Physics
- A biconvex lens has radii of curvature, 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is : Answer: 4. +5D Use the lens maker's formula: 1/f=(μ-1)(dfrac1R₁-dfrac1R₂) . For a biconvex lens R₁=+20 cm, R₂=-20 cm: 1/f=0.5(dfrac120+dfrac... 1
- If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is : Answer: 4. 1 : √(2) Speed of a transverse wave on a string is v=√(T/μ) , so v∝sqrt T . v_i/v_f=√(T/2T)=1/sqrt2 , i.e. 1:sqrt2 . Option (4) i... 1
- The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is : Answer: 3. √(2) : 1 Radius of gyration is k=√(I/M) . For a disc, about the normal axis through the centre I=1/2MR² ; about a diameter I=1/4M... 1
- Three resistors having resistances r_1, r_2 and r_3 are connected as shown in the given circuit. The ratio i_3/i_1 of currents in terms of resistances used in the circuit is : Answer: 2. r₂/r₂+r₃ Key idea: r₂ and r₃ are in parallel, and the current i₁ splits in inverse ratio of resistances (current divider). i₃=i₁·... 1
- A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit ? Answer: 1. 0.2 A Key idea: with no losses, input power = output power = 44 W. I_p=P/V_p=44/220=0.2 A. Option 1. (The secondary current is 44... 1
- A particle of mass 'm' is projected with a velocity v = kV_e (k < 1) from the surface of the earth. (V_e = escape velocity)The maximum height above the surface reached by the particle is : Answer: 4. Rk²/1-k² Key idea: conserve energy: 1/2mv²-dfracGMmR=-GMm/R+h , with V_e²=dfrac2GMR . 1/2m k²dfrac2GMR-dfracGMmR=-GMm/R+h⇒dfracGM... 1
- A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution.If this particle were projected with the same speed at an angle 'θ' to the horizontal, the maximum height attained by it equals 4R. The angle of… Answer: 4. θ = sin⁻¹(2gT²/π²R)^(1/2) Key idea: speed v=dfrac2π RT , and the maximum height is H=v²sin²θ/2g . 4R=4π²R²/T²·sin²θ/2g⇒sin²θ=8gRT... 1
- A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s^2) nearly : Answer: 2. 4.2 kg m/s Key idea: impulse equals change in momentum; the velocity reverses. Speed at the ground v=√(2gh)=√(200)≈14.1 m/s. Rebo... 1
- From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring i… Answer: 1. 3/4 Key idea: moment of inertia of a ring about its axis depends only on mass and radius, since every part is at distance R. Remo... 1
- A radioactive nucleus ^A_ZX undergoes spontaneous decay in the sequence^A_ZX → _Z-1B → _Z-3C → _Z-2D, where Z is the atomic number of element X. The possible decay particles in the sequence are : Answer: 3. β^+, α, β^- Key idea: α decay lowers Z by 2; β^- raises Z by 1; β^+ lowers Z by 1. Z→ Z-1 : β^+ . Z-1→ Z-3 : α . Z-3→ Z-2 : β^- .... 1
- The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be : Answer: 2. 1/2√(2) Key idea: activity decays as A=A₀(1/2)^t/T_(1/2) . dfrac AA₀=(1/2)^(150/100)=(1/2)^(3/2)=dfrac12sqrt2 . Option 2. 1
- A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is : Answer: 4. 216 MeV Key idea: gain in binding energy = total B.E. of fragments - B.E. of parent. 240×7.6=1824 MeV; fragments: 240×8.5=2040 Me... 1
- In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs ? Answer: 1. 60 cm Key idea: in a potentiometer, EMF ∝ balancing length. E₂/E₁=l₂/l₁⇒ l₂=36×2.5/1.5=60 cm. Option 1. 1
- The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 Ω. What will be the effective resistance if they are connected in series ? Answer: 4. 4 Ω Key idea: four equal resistors R: parallel gives R/4 , series gives 4R . R/4=0.25\ Ω⇒ R=1\ Ω . In series: 4R=4\ Ω . Option 4. 1
- The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3×10^-3 watt will be : (h = 6.6×10^-34 Js) Answer: 3. 10¹⁶ Key idea: number of photons per second N=P/hc/λ=Pλ/hc . N=dfrac3.3×10⁻³×600×10⁻⁹6.6×10⁻³⁴×3×10⁸=dfrac1.98×10⁻⁹1.98×10⁻²⁵=10¹... 1
- An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λ_d, then : Answer: 3. λ = (2mc/h)λ_d² Key idea: with negligible work function, the photoelectron's kinetic energy equals the photon energy. hc/λ=p²/2m⇒... 1
- An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance 'R' are connected in series to an ac source of potential difference 'V' volts as shown in figure.Potential difference across L, C and R is 40 V, 10 V and 40 V, res… Answer: 4. 5\ Ω Key idea: voltages in series LCR add as phasors: V=√(V_R²+(V_L-V_C)²) , and Z=V/I with both in the same (rms) form. V=√(40²+... 1
- A series LCR circuit containing 5.0 H inductor, 80 μF capacitor and 40 Ω resistor is connected to 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant an… Answer: 3. 46 rad/s and 54 rad/s Key idea: power falls to half of its resonance value when |X_L-X_C|=R , which happens at ω=ω₀±dfrac R2L (fo... 1
- Two conducting circular loops of radii R_1 and R_2 are placed in the same plane with their centres coinciding. If R_1 >> R_2, the mutual inductance M between them will be directly proportional to : Answer: 4. R₂²/R₁ Key idea: the large loop R₁ produces a nearly uniform field at its centre, which passes through the small loop. B=μ₀I/2R₁... 1
- For the given circuit, the input digital signals are applied at the terminals A, B and C. What would be the output at the terminal y ? Answer: 3. Output waveform (3): constant 5 V Key idea: from the circuit, y=A· B+B· C (AND of A,B ORed with NAND of B,C). IntervalABCA.BNAND(... 1
- The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them. Answer: 3. current in n-type > current in p-type. Key idea: I∝ n μ E , and the same concentration means the current depends on mobility. In... 1
- Polar molecules are the molecules : Answer: 4. having a permanent electric dipole moment. Polar molecules (such as H₂O , HCl) have centres of positive and negative charge that... 1
- Two charged spherical conductors of radius R_1 and R_2 are connected by a wire. Then the ratio of surface charge densities of the spheres (σ_1/σ_2) is : Answer: 2. R₂/R₁ Key idea: connected conductors are at the same potential. V=kQ/R=kσ4π R²/R=4π kσ R . Equal V gives σ₁R₁=σ₂R₂ , so σ₁/σ₂=R₂/... 1
- Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop. Answer: 4. 1980 V Key idea: total charge and volume are conserved when drops merge. Volume: 4/3π R³=27·4/3π r³⇒ R=3r . Charge: Q=27q . V_big... 1