MHT CET 2026 May 14 Shift 1 PCM Question Paper- Download PDF with Solutions

Step 1: Concept

For a binary salt of type BA the dissociation is \(\mathrm{BA}(s) \rightleftharpoons \mathrm{B}^{+}(aq) + \mathrm{A}^{-}(aq)\).
The solubility product is \(K_{sp} = S^{2}\), where \(S\) is the molar solubility.

Step 2: Meaning

We are given \(K_{sp} = 4.9 \times 10^{-13}\).
We need \(S\), the maximum amount of salt that dissolves in \(1\,\mathrm{dm}^{3}\) of solution.

Step 3: Analysis

Rearranging: \(S = \sqrt{K_{sp}} = \sqrt{4.9 \times 10^{-13}}\).
Rewrite as \(\sqrt{49 \times 10^{-14}}\) to simplify the square root.

Step 4: Conclusion
\(S = \sqrt{49 \times 10^{-14}} = 7 \times 10^{-7}\,\mathrm{mol\,dm}^{-3}\).


Final Answer: (A) Quick Tip: For AB-type salts, solubility is always the square root of \(K_{sp}\). Adjust the power of 10 to an even number for easy square-rooting.

Step 1: Concept

A strong dibasic acid (e.g.\ \(\mathrm{H_2SO_4}\)) dissociates completely,
releasing two \(\mathrm{H}^{+}\) ions per formula unit.
Hence \([\mathrm{H}^{+}] = 2 \times Molarity\).

Step 2: Meaning

The molarity is \(0.01\,\mathrm{M}\), so \([\mathrm{H}^{+}] = 2 \times 0.01 = 0.02\,\mathrm{M} = 2 \times 10^{-2}\,\mathrm{M}\).

Step 3: Analysis
\(\mathrm{pH} = -\log[\mathrm{H}^{+}] = -\log(2 \times 10^{-2}) = 2 - \log 2\).
Using \(\log 2 \approx 0.3010\).

Step 4: Conclusion
\(\mathrm{pH} = 2 - 0.3010 = 1.699 \approx 1.7\).


Final Answer: (D) Quick Tip: For dibasic acids, double the concentration before taking the log. \(\mathrm{pH} = n - \log(\mathrm{coeff})\) where the concentration is \(\mathrm{coeff} \times 10^{-n}\).

Step 1: Concept
\(\mathrm{PbCl_2}\) is an \(AB_2\)-type salt. \(\mathrm{PbCl_2} \rightleftharpoons \mathrm{Pb^{2+}} + 2\,\mathrm{Cl^{-}}\).
The solubility product expression is \(K_{sp} = (S)(2S)^{2} = 4S^{3}\).

Step 2: Meaning

Given \(K_{sp} = 3.2 \times 10^{-5}\), solve \(4S^{3} = 3.2 \times 10^{-5}\).

Step 3: Analysis
\(S^{3} = \dfrac{3.2 \times 10^{-5}}{4} = 0.8 \times 10^{-5} = 8 \times 10^{-6}\).

Step 4: Conclusion
\(S = \sqrt[3]{8 \times 10^{-6}} = 2 \times 10^{-2}\,\mathrm{mol\,dm}^{-3}\).


Final Answer: (B) Quick Tip: For \(AB_2\) or \(A_2B\) salts use \(K_{sp} = 4S^{3}\). Adjust the exponent to a multiple of 3 before taking the cube root.

Step 1: Concept

According to Brønsted--Lowry theory, an acid is a proton donor and a base
is a proton acceptor. Conjugate acid--base pairs differ by one proton.

Step 2: Meaning

Labelling the species: \(\underbrace{\mathrm{HCl}}_{\mathrm{Acid_1}} + \underbrace{\mathrm{H_2O}}_{\mathrm{Base_2}} \rightleftharpoons \underbrace{\mathrm{H_3O^+}}_{\mathrm{Acid_2}} + \underbrace{\mathrm{Cl^-}}_{\mathrm{Base_1}}\).

Step 3: Analysis

HCl (Acid\(_1\)) donates a proton to form \(\mathrm{Cl^-}\) (Base\(_1\)). \(\mathrm{H_2O}\) (Base\(_2\)) accepts the proton to form \(\mathrm{H_3O^+}\)
(Acid\(_2\)). The question asks for Base\(_1\), the conjugate base of HCl.

Step 4: Conclusion
\(\mathrm{Cl^-}\) is the conjugate base produced when HCl loses a proton,
making it the correct answer.


Final Answer: (C) Quick Tip: A Brønsted base is what remains after an acid loses \(\mathrm{H^+}\). \(\mathrm{HCl} - \mathrm{H^+} = \mathrm{Cl^-}\).

Step 1: Concept

For a weak base, \([\mathrm{OH^-}] = C\alpha\), where \(C\) is the molar
concentration and \(\alpha\) is the degree of dissociation. \(\mathrm{pOH} = -\log[\mathrm{OH^-}]\) and \(\mathrm{pH} = 14 - \mathrm{pOH}\).

Step 2: Meaning
\(C = 2 \times 10^{-3}\,\mathrm{M}\), \(\alpha = 0.05\). \([\mathrm{OH^-}] = (2 \times 10^{-3}) \times 0.05 = 1 \times 10^{-4}\,\mathrm{M}\).

Step 3: Analysis
\(\mathrm{pOH} = -\log(10^{-4}) = 4\).

Step 4: Conclusion
\(\mathrm{pH} = 14 - 4 = 10\).


Final Answer: (B) Quick Tip: Always check whether the question asks for pH or pOH. For bases, compute pOH first, then use \(\mathrm{pH} = 14 - \mathrm{pOH}\).

Step 1: Concept

This is a symmetric 3-D resistor network. Cubic symmetry allows us to use
current-distribution analysis to simplify the circuit.

Step 2: Meaning

A body diagonal connects two corners sharing no face. Injecting total
current \(6I\) at one corner, symmetry forces it to split into \(2I\) along
each of the three edges meeting there.

Step 3: Analysis

Tracing the potential drop along one representative path: \[V = (2I)(r) + (I)(r) + (2I)(r) = 5Ir.\]

Step 4: Conclusion
\(R_{\mathrm{eq}} = \dfrac{V}{I_{\mathrm{total}}} = \dfrac{5Ir}{6I} = \dfrac{5}{6}\,r\).


Final Answer: (A) Quick Tip: Cube resistance triad: edge diagonal \(= \tfrac{7}{12}r\), face diagonal \(= \tfrac{3}{4}r\), body diagonal \(= \tfrac{5}{6}r\).

Step 1: Concept

The field due to a long wire at distance \(d\) is \(B = \mu_0 I / (2\pi d)\).
Force on a moving charge: \(\vec{F} = q(\vec{v} \times \vec{B})\).

Step 2: Meaning

By the right-hand thumb rule, for current flowing upward, \(\vec{B}\) at the
proton's location points into the page.

Step 3: Analysis

With \(\vec{v}\) upward and \(\vec{B}\) into the page, \(\vec{v} \times \vec{B}\) points toward the wire.
Magnitude: \(F = ev\!\left(\dfrac{\mu_0 I}{2\pi d}\right) = \dfrac{\mu_0 eIv}{2\pi d}\).

Step 4: Conclusion

The force is directed toward the wire (attractive) with magnitude \(\dfrac{\mu_0 eIv}{2\pi d}\).


Final Answer: (A) Quick Tip: Like currents attract. A moving proton constitutes a current parallel to and in the same direction as the wire, so the force is attractive.

Step 1: Concept

Field at centre of a circular coil: \(B = \mu_0 NI/(2R)\).
Total wire length \(L = N \cdot 2\pi R\) is conserved on rewinding.

Step 2: Meaning

New radius \(R' = R/2\).
Conservation of length: \(N(2\pi R) = N'(2\pi R/2) \Rightarrow N' = 2N\).

Step 3: Analysis
\(B' = \dfrac{\mu_0 N' I}{2R'} = \dfrac{\mu_0 (2N) I}{2(R/2)} = \dfrac{4\mu_0 NI}{2R} = 4B\).

Step 4: Conclusion
\(B' : B = 4 : 1\).


Final Answer: (A) Quick Tip: When rewinding a coil, \(B \propto N/R\). Halving \(R\) doubles \(N\), giving a net factor of \(2 \times 2 = 4\).

Step 1: Concept

Magnetic dipole moment of a current loop: \(M = I \times A\), where \(I\) is
the equivalent current and \(A\) is the area enclosed.

Step 2: Meaning

Equivalent current due to the revolving electron: \(I = e/T = ef\) (since \(f = 1/T\)).

Step 3: Analysis

Area of the circular orbit: \(A = \pi r^{2}\).
Therefore \(M = ef \cdot \pi r^{2}\).

Step 4: Conclusion
\(M = \pi efr^{2}\).


Final Answer: (A) Quick Tip: \(M = IA\). Current \(= e \times f\) and area \(= \pi r^{2}\); multiply directly.

Step 1: Concept

Transformer efficiency: \(\eta = P_s/P_p = (V_s I_s)/(V_p I_p)\).

Step 2: Meaning
\(V_p = 220\,\mathrm{V}\), \(V_s = 110\,\mathrm{V}\), \(I_p = 5\,\mathrm{A}\), \(\eta = 0.80\).

Step 3: Analysis
\(I_s = \dfrac{\eta\,V_p\,I_p}{V_s} = \dfrac{0.80 \times 220 \times 5}{110} = 0.80 \times 2 \times 5 = 8\,\mathrm{A}\).

Step 4: Conclusion

The secondary current is \(8\,\mathrm{A}\).


Final Answer: (B) Quick Tip: Power out \(=\) efficiency \(\times\) power in. At 100% the secondary current would be 10\,A; at 80% it is \(10 \times 0.8 = 8\,\mathrm{A}\).

Step 1: Concept

Use \(\cos 2x = 2\cos^{2}x - 1\) and \(\sin^{3}x = \sin x(1 - \cos^{2}x)\)
to simplify the integrand.

Step 2: Meaning

Numerator: \(\sin x + \sin x(1 - \cos^{2}x) = \sin x(2 - \cos^{2}x)\). \[I = \int \frac{\sin x(2 - \cos^{2}x)}{2\cos^{2}x - 1}\,dx.\]

Step 3: Analysis

Substitute \(u = \cos x\), \(du = -\sin x\,dx\): \[I = -\!\int \frac{2 - u^{2}}{2u^{2} - 1}\,du = \int \frac{u^{2} - 2}{2u^{2} - 1}\,du = \int\!\left(\frac{1}{2} - \frac{3/2}{2u^{2}-1}\right)du.\]
Integrating via partial fractions on the second term: \[I = \tfrac{1}{2}u - \tfrac{3}{4\sqrt{2}}\log\!\left|\frac{\sqrt{2}\,u - 1}{\sqrt{2}\,u + 1}\right| + C.\]

Step 4: Conclusion

Re-substituting \(u = \cos x\) and comparing with the given form: \(P = \dfrac{1}{2}\), \(Q = \dfrac{-3}{4\sqrt{2}}\).


Final Answer: (B) Quick Tip: The substitution \(u = \cos x\) converts the integrand into a rational function, readily handled by partial fractions.

Step 1: Concept

If \(a,b,c\) are in A.P., then \(2b = a + c\).
By the Sine Rule this is equivalent to \(2\sin B = \sin A + \sin C\).

Step 2: Meaning

Using sum-to-product: \(4\sin\tfrac{B}{2}\cos\tfrac{B}{2} = 2\sin\tfrac{A+C}{2}\cos\tfrac{A-C}{2}\).
Since \(\tfrac{A+C}{2} = 90^{\circ} - \tfrac{B}{2}\), we get \(\cos\tfrac{A-C}{2} = 2\sin\tfrac{B}{2}\).

Step 3: Analysis
\[\cos A + \cos C + 2\cos B = 2\cos\tfrac{A+C}{2}\cos\tfrac{A-C}{2} + 2\!\left(1 - 2\sin^{2}\tfrac{B}{2}\right).\]
Substitute \(\cos\tfrac{A+C}{2} = \sin\tfrac{B}{2}\) and \(\cos\tfrac{A-C}{2} = 2\sin\tfrac{B}{2}\): \[= 4\sin^{2}\tfrac{B}{2} + 2 - 4\sin^{2}\tfrac{B}{2} = 2.\]

Step 4: Conclusion
\(\cos A + 2\cos B + \cos C = 2\).


Final Answer: (B) Quick Tip: Quick check: equilateral triangle (\(A = B = C = 60^{\circ}\)) gives \(0.5 + 1 + 0.5 = 2\). \(\checkmark\)

Step 1: Concept

Area \(= \tfrac{1}{2}ac\sin B\).
Expand: \(b^{2} - (c-a)^{2} = b^{2} - c^{2} - a^{2} + 2ac\).

Step 2: Meaning

Cosine Rule: \(b^{2} - c^{2} - a^{2} = -2ac\cos B\).
So the given area \(= 2ac(1 - \cos B)\).

Step 3: Analysis

Equating: \(\tfrac{1}{2}ac\sin B = 2ac(1 - \cos B) \Rightarrow \sin B = 4(1 - \cos B)\).
Half-angle substitution gives \(\tan\tfrac{B}{2} = \tfrac{1}{4}\).

Step 4: Conclusion
\[\tan B = \frac{2\tan(B/2)}{1 - \tan^{2}(B/2)} = \frac{2 \cdot \tfrac{1}{4}}{1 - \tfrac{1}{16}} = \frac{\tfrac{1}{2}}{\tfrac{15}{16}} = \frac{8}{15}.\]


Final Answer: (D) Quick Tip: When area is given in terms of sides, use the Cosine Rule to reduce the expression to a single trigonometric equation in one angle.

Step 1: Concept

Apply the Cosine Rule: \(c^{2} = a^{2} + b^{2} - 2ab\cos C\).

Step 2: Meaning
\(a = 1\), \(b = \sqrt{3}\), \(C = \dfrac{\pi}{6} = 30^{\circ}\),
so \(\cos C = \dfrac{\sqrt{3}}{2}\).

Step 3: Analysis
\[c^{2} = 1 + 3 - 2(1)(\sqrt{3})\cdot\frac{\sqrt{3}}{2} = 4 - 3 = 1.\]

Step 4: Conclusion
\(c = 1\).


Final Answer: (C) Quick Tip: For SAS (two sides and included angle), the Cosine Rule is the direct and only formula needed to find the third side.

Step 1: Concept

Rewrite: \((x-1)^{3/4}(x+2)^{5/4} = (x+2)^{2}\!\left(\dfrac{x-1}{x+2}\right)^{3/4}\).

Step 2: Meaning

Let \(t = \dfrac{x-1}{x+2}\).
Then \(dt = \dfrac{3}{(x+2)^{2}}\,dx\), so \(\dfrac{dx}{(x+2)^{2}} = \dfrac{dt}{3}\).

Step 3: Analysis
\[\int \frac{dx}{(x+2)^{2}\cdot t^{3/4}} = \frac{1}{3}\int t^{-3/4}\,dt = \frac{1}{3}\cdot\frac{t^{1/4}}{1/4} + C = \frac{4}{3}\,t^{1/4} + C.\]

Step 4: Conclusion

Re-substituting: \(I = \dfrac{4}{3}\!\left(\dfrac{x-1}{x+2}\right)^{1/4} + C\).


Final Answer: (C) Quick Tip: For integrands with fractional powers of two linear factors, try \(t = \dfrac{ax+b}{cx+d}\); this reduces the integral to a simple power function.