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Quiz 3
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Q.1
In Young’s double sit experiment the separation d between the slits is \(2\,mm\), the wavelength \(\lambda\) of the light used is 5896 Å and distance \(D\) between the screen and slits is 100 cm. it is found that the angular width of the fringes is \(0.20^{\circ}\). To increase the fringe angular width to \(0.21^{\circ}\) (with same \(\lambda\) and \(D\) ) the separation between the slits needs to be changed to
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1.8 mm
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1.9 mm
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2.1 mm
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1.7 mm
Explanation
Angular fringe width β = λ/d, so β is inversely proportional to d. β₁/β₂ = d₂/d₁ → 0.20/0.21 = d₂/2mm → d₂ = 2 × 0.20/0.21 ≈ 1.9 mm.
Q.2
A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is \(20\,\, m/s^2\) at a distance of 5 m from the mean position. The time period of oscillation is
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\(1 s\)
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\(2 s\)
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\(\pi s\)
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\(2\pi s\)
Explanation
For SHM, acceleration a = ω²x. Given a=20 m/s² at x=5m: ω² = 20/5 = 4 → ω = 2 rad/s T = 2π/ω = 2π/2 = π seconds.
Q.3
In the circuit shown in the figure, the input voltage \(V_{i}\) is \(20 V\), \(V_{BE}=0\) and \(V_{CE}=0\). The values of \(I_{B}\), \(I_C\) and \(\beta\) are given by
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\(I_B =\text{40 }\mu A, I_C=\text{10 }mA, \beta =250\)
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\(I_B =\text{25 }\mu A, I_C=\text{5 }mA, \beta =200\)
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\(I_B =\text{20 }\mu A, I_C=\text{5 }mA, \beta =250\)
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\(I_B =\text{40 }\mu A, I_C=\text{5 }mA, \beta =125\)
Explanation
I_B = (V_i − V_BE)/R_B = (20−0)/500kΩ = 40 μA. I_C = (V_CC − V_CE)/R_C = (20−0)/4kΩ = 5 mA. β = I_C/I_B = 5mA/40μA = 125.
Q.4
An electron falls from rest through a vertical distance \(h\) in a uniform and vertically upward directed electric field \(E\). The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance \(h\). The time of fall of the electron, in comparison to the time of fall of the proton is
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smaller
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equal
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10 times greater
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5 times greater
Explanation
Both electron and proton experience the same magnitude of electric force (F=eE), but a = F/m, and the electron's mass is about 1836 times smaller than the proton's — giving it a vastly larger acceleration. Using h=½at² → t=√(2h/a), a much larger acceleration means a much SMALLER time to fall the same distance h. So the electron's fall time is smaller than the proton's.
Q.5
The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is
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1: -1
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2: -1
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1: -2
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1: 1
Explanation
In the Bohr model, for any orbit, kinetic energy KE = −(total energy E) — i.e. KE and E always have equal magnitude but opposite sign. So the ratio KE:E = 1:(−1).
Q.6
A turning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of \(27^0C\) two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at \(27^0C\) is
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300 m/s
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350 m/s
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339 m//s
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330 m/s
Explanation
The distance between successive resonance positions in a tube corresponds to half a wavelength: λ/2 = 73 − 20 = 53 cm → λ = 106 cm = 1.06 m. v = fλ = 320 × 1.06 ≈ 339 m/s.
Q.7
An electron of mass m with an initial velocity \(\vec{V}=V_0 \hat{i} \left( V_0>0 \right) \) enters an electric field \(\vec{E}=-E_0 \vec{i} \left( E_0=constant>0 \right)\) at \(t=0\). If \(\lambda _0\) is its de-Broglie wavelength initially, then its de-Broglie wavelength at time \(t\) is
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\( \lambda _0\left( 1+\frac{eE_0}{mV_0}t \right) \)
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\( \frac{\lambda _0}{\left( 1+\frac{eE_0}{mV_0}t \right)} \)
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\( \lambda _0t\)
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\( \lambda _0\)
Explanation
Force on the electron: F = (−e)(−E₀î) = eE₀î — in the SAME direction as its initial velocity V₀î, so the electron speeds up: v(t) = V₀ + (eE₀/m)t. De Broglie wavelength λ = h/(mv), which shrinks as v grows: λ(t) = h/[m(V₀+(eE₀/m)t)] = [h/(mV₀)] / [1+(eE₀/(mV₀))t] = λ₀ / [1+(eE₀/(mV₀))t].
Q.8
A small sphere of radius \(r\) falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to
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\(r^3\)
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\(r^4\)
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\(r^2\)
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\(r^5\)
Explanation
At terminal velocity, the sphere moves at constant speed, so ALL the gravitational PE it loses converts directly to heat — the rate of heat production equals (weight) × (terminal velocity). Weight ∝ r³ (volume). Terminal velocity (Stokes' law) v_t ∝ r². Rate of heat production ∝ r³ × r² = r⁵.
Q.9
The power radiated by a black body is P and it radiates maximum energy at wavelength, \(\lambda _0\). If the temperature of the black body is now changed so that it radiates maximum energy at wavelength \(\frac{3}{4}\lambda _{0}\), the power radiated by it becomes \(nP\). The value of \(n\) is
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\( \frac{4}{3}\)
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\( \frac{256}{81}\)
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\( \frac{3}{4}\)
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\( \frac{81}{256}\)
Explanation
By Wien's law, λ_max ∝ 1/T. If λ_max shrinks to (3/4)λ₀, temperature increases to T_new = T_old × (4/3). By Stefan's law, P ∝ T⁴, so: n = P_new/P_old = (4/3)⁴ = 256/81.
Q.10
An inductor \(20 \,m H\), a capacitor \(100\mu F\) and a resistor \(50\Omega \) are connected in series across a source of emf, \(V=\text{10}\sin\text{314 }t\). The power loss in the circuit is
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0.79 W
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0.43 W
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1.13 W
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2.74 W
Explanation
Reactances: X_L = ωL = 314×0.020 = 6.28 Ω; X_C = 1/(ωC) = 1/(314×100×10⁻⁶) ≈ 31.85 Ω. Impedance: Z = √[R² + (X_L−X_C)²] = √[50² + (−25.57)²] ≈ 56.2 Ω. Peak current I₀ = V₀/Z = 10/56.2 ≈ 0.178 A → I_rms ≈ 0.126 A. Power loss = I_rms² × R ≈ (0.126)² × 50 ≈ 0.79 W.
Q.11
The similarity of bone structure in the forelimbs of many vertebrates is an example of
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Adaptive radiation
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Homology
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Analogy
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Convergent evolution
Explanation
Similar underlying bone structure in the forelimbs of different vertebrates (e.g. human arm, bat wing, whale flipper), despite serving very different functions, is the classic example of homology — structures with a shared evolutionary origin.
Q.12
The kinetic energies of a plant in an elliptical orbit about the sun, at positions A,B and C are \(k_A,k_B\,\,and\,\,k_C\), respectively. AC is the major axis and SB is perpendicular to AC at the position of the sun S as shown in the figure. Then
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\(K_A < K_B < K_C\)
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\(K_B < K_A < K_C\)
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\(K_A > K_B > K_C\)
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\(K_B > K_A > K_C\)
Explanation
By Kepler's second law (equal areas in equal times), a planet moves fastest when closest to the Sun and slowest when farthest. Since S sits closer to A than to C, A is the near point (highest speed, highest KE) and C is the far point (lowest speed, lowest KE), with B — at an intermediate distance — falling in between: K_A > K_B > K_C.
Q.13
A sample of 0.1 g water at \(100^0\text{C}\) and normal pressure \(\left( 1.013\times 10^5\text{Nm}^{-2} \right) \) requires 54 cal of heat energy to convert to steam at \(100^0\text{C}\). If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is
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84.5 J
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208.7 J
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104.3 J
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42.2 J
Explanation
Heat added: Q = 54 cal × 4.184 J/cal ≈ 226 J. Work done by the expanding steam: W = PΔV ≈ P × V_steam (the initial water volume is negligible in comparison) = (1.013×10⁵)(167.1×10⁻⁶) ≈ 16.9 J. Change in internal energy: ΔU = Q − W ≈ 226 − 16.9 ≈ 208.7 J.
Q.14
A block of mass \(m\) is placed on a smooth inclined wedge ABC of inclination \(\theta\) as shown in the figure. The wedge is given acceleration \(a\) towards the right. The relation between a and \(\theta\) for the block to remain stationary on the wedge is
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\(a=\frac{g}{\mathrm{cosec}\theta}\)
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\(a=g\tan\theta\)
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\(a=\frac{g}{\sin\theta}\)
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\(a=g\cos\theta\)
Explanation
In the wedge's own (accelerating) reference frame, a pseudo-force ma acts backward on the block, alongside gravity (mg, down) and the normal force (perpendicular to the frictionless incline). For the block to stay put, the component of the pseudo-force along the incline must balance gravity's component along the incline: ma·cosθ = mg·sinθ → a = g·tanθ.
Q.15
A carbon resistor of (47 + 4.7) is to be marked with rings of different colors for its identification. The color code sequence will be
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Yellow – Violet – Orange – Silver
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Yellow – Green – Violet – Gold
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Green – Orange – Violet – Gold
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Violet – Yellow – Orange – Silver
Explanation
(Treating the value as 47 kΩ ± 10%, since the ±4.7 tolerance figure only makes sense as 10% of a 47k value.) Standard resistor color code: first digit 4 = Yellow, second digit 7 = Violet, multiplier ×10³ = Orange, and a ±10% tolerance = Silver — giving Yellow–Violet–Orange–Silver.
Q.16
A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from
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the current source
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the magnetic field
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the induced electric field due to the changing magnetic field
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the lattice structure of the material of the rod
Explanation
Pushing a diamagnetic material out of a magnetic field, and lifting it against gravity, requires energy. As the rod moves, it slightly changes the flux through the electromagnet's coils, and by Lenz's law, extra work must be done by whatever maintains the current (the current source/power supply) to keep the field steady against this change — that extra electrical work is what ultimately supplies the mechanical energy lifting the rod.
Q.17
A metallic rod of mass per unit length \(\text{0.5Kgm}^{-1}\) is lying horizontally on a smooth inclined plane which makes an angle of with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction \(0.25\,\, T\) is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is
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7.14 A
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5.98 A
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14.76 A
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11.32 A
Explanation
(The incline angle, missing from the source text, works out to 30° based on the marked answer.) The horizontal magnetic force F = BIL on the horizontal current-carrying rod must have a component along the incline that balances the rod's weight component along the incline (mass per length × g × sinθ): BI·cosθ = (mass/length)·g·sinθ → I = (mass/length)·g·tanθ / B I = 0.5 × 9.8 × tan(30°) / 0.25 ≈ 0.5 × 9.8 × 0.577 / 0.25 ≈ 11.32 A.
Q.18
Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is
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\(40\Omega \)
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\(250\Omega \)
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\(25\Omega \)
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\(500 \Omega \)
Explanation
Current sensitivity S_i = deflection/current, voltage sensitivity S_v = deflection/voltage = deflection/(current×resistance) = S_i/G. G = S_i/S_v = (5 div/mA) / (20 div/V) = (5000 div/A) / (20 div/V) = 250 Ω.
Q.19
A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy \(\left( K_t \right) \) as well as rotational kinetic energy \(\left( K_r \right)\) simultaneously. The ratio \(K_t:\left( K_t+K_r \right)\) for the sphere is
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5: 7
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7: 10
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10: 7
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2: 5
Explanation
For a rolling solid sphere (I = (2/5)mr²): K_t = ½mv², K_r = ½Iω² = ½(2/5)mr²(v/r)² = (1/5)mv². K_t + K_r = ½mv² + (1/5)mv² = (7/10)mv². K_t / (K_t+K_r) = (1/2) / (7/10) = 5/7.
Q.20
When the light of frequency \(2v_0\) (where \(v_0\) is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is \(v_1\) . When the frequency of the incident radiation is increased to \(5v_0\), the maximum velocity of electrons emitted from the same plate is \(v_2\) The ratio of \(v_1\) to \(v_2\) is
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4: 1
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1: 2
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1: 4
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2: 1
Explanation
KE_max = h(ν−ν₀). At ν=2ν₀: ½mv₁² = h(2ν₀−ν₀) = hν₀ → v₁² ∝ 1×(hν₀/m) At ν=5ν₀: ½mv₂² = h(5ν₀−ν₀) = 4hν₀ → v₂² ∝ 4×(hν₀/m) v₁²/v₂² = 1/4 → v₁/v₂ = 1/2.
Q.21
An em wave is propagating in a medium with a velocity \(\vec{V} = V \hat{i}\). The instantaneous oscillating electric field of his em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along
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– y direction
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– x direction
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+ z direction
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– z direction
Explanation
For an EM wave, the direction of propagation follows E×B (right-hand rule). With E along +y (ĵ) and propagation along +x (î), we need ĵ×B̂ = î. Since ĵ×k̂ = î (a standard vector identity), B must point along +z.
Q.22
In a p-n junction diode, change in temperature due to heating
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affects only reverse resistance
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affects only forward resistance
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affects the overall V –I characteristics of p-n junction
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does not affect resistance of p-n junction
Explanation
Temperature affects the diode's charge-carrier behaviour throughout — both the forward conduction and (especially) the reverse leakage current change significantly with temperature — so it affects the overall V-I characteristic, not just one branch of it.
Q.23
A solid sphere is roasting freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?
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Angular velocity
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Moment of inertia
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Rotational kinetic energy
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Angular momentum
Explanation
With no external torque acting on the freely spinning sphere in space, its angular momentum L is conserved — even as the radius (and hence moment of inertia I = (2/5)MR²) changes, angular velocity ω and rotational KE (= L²/2I) both change to keep L itself constant.
Q.24
Which one is wrongly matched?
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Gemma cups – Marchantia
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Uniflagellate gametes – Polysiphonia
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Biflagellate zoospores – Brown algae
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Unicellular organism – Chlorella
Explanation
Polysiphonia is a red alga (Rhodophyta), and red algae are distinctively characterised by having NO flagellated stage at all in their life cycle — so "uniflagellate gametes" is the wrong pairing. (Gemma cups in Marchantia, biflagellate zoospores in brown algae, and Chlorella as a unicellular alga are all correctly matched.)
Q.25
According to Hugo de Vries, the mechanism of evolution is
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Minor mutations
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Saltation
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Phenotypic variations
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Multiple step mutations
Explanation
Hugo de Vries proposed that evolution proceeds through saltation — sudden, large-scale mutational jumps — rather than the slow, gradual accumulation of small variations that Darwin's natural selection describes.
Q.26
Which of the following pairs is wrongly matched?
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XO type sex determination : Grasshopper
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ABO blood grouping : Co-dominance
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T.H. Morgan : Linkage
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Starch synthesis in pea : Multiple alleles
Explanation
The classic Mendelian pea trait for starch/seed shape is controlled by a single gene with just two alleles (a simple dominant/recessive pair) — not by multiple alleles (a multiple-allele system, like ABO blood grouping, needs three or more alleles for one gene). So "Starch synthesis in pea : Multiple alleles" is the wrongly matched pair.
Q.27
The correct order of steps in Polymerase Chain Reaction (PCR) is
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Denaturation, Annealing, Extension
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Denaturation, Extension, Annealing
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Annealing, Extension, Denaturation
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Extension, Denaturation, Annealing
Explanation
The three core steps of a PCR cycle, in order, are: Denaturation (heating to separate the DNA strands), Annealing (primers bind to the single strands), and Extension (DNA polymerase synthesises the new complementary strand).
Q.28
Many ribosomes may associate with a single mRNA to form multiple copies of a polypeptide simultaneously. Such strings of ribosomes are termed as
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Polysome
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Plastidome
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Polyhedral bodies
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Nucleosome
Explanation
A cluster of multiple ribosomes simultaneously translating the same mRNA molecule (each producing its own copy of the polypeptide) is called a polysome (polyribosome).
Q.29
Use of bioresources by multinational companies and organizations without authorisation from the concerned country and its people is called
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Biopiracy
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Bio-infringemen
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Biodegradation
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Bioexploitation
Explanation
The unauthorised use of a country's bioresources (genetic material, traditional knowledge, etc.) by outside companies or organisations without permission or benefit-sharing is called biopiracy.
Q.30
Match the columns
Column I
Column II
a. Herbarium
(i) It is a place having a
collection of preserved
plants and animals
b. Key
(ii) A list that enumerates
methodically all the
species found in an area
with brief description
aiding identification
c. Museum
(iii) Is a place where dried
and pressed plant
specimens mounted on
sheets are kept
d. Catalogue
(iv) A booklet containing a
list of characters and
their alternates which
are helpful in
identification of various
taxa.
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a. -- (iii) b. -- (iv) c. -- (i) d. -- (ii)
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a. -- (ii) b. -- (iv) c. -- (iii) d. -- (i)
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a. -- (iii) b. -- (ii) c. -- (i) d. -- (iv)
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a. -- (i) b. -- (iv) c. -- (iii) d. -- (ii)
Explanation
Matching each term to its definition: Herbarium — a place with dried, pressed plant specimens mounted on sheets (iii). Key — a booklet of alternate characters used for identification (iv). Museum — a place with preserved plant and animal specimens (i). Catalogue — a methodical list of species found in an area with brief identifying descriptions (ii).
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