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NEET Previous Year Question Papers
Quiz 2
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Q.1
A physical quantity of the dimensions of length that can be formed out of \(c\), \(G\) and \(\frac{e^2}{4\pi \epsilon_0}\) is [\(c\) is velocity of light, \(G\) is universal constant of gravitation and \(e\) is charge]
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\(\frac{1}{c^2}\left[ G\frac{e^2}{4\pi \varepsilon _0} \right] ^{1/2}\)
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\(c^2\left[ G\frac{e^2}{4\pi \varepsilon_0} \right] ^{1/2}\)
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\(\frac{1}{c^2}\left[ \frac{e^2}{G4\pi \varepsilon_0} \right] ^{1/2}\)
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\(\frac{1}{c}G\frac{e^2}{4\pi \varepsilon_0}\)
Explanation
Checking dimensions: [G] = M⁻¹L³T⁻², and [e²/4πε₀] has dimensions of energy×length = ML³T⁻² (from Coulomb's law, F = ke²/r² rearranged). So G×(e²/4πε₀) has dimensions [M⁻¹L³T⁻²][ML³T⁻²] = L⁶T⁻⁴; its square root is L³T⁻². Dividing by c² (L²T⁻²) leaves just L — a length. So (1/c²)√[G·e²/(4πε₀)] has the dimensions of length.
Q.2
Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]
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350 Hz
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361 Hz
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411 Hz
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448 Hz
Explanation
With both source and observer approaching each other, the Doppler formula is f' = f(v + v_observer)/(v − v_source): f' = 400 × (340 + 16.5)/(340 − 22) = 400 × 356.5/318 ≈ 448 Hz.
Q.3
In a common emitter transistor amplifier the audio signal voltage across the collector is 3 V. The resistance of collector is \(3 K\Omega\). If current gain is 100 and the base resistance is \(2 K\Omega\), the voltage and power gain of the amplifier is
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200 and 1000
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15 and 200
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150 and 15000
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20 and 2000
Explanation
Voltage gain of a CE amplifier: A_v = β × (R_collector/R_base) = 100 × (3000/2000) = 150. Power gain = A_v × current gain = 150 × 100 = 15,000.
Q.4
The acceleration due to gravity at a height 1 km above the earth is the same as at a depth \(d\) below the surface of earth. Then
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\(d=\frac{1}{2}Km\)
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\(d=1Km\)
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\(d=\frac{3}{2}Km\)
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\(d=2Km\)
Explanation
Using the standard (small-height/depth) approximations: g at height h ≈ g(1 − 2h/R); g at depth d ≈ g(1 − d/R). Setting them equal: 2h/R = d/R → d = 2h = 2 × 1 km = 2 km.
Q.5
Which of the following statements are correct? (a) Centre of mass of a body always coincides with the centre of gravity of the body. (b) Centre of mass of a body is the point at which the total gravitational torque on the body is zero (c) A couple on a body produce both translational and rotational motion in a body. (d) Mechanical advantage greater than one means that small effort can be used to lift a large load.
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(b) and (d)
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(a) and (b)
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(b) and (c)
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(c) and (d)
Explanation
(b) is correct — the centre of gravity (and, in a uniform field, the centre of mass) is defined as the point about which the total gravitational torque on the body vanishes. (d) is correct — mechanical advantage = load/effort, so MA > 1 by definition means the load exceeds the effort needed, i.e. a small effort lifts a larger load. (a) is not universally true — centre of mass and centre of gravity only coincide in a UNIFORM gravitational field, not always. (c) is false — a pure couple (equal and opposite forces, zero net force) produces rotational motion only, with zero net translational effect.
Q.6
If \(\theta_1\) and \(\theta_2\) be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip \(\theta\) is given by
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\(\cot ^2\theta \,\,=\,\,\cot ^2\theta _1+\cot ^2\theta _2\)
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\(\tan ^2\theta \,\,=\,\,\tan ^2\theta _1+\tan ^2\theta _2\)
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\(\cot ^2\theta \,\,=\,\,\cot ^2\theta _1-\cot ^2\theta _2\)
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\(\tan ^2\theta =\,\,\tan ^2\theta _1-\tan ^2\theta _2\)
Explanation
The relation between the true angle of dip θ and the two apparent angles measured in perpendicular vertical planes is cot²θ = cot²θ₁ + cot²θ₂ — a standard result from resolving the horizontal component of Earth's magnetic field along the two perpendicular planes.
Q.7
An arrangement of three parallel straight wires placed perpendicular to the plane of paper carrying the same current \(I\) along the same direction is shown in Figure. The magnitude of force per unit length on the middle wire \(B\) is given by
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\(\frac{\mu _0I^2}{2\pi d}\)
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\(\frac{2\mu _0I^2}{\pi d}\)
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\(\frac{\sqrt{2}\mu _0I^2}{\pi d}\)
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\(\frac{\mu _0I^2}{\sqrt{2\pi d}}\)
Explanation
Wire A sits directly below B (distance d) and wire C sits directly to the right of B (distance d), forming a right angle at B — with all three currents in the same direction, each of A and C attracts B with force F = μ₀I²/(2πd), and these two forces are perpendicular to each other (one toward A, one toward C). Resultant = √(F² + F²) = √2 × F = √2 × μ₀I²/(2πd) = μ₀I²/(√2·πd).
Q.8
Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will:
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Keep floating at the same distance between them
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Move towards each other
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Move away from each other
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Will become stationary
Explanation
With no other forces acting, the astronauts' own mutual gravitational attraction — however small — is the only force present, and it's always attractive. Over time, that pulls them slowly toward each other.
Q.9
In an electromagnetic wave in free space the root mean square value of the electric field is \(E_{rms} = 6 V/m\). The peak value of the magnetic field is
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\(1.41 \times 10^{–8} T\)
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\(2.83 \times 10^{–8} T\)
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\(0.70 \times 10^{–8} T\)
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\(4.23 \times 10^{–8} T\)
Explanation
E₀ = E_rms × √2 = 6 × 1.414 ≈ 8.49 V/m. For an EM wave in free space, B₀ = E₀/c = 8.49 / (3×10⁸) ≈ 2.83×10⁻⁸ T.
Q.10
The bulk modulus of a spherical object is \(B\). If it is subjected to uniform pressure \(p\), the fractional decrease in radius is
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\(\frac{p}{B}\)
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\(\frac{B}{3p}\)
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\(\frac{3p}{B}\)
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\(\frac{p}{3B}\)
Explanation
Bulk modulus B = −p/(ΔV/V), so the fractional volume decrease is ΔV/V = p/B. For a sphere, V ∝ r³, so ΔV/V = 3(Δr/r) — meaning the fractional decrease in radius is (1/3) of the fractional volume decrease: Δr/r = p/(3B).
Q.11
The ratio of resolving powers of an optical microscope for two wavelengths \(\lambda_1 = 4000 \overset{\circ}{\mathrm {A}}\) and \(\lambda_2 = 6000 \overset{\circ}{\mathrm {A}}\) is
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8 : 27
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9 : 4
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3 : 2
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16 : 81
Explanation
Resolving power of a microscope is inversely proportional to wavelength (shorter wavelength resolves finer detail). So the ratio of resolving powers is the inverse ratio of wavelengths: RP₁/RP₂ = λ₂/λ₁ = 6000/4000 = 3/2.
Q.12
Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take g constant with a value \(10 m/s^2\) . The work done by the (i) gravitational force and the (ii) resistive force of air is
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(i) – 10 J (ii) –8.25 J
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(i) 1.25 J (ii) –8.25 J
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(i) 100 J (ii) 8.75 J
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(i) 10 J (ii) –8.75 J
Explanation
Work done by gravity = mgh = (0.001 kg)(10 m/s²)(1000 m) = 10 J. Actual kinetic energy gained = ½mv² = ½(0.001)(50²) = 1.25 J. By the work-energy theorem, W_gravity + W_resistive = ΔKE: 10 + W_resistive = 1.25 → W_resistive = 1.25 − 10 = −8.75 J.
Q.13
A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be
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225
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450
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1000
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1800
Explanation
Radiated power (Stefan-Boltzmann law) is proportional to surface area × T⁴, i.e. P ∝ r²T⁴. P₂/P₁ = (r₂/r₁)² × (T₂/T₁)⁴ = (1/2)² × (2)⁴ = (1/4)(16) = 4. P₂ = 450 × 4 = 1800 W.
Q.14
Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively
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\(g, \frac{g}{3}\)
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\( \frac{g}{3},g\)
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\(g,g\)
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\(\frac{g}{3}, \frac{g}{3}\)
Explanation
Just before the string is cut, the spring supports the full weight (3m+m)g = 4mg, so it's stretched to exert exactly that much force. The instant the string is cut: Block B loses its only support (the string) and becomes a free-falling body — its acceleration is g. Block A: a spring's force can't change instantaneously (it depends on extension, which hasn't had time to change yet), so the spring still pulls A upward with 4mg. A's own weight is only 3mg now (B no longer pulls down on it via the string). Net force on A = 4mg − 3mg = mg upward, so A's acceleration = mg/(3m) = g/3, directed upward. So A and B have accelerations g/3 and g respectively.
Q.15
Two Polaroids \(P_1\) and \(P_2\) are placed with their axis perpendicular to each other. Unpolarised light \(I_0\) is incident on \(P_1\) . A third polaroid \(P_3\) is kept in between \(P_1\) and \(P_2\) such that its axis makes an angle \(45^{\circ}\) with that of \(P_1\). The intensity of transmitted light through \(P_2\) is
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\(\frac{I_0}{2}\)
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\(\frac{I_0}{4}\)
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\(\frac{I_0}{8}\)
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\(\frac{I_0}{16}\)
Explanation
Unpolarised light through P₁: intensity drops to I₀/2. Through P₃ (at 45° to P₁): intensity becomes (I₀/2)cos²(45°) = (I₀/2)(1/2) = I₀/4. Through P₂ (at 90° to P₁, i.e. 45° to P₃): intensity becomes (I₀/4)cos²(45°) = (I₀/4)(1/2) = I₀/8.
Q.16
A long solenoid of diameter 0.1 m has \(2 \times 10^4\) turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to \(0 A\) from \(4 A\) in 0.05 s. If the resistance of the coil is \(10\pi^2 \omega\), the total charge flowing through the coil during this time is
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\(32\pi\, \mu C\)
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\(16 \mu C\)
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\(32\, \mu C\)
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\(16\pi\, \mu C\)
Explanation
Field inside the solenoid: B = μ₀nI. Change in field as current drops from 4A to 0: ΔB = μ₀ × (2×10⁴) × 4 = 8×10⁴μ₀. Since the coil (radius 0.01 m) is much smaller than the solenoid and sits at its centre, use the coil's own area for flux: A = π(0.01)² = π×10⁻⁴ m². Total charge q = N·ΔΦ/R = N·ΔB·A/R = 100 × (8×10⁴×4π×10⁻⁷) × (π×10⁻⁴) / (10π²) = 100 × 3.2π×10⁻² × π×10⁻⁴ / (10π²) = 3.2π²×10⁻⁴ / (10π²) × 100/100... simplifying directly: = 32×10⁻⁶ C = 32 μC.
Q.17
Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities \(\omega_1\) and \(\omega_2\) . They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is
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\(\frac{1}{2}I\left( \omega _1+\omega _2 \right) ^2\)
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\(\frac{1}{4}I\left( \omega _1-\omega _2 \right) ^2\)
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\(I\left( \omega _1-\omega _2 \right) ^2\)
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\(\frac{1}{8}I\left( \omega _1-\omega _2 \right) ^2\)
Explanation
Conservation of angular momentum: Iω₁ + Iω₂ = (2I)ω_f → ω_f = (ω₁+ω₂)/2. Initial KE = ½Iω₁² + ½Iω₂². Final KE = ½(2I)ω_f² = I(ω₁+ω₂)²/4. Energy lost = ½I(ω₁²+ω₂²) − I(ω₁+ω₂)²/4 = I[2(ω₁²+ω₂²) − (ω₁+ω₂)²]/4 = I(ω₁−ω₂)²/4 (after expanding and simplifying) = (1/4)I(ω₁−ω₂)².
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