A hollow conducting sphere of radius $R$ is given a positive charge $+Q$. The electric potential ($V$) and the electric field ($E$) at a point residing at a distance $r$ ($r
✅ Correct Answer: $V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$, $E=0$
Explanation: Inside a hollow conducting sphere, the net electric field $E$ is completely zero because there is no enclosed charge[cite: 392]. Because $E = -\frac{dV}{dr} = 0$, the electric potential $V$ must remain constant from the center all the way to the surface[cite: 393]. Its value inside is equal to its value on the surface: $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$[cite: 394].
Question 6
A uniform copper wire of resistance $R$ is stretched uniformly until its length increases by $0.1\%$. The percentage change in its electrical resistance will be:
✅ Correct Answer: $0.2\%$
Explanation: When a wire is stretched, its volume ($V = A \cdot l$) stays strictly constant[cite: 409]. The formula for resistance is $R = \rho\frac{l}{A} = \rho\frac{l^2}{V}$[cite: 411]. Since resistivity $\rho$ and volume $V$ are constant, $R \propto l^2$[cite: 412]. Using fractional errors for small percentage changes: $\frac{\Delta R}{R} \times 100 = 2 \left(\frac{\Delta l}{l} \times 100\right) = 2 \times 0.1\% = 0.2\%$[cite: 413, 414].
Question 7
A convex lens made of glass ($\mu_g=1.5$) has a focal length $f$ in air. When it is completely immersed in a liquid of refractive index $\mu_l=1.5$ its focal length becomes:
✅ Correct Answer: Infinite
Explanation: According to the Lens Maker's Formula: $\frac{1}{f_l} = \left(\frac{\mu_g}{\mu_l}-1\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$[cite: 429, 430]. Since the refractive index of the lens matches the liquid exactly ($\mu_g = \mu_l = 1.5$), the term $\left(\frac{\mu_g}{\mu_l}-1\right) = (1-1) = 0$[cite: 430]. Thus, $\frac{1}{f_l} = 0 \implies f_l = \infty$[cite: 431]. The lens behaves simply as a flat glass sheet and stops converging light[cite: 431].
Question 8
A solid sphere, a thin circular disc, and a hollow sphere, all possessing identical masses and radii, are released from rest at the top of an inclined plane. If they roll down cleanly without slipping, which object will reach the bottom with the maximum linear velocity?
✅ Correct Answer: Solid sphere
Explanation: Using the law of conservation of energy for pure rolling down an incline, the velocity at the bottom is given by $v = \sqrt{\frac{2gh}{1+\frac{I}{mR^2}}}$[cite: 446, 447]. The smaller the value of $\frac{I}{mR^2}$, the larger the velocity $v$[cite: 448]. For a solid sphere, $\frac{I}{mR^2} = 0.4$[cite: 449]. For a circular disc, $\frac{I}{mR^2} = 0.5$[cite: 450]. For a hollow sphere, $\frac{I}{mR^2} \approx 0.67$[cite: 451]. Since the solid sphere has the lowest value, it achieves the maximum linear velocity[cite: 452].
Question 9
Assertion (A): In an adiabatic expansion of an ideal gas, the internal energy of the system always decreases, causing a drop in temperature.
Reason (R): In an adiabatic process, there is no heat exchange between the system and its surroundings ($\Delta Q=0$).
✅ Correct Answer: Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: From the First Law of Thermodynamics, $\Delta Q = \Delta U + \Delta W$[cite: 467]. In an adiabatic process, $\Delta Q=0$, which gives $\Delta U = -\Delta W$[cite: 468]. During an expansion, the gas does positive work ($\Delta W>0$), meaning $\Delta U<0$[cite: 469]. Since the internal energy of an ideal gas depends solely on its temperature ($\Delta U = nC_v\Delta T$), a decrease in internal energy directly forces the temperature to drop[cite: 469].
Question 10
Assertion (A): When a tiny, opaque circular disc is placed directly in the path of light originating from a distant point source, a bright spot is observed right at the center of its geometric shadow.
Reason (R): Wavelets originating from the continuous circular edge of the obstacle travel equal path lengths to reach the center of the shadow, interfering constructively.
✅ Correct Answer: Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: This phenomenon describes Poisson's Spot (or Arago's Spot)[cite: 485]. Light diffracted from the outer circular rim of the opaque disc travels completely symmetric paths to the absolute center of the shadow[cite: 485]. Because the path difference is zero, they undergo constructive interference, forming a bright spot[cite: 486].
Question 11
Consider the following two statements regarding gravitational fields:
Statement I: The acceleration due to gravity ($g$) decreases monotonically as we move upwards above the Earth's surface, as well as when we move downwards towards its center.
Statement II: The gravitational potential energy of a two-body system is always negative and reaches its maximum value of zero at infinity.
✅ Correct Answer: Both Statement I and Statement II are correct.
Explanation: Statement I is correct: Above the surface, $g_h = g(1+h/R)^{-2}$, and below the surface, $g_d = g(1-d/R)$[cite: 504]. In both directions, $g$ decreases from its maximum value at the surface[cite: 505]. Statement II is correct: Since gravity is an attractive force, the potential energy formula is $U = -\frac{GMm}{r}$[cite: 506]. It remains negative everywhere and increases towards a maximum value of $0$ as $r \to \infty$[cite: 506].
Question 12
Consider the following two statements regarding Bohr's atomic model:
Statement I: The angular momentum of an electron orbiting in any permitted stationary orbit is an integral multiple of $h/2\pi$.
Statement II: When an electron transitions from a higher energy orbit to a lower energy orbit, the frequency of the emitted photon depends on the target orbit but is completely independent of the initial orbit.
✅ Correct Answer: Statement I is correct but Statement II is incorrect.
Explanation: Statement I is correct: It represents Bohr's famous quantization condition, $mvr = \frac{nh}{2\pi}$[cite: 523]. Statement II is incorrect: The energy (and frequency) of the emitted photon is determined by the difference between both levels: $h\nu = E_{\text{initial}} - E_{\text{final}}$[cite: 524]. Therefore, it depends on both orbits[cite: 525].
Question 13
A small body of mass $m$ starts from rest and moves along a straight line under the influence of an engine delivering a constant power $P$. The instantaneous velocity of the body at time $t$ is proportional to:
✅ Correct Answer: $t^{1/2}$
Explanation: Power is defined as $P = F \cdot v$[cite: 540]. We can express force as mass times acceleration: $P = \left(m \frac{dv}{dt}\right) v \implies v \, dv = \left(\frac{P}{m}\right) dt$[cite: 540, 541]. Integrating both sides with respect to time from rest: $\int_0^v v \, dv = \frac{P}{m} \int_0^t dt \implies \frac{v^2}{2} = \frac{P}{m}t \implies v = \sqrt{\frac{2P}{m}} \cdot t^{1/2}$[cite: 542]. Since $P$ and $m$ are constants, $v \propto t^{1/2}$[cite: 543].
Question 14
A straight conducting rod of length $L$ is rotated with a constant angular velocity $\omega$ about an axis passing through one of its ends and perpendicular to its length. A uniform, constant magnetic field $B$ acts parallel to the axis of rotation. The induced electromotive force (emf) developed between the two ends of the rod is:
✅ Correct Answer: $\frac{1}{2}BL^2\omega$
Explanation: Consider a small element of length $dx$ at a distance $x$ from the center of rotation[cite: 559]. The motional emf induced across this small element is $d\varepsilon = B \cdot v \cdot dx$[cite: 560]. Since $v = x\omega$, we have $d\varepsilon = B(x\omega)dx$[cite: 561, 562]. Integrating from $x=0$ to $x=L$: $\varepsilon = \int_0^L B\omega x \, dx = B\omega \left[\frac{x^2}{2}\right]_0^L = \frac{1}{2}BL^2\omega$[cite: 563, 564].
Question 15
A projectile is launched from the ground with an initial velocity $\vec{u} = u_x\hat{i} + u_y\hat{j}$, where $\hat{i}$ points horizontally along the ground and $\hat{j}$ points vertically upwards. At the highest point of its trajectory, the velocity and acceleration vectors of the projectile are:
✅ Correct Answer: $\vec{v} = u_x\hat{i}$, $\vec{a} = -g\hat{j}$
Explanation: Throughout a projectile's flight, there is no horizontal force acting on it ($a_x=0$), meaning the horizontal component of velocity remains constant at $v_x=u_x$[cite: 579]. At the absolute maximum height, the vertical component of velocity drops to zero ($v_y=0$)[cite: 580]. The only acceleration acting on the projectile is gravity, which pulls downward at all times ($\vec{a} = -g\hat{j}$)[cite: 581].
Question 16
In a potentiometer arrangement, a cell of emf $1.5\text{ V}$ gives a balance point at $30\text{ cm}$ length of the wire. If this cell is replaced by another cell of unknown emf and the balance point shifts to $40\text{ cm}$, the emf of the second cell is:
✅ Correct Answer: $2.0\text{ V}$
Explanation: The basic operating principle of a potentiometer states that the open-circuit potential drop across a balanced length of wire is directly proportional to the length itself: $E \propto l$[cite: 596]. Therefore: $\frac{E_1}{E_2} = \frac{l_1}{l_2} \implies \frac{1.5}{E_2} = \frac{30}{40}$[cite: 598]. Solving gives $E_2 = 1.5 \times \frac{40}{30} = 2.0\text{ V}$[cite: 598].
Question 17
When an inductor $L$ is connected in series with a resistor $R$ to an AC source, the phase difference between the voltage and current is $\pi/3$. If instead a capacitor $C$ is connected in series with the same resistor $R$ to the same source, the phase difference is again $\pi/3$. If $L$, $C$, and $R$ are all connected in series with the same AC source, the power factor of the circuit will be:
✅ Correct Answer: $1.0$
Explanation: From the individual circuits, $\tan(\pi/3) = \frac{X_L}{R} \implies X_L = R\sqrt{3}$ and $\tan(\pi/3) = \frac{X_C}{R} \implies X_C = R\sqrt{3}$[cite: 614, 615]. This shows that $X_L = X_C$[cite: 616]. When all three are combined in series, the circuit enters electrical resonance[cite: 616]. The net reactance becomes zero ($X = X_L - X_C = 0$), and the total impedance equals the resistance ($Z = R$)[cite: 617]. The power factor is $\cos\phi = \frac{R}{Z} = \frac{R}{R} = 1.0$[cite: 618].
Question 18
One mole of an ideal monatomic gas ($\gamma=5/3$) is mixed with one mole of a diatomic gas ($\gamma=7/5$). The explicit value of adiabatic exponent $\gamma$ for the resulting gas mixture is approximately:
✅ Correct Answer: $1.50$
Explanation: We use the formula: $n_{\text{mix}} C_{v_{\text{mix}}} = n_1 C_{v_1} + n_2 C_{v_2}$[cite: 633, 634]. For monatomic gas, $C_{v_1} = \frac{3}{2}R$. For diatomic gas, $C_{v_2} = \frac{5}{2}R$[cite: 635]. Given $n_1=1$ and $n_2=1$: $2 \cdot C_{v_{\text{mix}}} = 1 \left(\frac{3}{2}R\right) + 1 \left(\frac{5}{2}R\right) = 4R \implies C_{v_{\text{mix}}} = 2R$[cite: 636, 637, 638]. Using $C_{p_{\text{mix}}} = C_{v_{\text{mix}}} + R = 3R$[cite: 639]. The new adiabatic exponent is $\gamma_{\text{mix}} = \frac{C_{p_{\text{mix}}}}{C_{v_{\text{mix}}}} = \frac{3R}{2R} = 1.50$[cite: 640].
Question 19
A parallel plate air capacitor is charged by a DC battery to a potential difference $V_0$ and then disconnected from the source. A dielectric slab of dielectric constant $K$ is now inserted to completely fill the space between the plates. Which of the following statements describes the final state of the capacitor?
✅ Correct Answer: The charge remains unchanged, while the potential difference decreases by a factor of $K$.
Explanation: Because the battery is disconnected before inserting the dielectric, the net charge ($Q_0$) on the plates is isolated and cannot change ($Q=Q_0$)[cite: 656]. Inserting the dielectric increases the capacitance of the system to $C = KC_0$[cite: 657]. Using the relationship $V = \frac{Q}{C} = \frac{Q_0}{KC_0} = \frac{V_0}{K}$, the potential difference across the plates drops by a factor of $K$[cite: 658].
Question 20
In an experiment, the physical quantities $a, b, c$ and $d$ are measured with percentage errors of $1\%, 2\%, 3\%,$ and $4\%$ respectively. A calculated quantity $P$ is given by $P = \frac{a^3b^2}{cd}$. The maximum percentage error possible in the determination of $P$ is:
✅ Correct Answer: $14\%$
Explanation: The maximum relative error in a calculated quantity is found by summing the individual fractional errors multiplied by their respective exponents: $\frac{\Delta P}{P} = 3\left(\frac{\Delta a}{a}\right) + 2\left(\frac{\Delta b}{b}\right) + 1\left(\frac{\Delta c}{c}\right) + 1\left(\frac{\Delta d}{d}\right)$[cite: 674, 675, 676]. Converting this to percentage errors: $\%$ error in $P = 3(1\%) + 2(2\%) + 1(3\%) + 1(4\%) = 3\% + 4\% + 3\% + 4\% = 14\%$[cite: 678].
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