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Physics - 4 — MH-CET Full Length Paper MCQs with Solutions

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Sample questions with solutions

Q1 — easy
The flow of a liquid through a pipe of diameter $D$ is turbulent if the Reynolds number $R_e$ is:
A. $< 1000$
B. $1000 < R_e < 2000$
C. $> 2000$  ✓ Correct
D. Zero
Solution: Flow is laminar for $R_e < 1000$, unstable between 1000 and 2000, and fully turbulent when $R_e > 2000$.
Q2 — easy
The mean free path $\lambda$ of gas molecules having molecular diameter $d$ and number density $n$ is given by:
A. $\frac{1}{\sqrt{2}\pi n d^2}$  ✓ Correct
B. $\frac{\sqrt{2}}{\pi n d^2}$
C. $\frac{1}{\sqrt{2}\pi n^2 d}$
D. $\frac{1}{2\pi n d^2}$
Solution: Standard Clausius-Maxwell kinetic theory expression: $\lambda = \frac{1}{\sqrt{2}\pi n d^2}$.
Q3 — easy
The capacitance of a parallel-plate air capacitor is $8\,\mu\text{F}$. If the space between plates is completely filled with mica of dielectric constant $K = 6$, the capacitance becomes:
A. $1.33\,\mu\text{F}$
B. $24\,\mu\text{F}$
C. $48\,\mu\text{F}$  ✓ Correct
D. $14\,\mu\text{F}$
Solution: With dielectric: $C' = KC_0 = 6 \times 8\,\mu\text{F} = 48\,\mu\text{F}$.
Q4 — easy
The electric field intensity at an axial point distance $r$ from an electric dipole of moment $p$ ($r \gg$ dipole length) is proportional to:
A. $1/r$
B. $1/r^2$
C. $1/r^3$  ✓ Correct
D. $1/r^4$
Solution: Axial electric field: $E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3} \propto 1/r^3$.
Q5 — easy
Ampere's circuital law around any closed path enclosing net steady current $I_{\text{enc}}$ is stated mathematically as:
A. $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$  ✓ Correct
B. $\oint \vec{B}\times d\vec{l} = \mu_0 I_{\text{enc}}$
C. $\oint \vec{B}\cdot d\vec{A} = \mu_0 I_{\text{enc}}$
D. $\oint \vec{B}\cdot d\vec{l} = 0$
Solution: Standard Ampere circuital integral: $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$.
Q6 — easy
A charged particle enters a transverse magnetic field with pitch angle $90^\circ$. The work done by the magnetic Lorentz force on the charged particle along its path is always:
A. Positive
B. Negative
C. Zero  ✓ Correct
D. Dependent on particle speed
Solution: Magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ is always perpendicular to $\vec{v}$. Work done $= \vec{F}_m \cdot d\vec{s} = 0$.
Q7 — easy
The curie temperature $T_c$ of a ferromagnetic material represents the transition temperature above which the material behaves as:
A. Diamagnetic
B. Paramagnetic  ✓ Correct
C. Superconducting
D. Ferrimagnetic
Solution: Above the Curie temperature, thermal agitation destroys domain alignment, transforming ferromagnets into paramagnets.
Q8 — easy
The magnetic energy stored per unit volume in a uniform magnetic field $B$ in vacuum is given by:
A. $\frac{B^2}{2\mu_0}$  ✓ Correct
B. $\frac{1}{2}\mu_0 B^2$
C. $\frac{B}{2\mu_0}$
D. $\frac{B^2}{\mu_0}$
Solution: Energy density: $u_B = \frac{B^2}{2\mu_0}$.
Q9 — easy
In a series LCR alternating current circuit, the phase angle $\phi$ between applied voltage and resulting current is zero when:
A. $X_L > X_C$
B. $X_L < X_C$
C. $X_L = X_C$  ✓ Correct
D. $R = 0$
Solution: At resonance: $X_L = X_C$, impedance $Z = R$, phase angle $\phi = 0$.
Q10 — easy
According to Bohr's postulate, the orbital angular momentum of an electron revolving in the $n^{\text{th}}$ stationary orbit is quantized as:
A. $\frac{nh}{2\pi}$  ✓ Correct
B. $\frac{2\pi n}{h}$
C. $\frac{n^2 h}{2\pi}$
D. $\frac{h}{2\pi n}$
Solution: Bohr quantization: $L = \frac{nh}{2\pi}$.
Q11 — easy
In a common emitter (CE) NPN transistor amplifier, the phase difference between the input AC signal and the output amplified signal is:
A. $0^\circ$
B. $90^\circ$
C. $180^\circ$  ✓ Correct
D. $270^\circ$
Solution: In CE configuration, output is $180^\circ$ out of phase with the input signal.
Q12 — easy
The decimal equivalent of binary number $(1101)_2$ is:
A. 11
B. 13  ✓ Correct
C. 15
D. 9
Solution: $(1101)_2 = 1(2^3) + 1(2^2) + 0(2^1) + 1(2^0) = 8 + 4 + 0 + 1 = 13$.
Q13 — easy
During an isothermal expansion of an ideal gas, the change in its internal energy ($\Delta U$) is:
A. Positive
B. Negative
C. Zero  ✓ Correct
D. Proportional to work done
Solution: Internal energy of an ideal gas depends only on temperature. Isothermal ($\Delta T = 0$) means $\Delta U = 0$.
Q14 — easy
Which of the following pairs of physical quantities have identical dimensional formulas?
A. Work and Torque  ✓ Correct
B. Force and Pressure
C. Momentum and Energy
D. Power and Force
Solution: Work $= [ML^2T^{-2}]$ and Torque $= [L][MLT^{-2}] = [ML^2T^{-2}]$. Same dimensions.
Q15 — easy
In an intrinsic semiconductor, the Fermi energy level at absolute zero temperature lies:
A. In the middle of the forbidden energy gap  ✓ Correct
B. Inside the conduction band
C. Inside the valence band
D. At the top edge of conduction band
Solution: In an intrinsic semiconductor at $T = 0\text{ K}$, the Fermi level lies precisely at the midpoint of the forbidden bandgap.
Q16 — medium
A motor cyclist loops a vertical globe of death of diameter $10\text{ m}$. The minimum speed he must maintain at the topmost point to avoid falling off is ($g = 10\text{ m/s}^2$):
A. $5\text{ m/s}$
B. $7.07\text{ m/s}$  ✓ Correct
C. $10\text{ m/s}$
D. $14.14\text{ m/s}$
Solution: Diameter $D = 10\text{ m} \implies$ radius $r = 5\text{ m}$. Critical velocity at apex: $v_{\text{top}} = \sqrt{gr} = \sqrt{10 \times 5} = \sqrt{50} \approx 7.07\text{ m/s}$.
Q17 — medium
The moment of inertia of a uniform circular disc of mass $M$ and radius $R$ about a tangent in its plane is:
A. $\frac{1}{4}MR^2$
B. $\frac{1}{2}MR^2$
C. $\frac{5}{4}MR^2$  ✓ Correct
D. $\frac{3}{2}MR^2$
Solution: About a diameter: $I_d = \frac{1}{4}MR^2$. By parallel axis theorem, tangent in plane is at distance $R$: $I = I_d + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$.
Q18 — medium
A constant external torque of $20\text{ N}\cdot\text{m}$ acts on a flywheel of moment of inertia $5\text{ kg}\cdot\text{m}^2$ initially at rest. The kinetic energy acquired by the flywheel in $4\text{ seconds}$ is:
A. $320\text{ J}$
B. $640\text{ J}$  ✓ Correct
C. $160\text{ J}$
D. $1280\text{ J}$
Solution: Angular acceleration $\alpha = \tau / I = 20/5 = 4\text{ rad/s}^2$. At $t=4\text{ s}$: $\omega = \alpha t = 16\text{ rad/s}$. KE $= \frac{1}{2}I\omega^2 = \frac{1}{2}(5)(256) = 640\text{ J}$.
Q19 — medium
A U-tube contains water and kerosene separated by mercury. If mercury levels in both limbs are equal, the ratio of column heights of water and kerosene ($h_w / h_k$) is ($\rho_w = 1.0\text{ g/cm}^3$, $\rho_k = 0.8\text{ g/cm}^3$):
A. 0.8  ✓ Correct
B. 1.25
C. 1.0
D. 0.64
Solution: Equal hydrostatic pressure at mercury interface: $\rho_w g h_w = \rho_k g h_k \implies \frac{h_w}{h_k} = \frac{\rho_k}{\rho_w} = \frac{0.8}{1.0} = 0.8$.
Q20 — medium
A needle of length $5\text{ cm}$ floats on the surface of water. The minimum downward force required to pull the needle off the surface is ($T_{\text{water}} = 0.07\text{ N/m}$):
A. $3.5 \times 10^{-3}\text{ N}$
B. $7.0 \times 10^{-3}\text{ N}$  ✓ Correct
C. $1.4 \times 10^{-2}\text{ N}$
D. $7.0\text{ N}$
Solution: Two boundary contact lines: total length $L_{\text{eff}} = 2l$. Force $F = 2lT = 2(0.05)(0.07) = 7.0 \times 10^{-3}\text{ N}$.
Q21 — medium
In an isobaric heating process of an ideal diatomic gas, the fraction of total heat absorbed that goes into performing external work ($\Delta W / Q$) is:
A. $2/7$  ✓ Correct
B. $5/7$
C. $2/5$
D. $3/5$
Solution: At constant pressure: $\frac{\Delta W}{Q} = \frac{R}{C_p} = \frac{R}{\frac{7}{2}R} = \frac{2}{7}$.
Q22 — medium
A black body radiator at $27^\circ\text{C}$ radiates heat at a rate of $5\text{ W/cm}^2$. The rate of heat radiation by the same body at $327^\circ\text{C}$ is:
A. $10\text{ W/cm}^2$
B. $40\text{ W/cm}^2$
C. $80\text{ W/cm}^2$  ✓ Correct
D. $160\text{ W/cm}^2$
Solution: Stefan's law: $E \propto T^4$. $T_1 = 300\text{ K}$, $T_2 = 600\text{ K}$. Ratio $= (600/300)^4 = 16$. $E_2 = 16 \times 5 = 80\text{ W/cm}^2$.
Q23 — medium
A mass $m$ is suspended simultaneously from two identical vertical springs each of force constant $k$ connected in parallel. The time period of vertical oscillation is:
A. $2\pi\sqrt{\frac{m}{k}}$
B. $2\pi\sqrt{\frac{m}{2k}}$  ✓ Correct
C. $2\pi\sqrt{\frac{2m}{k}}$
D. $\pi\sqrt{\frac{m}{k}}$
Solution: Parallel springs: $k_{\text{eq}} = 2k$. Time period $T = 2\pi\sqrt{\frac{m}{k_{\text{eq}}}} = 2\pi\sqrt{\frac{m}{2k}}$.
Q24 — medium
A simple harmonic oscillator has total mechanical energy $E$. At half the maximum amplitude ($x = A/2$), its potential energy is:
A. $E/2$
B. $E/4$  ✓ Correct
C. $3E/4$
D. $E/\sqrt{2}$
Solution: Total energy $E = \frac{1}{2}kA^2$. PE at $x = A/2$: $U = \frac{1}{2}k(A/2)^2 = \frac{1}{4} \cdot \frac{1}{2}kA^2 = E/4$.
Q25 — medium
Two tuning forks when sounded together produce 5 beats per second. When the prong of the fork of higher frequency is loaded with a little wax, the beat frequency decreases to 2 beats/s. If the other fork has frequency $256\text{ Hz}$, the original frequency of the loaded fork was:
A. $251\text{ Hz}$
B. $261\text{ Hz}$  ✓ Correct
C. $258\text{ Hz}$
D. $254\text{ Hz}$
Solution: Loading with wax decreases frequency and decreases beat frequency from 5 to 2. So $n_A > n_B$: $n_A = 256 + 5 = 261\text{ Hz}$.
Q26 — medium
The velocity of sound in a gas is $v$. If the absolute temperature of the gas is made 4 times and the pressure is doubled, the new speed of sound is:
A. $v$
B. $2v$  ✓ Correct
C. $4v$
D. $v\sqrt{2}$
Solution: Speed of sound $v = \sqrt{\frac{\gamma RT}{M}}$, depends only on temperature (independent of pressure). If $T' = 4T$, $v' = \sqrt{4}v = 2v$.
Q27 — medium
In a single-slit Fraunhofer diffraction pattern, the first diffraction minimum for light of wavelength $600\text{ nm}$ occurs at an angle of $30^\circ$. The width of the slit is:
A. $1.2\,\mu\text{m}$  ✓ Correct
B. $0.6\,\mu\text{m}$
C. $2.4\,\mu\text{m}$
D. $0.3\,\mu\text{m}$
Solution: First minimum: $a \sin\theta = \lambda \implies a = \frac{600 \times 10^{-9}}{\sin 30^\circ} = \frac{600 \times 10^{-9}}{0.5} = 1.2 \times 10^{-6}\text{ m} = 1.2\,\mu\text{m}$.
Q28 — medium
Two coherent sources of monochromatic light produce an interference pattern on a screen. If the ratio of maximum to minimum intensity in the pattern is $9:1$, the ratio of their individual amplitudes is:
A. 2 : 1  ✓ Correct
B. 3 : 1
C. 9 : 1
D. 4 : 1
Solution: $\frac{I_{\max}}{I_{\min}} = \left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2 = 9 \implies \frac{A_1+A_2}{A_1-A_2} = 3 \implies A_1/A_2 = 2/1$.
Q29 — medium
A uniform copper wire of resistance $20\,\Omega$ is bent into a closed circular loop. The effective electrical resistance between two diametrically opposite points of this loop is:
A. $20\,\Omega$
B. $10\,\Omega$
C. $5\,\Omega$  ✓ Correct
D. $2.5\,\Omega$
Solution: Two parallel semicircles each of resistance $10\,\Omega$. $R_{\text{eff}} = \frac{10 \times 10}{10 + 10} = 5\,\Omega$.
Q30 — medium
In a meter bridge experiment, balance is obtained at $20\text{ cm}$ from the left end when unknown resistance $X$ is placed in the left gap and standard $12\,\Omega$ resistor in the right gap. The value of $X$ is:
A. $2\,\Omega$
B. $3\,\Omega$  ✓ Correct
C. $4\,\Omega$
D. $6\,\Omega$
Solution: $\frac{X}{R} = \frac{l}{100-l} \implies \frac{X}{12} = \frac{20}{80} = \frac{1}{4} \implies X = 3\,\Omega$.