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Physics - 2 — MH-CET Full Length Paper MCQs with Solutions

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Sample questions with solutions

Q1 — easy
A uniform rod of mass $M$ and length $L$ is rotated about an axis perpendicular to its length passing through one end. Its moment of inertia is:
A. $\frac{1}{12}ML^2$
B. $\frac{1}{3}ML^2$  ✓ Correct
C. $\frac{1}{2}ML^2$
D. $\frac{2}{3}ML^2$
Solution: For a thin rod rotating about an end axis: $I = \frac{1}{3}ML^2$
Q2 — easy
Two stars $S_1$ and $S_2$ radiate maximum energy at wavelengths $400\text{ nm}$ and $600\text{ nm}$ respectively. According to Wien's law, the ratio of their absolute temperatures $T_1 / T_2$ is:
A. 2 : 3
B. 3 : 2  ✓ Correct
C. 4 : 9
D. 9 : 4
Solution: Wien's law: $\lambda_{\max} T = b \implies T_1/T_2 = \lambda_2/\lambda_1 = 600/400 = 3/2$
Q3 — easy
The average kinetic energy associated per translational degree of freedom of a gas molecule at absolute temperature $T$ is ($k_B$ is Boltzmann's constant):
A. $\frac{1}{2}k_B T$  ✓ Correct
B. $\frac{3}{2}k_B T$
C. $k_B T$
D. $\frac{5}{2}k_B T$
Solution: By the law of equipartition of energy, each degree of freedom carries average energy $\frac{1}{2}k_B T$
Q4 — easy
A uniform spring of force constant $k$ is cut into two equal halves. The spring constant of each individual half is:
A. $k/2$
B. $k$
C. $2k$  ✓ Correct
D. $4k$
Solution: Since $k \propto 1/L$, cutting the length in half doubles the spring constant: $k' = 2k$
Q5 — easy
The limit of resolution of an optical telescope having objective diameter $D$ for light of wavelength $\lambda$ is given by:
A. $\frac{1.22\lambda}{D}$  ✓ Correct
B. $\frac{D}{1.22\lambda}$
C. $\frac{0.61\lambda}{D}$
D. $\frac{2.44\lambda}{D}$
Solution: By Airy's diffraction formula, the angular limit of resolution for a circular aperture is $\Delta \theta = \frac{1.22\lambda}{D}$
Q6 — easy
An infinite plane sheet of charge carries uniform surface charge density $\sigma$. The electric field intensity at distance $r$ from the sheet is proportional to:
A. $r$
B. $1/r$
C. $1/r^2$
D. $r^0$ (independent of $r$)  ✓ Correct
Solution: By Gauss's law, $E = \frac{\sigma}{2\varepsilon_0}$, which is uniform and independent of distance
Q7 — easy
A cell of EMF $E$ and internal resistance $r$ is connected across an external variable load $R$. The maximum power is transferred to $R$ when:
A. $R = 2r$
B. $R = r$  ✓ Correct
C. $R = r/2$
D. $R \to \infty$
Solution: Maximum power transfer theorem requires that the load resistance match internal resistance: $R = r$
Q8 — easy
The core of a transformer is laminated specifically in order to minimize:
A. Hysteresis loss
B. Eddy current loss  ✓ Correct
C. Copper loss (Joule heating)
D. Flux leakage
Solution: Laminating the core breaks continuous circulating paths of induced currents, reducing eddy current losses
Q9 — easy
If the momentum of a free particle is doubled, its de Broglie wavelength becomes:
A. Double
B. Four times
C. One-half  ✓ Correct
D. One-fourth
Solution: $\lambda = h/p$. Doubling momentum halves the wavelength
Q10 — easy
The binding energy per nucleon of ${}^{56}_{26}\text{Fe}$ is approximately:
A. $1.1\text{ MeV}$
B. $7.6\text{ MeV}$
C. $8.8\text{ MeV}$  ✓ Correct
D. $12.4\text{ MeV}$
Solution: Iron-56 has the highest binding energy per nucleon on the nuclear curve, approximately 8.8 MeV/nucleon
Q11 — easy
A transistor common-emitter amplifier has a current gain $\beta = 49$. The value of common-base current gain $\alpha$ is:
A. 0.96
B. 0.98  ✓ Correct
C. 0.99
D. 1.02
Solution: $\alpha = \frac{\beta}{1 + \beta} = \frac{49}{50} = 0.98$
Q12 — easy
A photodiode operates efficiently when it is connected in:
A. Forward bias mode
B. Reverse bias mode  ✓ Correct
C. Unbiased equilibrium
D. Alternating polarity mode
Solution: Photodiodes are operated in reverse bias because the fractional change in reverse saturation current upon illumination is easier to detect
Q13 — easy
The Boolean expression $Y = \overline{A + B}$ represents which universal logic gate?
A. NAND
B. NOR  ✓ Correct
C. XOR
D. AND
Solution: $Y = \overline{A+B}$ represents the inverted OR operation, which is the NOR gate
Q14 — easy
A thin uniform circular ring and a solid disc have equal mass and equal radius. The ratio of their moments of inertia about their respective central transverse axes is:
A. 1 : 1
B. 2 : 1  ✓ Correct
C. 1 : 2
D. 4 : 1
Solution: $I_{\text{ring}} = MR^2$, $I_{\text{disc}} = \frac{1}{2}MR^2$. Ratio = 2:1
Q15 — easy
A stationary sound wave in an air column open at both ends has its third harmonic at $900\text{ Hz}$. The fundamental frequency of this open pipe is:
A. $150\text{ Hz}$
B. $300\text{ Hz}$  ✓ Correct
C. $450\text{ Hz}$
D. $600\text{ Hz}$
Solution: Open pipe supports all harmonics: $f_m = m \cdot f_1$. $900 = 3f_1 \implies f_1 = 300\text{ Hz}$
Q16 — easy
In pure intrinsic silicon crystal at absolute zero temperature ($T = 0\text{ K}$), the material acts as a perfect:
A. Superconductor
B. Conductor
C. Semiconductor
D. Insulator  ✓ Correct
Solution: At 0 K, all valence electrons are locked in covalent bonds; the conduction band is empty, making it a perfect insulator
Q17 — hard
A spherical raindrop of radius $r$ falls through air with terminal speed $v_t$. If two identical raindrops coalesce into a single drop, the new terminal velocity in the same medium will be:
A. $2^{1/3} v_t$
B. $2^{2/3} v_t$  ✓ Correct
C. $2 v_t$
D. $4 v_t$
Solution: Volume conserved: $R = 2^{1/3}r$. Terminal velocity $v_t \propto r^2$, so $v_t' = (R/r)^2 v_t = (2^{1/3})^2 v_t = 2^{2/3} v_t$
Q18 — hard
Water flows through a horizontal tube of non-uniform cross-section. At two points $P_1$ and $P_2$, the cross-sectional areas are $A$ and $2A$. If the speed at $P_1$ is $4\text{ m/s}$, the pressure difference $(P_1 - P_2)$ for water ($\rho = 1000\text{ kg/m}^3$) is:
A. $-6000\text{ Pa}$  ✓ Correct
B. $+6000\text{ Pa}$
C. $-8000\text{ Pa}$
D. $+8000\text{ Pa}$
Solution: Continuity: $v_2 = 2\text{ m/s}$. Bernoulli: $P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}(1000)(4 - 16) = -6000\text{ Pa}$
Q19 — hard
A rectangular loop of area $0.04\text{ m}^2$ rotates at $50\text{ rev/s}$ in a uniform magnetic field $B = 0.5\text{ T}$ perpendicular to the rotation axis. The maximum induced EMF in a coil of 100 turns is:
A. $314\text{ V}$
B. $628\text{ V}$  ✓ Correct
C. $157\text{ V}$
D. $62.8\text{ V}$
Solution: $\varepsilon_0 = NBA\omega = 100 \times 0.5 \times 0.04 \times (2\pi \times 50) = 200\pi \approx 628\text{ V}$
Q20 — hard
Two capillary tubes of equal length having radii in the ratio 1:2 are connected in series. The ratio of pressure drops across them for steady laminar flow is:
A. 16 : 1  ✓ Correct
B. 8 : 1
C. 4 : 1
D. 1 : 16
Solution: Poiseuille's law: $\Delta P \propto 1/r^4$. Ratio = $(2/1)^4 = 16:1$
Q21 — hard
During an adiabatic expansion of an ideal monatomic gas ($\gamma = 5/3$), the volume increases by a factor of 8. The temperature decreases by a factor of:
A. 2
B. 4  ✓ Correct
C. 8
D. 16
Solution: $TV^{\gamma-1} = \text{const}$. $T_2 = T_1 (1/8)^{2/3} = T_1 (1/4)$. Temperature decreases by factor 4
Q22 — hard
A parallel plate air capacitor has capacitance $C$. If a dielectric slab of dielectric constant $K = 4$ and thickness $d/2$ is inserted ($d$ is plate separation), the new capacitance is:
A. $\frac{8}{5}C$  ✓ Correct
B. $\frac{5}{8}C$
C. $2C$
D. $\frac{4}{3}C$
Solution: $C' = \frac{\varepsilon_0 A}{d - t + t/K} = \frac{\varepsilon_0 A}{d/2 + d/8} = \frac{\varepsilon_0 A}{5d/8} = \frac{8}{5}C$
Q23 — hard
A stone is dropped from a cliff of height $H$. It travels a distance of $25\text{ m}$ in its final second before hitting the ground. Taking $g = 10\text{ m/s}^2$, $H$ is:
A. $30\text{ m}$
B. $45\text{ m}$  ✓ Correct
C. $60\text{ m}$
D. $80\text{ m}$
Solution: Distance in nth second: $s_n = u + \frac{g}{2}(2n-1)$. $25 = 5(2n-1) \implies n = 3\text{ s}$. $H = \frac{1}{2}(10)(9) = 45\text{ m}$
Q24 — hard
Two equal point charges $+q$ are placed at distance $2a$ apart. A third charge $-2q$ is placed at the exact midpoint between them. The net electrostatic potential energy of the three-charge system is:
A. Zero
B. $-\frac{7q^2}{4\pi\varepsilon_0 a}$  ✓ Correct
C. $-\frac{8q^2}{4\pi\varepsilon_0 a}$
D. $-\frac{15q^2}{8\pi\varepsilon_0 a}$
Solution: $U = \frac{1}{4\pi\varepsilon_0}[\frac{(q)(-2q)}{a} + \frac{(-2q)(q)}{a} + \frac{(q)(q)}{2a}] = \frac{q^2}{4\pi\varepsilon_0 a}[-2 - 2 + 1/2] = -\frac{7q^2}{2 \cdot 4\pi\varepsilon_0 a}$
Q25 — medium
A curved highway of radius $100\text{ m}$ is banked for a design speed of $72\text{ km/h}$. Taking $g = 10\text{ m/s}^2$, the optimum angle of banking $\theta$ is:
A. $\tan^{-1}(0.2)$
B. $\tan^{-1}(0.4)$  ✓ Correct
C. $\tan^{-1}(0.5)$
D. $\tan^{-1}(0.8)$
Solution: $v = 72\text{ km/h} = 20\text{ m/s}$. $\tan\theta = \frac{v^2}{rg} = \frac{400}{100 \times 10} = 0.4 \implies \theta = \tan^{-1}(0.4)$
Q26 — medium
A ballet dancer spins with angular velocity $\omega$ with arms outstretched. When she pulls her arms inward, her moment of inertia reduces by 40%. Her new angular velocity is:
A. $1.4\omega$
B. $1.67\omega$  ✓ Correct
C. $2.5\omega$
D. $0.6\omega$
Solution: By conservation of angular momentum: $I_1 \omega_1 = I_2 \omega_2$. $I_2 = 0.6 I_1 \implies \omega_2 = \frac{I_1 \omega}{0.6 I_1} = \frac{5}{3}\omega \approx 1.67\omega$
Q27 — medium
Work done in blowing a spherical soap bubble from radius $R$ to $2R$ in air (surface tension $T$) is:
A. $12\pi T R^2$
B. $24\pi T R^2$  ✓ Correct
C. $8\pi T R^2$
D. $6\pi T R^2$
Solution: Soap bubble has two surfaces. $\Delta S = 2[4\pi (2R)^2] - 2[4\pi R^2] = 32\pi R^2 - 8\pi R^2 = 24\pi R^2$. $W = T \cdot \Delta S = 24\pi T R^2$
Q28 — medium
An ideal gas expands isothermally from volume $V_0$ to $2V_0$, doing work $W_1$. If the same gas is expanded adiabatically from $V_0$ to $2V_0$, doing work $W_2$, then:
A. $W_1 > W_2$  ✓ Correct
B. $W_1 < W_2$
C. $W_1 = W_2$
D. $W_1 = 0$
Solution: On a P-V diagram, the isothermal curve lies above the adiabatic curve from the same initial state. Area under isothermal is greater, so $W_1 > W_2$
Q29 — medium
A particle executes S.H.M. represented by $x = 0.05 \sin(100t + \pi/6)\text{ m}$. The maximum velocity of the particle is:
A. $5\text{ m/s}$  ✓ Correct
B. $0.5\text{ m/s}$
C. $50\text{ m/s}$
D. $2.5\text{ m/s}$
Solution: $A = 0.05\text{ m}$, $\omega = 100\text{ rad/s}$. $v_{\max} = A\omega = 0.05 \times 100 = 5\text{ m/s}$
Q30 — medium
The fundamental frequency of a stretched string of fixed length under tension $T$ is $n$. If the tension is increased by 69%, the percentage increase in its fundamental frequency is:
A. 30%  ✓ Correct
B. 69%
C. 13%
D. 25%
Solution: Frequency $n \propto \sqrt{T}$. $T' = 1.69T \implies n' = \sqrt{1.69} n = 1.30n$, a 30% increase