MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
NEET
NEET Previous Year Question Papers
Quiz 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Q.1
A spring of force constant k is cut into lengths of ratio 1 : 2 :They are connected in series and the new force constant is \(K'\). Then they are connected in parallel and force constant is \(K''\). Then \(K' : K''\) is
0%
1 : 6
0%
1 : 9
0%
1 : 11
0%
1 : 14
Explanation
(The ratio, cut off in the source text, is the well-known 1:2:3 version of this question.) Cutting spring constant k into pieces of length ratio 1:2:3 (i.e. 1/6, 2/6, 3/6 of the total length) gives individual constants k₁=6k, k₂=3k, k₃=2k (a piece's constant scales inversely with its length fraction). Series (recombining ALL the cut pieces in series always reconstructs the original spring): 1/K' = 1/6k + 1/3k + 1/2k = 6/6k = 1/k → K' = k. Parallel: K'' = k₁+k₂+k₃ = 6k+3k+2k = 11k. K':K'' = k : 11k = 1:11.
Q.2
The given electrical network is equivalent to
0%
AND gate
0%
OR gate
0%
NOR gate
0%
NOT gate
Explanation
Tracing the logic: the first NOR gate computes Q1 = NOR(A,B) = NOT(A OR B). That same signal feeds both inputs of the second NOR gate, giving Q2 = NOR(Q1,Q1) = NOT(Q1) = (A OR B) (since ORing a signal with itself just returns the signal, and NOR then inverts it). The final NOT gate then inverts Q2, giving Y = NOT(A OR B) = NOR(A,B) — so overall, this whole arrangement is equivalent to a single NOR gate.
Q.3
A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F., because the method involves :
0%
Cells
0%
Potential gradients
0%
A condition of no current flow through the galvanometer
0%
A combination of cells, galvanometer and resistances
Explanation
A potentiometer's accuracy comes from the fact that at the balance point, no current flows through the galvanometer at all — unlike a voltmeter, which always draws some current from the circuit and slightly disturbs the very voltage it's measuring.
Q.4
A gas mixture consists of 2 moles of \(O_2\) and 4 moles of \(Ar\) at temperature \(T\). Neglecting all vibrational modes, the total internal energy of the system is
0%
\(4RT\)
0%
\(15 RT\)
0%
\(9RT\)
0%
\(11RT\)
Explanation
Internal energy per mole: monatomic = (3/2)RT (3 translational degrees of freedom), diatomic (ignoring vibration) = (5/2)RT (3 translational + 2 rotational). O₂ (diatomic, 2 mol): 2 × (5/2)RT = 5RT Ar (monatomic, 4 mol): 4 × (3/2)RT = 6RT Total = 5RT + 6RT = 11RT.
Q.5
Radioactive material \(A\) has a decay constant \(8\lambda\) and material \(B\) has a decay constant \(8\lambda\). Initially, they have the same number of nuclei. After what time, the ratio of number of nuclei of material \(B\) to that \(A\) will be \(\frac{1}{e}\)
0%
\(\frac{1}{\lambda}\)
0%
\(\frac{1}{7\lambda}\)
0%
\(\frac{1}{8\lambda}\)
0%
\(\frac{1}{9\lambda}\)
Explanation
(The source text lists both decay constants as "8λ" — a scraping duplication. The famous original version of this question has material A with decay constant λ and material B with decay constant 8λ, which is what the marked answer is based on.) N_A = N₀e^(−λt), N_B = N₀e^(−8λt) N_B/N_A = e^(−8λt+λt) = e^(−7λt) Setting this equal to 1/e: e^(−7λt) = e^(−1) → 7λt = 1 → t = 1/(7λ).
Q.6
A U tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile, the water rises by 65 mm from its original level (see diagram). The density of the oil is
0%
\(650\,\,kg\,\,m^{-3}\)
0%
\(425\,\,kg\,\,m^{-3}\)
0%
\(800\,\,kg\,\,m^{-3}\)
0%
\(928\,\,kg\,\,m^{-3}\)
Explanation
Water rises 65 mm on the water-only side, so (by conservation of water volume in equal-bore tubes) it drops 65 mm on the oil side — a net water-level difference of 65+65 = 130 mm between the two arms. The oil's own surface sits 10 mm above the FINAL (risen) water level on the other side, so the oil column's total height is 130 + 10 = 140 mm. Balancing pressure at the level of the lower (oil-side) water surface: the oil column's pressure must equal the pressure of the 130 mm net water-level difference: ρ_oil × g × 140 = ρ_water × g × 130 ρ_oil = 1000 × 130/140 ≈ 928 kg/m³.
Q.7
A 250-Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of \(85 \mu A\) and subjected to a magnetic field of strength \(0.85 T\). Work done for rotating the coil by \(180^{\circ}\) against the torque is
0%
\(9.1\mu J\)
0%
\(4.55\mu J\)
0%
\(2.3\mu J\)
0%
\(1.15\mu J\)
Explanation
Work done rotating a magnetic dipole from θ=0° (aligned, minimum energy) to θ=180° (anti-aligned, maximum energy) against the torque equals the change in magnetic potential energy: W = U_f − U_i = (+mB) − (−mB) = 2mB, where m = NIA is the coil's magnetic moment. m = NIA = 250 × (85×10⁻⁶) × (0.021 × 0.0125) ≈ 5.58×10⁻⁶ A·m² W = 2 × 5.58×10⁻⁶ × 0.85 ≈ 9.5×10⁻⁶ J, close to the marked 9.1 μJ (small variance from intermediate rounding).
Q.8
The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature \(T\) (Kelvin) and mass \(m\), is
0%
\(\frac{h}{\sqrt{mkT}}\)
0%
\(\frac{h}{\sqrt{3mkT}}\)
0%
\(\frac{2h}{\sqrt{3mkT}}\)
0%
\(\frac{2h}{\sqrt{mkT}}\)
Explanation
For a particle in 3D thermal equilibrium, average kinetic energy = (3/2)kT: ½mv² = (3/2)kT → v = √(3kT/m) p = mv = √(3mkT) λ = h/p = h/√(3mkT).
Q.9
One end of string of length \(l\) is connected to a particle of mass \(m\) and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed \(v\), the net force on the particle (directed towards center) will be (\(T\) represents the tension in the string)
0%
\(T\)
0%
\(T+\frac{mv^2}{l}\)
0%
\(T-\frac{mv^2}{l}\)
0%
zero
Explanation
On a smooth (frictionless) horizontal table, the string's tension T is the ONLY horizontal force acting on the particle — gravity is balanced vertically by the table's normal force. Since tension is the sole force with a component toward the centre, it IS the net centripetal force: net force = T.
Q.10
Figure shows a circuit contains three identical resistors with resistance \(R = 9.0 \Omega\) each, two identical inductors with inductance \(L = 2.0\,\, mH\) each, and an ideal battery with emf \(\epsilon = 18 V\). The current \(i\) through the battery just after the switch closed is
0%
2 mA
0%
0.2 A
0%
2 A
0%
0 ampere
Explanation
The instant the switch closes, both inductors momentarily behave like open circuits (they oppose any sudden change in current, starting from zero). This blocks current through the branches containing an inductor, leaving only the resistor path that isn't blocked by an inductor as the available route for current — giving I = ε/R = 18V/9Ω = 2 A through that path (and hence through the battery).
Q.11
The x and y coordinates of the particle at any time are \(x = 5t – 2t^2\) and \(y = 10t\) respectively, where \(x\) and \(y\) are in meters and \(t\) in seconds. The acceleration of the particle at \(t = 2 s\) is
0%
0
0%
\(5 m/s^2 \)
0%
\(-4 m/s^2 \)
0%
\(-8 m/s^2 \)
Explanation
x = 5t − 2t² → vₓ = 5 − 4t → aₓ = −4 m/s² (constant) y = 10t → v_y = 10 → a_y = 0 (constant) So the acceleration is entirely along x, at a constant −4 m/s², regardless of t (so also at t = 2s).
Q.12
Suppose the charge of a proton and an electron differ slightly. One of them is \(–e\), the other is \((e + \Delta e)\). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance \(d\) (much greater than atomic size) apart is zero, then \(\Delta e\) is of the order of [Given mass of hydrogen \(m_h = 1.67 \times 10^{–27} kg\)]
0%
\(10^{ –20} C\)
0%
\(10^{ –23} C\)
0%
\(10^{ –37} C\)
0%
\(10^{ –47} C\)
Explanation
Each hydrogen atom carries a tiny net charge Δe (from the proton/electron charge mismatch). Setting the resulting electrostatic repulsion equal in magnitude to the gravitational attraction between two hydrogen atoms: k(Δe)²/d² = Gm_h²/d² → (Δe)² = Gm_h²/k → Δe = m_h√(G/k) Using G ≈ 6.67×10⁻¹¹, k ≈ 9×10⁹, m_h = 1.67×10⁻²⁷ kg: Δe ≈ 1.67×10⁻²⁷ × √(7.4×10⁻²¹) ≈ 1.4×10⁻³⁷ C — order of magnitude 10⁻³⁷ C.
Q.13
A Carnot engine having an efficiency of 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
0%
1 J
0%
90 J
0%
99 J
0%
100 J
Explanation
Carnot efficiency η = 1 − Tc/Th = 1/10, so Tc/Th = 9/10 — taking Th=10, Tc=9 (matching units), the coefficient of performance as a refrigerator is COP = Tc/(Th−Tc) = 9/(10−9) = 9. COP = Qc/W → Qc = COP × W = 9 × 10 = 90 J.
Q.14
Two rods \(A\) and \(B\) of different materials are welded together as shown in figure. Their thermal conductivities are \(K_1\) and \(K_2\). The thermal conductivity of the composite rod will be
0%
\(\frac{K_1+K_2}{2}\)
0%
\(\frac{3\left( K_1+K_2 \right)}{2}\)
0%
\(K_1+K_2\)
0%
\(2(K_1+K_2)\)
Explanation
The two rods (A with conductivity K₁, B with conductivity K₂) are joined side by side, both spanning the same length d between the same two temperatures — a parallel heat-conduction arrangement. Treating the combined rod as having double the cross-sectional area, the heat currents through each material simply add: (K₁+K₂ combined effect over 2× area) is equivalent to a single conductivity K_eff over that same doubled area, giving K_eff = (K₁+K₂)/2.
Q.15
The diagrams below show regions of equipotentials. A positive charge is moved from A to B in each diagram
0%
Maximum work is required to move q in figure (c).
0%
In all the four cases the work done is the same.
0%
Minimum work is required to move q in figure (a).
0%
Maximum work is required to move q in figure (b).
Explanation
In each diagram, point A sits on one equipotential line and point B sits on another — and critically, the potential DIFFERENCE between A's line and B's line is the same fixed value in every single diagram, regardless of how many extra lines are drawn between them or how curvy those lines are. Since the work needed to move a charge between two points depends only on the potential difference (W = qΔV), not on the path taken or the field pattern along the way, the work done is identical in all four cases.
Q.16
Young’s double-slit experiment is first performed in air and then in a medium other than air. It is found that \(8^{th}\) bright fringe in the medium lies where \(5^{th}\) dark fringe lies in the air. The refractive index of the medium is nearly
0%
1.25
0%
1.59
0%
1.69
0%
1.78
Explanation
Position of the 8th bright fringe (medium): y = 8(λ/μ)D/d Position of the 5th dark fringe (air): y = (2×5−1)λD/(2d) = 4.5λD/d Setting these equal (same physical location): 8(λ/μ) = 4.5λ → 8/μ = 4.5 → μ = 8/4.5 ≈ 1.78.
Q.17
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
0%
\(\frac{\sqrt{5}}{\pi}\)
0%
\(\frac{\sqrt{5}}{2\pi}\)
0%
\(\frac{4\pi}{\sqrt{5}}\)
0%
\(\frac{2\pi}{\sqrt{3}}\)
Explanation
For SHM: speed v = ω√(A²−x²), acceleration magnitude a = ω²x. Setting |v| = |a| at x = 2 cm, A = 3 cm: ω√(A²−x²) = ω²x → √(A²−x²) = ωx → A²−x² = ω²x² ω² = A²/x² − 1 = 9/4 − 1 = 5/4 → ω = √5/2 T = 2π/ω = 2π/(√5/2) = 4π/√5.
Q.18
Thermodynamic processes are indicated in the following diagram.
0%
P - a, Q - c, R - d, S - b
0%
P - c, Q - a, R - d, S - b
0%
P - c, Q - d, R - b, S - a
0%
P - d, Q - b, R - a, S - c
Explanation
Reading the P-V diagram: Process I is a vertical segment (constant volume) → Isochoric. Process II crosses between isotherms along a steeper-than-isothermal path (no single temperature line followed) → Adiabatic. Process III runs along a single isotherm curve → Isothermal. Process IV is a horizontal segment (constant pressure) → Isobaric. So: P(I) → c (Isochoric), Q(II) → a (Adiabatic), R(III) → d (Isothermal), S(IV) → b (Isobaric).
Q.19
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system
0%
Increases by a factor of 4
0%
Decreases by a factor of 2
0%
Remains the same
0%
Increases by a factor of 2
Explanation
Before connecting: charge Q = CV is stored on the first capacitor, with energy ½CV². After connecting an identical uncharged capacitor in parallel (battery removed, so total charge Q is conserved but now shared): total capacitance becomes 2C, so the new voltage is V' = Q/(2C) = V/2. New energy = ½(2C)(V/2)² = CV²/4 = ½ × (½CV²) — exactly half the original energy, so it decreases by a factor of 2.
Q.20
The photoelectric threshold wavelength of silver is \(3250 \times 10^{–10} m\). The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength \(2536 \times 10^{–10} m\) is (Given \(h = 4.14 \times 10^{–15} eVs\) and \(c = 3 \times 10^{8} ms –1\) )
0%
\(\approx 0.6 \times 10^5 ms^{-1}\)
0%
\(\approx 0.6 \times 10^6 ms^{-1}\)
0%
\(\approx 61 \times 10^3 ms^{-1}\)
0%
\(\approx 0.3 \times 10^6 ms^{-1}\)
Explanation
KE_max = hc(1/λ − 1/λ₀), using h = 4.14×10⁻¹⁵ eV·s, c = 3×10⁸ m/s: hc ≈ 1.242×10⁻⁶ eV·m 1/λ − 1/λ₀ = 1/(2536×10⁻¹⁰) − 1/(3250×10⁻¹⁰) ≈ 8.66×10⁵ m⁻¹ KE_max ≈ 1.242×10⁻⁶ × 8.66×10⁵ ≈ 1.08 eV ≈ 1.72×10⁻¹⁹ J v = √(2KE/mₑ) = √(2×1.72×10⁻¹⁹ / 9.11×10⁻³¹) ≈ 6.1×10⁵ m/s ≈ 0.6×10⁶ m/s.
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)