12th Science Maharashtra Board
In this category students supposed to ask questions which belong to std 12th maharashtra board Science
- The order and degree of the differential equation \(\left(\frac{d^2y}{dx^2}\right)^3+\left(\frac{dy}{dx}\right)^4+y=0\) are respectively Answer: B. 2 and 3 Highest derivative d²y/dx² (order 2), raised to power 3 (degree 3). 1
- If \(y=x^x\), then \(\frac{dy}{dx}\) at \(x=e\) is Answer: B. 2e^e dy/dx=x^x(1+ln x) ; at x=e : e^e(1+1)=2e^e . 1
- If \(x=a\cos^3t\) and \(y=a\sin^3t\), then \(\frac{dy}{dx}=\) Answer: A. -tan t dy/dx=3asin²tcos t/-3acos²tsin t=-tan t . 1
- If \(x^2+y^2=25\), then \(\frac{d^2y}{dx^2}\) at the point \((3,4)\) is Answer: D. -25/64 y'=-x/y=-3/4 ; y''=-y-xy'/y²=-4+9/4/16=-25/64 . 1
- \(\int\frac{dx}{x^2+4x+13}=\) Answer: A. 1/3tan⁻¹(x+2/3)+C x²+4x+13=(x+2)²+3² ; use ∫dx/u²+a²=1/atan⁻¹u/a . 1
- \(\int e^x\left(\frac1x-\frac1{x^2}\right)dx=\) Answer: A. e^x/x+C Form ∫ e^x(f+f')dx=e^xf+C with f=1/x . 1
- \(\int\frac{dx}{1+e^x}=\) Answer: C. x-ln(1+e^x)+C 1/1+e^x=1-e^x/1+e^x ; integrate term by term. 1
- A solid sphere rolls without slipping on a horizontal surface. The fraction of its total kinetic energy that is rotational is Answer: B. 2/7 KE_rot=1/2·2/5 mR²ω²=1/5 mv² , KE_total=1/2mv²+1/5mv²=7/10mv² . Fraction =1/5/7/10=2/7 . 1
- A flywheel of moment of inertia 2 kg m2 is speeded up from 10 rad/s to 20 rad/s. The work done on it is Answer: D. 300 J W=Δ KE=1/2 I(ω₂²-ω₁²)=1/2(2)(400-100)=300 J. 1
- A curved road of radius 50 m is to be banked so that a car moving at 10 m/s needs no friction. The angle of banking is (g = 10 m/s2) Answer: C. tan⁻¹(0.2) tanθ=v²/rg=100/50×10=0.2 . 1
- A small stone tied to a string of length 0.9 m is whirled in a vertical circle. The minimum speed it must have at the lowest point to just complete the circle is (g = 10 m/s2) Answer: A. 6.7 m/s v_min=√(5gr)=√(5×10×0.9)=√(45)≈6.7 m/s. 1
- A circular coil of 100 turns and radius 10 cm carries a current of 5 A. The magnetic field at its centre is (\(\mu_0=4\pi\times10^{-7}\) T m/A) Answer: D. π×10⁻³ T B=μ₀NI/2R=frac4π×10⁻⁷×100×50.2=π×10⁻³ T. 1
- A proton and an α-particle having the same kinetic energy enter a uniform magnetic field perpendicular to it. The ratio of the radii of their circular paths \(r_p:r_\alpha\) is Answer: A. 1 : 1 r=√(2mK)/qB . For α: mass 4m , charge 2e gives √(8mK)/2eB=√(2mK)/eB , same as proton. 1
- A rectangular coil of 50 turns and area 0.02 m2 carries 2 A in a uniform field of 0.5 T. The torque on it when its plane is parallel to the field is Answer: B. 1 N m Plane parallel to field means the normal is perpendicular to B : τ=NIAB=50×2×0.02×0.5=1 N m. 1
- The magnetic susceptibility of a paramagnetic salt is \(\chi\) at 300 K. Its susceptibility at 200 K will be Answer: D. 1.5chi Curie's law: chi∝1/T⇒chi'=chi×300/200=1.5chi . 1
- A conducting rod of length 1 m moves at 5 m/s perpendicular to a uniform magnetic field of 0.4 T on frictionless rails connected through a 4 Ω resistor (other resistances negligible). The external force needed to keep it moving at constant speed is Answer: D. 0.2 N e=Blv=2 V, I=0.5 A, F=BIl=0.4×0.5×1=0.2 N. 1
- When the current in a coil falls uniformly from 2 A to zero in 0.05 s, an emf of 8 V is induced in it. The self-inductance of the coil is Answer: A. 0.2 H L=e/dI/dt=8/2/0.05=0.2 H. 1
- In a series LCR circuit, \(L=\frac{1}{\pi}\) H and \(C=\frac{1}{\pi}\) μF. The resonant frequency is Answer: A. 500 Hz f=1/2π√(LC)=frac12π·frac10⁻³π=500 Hz. 1
- In a series RL circuit, R = 30 Ω and \(X_L\) = 40 Ω. The power factor of the circuit is Answer: D. 0.6 Z=√(30²+40²)=50\ Ω , cosφ=R/Z=0.6 . 1
- Light of wavelength 310 nm falls on a metal of work function 2.5 eV. The stopping potential is (take \(hc=1240\) eV nm) Answer: C. 1.5 V E=1240/310=4 eV; eV₀=4-2.5=1.5 eV, so V₀=1.5 V. 1
- The de Broglie wavelength of an electron accelerated from rest through a potential difference of 100 V is about Answer: B. 1.227 Å λ=12.27/sqrt V Å =12.27/10=1.227 Å. 1
- The ratio of the wavelength of the first line of the Balmer series to that of the first line of the Lyman series in hydrogen is Answer: D. 27/5 frac1λ_B∝1/4-1/9=5/36 , frac1λ_L∝1-1/4=3/4 . λ_B/λ_L=3/4/5/36=27/5 . 1
- \({}^{238}_{92}\mathrm{U}\) decays to \({}^{206}_{82}\mathrm{Pb}\). The numbers of α and β− particles emitted are Answer: D. 8 α and 6 β α count =238-206/4=8 , lowering Z by 16 to 76. To reach 82, 6 β⁻ are needed. 1
- A 6 V Zener diode is used with a series resistance of 200 Ω across a 10 V supply. A load of 1 kΩ is connected across the Zener. The current through the Zener diode is Answer: C. 14 mA Series current =10-6/200=20 mA; load current =6/1000=6 mA; Zener current =14 mA. 1