JAMB UTME - Physics - 2021

Bayanin Amsa

Given: Inductance (𝐿L) = 2 H Frequency (𝑓f) = 50π Hz

The reactance of the inductor (𝑋𝐿XL​) can be calculated using the formula:

𝑋𝐿=2𝜋𝑓𝐿XL​=2πfL

Substituting the given values:

𝑋𝐿=2𝜋×50𝜋×2XL​=2π×50π×2

𝑋𝐿=200𝜋2XL​=200π2

So, the reactance of the inductor is 200𝜋2200π2 ohms.

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The slope of the straight line displacement-time graph indicates uniform velocity.

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The relationship between the angle of incidence and the angle of refraction can be determined using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of light in the two media. We can use this relationship to find the sine of the angle of incidence: sin(incidence) / sin(refraction) = velocity of light in air / velocity of light in glass sin(incidence) / sin(30) = 3 x 10^8 / 2 x 10^8 sin(incidence) = sin(30) x (3 / 2) sin(incidence) = 0.5 x (3 / 2) sin(incidence) = 0.75 So, the sine of the angle of incidence is 0.75.

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Temperature is not a vector quantity. A vector quantity has both magnitude and direction, whereas temperature is a scalar quantity and only has magnitude. Momentum, force, and displacement are all vector quantities as they have both magnitude and direction.

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AT resonance; XL ${}_L$ = Xc ${}_c$

∴ $\therefore$ = VL ${}_L$ = Vc ${}_c$ = 40V

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P = mv2t=800×25×2520 $\frac{m{v}^2}{t}=\frac{800\times 25\times 25}{20}$

2.5 x 104 ${}^4$W

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Given amount remaining = 164 $\frac{1}{64}$

= 164=1(2)6 $\frac{1}{64}=\frac{1}{(2{)}^6}$

∴ $\therefore$ = 120years6 $\frac{120\text{years}}{6}$ = 20 years

⇒ $\Rightarrow$ T12 $\frac{1}{2}$ = 20 years

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Surface tension of a liquid decreases when its temperature increases.

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An overtone is a higher frequency harmonic that is an integer multiple of the fundamental frequency of a vibrating system, such as a stretched string. The first overtone is called the first harmonic, the second overtone is called the second harmonic, and so on. The frequency of the nth harmonic is given by the equation f_n = n * f_1, where f_1 is the fundamental frequency and n is an integer greater than or equal to 1. In this case, the fundamental frequency is 40 Hz, so the frequency of the first overtone is 2 * 40 = 80 Hz. The frequency of the second overtone is 3 * 40 = 120 Hz, and the frequency of the third overtone is 4 * 40 = 160 Hz. Therefore, there are 3 overtones between 40 Hz and 180 Hz, which corresponds to: 3.

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75.0 Nm^-2

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Young's modulus stressstrain $\frac{stress}{strain}$

Strain = el=0.0510=1070.005 $\frac{e}{l}=\frac{0.05}{10}=\frac{{10}^7}{0.005}$

2 x 109 ${}^9$Nm−2

Bayanin Amsa

Givn
m = 0.1kg
u = 10ms−1 ${}^{-1}$
h = 10m
g = 10ms−2 ${}^{-2}$

At hmax ${}_{max}$; v = 0

Using
v̸ $\mathrm{v̸}$ = u̸ $\mathrm{u̸}$ = g̸t $\mathrm{g̸}t$
/0 $\text{⧸}0$ = w̸ $\mathrm{w̸}$ - /10t $\text{⧸}10t$

v2 ${}^2$ = u2 ${}^2$ - 2gh
02 ${}^2$ = 102 ${}^2$ - 2 x 10h
20h = 100
h = 5m
Total height = 10 + 5
= 15m
Total energy = mghr = 0.1 x 10 x1 5
= 15J

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Concave Mirror: It is a spherical mirror whose outer surface is polished and inner or concave side is reflecting surface.

Convex Mirror: It is a spherical mirror whose inner surface is polished and outer side or convex side is the reflecting surface.

f = m4 = f = d4

f = 204 = 5cm

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The distance between a person and their image in a plane mirror is equal to twice the distance from the person to the mirror. Before the mirror was moved, the distance between the man and his image was: 2 * 4m = 8m After the mirror was moved 1m towards the man, the new distance from the man to the mirror is: 4m - 1m = 3m So, the new distance between the man and his image is: 2 * 3m = 6m So, the distance between the man and his new image is 6m

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A normal arrangement is an arrangement where the principal focus of objective coincides with the eyepiece so that the final image is at infinity

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In simple harmonic motion, the acceleration of the particle is directed towards its mean position and directly proportional to its displacement.

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A. I, II, and III only.

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p1 ${}_1$ = P0 ${}_0$, V1 ${}_1$ = V0 ${}_0$, V2 ${}_2$ = 15 $\frac{1}{5}$V0 ${}_0$ P2 ${}_2$ = ?

P1 ${}_1$V1 ${}_1$ = P2 ${}_2$V2 ${}_2$

P2 ${}_2$ = P1V1V1 $\frac{P_1{V}_1}{V_1}$

= P̸oVo15Po $\frac{{\mathrm{P̸}}_o{V}_o}{\frac{1}{5}{P}_o}$ = 5P̸o

Bayanin Amsa

Using the law of conservation of energy

mc?? $??$ = mgh

?? $??$ = ghc=10×20450=49 $\frac{gh}{c}=\frac{10\times 20}{450}=\frac{4}{9}$ºC

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The inner diameter of a test tube can be measured accurately using a pair of vernier calipers.

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Only 'mass' is a fundamental quantity 

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Given 
m = 10−2 ${}^{-2}$
w = 100rads−1 ${}^{-1}$
F = mw2 ${}^2$r
r = 0.2m
F = ?

= 10−2 ${}^{-2}$ x (100)2 ${}^2$ x 0.2

= 20N

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The period of vibration is the time it takes for one complete cycle of a wave to pass a fixed point. It is related to the wavelength and speed of the wave by the equation: T = λ / v where T is the period, λ is the wavelength, and v is the speed of the wave. Given the wavelength of the wave in air is 1.65 m and the speed of sound in air is 330 m/s, we can substitute these values in the above equation to get: T = λ / v = 1.65 m / 330 m/s = 0.005 s So, the period of vibration in air is 0.005 s.

Bayanin Amsa

Given 
m1 ${}_1$ = 0.05kg, u1 ${}_1$ = 200ms−1 ${}^{-1}$, m2 ${}_2$ = 0.95kg

K.E = 12 $\frac{1}{2}$mr ${}_r$v2 ${}^2$

m1 ${}_1$u1 ${}_1$ = v(m1 ${}_1$ + m2 ${}_2$) [law of conversation of momentum]

v = 0.05×2000.05+95 $\frac{0.05\times 200}{0.05+95}$ = 10ms−1 ${}^{-1}$

K.E = 12 $\frac{1}{2}$(1)102 ${}^2$ = 50J

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The ratio of forces experienced by two bodies can be determined using Newton's second law, which states that the force acting on an object is equal to its mass times its acceleration. If two bodies have masses m1 and m2 and are subjected to forces F1 and F2, respectively, such that they experience accelerations a1 and a2, then we have the following two equations: F1 = m1 * a1 F2 = m2 * a2 Dividing these two equations, we get: F1 / F2 = (m1 * a1) / (m2 * a2) = (m1 / m2) * (a1 / a2) Given that the ratio of masses is 3:1 and the ratio of accelerations is 2:9, we can substitute these values in the above equation to get: F1 / F2 = (3 / 1) * (2 / 9) = 2 / 3 So, the ratio of forces experienced by the two masses is 2:3.

Bayanin Amsa

V = 2dt $\frac{2d}{t}$ 
v = 340ms−1 ${}^{-1}$
t = 0.5s
d= ?

d = 2d2=340×0.52 $\frac{2d}{2}=\frac{340\times 0.5}{2}$ = 85m

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R.D = weigh in air - weigh in liquid Xweigh in air - weigh in water $\frac{\text{weigh in air - weigh in liquid X}}{\text{weigh in air - weigh in water}}$

= 10?710?6=34

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The relative humidity of air can be measured using a hygrometer. A hygrometer is an instrument that measures the amount of moisture or water vapor in the air. The other instruments listed are used for different purposes: - A hydrometer is an instrument used to measure the specific gravity of a liquid. - A manometer is an instrument used to measure pressure. - A hypsometer is an instrument used to measure height or altitude. So, the instrument that may be used to measure relative humidity is a hygrometer

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Heat involved (H) = 88,000J, Mass(M)= 0.5kg,  specific heat capacity of water(C) = 4200JkgºC,

Specific latent heat of fusion(L) = ?, Temperature change(Δθ) =  θ2 -  θ1 =  (10 - 0)° = 10°

H = MCΔθ + ML 

or

H = M(CΔθ + L) -->HM $\frac{H}{M}$ = CΔθ + L 

: L = HM $\frac{H}{M}$ -  CΔθ = 88,0000.5 $\frac{88,000}{0.5}$ - 4200 × $\times$ 10

L = 176,000 - 42000 = 134,000

L = 134,000 or 134kj/kg

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When the brakes in a car are applied, the frictional force on the tires is in the opposite direction of the motion of the car. This frictional force slows down the car and eventually brings it to a stop. So, the frictional force on the tires during braking is an advantage because it helps stop the car. Therefore, the answer is: an advantage because it is in the opposite direction of the motion of the car.

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The final velocity of the two balls after the collision can be calculated using the principle of conservation of momentum. In an isolated system, the total momentum before the collision is equal to the total momentum after the collision. Let's call the initial velocity of the first ball v1, and the final velocity of the two balls vf. Then, the momentum of the first ball before the collision is given by m1v1, and the total momentum of the two balls after the collision is given by (m1 + m2)vf, where m1 and m2 are the masses of the balls. Since the second ball is initially at rest, its initial velocity is zero, and its mass is 0.5 kg. So, the conservation of momentum equation is: m1 * v1 = (m1 + m2) * vf 0.5 * 10 = (0.5 + 0.5) * vf 5 = 1 * vf vf = 5 m/s So, the common velocity of the two balls after the collision is 5 m/s

Bayanin Amsa

Using y = A sin (wt + 0)

Given y = 0.25 x 10?3 ${}^{?3}$sin (500t - 0.025x)

By comparing, w = 5.00 x 102 ${}^2$rads?1

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Pressure varies inversely as A

i.e P ∝ $\propto$ 1A

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The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20ºC and 80ºC respectively is 3.0 x 10^3 K/m.

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mcΔθ $\mathrm{\Delta }\theta$ = mλ $\lambda$f

2 x 400 (100 - 0) = m(3.3 x 105 ${}^5$)

m = 2×400×1003.3×105=833

Bayanin Amsa

Given f = 1.6 x1016 ${}^{16}$H3 ${}_3$, h = 6.63 x 10−39 ${}^{-39}$Js

e = 1.6 x 10−19 ${}^{-19}$C, V = ?

V = we=hfe $\frac{w}{e}=\frac{hf}{e}$ = 1.6×1016×6.63×10341.6×10−19 $\frac{1.6\times {10}^{16}\times 6.63\times {10}^{34}}{1.6\times {10}^{-19}}$

= 66.3V

Bayanin Amsa

Radioactive radiation that is deflected by both magnetic and electric fields is called beta radiation. Beta radiation is a type of high-energy, high-speed electron that is emitted from the nucleus of an atom. Gamma rays are high-energy photons that are not deflected by electric or magnetic fields. X-rays are also high-energy photons that are not deflected by electric or magnetic fields. Ultra-violet radiation is a form of electromagnetic radiation that has a higher frequency than visible light, but lower frequency than x-rays. So, the radiation that is deflected by both magnetic and electric fields is beta radiation

Bayanin Amsa

H = mc?? $??$ + mL

H = ?

m = 50g, c = 4.23J/gk,L = 2260, ?? $??$ = 80ºC

H = 50 [4.23 x 80 + 2260]

= 129920 ? $?$ 130,000J

Bayanin Amsa

The collision between the gases is perfectly elastic