MH CET 2025 - 19 April — MH-CET Previous Year Question Papers MCQs with Solutions

Solution: Three equal inductors in series: $L_e=L_1+L_2+L_3=L+L+L=3L$. Energy stored $U=\dfrac{1}{2}L_e I^2=\dfrac{3LI^2}{2}$.

Solution: Apparent frequency for a source moving towards a stationary observer: $n'=n\left[\dfrac{v}{v-v_s}\right]$. As the source moves towards the observer, frequency increases, hence wavelength decreases.

Solution: In a reverse-biased p-n junction, the width of the depletion region increases, which leads to an increase in the electric field strength. Thus the electric field strength is maximum in the depletion layer.

Solution: Since the process is isothermal, there is no change in internal energy. Hence the supplied energy is used to perform work on the surroundings, so the gas does positive work.

Solution: The potential energy of a dipole in an electric field is $W=-pE\cos\theta$. When $\theta=0^\circ$, $W_1=-pE\cos 0^\circ=-pE$. When $\theta=90^\circ$, $W_2=-pE\cos 90^\circ=0$. Work done $=W_2-W_1=0-(-pE)=pE$.

Solution: A triglyceride (glyceryl ester of three fatty acids $R_1,R_2,R_3$) on hydrolysis with $3\ce{H2O}$ yields glycerol ($\ce{CH2OH-CHOH-CH2OH}$) and three free fatty acids ($\ce{R1COOH}$, $\ce{R2COOH}$, $\ce{R3COOH}$).

Solution: Transition elements have partially filled d-orbitals. $\text{Hg} = [\text{Xe}]4f^{14}5d^{10}6s^2$. Among the given options, Hg has a completely filled d-orbital in its ground state, hence it is not regarded as a transition element.

Solution: Glass exhibits the same magnitude for all properties such as refractive index, conductivity, etc., in every direction. Hence glass is isotropic in nature.

Solution: The duration of half-life ($t_{1/2}$) for a first order reaction is independent of the initial reactant concentration.

Solution: 1 mole of $\ce{CO2}$ at STP occupies $22.4\ \text{dm}^3$. Therefore 0.5 mole of $\ce{CO2}$ occupies $= 0.5\ \text{mol} \times 22.4\ \text{dm}^3\ \text{mol}^{-1} = 11.2\ \text{dm}^3$.

Solution: On heating, lithium carbonate decomposes: $\ce{Li2CO3 ->[\Delta] Li2O + CO2}$.

Solution: The addition polymerization of tetrafluoroethene with a free radical or desulphated catalyst at high pressure gives Teflon: $n\,\ce{CF2=CF2} ->[\text{catalyst, high pressure}] \ce{-[CF2-CF2]_n-}$. Teflon (PTFE) is poly(1,1,2,2-tetrafluoroethene).

Solution: Ozonolysis is a reaction in which ozone $\ce{O3}$ reacts with unsaturated compounds containing double bonds to form addition products called ozonides. These ozonides are then decomposed by water or dilute acids, producing aldehydes or ketones.

Solution: The equation of a plane through A(7,8,6) is $a(x-7) + b(y-8) + c(z-6) = 0$ ...(i). Since the required plane is parallel to the XY-plane, its direction ratios are 0, 0, 1. So the equation is $0(x-7) + 0(y-0) + 1(z-6) = 0 \Rightarrow z - 6 = 0 \Rightarrow z = 6$.

Solution: Given $\left(\sqrt{x^{2}+y^{2}}\right)^{2}+\left(\sqrt{y^{2}+z^{2}}\right)^{2}+\left(\sqrt{z^{2}+x^{2}}\right)^{2}=242$, i.e. $2(x^{2}+y^{2}+z^{2})=242$ so $x^{2}+y^{2}+z^{2}=121$. The distance from origin $=\sqrt{x^{2}+y^{2}+z^{2}}=\sqrt{121}=11$.

Solution: For the given structure the molecular formula is $\ce{C2H6O}$. Molar mass $=(12\times2)+(1\times6)+(16\times1)=24+6+16=46\ \text{g mol}^{-1}$.

Solution: Using conservation of energy, $\dfrac{1}{2}mv_1^2=\dfrac{1}{2}mv_2^2+2mgL$, so $v_1^2=v_2^2+4gL$ ...(i). Maximum tension $T_{max}=\dfrac{mv_1^2}{L}+mg$ and minimum tension $T_{min}=\dfrac{mv_2^2}{L}-mg$. Thus $\dfrac{T_{max}}{T_{min}}=\dfrac{v_1^2+gL}{v_2^2-gL}=\dfrac{v_2^2+5gL}{v_2^2-gL}$ (from i). Given ratio $=3$: $3(v_2^2-gL)=v_2^2+5gL\Rightarrow 2v_2^2=8gL\Rightarrow v_2^2=4gL$, so $v_2=\sqrt{4gL}$.

Solution: Magnetic fields from the two wires: $B_1=\dfrac{\mu_0 I_1}{2\pi X}$ and $B_2=\dfrac{\mu_0 I_2}{2\pi X}$. Since they are perpendicular, $B_{net}=\sqrt{B_1^2+B_2^2}=\dfrac{\mu_0}{2\pi X}\sqrt{I_1^2+I_2^2}$. By Pythagoras, $2X^2=5\times5$ so $X=\dfrac{5}{\sqrt{2}}$ cm. Then $B_{net}=\dfrac{4\pi\times10^{-7}}{2\pi\times\dfrac{5}{\sqrt{2}}\times10^{-2}}\sqrt{4^2+3^2}=2\sqrt{2}\times10^{-5}\ \text{T}$.

Solution: Using Rydberg's formula $\dfrac{1}{\lambda}=R\left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right]$. For Balmer ($n_1=2,n_2=3$): $\dfrac{1}{\lambda_1}=R\left[\dfrac{1}{4}-\dfrac{1}{9}\right]=\dfrac{5R}{36}$. For Paschen ($n_1=3,n_2=4$): $\dfrac{1}{\lambda_2}=R\left[\dfrac{1}{9}-\dfrac{1}{16}\right]=\dfrac{7R}{144}$. So $\dfrac{\lambda_2}{\lambda_1}=\dfrac{5R}{36}\times\dfrac{144}{7R}=\dfrac{20}{7}$. Since $f=\dfrac{c}{\lambda}$, $\dfrac{f_1}{f_2}=\dfrac{\lambda_2}{\lambda_1}=\dfrac{20}{7}$.

Solution: Given $V_2=2V_1$, $\gamma=\dfrac{3}{2}$. Gas Y (adiabatic): $P_1 V_1^\gamma=P_2 V_2^\gamma\Rightarrow \dfrac{P_1}{P_2}=\left(\dfrac{V_2}{V_1}\right)^\gamma=2^{3/2}=2\sqrt{2}$, so $P_1=2\sqrt{2}P_2$. Gas Z (isobaric): $P_1=P_2$. Gas X (isothermal): $P_1 V_1=P_2 V_2\Rightarrow \dfrac{P_1}{P_2}=2$, so $P_1=2P_2$. Ratio of initial pressures $=2:2\sqrt{2}:1$.

Solution: Initially $C_1=\dfrac{A\varepsilon_0}{d}$. After inserting the dielectric, $C_2=\dfrac{A\varepsilon_0}{d-t+\dfrac{t}{k}}=\dfrac{A\varepsilon_0}{d-t+\dfrac{t}{3}}$. Given $C_2=C_1+50\%\ \text{of}\ C_1=\dfrac{3}{2}C_1$. So $\dfrac{A\varepsilon_0}{\dfrac{3d-3t+t}{3}}=\dfrac{3}{2}\dfrac{A\varepsilon_0}{d}\Rightarrow\dfrac{3}{3d-2t}=\dfrac{3}{2d}\Rightarrow 2d=3d-2t\Rightarrow 2t=d$, so $t=\dfrac{d}{2}$.

Solution: Gravitational force provides centripetal force: $\dfrac{Gm^2}{(2r)^2}=\dfrac{mv^2}{r}$ or $\dfrac{Gm}{4r}=v^2$. Therefore $v=\sqrt{\dfrac{Gm}{4r}}$.

Solution: Mutual inductance of two concentric coplanar circular coils: $M=\dfrac{\mu_0\pi N_1 N_2 R_2^2}{2R_1}$. Therefore $M\propto\dfrac{R_2^2}{R_1}$.

Solution: Given $T=0.1\ \text{s}$. $T=2\pi\sqrt{\dfrac{m}{K}}$ with $Kx=mg$ (x = amplitude), so $\dfrac{m}{K}=\dfrac{x}{g}$ and $T=2\pi\sqrt{\dfrac{x}{g}}$. Then $0.1=2\pi\sqrt{\dfrac{x}{g}}\Rightarrow\dfrac{1}{100}=4\pi^2\dfrac{x}{g}\Rightarrow x=\dfrac{g}{400\pi^2}=\dfrac{10}{400\pi^2}=\dfrac{1}{40\pi^2}$. Then $V_{max}=x\omega=x\times2\pi f=\dfrac{1}{40\pi^2}\times2\pi\times10=\dfrac{1}{2\pi}\ \text{m/s}$ (using $f=\dfrac{1}{T}$).

Solution: The string vibrates in three segments, so $l=\dfrac{3\lambda}{2}$, giving $\lambda=\dfrac{2l}{3}=\dfrac{2(0.9)}{3}=0.6\ \text{m}$. Wave speed $v=\sqrt{\dfrac{T}{m}}=\sqrt{\dfrac{40}{0.1}}=20\ \text{m/s}$. Frequency $n=\dfrac{v}{\lambda}=\dfrac{20}{0.6}=\dfrac{100}{3}\ \text{Hz}$. Amplitude of particle velocity $=(a_{\max})\omega=(a_{\max})(2\pi n)=0.3\times10^{-2}\times2\pi\times\dfrac{100}{3}=\dfrac{\pi}{5}\ \text{m/s}$.

Solution: Torque of the rod's weight about the pivot at angle $\theta$ is $\tau=Mg\sin\theta\,\dfrac{L}{2}$. Also $\tau=I\alpha$, so $I\alpha=Mg\sin\theta\,\dfrac{L}{2}$. For a thin uniform rod about an axis through one end, $I=\dfrac{ML^2}{3}$, giving $\dfrac{ML^2}{3}\alpha=Mg\sin\theta\,\dfrac{L}{2}$, hence $\dfrac{L\alpha}{2}=g\dfrac{\sin\theta}{2}$ and $\alpha=\dfrac{3g\sin\theta}{2L}$.

Solution: Let $a$ be the distance between $q_2$ and the proton p, so the distance between $q_1$ and p is $L+a$. For equilibrium, the net field is zero: $\dfrac{kq_1}{r_1^2}+\dfrac{kq_2}{r_2^2}=0\Rightarrow\dfrac{+6q}{(L+a)^2}+\dfrac{-3q}{a^2}=0\Rightarrow\dfrac{6q}{(L+a)^2}=\dfrac{3q}{a^2}\Rightarrow\dfrac{2}{(L+a)^2}=\dfrac{1}{a^2}$. Taking the square root: $\dfrac{\sqrt{2}}{L+a}=\dfrac{1}{a}\Rightarrow\sqrt{2}a=L+a\Rightarrow a=\dfrac{L}{\sqrt{2}-1}$. Distance between $q_1$ and p $=L+a=L+\dfrac{L}{\sqrt{2}-1}=\dfrac{\sqrt{2}L}{\sqrt{2}-1}$.

Solution: $3E=n\times(4\pi r^2\times T)-(4\pi R^2\times T)$. With $3\times(4\pi R^2\times T)=n\times(4\pi r^2\times T)-(4\pi R^2\times T)$, this gives $n=\dfrac{4R^2}{r^2}$.

Solution: For a spectral series $\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$. For the 2nd line of the Balmer series ($n_1=2,\,n_2=\infty$): $\dfrac{1}{\lambda_1}=R\left(\dfrac{1}{2^2}-\dfrac{1}{\infty^2}\right)=\dfrac{R}{4}$, so $\lambda_1=\dfrac{4}{R}$. For the last line ($n_1=3,\,n_2=\infty$): $\dfrac{1}{\lambda_2}=R\left(\dfrac{1}{3^2}\right)=\dfrac{R}{9}$, so $\lambda_2=\dfrac{9}{R}$. Therefore $\dfrac{\lambda_2}{\lambda_1}=\dfrac{9}{R}\times\dfrac{R}{4}=\dfrac{9}{4}$.