A-4} \implies K_\alpha = \left(\frac{A-4}{A}\right) Q$.
Question 18
Two ideal thin convex lenses of focal lengths $f_1=15\text{ cm}$ and $f_2=30\text{ cm}$ are placed coaxially in contact with each other. This combined lens system is now used to form an image of an object placed infinitely far away. The effective focal length ($F$) of the combination is:
✅ Correct Answer: $10\text{ cm}$
Explanation: When two thin lenses are in contact, the equivalent focal length $F$ of the lens combination is given by:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
$\frac{1}{F} = \frac{1}{15} + \frac{1}{30} = \frac{2+1}{30} = \frac{3}{30} = \frac{1}{10} \implies F = 10\text{ cm}$.
Question 19
An electromagnetic wave is propagating through vacuum along the positive z-direction ($+\hat{k}$). At a specific point in space and time, the instantaneous electric field vector is directed along the positive x-axis ($\vec{E} = E_0 \hat{i}$). The direction of the corresponding magnetic field vector ($\vec{B}$) at that same instant must be pointing along:
✅ Correct Answer: The positive y-axis ($+\hat{j}$)
Explanation: NCERT EM Waves states that the direction of propagation of an electromagnetic wave is given by the cross product of the electric field vector and the magnetic field vector ($\vec{E} \times \vec{B}$).
Direction of wave propagation = $+\hat{k}$. Given $\vec{E}$ direction = $+\hat{i}$.
We know that $\hat{i} \times \hat{j} = \hat{k}$. Therefore, the magnetic field vector $\vec{B}$ must point along the positive y-axis ($+\hat{j}$).
Question 20
In an experimental setup studying the photoelectric effect, a student plots a graph of the stopping potential ($V_0$) on the vertical axis against the frequency ($\nu$) of the incident radiation on the horizontal axis. The mathematical slope of this line is a universal ratio given by:
✅ Correct Answer: $h/e$
Explanation: Einstein's photoelectric equation is $h\nu = \phi_0 + eV_0$.
Rearranging the formula to isolate the stopping potential ($V_0$) on the y-axis:
$V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$
Comparing this with the equation of a straight line ($y = mx + c$), the slope $m$ is exactly equal to $\frac{h}{e}$.
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