Tag: JAMB Physics 2026

2026 JAMB Physics Likely Questions and Answers with details explanations 2026 JAMB Physics Likely Questions and Answers If you are preparing for the 2026 JAMB Physics exam, studying 2026 JAMB Physics likely questions and answers is one of the most effective ways to boost your score. Physics is one of the major objective subjects in JAMB, and mastering likely questions helps you anticipate the exam pattern, understand key concepts, and perform with confidence on test day. This post provides predicted 2026 JAMB Physics likely questions and answers, detailed explanations, and a free PDF download you can use for offline practice and revision. InstructorMA COACH Last updatedtoday Total Viewers 1k+ ExamJAMB 2026 JAMB Physics Likely Questions and Answers Join Our Social Media Join Our Social media for latest updates and reading classes WhatsApp Community WhatsApp Channel Telegram Channel TikTok YouTube FaceBook Join Our Social media for latest updates and reading classes WhatsApp Community WhatsApp Channel Telegram Channel TikTok YouTube FaceBook Question 1Question: Which types of motion does a coin undergo as it moves down an inclined plane?(a) rotational and translational(b) circular and random(c) translational(d) rotationalTopic: Types of MotionCorrect Option: (a)Detailed Explanation:When a coin rolls down an inclined plane, it exhibits two simultaneous motions:Translational Motion: The center of mass of the coin moves in a straight line down the slope.Rotational Motion: The coin spins around its own center/axis as it moves.Therefore, the combined motion is rotational and translational.Question 2Question: An object of mass 20kg is released from a height of 5m above the ground level. The kinetic energy of the object just before hitting the ground (a)(a) 1000J(b) 2000J(c) 1500J(d) 4000JTopic: Work, Energy, and Power (Conservation of Energy)Correct Option: (a)Detailed Explanation:According to the law of conservation of energy, the Potential Energy (P.E.) at the top equals the Kinetic Energy (K.E.) just before hitting the ground (assuming negligible air resistance).P.E.=mghP.E. = mghP.E.=mgh m=20kg,g=10m/s2,h=5mm = 20text{kg}, quad g = 10text{m/s}^2, quad h = 5text{m}m=20kg,g=10m/s2,h=5m P.E.=20×10×5=1000JP.E. = 20 times 10 times 5 = 1000text{J}P.E.=20×10×5=1000JTherefore, K.E. at the bottom = 1000J. Question 3Question: The small droplets of water that form on the grass in the early hours of the morning is(a) dew(b) fog(c) hail(d) mistTopic: Change of State / MeteorologyCorrect Option: (a)Detailed Explanation:Dew occurs when atmospheric vapor condenses directly onto cool surfaces (like grass) at night.Fog and Mist are water droplets suspended in the air.Hail is frozen precipitation.Question 4Question: The equation PaVbTc=constantP^a V^b T^c = text{constant}PaVbTc=constant reduces to Boyle's law if (a) a = 0, b = 1, c = -1(b) a = 1, b = 1, c = 0(c) a = 1, b = 1, c = 1(d) a = 1, b = 1, c = -1Topic: Gas LawsCorrect Option: (b)Detailed Explanation:Boyle's Law states that for a fixed mass of gas at constant temperature, Pressure is inversely proportional to Volume. The formula is PV=constantPV = text{constant}PV=constant.Looking at PaVbTcP^a V^b T^cPaVbTc: We need P1P^1P1 and V1V^1V1 (so a=1,b=1a=1, b=1a=1,b=1). Temperature (TTT) does not appear in the Boyle's law equation (it is constant), which mathematically is equivalent to T0T^0T0 (since anything to the power of 0 is 1). So c=0c=0c=0.Matches option (b). Question 5Question: The distance between two successive troughs of a wave is 30cm and the velocity is 300ms⁻¹. Calculate the frequency.(a) 100Hz(b) 2250Hz(c) 1000Hz(d) 225HzTopic: WavesCorrect Option: (c)Detailed Explanation:Distance between successive troughs = Wavelength (λlambdaλ) = 30cm = 0.3m. Velocity (vvv) = 300 m/s. Formula: v=fλv = f lambdav=fλ f=vλ=3000.3=30003=1000Hzf = frac{v}{lambda} = frac{300}{0.3} = frac{3000}{3} = 1000text{Hz}f=λv=0.3300=33000=1000Hz. Question 6Question: In a compound microscope, the objective and the eyepiece focal lengths are(a) at infinity(b) long(c) the same(d) shortTopic: Optical InstrumentsCorrect Option: (d)Detailed Explanation:To achieve high magnification, both lenses in a compound microscope must have short focal lengths.Magnifying power is inversely proportional to focal length.The objective lens has a very short focal length (to form a real, magnified image).The eyepiece has a short focal length (to act as a magnifier for the first image).Question 7Question: What is the best method of demagnetizing a steel bar magnet?(a) solenoid method(b) hammering it(c) heating it(d) rough handling itTopic: MagnetismCorrect Option: (a)Detailed Explanation:While heating and hammering can demagnetize a magnet, the electrical method (solenoid) is the most effective and controlled "scientific" method. Placing the magnet in a solenoid carrying an Alternating Current (AC) and slowly withdrawing it in an East-West direction destroys the magnetic domains completely.Question 8Question: The magnitude of the angle of dip at the equator is(a) 180⁰(b) 0⁰(c) 90⁰(d) 60⁰Topic: Earth's MagnetismCorrect Option: (b)Detailed Explanation:At the magnetic Equator, the earth's magnetic field lines run parallel to the horizontal. Thus, the angle of dip is 0⁰.At the magnetic Poles, the field lines are vertical, so the dip is 90⁰.Question 9Question: A glass bottle of initial volume 2×104cm32 times 10^4 text{cm}^32×104cm3 is heated from 20°C to 50°C. If the linear expansivity of glass is 9×10−6°C−19 times 10^{-6} text{°C}^{-1}9×10−6°C−1, the volume of the bottle at 50°C is (a) 20013.5cm³(b) 20016.2cm³(c) 2005.4cm³(d) 20008.1cm³Topic: Thermal ExpansionCorrect Option: (b)Detailed Explanation:Linear expansivity (αalphaα) = 9×10−6K−19 times 10^{-6} text{K}^{-1}9×10−6K−1. Cubic (Volume) expansivity (γgammaγ) = 3α=3×9×10−6=27×10−6K−13alpha = 3 times 9 times 10^{-6} = 27 times 10^{-6} text{K}^{-1}3α=3×9×10−6=27×10−6K−1. Change in temp (ΔθDelta thetaΔθ) = 50−20=30°C50 - 20 = 30text{°C}50−20=30°C. Original Volume (V1V_1V1) = 20,000cm320,000 text{cm}^320,000cm3. Change in Volume (ΔVDelta VΔV) = V1γΔθV_1 gamma Delta thetaV1γΔθΔV=20,000×(27×10−6)×30Delta V = 20,000 times (27 times 10^{-6}) times 30ΔV=20,000×(27×10−6)×30ΔV=2×104×27×10−6×3×101Delta V = 2 times 10^4 times 27 times 10^{-6} times 3 times 10^1ΔV=2×104×27×10−6×3×101ΔV=162×10−1=16.2cm3Delta V = 162 times 10^{-1} = 16.2 text{cm}^3ΔV=162×10−1=16.2cm3 New Volume = 20,000+16.2=20016.2 cm320,000 + 16.2 = textbf{20016.2 cm}^320,000+16.2=20016.2 cm3. Question 10Question: When the linear momentum of a body is constant, the net force acting on it (a) is(a) zero(b) decreases(c) increases(d) remains constantTopic: Newton's Second LawCorrect Option: (a)Detailed Explanation:Newton's Second Law states that Force is the rate of change of momentum (F=ΔptF = frac{Delta p}{t}F=tΔp).If momentum (ppp) is constant, the change in momentum (ΔpDelta pΔp) is zero.Therefore, Force = 0. Question 11Question: A device that converts mechanical energy into electrical energy is (a)(a) transformer(b) dynamo(c) an electric motor(d) an induction coilTopic: Electromagnetic InductionCorrect Option: (b)Detailed Explanation:A Dynamo (or Generator) converts Mechanical Energy →rightarrow→ Electrical Energy. An Electric Motor converts Electrical →rightarrow→ Mechanical. Question 12Question: A quantity of water at 20°C is mixed with another quantity of water at 70°C. The final steady temperature of the mixture is 40°C. Determine the ratio of the mass of the hot water to that of the cold water.(a) 2:1(b) 2:3(c) 5:2(d) 2:7Topic: Heat Energy (Method of Mixtures)Correct Option: (b)Detailed Explanation:Heat lost by hot water = Heat gained by cold water.Let mhm_hmh be mass of hot water and mcm_cmc be mass of cold water. Specific heat capacity ccc cancels out.mhc(70−40)=mcc(40−20)m_h c (70 - 40) = m_c c (40 - 20)mhc(70−40)=mcc(40−20) mh(30)=mc(20)m_h (30) = m_c (20)mh(30)=mc(20) mhmc=2030=23frac{m_h}{m_c} = frac{20}{30} = frac{2}{3}mcmh=3020=32Ratio is 2:3. Question 13Question: The quantity of a note depends on its(a) pitch(b) frequency(c) amplitude(d) overtoneTopic: Sound WavesCorrect Option: (d)Detailed Explanation:Quality (or Timbre) allows us to distinguish between different instruments playing the same note. It depends on the number and intensity of overtones (harmonics).Pitch depends on frequency.Loudness depends on amplitude.Question 14Question: An ammeter can be adapted to measure potential difference by using a (a)(a) rheostat(b) multiplier(c) shunt(d) resistance boxTopic: Electrical InstrumentationCorrect Option: (b)Detailed Explanation:To measure potential difference (act as a voltmeter), a galvanometer/ammeter needs a high resistance connected in series. This high resistance is called a multiplier.(Note: A shunt is a low resistance in parallel used to convert a galvanometer to an ammeter).Question 15Question: A transformer has 800 turns of wire in the primary coil and 80 turns in the secondary coil. If the input voltage is 150V, calculate the magnitude of the output voltage.(a) 15V(b) 150V(c) 36V(d) 55VTopic: TransformersCorrect Option: (a)Detailed Explanation:Formula: VsVp=NsNpfrac{V_s}{V_p} = frac{N_s}{N_p}VpVs=NpNs Vs=Vp×NsNpV_s = V_p times frac{N_s}{N_p}Vs=Vp×NpNs Vs=150×80800V_s = 150 times frac{80}{800}Vs=150×80080 Vs=150×110=15VV_s = 150 times frac{1}{10} = 15text{V}Vs=150×101=15V. Question 16Question: An element of nucleon number P and atomic number Q emits an alpha particle from its nucleus. The resultant numbers of the new element formed are respectively(a) P+2 and Q-2(b) P-4 and Q+2(c) P-4 and Q-2(d) P+2 and Q+2Topic: RadioactivityCorrect Option: (c)Detailed Explanation:An alpha particle (αalphaα) is a Helium nucleus (24He^4_2text{He}24He). It has a mass number of 4 and an atomic number of 2.When emitted: Nucleon number (Mass) decreases by 4 (P→P−4P rightarrow P-4P→P−4). Atomic number (Proton) decreases by 2 (Q→Q−2Q rightarrow Q-2Q→Q−2). Question 17Question: Calculate the total distance covered by a train before coming to rest if its initial speed is 30ms−130text{ms}^{-1}30ms−1 with a constant retardation of 0.2ms−20.2text{ms}^{-2}0.2ms−2. (a) 1500m(b) 2750m(c) 2100m(d) 2250mTopic: Equations of MotionCorrect Option: (d)Detailed Explanation:Use the equation: v2=u2+2asv^2 = u^2 + 2asv2=u2+2as Final velocity v=0v = 0v=0 (comes to rest) Initial velocity u=30u = 30u=30 Acceleration a=−0.2a = -0.2a=−0.2 (retardation)0=302+2(−0.2)s0 = 30^2 + 2(-0.2)s0=302+2(−0.2)s0=900−0.4s0 = 900 - 0.4s0=900−0.4s0.4s=9000.4s = 9000.4s=900s=9000.4=90004=2250ms = frac{900}{0.4} = frac{9000}{4} = 2250text{m}s=0.4900=49000=2250m. Question 18Question: An object of volume 1m31text{m}^31m3 and mass 2kg is fully immersed in a liquid of density 1kgm−31text{kgm}^{-3}1kgm−3. Calculate its apparent weight. (a) 1N(b) 2N(c) 10N(d) 20NTopic: Archimedes' PrincipleCorrect Option: (c)Detailed Explanation:Real Weight ( WWW ): mg=2×10=20Nmg = 2 times 10 = 20text{N}mg=2×10=20N. Upthrust ( UUU ): Weight of liquid displaced = Volume×ρliquid×gtext{Volume} times rho_{text{liquid}} times gVolume×ρliquid×gU=1×1×10=10NU = 1 times 1 times 10 = 10text{N}U=1×1×10=10N. Apparent Weight: W−U=20−10=10NW - U = 20 - 10 = 10text{N}W−U=20−10=10N. Question 19Question: Which of the following is not a factor that can increase the rate of evaporation of water in a lake?(a) increase in the kinetic energy of the molecules of water(b) rise in temperature(c) increase in the pressure of the atmosphere(d) increase in the average speed of the molecules of waterTopic: Heat (Evaporation)Correct Option: (c)Detailed Explanation:Evaporation is the escape of energetic molecules from the surface.High Temp, High KE, and Wind speed increase evaporation.High Atmospheric Pressure pushes down on the liquid surface, making it harder for molecules to escape, thus decreasing the rate.Question 20Question: Where can a man place his face to obtain an enlarged image when using a concave mirror to shave?(a) at infinity(b) at the principal focus(c) between the centre of curvature and the principal focus(d) between the principal focus and the poleTopic: Optics (Mirrors)Correct Option: (d)Detailed Explanation:Concave mirrors produce a Virtual, Erect, and Enlarged image only when the object is placed between the Focus (F) and the Pole (P). This is the principle used in shaving mirrors.Question 21Question: A particle carrying a charge of 1×10−8C1 times 10^{-8}text{C}1×10−8C enters a magnetic field... at right angles to the field. If the force on this particle is 1.8×10−8N1.8 times 10^{-8}text{N}1.8×10−8N, what is the magnitude of the field? (Velocity is 3×102ms−13 times 10^2 text{ms}^{-1}3×102ms−1). (a) 6.0×10−1T6.0 times 10^{-1}text{T}6.0×10−1T (b) 6×103T6 times 10^3text{T}6×103T (c) 6×10−3T6 times 10^{-3}text{T}6×10−3T (d) 6×104T6 times 10^4text{T}6×104T Topic: ElectromagnetismCorrect Option: (c)Detailed Explanation:Formula: F=qvBsin⁡θF = qvB sinthetaF=qvBsinθ (where θ=90∘theta = 90^circθ=90∘, so sin⁡90=1sin90 = 1sin90=1)B=FqvB = frac{F}{qv}B=qvF B=1.8×10−8(1×10−8)×(3×102)B = frac{1.8 times 10^{-8}}{(1 times 10^{-8}) times (3 times 10^2)}B=(1×10−8)×(3×102)1.8×10−8Powers of 10−810^{-8}10−8 cancel out.B=1.8300=183000=61000=6×10−3TB = frac{1.8}{300} = frac{18}{3000} = frac{6}{1000} = 6 times 10^{-3}text{T}B=3001.8=300018=10006=6×10−3T. Question 22Question: A glass block of thickness 10cm and refractive index 3/2 is placed on an object. If an observer views the object vertically, the displacement of the object is(a) 8.5cm(b) 3.33cm(c) 5.00cm(d) 6.67cmTopic: RefractionCorrect Option: (b)Detailed Explanation:Vertical displacement (Lateral displacement) d=t(1−1n)d = t (1 - frac{1}{n})d=t(1−n1) d=10(1−11.5)=10(1−23)d = 10 (1 - frac{1}{1.5}) = 10 (1 - frac{2}{3})d=10(1−1.51)=10(1−32) d=10(13)=3.33cmd = 10 (frac{1}{3}) = 3.33text{cm}d=10(31)=3.33cm. Question 23Question: A girl stands on a scale in a lift. If the reading on the scale is less than her weight, then the lift is moving(a) downwards with uniform acceleration(b) downwards with uniform speed(c) upwards with uniform speed(d) upwards with uniform accelerationTopic: Mechanics (Lift Motion)Correct Option: (a)Detailed Explanation:Scale reading (RRR) is the Normal Reaction. When accelerating downwards, the net force is down (mg−R=mamg - R = mamg−R=ma). So, R=m(g−a)R = m(g - a)R=m(g−a). Since R<mgR < mgR<mg (reading is less than weight), the lift must be accelerating downwards. Question 24Question: In the formation of sea breeze, wind blows from (a)(a) sea to land(b) land to sea(c) sky to land(d) sea to skyTopic: Heat Transfer (Convection)Correct Option: (a)Detailed Explanation:Sea breeze occurs during the day. The land heats up faster than the sea. Hot air over the land rises (low pressure), and cool air from the sea blows towards the land to replace it. Hence, Sea to Land.Question 25Question: A simple pendulum of length 0.4m has a period of 2s. What is the period of a similar pendulum of length 0.8m at the same place?(a) 2ssqrt{2}text{s}2s (b) 22s2sqrt{2}text{s}22s (c) 4s4text{s}4s (d) 8s8text{s}8s Topic: Simple Harmonic MotionCorrect Option: (b)Detailed Explanation:Period T∝LT propto sqrt{L}T∝L.T2T1=L2L1frac{T_2}{T_1} = sqrt{frac{L_2}{L_1}}T1T2=L1L2 T22=0.80.4=2frac{T_2}{2} = sqrt{frac{0.8}{0.4}} = sqrt{2}2T2=0.40.8=2 T2=22sT_2 = 2sqrt{2}text{s}T2=22s. Question 26Question: An object is moving with a velocity of 10ms−110text{ms}^{-1}10ms−1. At what height must a similar body be situated to have a potential energy equal in value to the kinetic energy of the moving body? (a) 5m(b) 2.5m(c) 10m(d) 3mTopic: Mechanical EnergyCorrect Option: (a)Detailed Explanation:Set P.E.=K.E.P.E. = K.E.P.E.=K.E. mgh=12mv2mgh = frac{1}{2}mv^2mgh=21mv2 gh=v22gh = frac{v^2}{2}gh=2v2 10×h=102210 times h = frac{10^2}{2}10×h=2102 10h=50 ⟹ h=5m10h = 50 implies h = 5text{m}10h=50⟹h=5m. Question 27Question: A wave that travels through stretched strings is known as (a)(a) microwave(b) seismic wave(c) mechanical wave(d) electromagnetic waveTopic: WavesCorrect Option: (c)Detailed Explanation:Waves on strings require a material medium (the string) to travel. Therefore, they are Mechanical waves. Specifically, they are transverse mechanical waves.Question 28Question: What is the velocity of sound at 100°C if the velocity of sound at 0°C is 340ms−1340text{ms}^{-1}340ms−1? (a) 440ms−1440text{ms}^{-1}440ms−1 (b) 397ms−1397text{ms}^{-1}397ms−1 (c) 497ms−1497text{ms}^{-1}497ms−1 (d) 240ms−1240text{ms}^{-1}240ms−1 Topic: Sound WavesCorrect Option: (b)Detailed Explanation:Velocity of sound is proportional to the square root of absolute temperature (Kelvin).v∝Tv propto sqrt{T}v∝T v2v1=T2T1frac{v_2}{v_1} = sqrt{frac{T_2}{T_1}}v1v2=T1T2 T1=0+273=273KT_1 = 0 + 273 = 273text{K}T1=0+273=273K.T2=100+273=373KT_2 = 100 + 273 = 373text{K}T2=100+273=373K.v2=340×373273≈340×1.366≈340×1.169≈397ms−1v_2 = 340 times sqrt{frac{373}{273}} approx 340 times sqrt{1.366} approx 340 times 1.169 approx 397text{ms}^{-1}v2=340×273373≈340×1.366≈340×1.169≈397ms−1. Question 29Question: The angle of deviation of light of various colours passing through a triangular prism increases in the order(a) red →rightarrow→ green →rightarrow→ blue (b) blue →rightarrow→ green →rightarrow→ red (c) green →rightarrow→ violet →rightarrow→ blue (d) blue →rightarrow→ red →rightarrow→ green Topic: Dispersion of LightCorrect Option: (a)Detailed Explanation:Red light has the longest wavelength and deviates (bends) the least. Violet/Blue light has the shortest wavelength and deviates the most.Order of increasing deviation: Red < Orange < Yellow < Green < Blue < Indigo < Violet.Option (a) follows this order correctly.Question 30Question: Capacitors are used in the induction coil to (a)(a) dissipate energy(b) control circuits(c) prevent electric sparks(d) distort the electric fieldsTopic: ElectronicsCorrect Option: (c)Detailed Explanation:In an induction coil, a capacitor is connected across the "make-and-break" contact points. Its function is to absorb the induced current when the circuit breaks, thereby preventing sparking at the contact points and ensuring a rapid collapse of the magnetic field.Question 31Question: If a current of 2.5A flows through electrolyte for 3 hours and 1.8g of a substance [is deposited]... what mass will be deposited if a current of 4A flows through it for 4.8 hours?(a) 4.8g(b) 3.2g(c) 2.4g(d) 4.6gTopic: Electrolysis (Faraday's First Law)Correct Option: (d)Detailed Explanation:m=ZItm = ZItm=ZIt.Since it is the same substance, Z is constant.m2m1=I2t2I1t1frac{m_2}{m_1} = frac{I_2 t_2}{I_1 t_1}m1m2=I1t1I2t2 m2=1.8×4×4.82.5×3=1.8×19.27.5m_2 = 1.8 times frac{4 times 4.8}{2.5 times 3} = 1.8 times frac{19.2}{7.5}m2=1.8×2.5×34×4.8=1.8×7.519.2 m2=1.8×2.56=4.608gm_2 = 1.8 times 2.56 = 4.608text{g}m2=1.8×2.56=4.608g.Approx 4.6g. Question 32Question: Which of the following units is equivalent to watt?(a) kgm2s−1kg m^2 s^{-1}kgm2s−1 (b) kgms−2kg m s^{-2}kgms−2 (c) kgms−2kg m s^{-2}kgms−2 (d) kgm2s−3kg m^2 s^{-3}kgm2s−3 Topic: Units and DimensionsCorrect Option: (d)Detailed Explanation:Watt is Power.Power = WorkTime=Force×DistanceTimefrac{text{Work}}{text{Time}} = frac{text{Force} times text{Distance}}{text{Time}}TimeWork=TimeForce×DistanceForce = Mass ×times× Acc = kg⋅m/s2kg cdot m/s^2kg⋅m/s2Power = (kg⋅m/s2)⋅ms=kg⋅m2⋅s−3frac{(kg cdot m/s^2) cdot m}{s} = kg cdot m^2 cdot s^{-3}s(kg⋅m/s2)⋅m=kg⋅m2⋅s−3. Question 33Question: A diverging lens of focal length 30cm produces an image 20cm from the lens. Determine the object distance.(a) 10cm(b) 12cm(c) 60cm(d) 50cmTopic: Light (Lenses)Correct Option: (c)Detailed Explanation:Lens formula: 1f=1v+1ufrac{1}{f} = frac{1}{v} + frac{1}{u}f1=v1+u1Diverging lens: fff is negative (−30-30−30). Image is always virtual, so vvv is negative (−20-20−20).1−30=1−20+1ufrac{1}{-30} = frac{1}{-20} + frac{1}{u}−301=−201+u1 1u=120−130=3−260=160frac{1}{u} = frac{1}{20} - frac{1}{30} = frac{3 - 2}{60} = frac{1}{60}u1=201−301=603−2=601 u=60cmu = 60text{cm}u=60cm. Question 34Question: A string under tension produces a note of frequency 14Hz. Determine the frequency when the tension is quadrupled.(a) 56Hz(b) 28Hz(c) 18Hz(d) 14HzTopic: Waves in StringsCorrect Option: (b)Detailed Explanation:Frequency f∝Tf propto sqrt{T}f∝T.If Tension TTT becomes 4T4T4T, the new frequency f2∝4T=2Tf_2 propto sqrt{4T} = 2sqrt{T}f2∝4T=2T.The frequency doubles.14×2=28Hz14 times 2 = 28text{Hz}14×2=28Hz. Question 35Question: The work function of a metal is 2.56×10−19J2.56 times 10^{-19}text{J}2.56×10−19J. Calculate the frequency of a photon whose energy is required to eject from the metal an electron with kinetic energy of 3.0eV. (a) 1.12×1015Hz1.12 times 10^{15}text{Hz}1.12×1015Hz (b) 2.2×1014Hz2.2 times 10^{14}text{Hz}2.2×1014Hz (c) 1.12×105Hz1.12 times 10^{5}text{Hz}1.12×105Hz (d) 4.4×1014Hz4.4 times 10^{14}text{Hz}4.4×1014Hz Topic: Photoelectric EffectCorrect Option: (a)Detailed Explanation:E=hf=W+K.E.E = hf = W + K.E.E=hf=W+K.E. W=2.56×10−19JW = 2.56 times 10^{-19}text{J}W=2.56×10−19J.K.E.=3.0eV=3.0×1.6×10−19J=4.8×10−19JK.E. = 3.0text{eV} = 3.0 times 1.6 times 10^{-19}text{J} = 4.8 times 10^{-19}text{J}K.E.=3.0eV=3.0×1.6×10−19J=4.8×10−19J.Total Energy E=(2.56+4.8)×10−19=7.36×10−19JE = (2.56 + 4.8) times 10^{-19} = 7.36 times 10^{-19}text{J}E=(2.56+4.8)×10−19=7.36×10−19J.f=Eh=7.36×10−196.6×10−34≈1.115×1015Hzf = frac{E}{h} = frac{7.36 times 10^{-19}}{6.6 times 10^{-34}} approx 1.115 times 10^{15}text{Hz}f=hE=6.6×10−347.36×10−19≈1.115×1015Hz. Question 36Question: When a red rose flower is observed in blue light, what colour does the observer see?(a) blue(b) red(c) black(d) yellowTopic: Colour of ObjectsCorrect Option: (c)Detailed Explanation:A red rose appears red because it reflects red light and absorbs all other colors. When illuminated by blue light, the rose absorbs the blue light. Since there is no red light to reflect, no light reaches the observer's eye, and it appears black.Question 37Question: The density ρrhoρ of a spherical ball of diameter ddd and mass mmm is given by (a) ρ=…rho = dotsρ=… (a) …dots… (b) …dots… (c) …dots… (d) ρ=6mπd3rho = frac{6m}{pi d^3}ρ=πd36m Topic: DensityCorrect Option: (d)Detailed Explanation:Density ρ=mVrho = frac{m}{V}ρ=Vm.Volume of sphere V=43πr3V = frac{4}{3} pi r^3V=34πr3.Since r=d2r = frac{d}{2}r=2d, V=43π(d2)3=43πd38=πd36V = frac{4}{3} pi (frac{d}{2})^3 = frac{4}{3} pi frac{d^3}{8} = frac{pi d^3}{6}V=34π(2d)3=34π8d3=6πd3.ρ=mπd36=6mπd3rho = frac{m}{frac{pi d^3}{6}} = frac{6m}{pi d^3}ρ=6πd3m=πd36m. Question 38Question: A coin is pushed from the edge of a laboratory bench with a horizontal velocity of 15ms−115text{ms}^{-1}15ms−1. If the height of the bench from the floor is 1.5m, calculate the distance from the foot of the bench to the point of impact with the floor. (a) 8.22m(b) 2.25m(c) 1.5m(d) 1.41mTopic: Projectile MotionCorrect Option: (a)Detailed Explanation:Find time of flight (ttt): h=12gt2h = frac{1}{2}gt^2h=21gt21.5=0.5(10)t2 ⟹ 1.5=5t21.5 = 0.5(10)t^2 implies 1.5 = 5t^21.5=0.5(10)t2⟹1.5=5t2t2=0.3 ⟹ t=0.3≈0.5477st^2 = 0.3 implies t = sqrt{0.3} approx 0.5477text{s}t2=0.3⟹t=0.3≈0.5477s. Horizontal Distance (Range) R=u×tR = u times tR=u×tR=15×0.5477=8.2155mR = 15 times 0.5477 = 8.2155text{m}R=15×0.5477=8.2155m.Approx 8.22m. Question 39Question: Two sound waves have frequencies of 12Hz and 10Hz. Calculate their beats period.(a) 2.0s(b) 1.2s(c) 1.0s(d) 0.5sTopic: Sound (Beats)Correct Option: (d)Detailed Explanation:Beat Frequency (fbf_bfb) = f1−f2=12−10=2Hzf_1 - f_2 = 12 - 10 = 2text{Hz}f1−f2=12−10=2Hz.Beat Period (TTT) = 1fb=12=0.5sfrac{1}{f_b} = frac{1}{2} = 0.5text{s}fb1=21=0.5s. Question 40Question: A lens that is thinner at the middle and thicker at the edges is (a)(a) diverging(b) converging(c) plano-convex(d) converging meniscusTopic: LensesCorrect Option: (a)Detailed Explanation:This is the physical definition of a Concave lens. Concave lenses diverge light rays, so they are also called Diverging lenses. 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