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Quiz 2
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Q.1
The compound \(C_7H_8\) undergoes the following reactions : \[C_7H_8\xrightarrow{3Cl_2/\Delta}A\xrightarrow{Br_2/Fe}B\xrightarrow{Zn/HCl}C\] The product ‘C’ is
0%
o-bromotoluene
0%
m-bromotoluene
0%
3-bromo-2, 4, 6,-trichlorotoluene
0%
p-bromotoluene
Explanation
Toluene + 3Cl₂/heat (radical substitution on the methyl group) gives PhCCl₃ (A). PhCCl₃'s CCl₃ group is electron-withdrawing and meta-directing, so bromination (Br₂/Fe) places Br at the meta position (B). Zn/HCl then reduces the CCl₃ group back to CH₃, giving m-bromotoluene (C).
Q.2
The solubility of \(BaSO_4\) in water is \(2.42 \times 10^{-3}gL^{-1}\) at 298 K. The value of its solubility product \( \left ( K_{sp} \right ) \) will be (Given molar mass of \(BaSO_4 = 233 g.mol^{-1}\)
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\(1.08\times \,\,10^{-14}mol^2L^{-2}\)
0%
\(1.08\times 10^{-10}mol^2L^{-2}\)
0%
\(1.08\times 10^{-8}mol^2L^{-2}\)
0%
\(1.08\times 10^{-12}mol^2L^{-2}\)
Explanation
Molar solubility S = (2.42×10⁻³ g/L) / (233 g/mol) ≈ 1.039×10⁻⁵ mol/L. Ksp = [Ba²⁺][SO₄²⁻] = S² = (1.039×10⁻⁵)² ≈ 1.08×10⁻¹⁰ mol²/L².
Q.3
In the reaction The electrophile involved is
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dichloromethyl anion
0%
dichlorocarbene
0%
formyl cation
0%
dichloromethyl cation
Explanation
This is the Reimer-Tiemann reaction: phenol reacts with CHCl₃ and NaOH to give salicylaldehyde. NaOH deprotonates CHCl₃ to CCl₃⁻, which then undergoes α-elimination to form :CCl₂ (dichlorocarbene) — this electron-deficient carbene is the actual electrophile that attacks the electron-rich phenoxide ring.
Q.4
Considering Ellingham diagram, which of the following metals can be used to reduce alumina?
0%
Fe
0%
Zn
0%
Cu
0%
Mg
Explanation
On an Ellingham diagram, magnesium's oxide-formation line lies below aluminium's (more negative ΔG° for MgO formation) — meaning Mg is a stronger reducing agent than Al under these conditions and can reduce Al₂O₃ to metallic aluminium. Fe, Zn, and Cu are all weaker reducing agents (higher up the diagram) and can't do this.
Q.5
Which of the following oxides is most acidic in nature?
0%
BaO
0%
BeO
0%
CaO
0%
MgO
Explanation
Among alkaline earth oxides, basicity increases going down the group (larger, more electropositive metal ions), so BeO — the smallest, most covalent, least basic — is the most acidic (in fact, amphoteric) of the group, while BaO is the most basic.
Q.6
The compound A on treatment with Na gives B, and with \(\text{PC}1_5\) gives C. B and C react together to give diethyl ether. A, B and C are in the order
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\( C_2H_5OH,C_2H_5C\text{1,}C_2H_5ONa\)
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\( C_2H_5OH,C_2H_6,C_2H_5C1\)
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\( C_2H_5OH,C_2H_5ONa,C_2H_5C1\)
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\( C_2H_5C\text{1,}C_2H_5OH\)
Explanation
A = C₂H₅OH (ethanol). With Na: 2C₂H₅OH + 2Na → 2C₂H₅ONa (B, sodium ethoxide) + H₂. With PCl₅: C₂H₅OH + PCl₅ → C₂H₅Cl (C, ethyl chloride) + POCl₃ + HCl. B and C then combine via the Williamson ether synthesis: C₂H₅ONa + C₂H₅Cl → C₂H₅-O-C₂H₅ (diethyl ether) + NaCl.
Q.7
In the structure of \(ClF_3\), the number of lone pairs of electrons on central atom ‘Cl’ is
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two
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four
0%
one
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three
Explanation
ClF₃ has 5 electron domains around the central Cl atom: 3 bonding pairs (to the 3 F atoms) and 2 lone pairs — giving a T-shaped molecular geometry (trigonal bipyramidal electron geometry) with 2 lone pairs.
Q.8
For the redox reaction \[MnO_{4}^{-}+C_2O_{4}^{2-}+H^+\rightarrow Mn^{2+}+CO_2+H_2O\] The correct coefficients of the reactants for the balanced equation are
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\(MnO_{4}^{-}--\text{16 }C_2O_{4}^{2-}--\text{5 }H^+--2\)
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\(MnO_{4}^{-}--\text{5 }C_2O_{4}^{2-}--\text{16 }H^+--2\)
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\(MnO_{4}^{-}--\text{2 }C_2O_{4}^{2-}--\text{5 }H^+--16\)
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\(MnO_{4}^{-}--\text{2 }C_2O_{4}^{2-}--\text{16 }H^+--5\)
Explanation
The balanced redox equation is: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O (Mn goes from +7 to +2, a 5-electron gain per Mn; each C₂O₄²⁻ loses 2 electrons going to 2CO₂ — balancing electrons requires 2 MnO₄⁻ for every 5 C₂O₄²⁻, with 16 H⁺ needed to balance the oxygens/charge.)
Q.9
Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity?
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\( NO_2\)
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\( N_2O\)
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\( NO\)
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\( N_2O_5\)
Explanation
NO₂ is predominantly a secondary pollutant, formed mainly by the atmospheric oxidation of NO (itself largely from combustion) — rather than being commonly and directly released in significant quantities from major NATURAL sources the way NO (lightning, soil bacteria) and N₂O (soil nitrification/denitrification) are.
Q.10
Regarding cross-linked or network polymers, which of the following statements is incorrect?
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They contain strong covalent bonds in their polymer chains.
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They contain covalent bonds between various linear polymer chains
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Examples are Bakelite and melamine.
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They are formed from bi-and tri-functional monomers.
Explanation
The defining feature of cross-linked/network polymers is covalent bonding BETWEEN separate linear polymer chains (the cross-links), not merely within a single chain — so describing them simply as having "strong covalent bonds in their polymer chains" misses this defining cross-linking characteristic. (Bakelite and melamine as examples, formation from bi/tri-functional monomers, and covalent bonds connecting different chains are all correctly stated.)
Q.11
Consider the following species : \(CN^+\), \(CN^-\), \(NO\) and \(CN\) Which one of these will have the highest bond order?
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\(CN^+\)
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\(CN\)
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\(NO \)
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\(CN^-\)
Explanation
CN⁻ has 14 electrons total (6 from C + 7 from N + 1 for the negative charge) — isoelectronic with N₂, giving it the same very stable bond order of 3 (a full triple bond). CN, NO, and CN⁺ all have different electron counts (13, 15, and 12 respectively) giving lower bond orders (2.5, 2.5, and 2).
Q.12
Which part of poppy plant is used to obtain the drug “Smack”?
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Roots
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Flowers
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Latex
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Leaves
Explanation
"Smack" (a crude form of heroin) is derived from the latex (milky sap) exuded by the unripe seed pods of the poppy plant, which contains opium alkaloids.
Q.13
At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere? Given that Mass of oxygen molecule \(m = 2.76\times 10^{-26}kg\) Boltzmann's constant \(k_B = 1.38\times 10^{-23} JK^{-1}\)
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\(1.254 \times 10^4\, K\)
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\(2.508 \times 10^4\, K\)
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\(5.016 \times 10^4\, K\)
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\(8.360 \times 10^4\, K\)
Explanation
Setting rms speed equal to escape velocity (11.2 km/s = 11200 m/s): (3kT/m) = v_esc² → T = m·v_esc²/(3k) T = (2.76×10⁻²⁶)(11200²) / (3×1.38×10⁻²³) ≈ 8.36×10⁴ K.
Q.14
A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. the main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of – 0.004 cm, the correct diameter of the ball is
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0.053 cm
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0.529 cm
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0.525 cm
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0.521 cm
Explanation
Reading = main scale (5 mm = 0.5 cm) + circular scale (25 divisions × 0.001 cm) = 0.5 + 0.025 = 0.525 cm. Correcting for the zero error (subtracting a negative zero error means adding its magnitude): 0.525 − (−0.004) = 0.529 cm.
Q.15
Three objects, A : (a solid sphere), B : (a thin circular disk) and C : (a circular ring), each have the same mass \(M\) and radius \(R\), They all spin with the same angular speed \(\omega \) about their own symmetry axes. The amounts of work \(W\) required to bring them to rest, would satisfy the relation
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\(W_B>W_A>W_C\)
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\(W_C>W_B>W_A\)
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\(W_A>W_C>W_B\)
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\(W_A>W_B>W_C\)
Explanation
Work needed to stop a spinning object equals its rotational KE, ½Iω² — larger moment of inertia I means more work needed (same M, R, ω for all three). I_sphere = (2/5)MR² (smallest), I_disk = (1/2)MR², I_ring = MR² (largest, since all its mass sits at the rim). So W_ring > W_disk > W_sphere, i.e. W_C > W_B > W_A.
Q.16
The moment of the force, \(\vec{F} = 4 \hat{i} + 5\hat{ j} – 6\hat{ k}\) at \((2,0,-3)\), about the point \((2,-2, -2)\), is given by
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\(– 7 \hat{i} –4 \hat{j} -8 \hat{k}\)
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\(– 7 \hat{i} -8 \hat{j} -4 \hat{k}\)
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\(– 8 \hat{i} –4 \hat{j} -7 \hat{k}\)
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\(– 4 \hat{i} – \hat{j} -8 \hat{k}\)
Explanation
Position vector from the pivot to the point of application: r = (2,0,−3) − (2,−2,−2) = (0,2,−1). Moment = r × F = (0,2,−1) × (4,5,−6): i: (2)(−6) − (−1)(5) = −12+5 = −7 j: −[(0)(−6) − (−1)(4)] = −[0+4] = −4 k: (0)(5) − (2)(4) = −8 Moment = −7i − 4j − 8k.
Q.17
In the combination of the following gates the output Y . can be written in terms of inputs A and B as
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\( \overline{A\cdot B}\)
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\( \overline{A\cdot B}+A\cdot B \)
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\( \overline{A+B}\)
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\( A\cdot \overline{B}+\overline{A}\cdot B \)
Explanation
Tracing the circuit: A goes directly to the top AND gate, while B is inverted (NOT gate) and also feeds that same top AND gate — giving A·B̄. Symmetrically, B goes directly to the bottom AND gate, while A is inverted and feeds it too — giving Ā·B. The final OR gate combines them: Y = A·B̄ + Ā·B (this is the XOR function).
Q.18
The refractive index of the material of a prism is \(\sqrt{2}\) and the angle of the prism is \(30^0\) . One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is
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\(45^0\)
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\(60^0\)
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\(zero\)
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\(30^0\)
Explanation
For the ray to retrace its path after hitting the silvered face, it must strike that internal face exactly along the normal (perpendicular incidence there) — geometrically, this requires the angle of refraction at the first surface to equal the prism angle A = 30°. By Snell's law: sin(i) = μ·sin(30°) = √2 × 0.5 = √2/2 = 1/√2 → i = 45°.
Q.19
An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be
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36 cm away from the mirror
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36 cm towards the mirror
0%
30 cm away from the mirror
0%
30 cm towards the mirror
Explanation
Using the mirror formula 1/v + 1/u = 1/f (f=15 cm): At u=40cm: 1/v = 1/15 − 1/40 = (8−3)/120 = 5/120 → v = 24 cm. After moving 20cm closer, u=20cm: 1/v = 1/15 − 1/20 = (4−3)/60 → v = 60 cm. Image displacement = 60 − 24 = 36 cm, and since v increased (image moved farther), it's 36 cm AWAY from the mirror.
Q.20
The volume \(V\) of a monatomic gas varies with its temperature \(T\), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state \(A\) to state \(B\), is
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\(\frac{2}{3}\)
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\(\frac{2}{5}\)
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\(\frac{1}{3}\)
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\(\frac{2}{7}\)
Explanation
The straight V-T line through the origin means V/T is constant — by the ideal gas law PV=nRT, this means P stays constant too (an isobaric process). For an isobaric process: Q = nCpΔT, and work done W = PΔV = nRΔT. W/Q = R/Cp. For a monatomic gas, Cp = (5/2)R, so W/Q = R/((5/2)R) = 2/5.
Q.21
A moving block having mass \(m\), collides with another stationary block having mass \(4m\). The lighter block comes to rest after collision. When the initial velocity of the lighter block is \(v\), then the value of coefficient of restitution \(e\) will be
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0.8
0%
0.5
0%
0.4
0%
0.25
Explanation
By momentum conservation, mv = 4m·v' (v' = final speed of the 4m block, since the m block stops): v' = v/4. Coefficient of restitution e = (relative separation speed)/(relative approach speed) = (v/4 − 0)/(v − 0) = 1/4 = 0.25.
Q.22
A set of \(n\) equal resistors, of value \(R\) each, are connected in series to a battery of emf \(E\) and internal resistance \(R\). The current drawn is \(I\). Now, the \(n\) resistors are connected in parallel to the same battery. Then the current drawn from battery becomes \(10\, I\). The value of \(n\) is
0%
20
0%
9
0%
11
0%
10
Explanation
Series: total resistance = nR + R (internal), so I = E/[R(n+1)]. Parallel: n resistors in parallel give R/n, so total resistance = R/n + R = R(1+n)/n, giving I_parallel = En/[R(n+1)]. Setting I_parallel = 10I: En/[R(n+1)] = 10E/[R(n+1)] → n = 10.
Q.23
Unpolarised light is incident from air on a plane surface of material of refractive index \(\mu\). At a particular angle of incidence \(i\), it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation ?
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Reflected light is polarized with its electric vector parallel to the plane of incidence.
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\(i=\tan ^{-1}\left( \frac{1}{\mu} \right) \)
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Reflected light is polarized with its electric vector perpendicular to the plane of incidence.
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\(i=\sin ^{-1}\left( \frac{1}{\mu} \right) \)
Explanation
This is Brewster's angle condition (reflected and refracted rays perpendicular). At Brewster's angle, the reflected ray is completely plane-polarised, with its electric vector oscillating perpendicular to the plane of incidence — a standard, well-established result of Brewster's law.
Q.24
For a radioactive material, half-life is 10 minutes. If initially there are 600 numbers of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
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10
0%
20
0%
30
0%
15
Explanation
450 out of 600 nuclei have decayed, leaving 150 — that's 150/600 = 1/4 = (1/2)² of the original amount, meaning exactly 2 half-lives have passed. Time = 2 × 10 minutes = 20 minutes.
Q.25
A body initially at rest and sliding along a friction less track from a height \(h\) (as shown in the figure) just completes a vertical circle of diameter \(AB = D\). The height \(h\) is equal to
0%
\(\frac{7}{5}D\)
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\(D\)
0%
\(\frac{3}{2}D\)
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\(\frac{5}{4}D\)
Explanation
At the top of the loop (radius D/2), the minimum condition for completing the circle is that gravity alone supplies the centripetal force: v_top² = g(D/2). Energy conservation from the starting height h to the top of the loop (height D above the ground): gh = gD + ½v_top² = gD + ½·g(D/2) = gD + gD/4 h = D + D/4 = 5D/4.
Q.26
Two wires are made of the same volume. The first wire has cross-sectional area \(A\) and the second wire has cross-sectional area \(3A\). if the length of the first wire is increased by \(\Delta l\) on applying a force \(F\), how much force is needed to stretch the second wire by the same amount ?
0%
\(F\)
0%
\(9 F\)
0%
\(4 F\)
0%
\(6 F\)
Explanation
Same volume means L₁A = L₂(3A) → L₂ = L₁/3. From Young's modulus, F = YAΔl/L for a given stretch Δl. F₁ = YAΔl/L₁ (given as F). F₂ = Y(3A)Δl/L₂ = Y(3A)Δl/(L₁/3) = 9 × YAΔl/L₁ = 9F.
Q.27
The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is
0%
8 cm
0%
16 cm
0%
12.5 cm
0%
13.2 cm
Explanation
A closed pipe supports only odd harmonics, so its "3rd harmonic" is 3 times its fundamental: f = 3v/(4L_closed) = 3v/(4×20) = 3v/80. An open pipe's fundamental is f = v/(2L_open). Setting equal: v/(2L_open) = 3v/80 → L_open = 80/(2×3) = 80/6 ≈ 13.3 cm (close to the marked 13.2 cm).
Q.28
An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of
0%
Large focal length and small diameter
0%
Small focal length and large diameter
0%
Large focal length and large diameter
0%
Small focal length and small diameter
Explanation
Angular magnification of a refracting telescope increases with the objective's focal length (M = f_objective/f_eyepiece), while angular resolution improves with a larger objective diameter/aperture (resolving power ∝ aperture size) — so both benefit from a large focal length AND a large diameter.
Q.29
The magnetic potential energy stored in a certain inductor is \(25\, mJ\) , When the current in the inductor is \(60\, mA\). This inductor is of inductance
0%
138.88 H
0%
0.138 H
0%
1.389 H
0%
13.89 H
Explanation
Energy stored in an inductor: U = ½LI² → L = 2U/I². L = 2(0.025) / (0.06)² = 0.05/0.0036 ≈ 13.89 H.
Q.30
The electrostatic force between the metal plates of an isolated parallel plate capacitor \(C\) having a charge \(Q\) and area \(A\), is
0%
Proportional to the square root of the distance between the plates.
0%
inversely proportional to the distance between the plates.
0%
Linearly proportional to the distance between the plates
0%
Independent of the distance between the plates.
Explanation
The force between the plates of an isolated (constant charge Q) parallel-plate capacitor is F = Q²/(2ε₀A) — this expression involves only the charge and plate area, with no dependence on the plate separation at all.
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