MHT CET 2026 May 14 Shift 2 PCM Question Paper- Download PDF with Solutions

Step 1: Concept
To evaluate the integral, we use trigonometric substitution to simplify the square root expression. A common substitution for \(\sqrt{\frac{a-x}{a+x}}\) is \(x = a \cos \theta\).

Step 2: Meaning
Let \(x = 4 \cos \theta\). Then \(dx = -4 \sin \theta d\theta\). When \(x = 0\), \(\theta = \pi/2\); when \(x = 4\), \(\theta = 0\).

Step 3: Analysis
Substituting these into the integral:
\(\int_{\pi/2}^{0} \sqrt{\frac{4(1-\cos \theta)}{4(1+\cos \theta)}} (-4 \sin \theta) d\theta = \int_{0}^{\pi/2} \sqrt{\frac{2 \sin^2(\theta/2)}{2 \cos^2(\theta/2)}} (4 \sin \theta) d\theta\).
This simplifies to
\(\int_{0}^{\pi/2} \tan(\theta/2) \cdot 8 \sin(\theta/2) \cos(\theta/2) d\theta = \int_{0}^{\pi/2} 8 \sin^2(\theta/2) d\theta
= \int_{0}^{\pi/2} 4(1 - \cos \theta) d\theta\).

Step 4: Conclusion
Integrating: \(4[\theta - \sin \theta]_{0}^{\pi/2} = 4(\pi/2 - 1) = 2\pi - 4 = 2(\pi - 2)\).


Final Answer: (B) Quick Tip: For integrals involving \(\sqrt{\frac{a-x}{a+x}}\), rationalizing the numerator or using \(x = a \cos \theta\) are effective strategies.

Step 1: Concept
We determine the range by analyzing the composite function \(f(g(x))\) where \(g(x) = \sin x\) and \(f(u) = \log u\).

Step 2: Meaning
Since the domain is restricted to \(\sin x > 0\), we know that for any real \(x\), the value of the sine function is bounded such that \(0 < \sin x \leq 1\).

Step 3: Analysis
The natural logarithm function \(y = \log u\) is monotonically increasing. As the argument \(u\) ranges from values approaching \(0\) up to \(1\), \(\log u\) ranges from \(-\infty\) to \(\log(1) = 0\).

Step 4: Conclusion
Therefore, the outputs of the function cover all values from negative infinity up to and including zero, making the range \((-\infty, 0]\).


Final Answer: (C) Quick Tip: The log of a value between 0 and 1 is always negative. Since \(\sin x\) never exceeds 1, its log can never be positive.

Step 1: Concept
The second derivative \(\frac{d^2y}{dx^2} = 0\) indicates that the rate of change of the slope is zero, meaning the relation between \(x\) and \(y\) is linear.

Step 2: Meaning
Integrating the condition \(\frac{d^2y}{dx^2} = 0\) twice with respect to \(x\) provides the functional form of the relationship.

Step 3: Analysis
First integration: \(\frac{dy}{dx} = m\) (where \(m\) is a constant slope). Second integration: \(y = mx + c\) (where \(c\) is a constant).

Step 4: Conclusion
Among the given options, \(y = ax + b\) is the only expression that represents the standard general equation of a straight line.


Final Answer: (C) Quick Tip: Whenever the second derivative is zero, the curve is a straight line.

Step 1: Concept
Use the double angle formula for sine, \(\sin 2x = 2 \sin x \cos x\), to simplify the product of trigonometric terms.

Step 2: Meaning
The given equation \(\sin x \cos x = 1/4\) can be multiplied by 2 on both sides to give \(2 \sin x \cos x = 1/2\), which simplifies to \(\sin 2x = 1/2\).

Step 3: Analysis
The general solution for \(\sin \theta = \sin \alpha\) is \(\theta = n\pi + (-1)^n \alpha\). Since \(\sin(\pi/6) = 1/2\), we substitute \(\theta = 2x\) and \(\alpha = \pi/6\) to get \(2x = n\pi + (-1)^n (\pi/6)\).

Step 4: Conclusion
Dividing the entire equation by 2 yields the general solution \(x = \frac{n\pi}{2} + (-1)^n \frac{\pi}{12}\).


Final Answer: (B) Quick Tip: Multiplying by 2 to create \(\sin 2x\) is the standard first step for trigonometric products of sine and cosine.

Step 1: Concept
Simplify the complex terms \((1-i)^2\) and \((1+i)^2\) using algebraic expansion before applying the power \(n\).

Step 2: Meaning
Note that \((1-i)^2 = 1 + i^2 - 2i = 1 - 1 - 2i = -2i\) and \((1+i)^2 = 1 + i^2 + 2i = 1 - 1 + 2i = 2i\).

Step 3: Analysis
The expression becomes \(\frac{2^n}{(-2i)^n} + \frac{(2i)^n}{2^n} = \frac{1}{(-i)^n} + i^n = i^n + \frac{1}{(-i)^n}\). Using \(1/(-i) = i\), this simplifies to \(i^n + i^n\) is not correct; rather \(i^n + (-i)^{-n} = e^{in\pi/2} + e^{-in\pi/2}\).

Step 4: Conclusion
Using the identity \(2 \cos \theta = e^{i\theta} + e^{-i\theta}\), where \(\theta = n\pi/2\), the result is \(2 \cos(n\pi/2)\).


Final Answer: (C) Quick Tip: The squares \((1\pm i)^2 = \pm 2i\) are extremely useful for simplifying powers of these complex numbers.

Step 1: Concept
A chemical test that provides distinct observable results for \(1^\circ\), \(2^\circ\), and \(3^\circ\) alcohols based on their reactivity.

Step 2: Meaning
Lucas reagent is a solution of anhydrous zinc chloride (\(ZnCl_2\)) in concentrated hydrochloric acid (\(HCl\)). It reacts with alcohols to form alkyl chlorides, which are insoluble and create cloudiness (turbidity) in the solution.

Step 3: Analysis
The rate of this reaction depends on the stability of the carbocation formed:

* Tertiary (\(3^\circ\)) alcohols react immediately, producing turbidity instantly.
* Secondary (\(2^\circ\)) alcohols react within 5 to 10 minutes.
* Primary (\(1^\circ\)) alcohols do not produce turbidity at room temperature.

Step 4: Conclusion
Because each class of alcohol reacts at a significantly different speed, the Lucas Test reagent is the standard choice for distinguishing them.


Final Answer: (A) Quick Tip: Remember the 3-2-1 rule for Lucas reagent: 3 (Instant), 2 (5-10 mins), 1 (No reaction at room temp).

Step 1: Concept
The boiling point of a compound depends on the strength of the intermolecular forces (IMF) present between its molecules.

Step 2: Meaning
Stronger intermolecular forces require more energy (higher temperature) to overcome and transition the substance from a liquid to a gas.

Step 3: Analysis
Propane and ethane are non-polar hydrocarbons with weak London dispersion forces. Ethyl chloride is polar with dipole-dipole interactions. Ethyl alcohol (\(C_{2}H_{5}OH\)) contains an \(-OH\) group, which enables strong intermolecular hydrogen bonding.

Step 4: Conclusion
Hydrogen bonding is significantly stronger than dipole-dipole or dispersion forces; therefore, ethyl alcohol has the highest boiling point among the given options.


Final Answer: (C) Quick Tip: Hydrogen bonding (found in alcohols, carboxylic acids, and amines) usually results in a much higher boiling point than other organic compounds of similar mass.

Step 1: Concept
The Williamson Ether Synthesis is the standard method for preparing ethers, involving an \(S_{N}2\) reaction between a nucleophilic alkoxide or phenoxide ion and a primary alkyl halide.

Step 2: Meaning
To prepare anisole (\(C_{6}H_{5}OCH_{3}\)), you need a phenoxide source to provide the phenyl group and a methyl source to provide the alkyl group.

Step 3: Analysis
Sodium phenoxide (\(C_{6}H_{5}ONa\)) acts as the nucleophile and attacks the methyl group of methyl iodide (\(CH_{3}I\)), displacing the iodide ion.

Step 4: Conclusion
This specific combination follows the \(S_{N}2\) mechanism perfectly to yield methyl phenyl ether (anisole).


Final Answer: (A) Quick Tip: Williamson Synthesis: Use the larger part as the alkoxide and the smaller part as the alkyl halide for the best yield.

Step 1: Concept
The degree of dissociation (\(\alpha\)) of a weak electrolyte is the ratio of its molar conductivity (\(\Lambda_{m}\)) to its limiting molar conductivity (\(\Lambda_{m}^{0}\)), given by \(\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{0}}\).

Step 2: Meaning
The concentration of \(H^{+}\) ions in a weak acid is calculated using the formula \([H^{+}] = C\alpha\), where \(C\) is the initial concentration.

Step 3: Analysis
First, find \(\alpha = \frac{5}{390} \approx 0.01282\). Then, calculate the concentration: \([H^{+}] = 0.1 \times 0.01282 = 0.001282\) M.

Step 4: Conclusion
Converting the result to scientific notation gives \(1.28 \times 10^{-3}\) M.


Final Answer: (A) Quick Tip: For weak acids: Concentration of ions is simply the product of the total concentration and the degree of ionization.

Step 1: Concept
Orthophosphoric acid has the chemical formula \(H_{3}PO_{4}\). The sum of oxidation states in a neutral molecule must be zero.

Step 2: Meaning
We assign standard oxidation states: \(H = +1\) and \(O = -2\). Let the oxidation state of phosphorus be \(x\).

Step 3: Analysis
Setting up the equation: \(3(+1) + x + 4(-2) = 0\). This simplifies to \(3 + x - 8 = 0\), leading to \(x - 5 = 0\).

Step 4: Conclusion
Solving for \(x\) gives \(x = +5\). Therefore, phosphorus is in the \(+5\) oxidation state.


Final Answer: (C) Quick Tip: "Ic" acids of non-metals usually feature the element in its highest common oxidation state (e.g., \(H_{2}SO_{4}\) is \(+6\), \(H_{3}PO_{4}\) is \(+5\)).

Step 1: Concept
For pure rolling on an incline, the friction force \(f\) must be less than or equal to the maximum static friction \(\mu_{s} N\).
The acceleration is \(a = \frac{g \sin \theta}{1 + I/MR^{2}}\).

Step 2: Meaning
For a solid sphere, the moment of inertia about its center is \(I = \frac{2}{5} MR^{2}\).
The friction force required to provide torque for rolling is \(f = \frac{I a}{R^{2}}\).

Step 3: Analysis
Substituting \(a\) and \(I\): \(f = M g \sin \theta \left( \frac{I}{MR^{2} + I} \right)\).
For a solid sphere, this becomes \(f = \frac{2}{7} Mg \sin \theta\).
Since \(f \leq \mu_{s} Mg \cos \theta\), we get \(\mu_{s} \geq \frac{2}{7} \tan \theta\).

Step 4: Conclusion
The minimum value required to prevent slipping is exactly \(\frac{2}{7} \tan \theta\).


Final Answer: (A) Quick Tip: The general formula for \(\mu_{min}\) for any rolling object is \(\frac{\tan \theta}{1 + \frac{MR^{2}}{I}}\).

Step 1: Concept
Since no external torque acts on the system about the rotation axis, the total angular momentum is conserved (\(L_{initial} = L_{final}\)).

Step 2: Meaning
Angular momentum is \(L = I \omega\). The initial moment of inertia of the disc is \(I_{1} = \frac{1}{2} MR^{2}\).

Step 3: Analysis
When mass \(m\) is placed at the edge, the new moment of inertia is \(I_{2} = \frac{1}{2} MR^{2} + mR^{2} = R^{2}(\frac{M}{2} + m)\).
Conservation gives \(I_{1} \omega = I_{2} \omega'\). Thus, \(\frac{1}{2} MR^{2} \omega = R^{2}(\frac{M + 2m}{2}) \omega'\).

Step 4: Conclusion
Solving for \(\omega'\) gives \(\omega' = \frac{M\omega}{M + 2m}\).


Final Answer: (A) Quick Tip: "Gently placed" implies no external impulse/torque, making angular momentum conservation your go-to tool.

Step 1: Concept
The radius of gyration \(k\) is defined by \(I = Mk^{2}\). First, we must find the moment of inertia \(I\) about the tangent.

Step 2: Meaning
According to the parallel axis theorem, \(I_{tangent} = I_{cm} + Md^{2}\). For a hollow sphere, \(I_{cm} = \frac{2}{3} MR^{2}\) and the distance to the tangent is \(d = R\).

Step 3: Analysis
\(I = \frac{2}{3} MR^{2} + MR^{2} = \frac{5}{3} MR^{2}\). Equating this to \(Mk^{2}\), we get \(k^{2} = \frac{5}{3} R^{2}\).

Step 4: Conclusion
Taking the square root, \(k = \sqrt{\frac{5}{3}} R\).


Final Answer: (A) Quick Tip: Always remember the parallel axis theorem: \(I = I_{cm} + Mh^{2}\). It's the bridge between center-of-mass properties and any other axis.

Step 1: Concept
According to Jurin's Law, the height \(h\) to which a liquid rises in a capillary tube is inversely proportional to the radius \(r\) of the tube (\(h \propto 1/r\)).

Step 2: Meaning
The formula is \(h = \frac{2T \cos \theta}{\rho g r}\). For the same liquid and conditions, all terms except \(h\) and \(r\) are constant.

Step 3: Analysis
Since \(h_{1} r_{1} = h_{2} r_{2}\) (as \(hr = constant\)), we can rearrange the equation to find the ratio of the heights.

Step 4: Conclusion
The ratio \(\frac{h_{1}}{h_{2}} = \frac{r_{2}}{r_{1}}\).


Final Answer: (B) Quick Tip: Thinner tube = Higher rise. The inverse relationship means the ratio of heights is the inverse ratio of radii.

Step 1: Concept
The root mean square velocity (\(v_{rms}\)) is given by the formula \(v_{rms} = \sqrt{\frac{3RT}{M}}\).

Step 2: Meaning
Using the ideal gas equation in terms of density (\(PM = dRT\)), we can substitute \(RT/M = P/d\).

Step 3: Analysis
Substituting this into the \(v_{rms}\) formula gives \(v_{rms} = \sqrt{\frac{3P}{d}}\). At constant pressure \(P\), \(v_{rms}\) depends only on density.

Step 4: Conclusion
Therefore, \(v_{rms} \propto \frac{1}{\sqrt{d}}\).


Final Answer: (D) Quick Tip: \(v_{rms} = \sqrt{3P/d}\). Heavier (denser) gas molecules move slower at the same pressure.