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**Question 1**
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A boy standing some distance from the foot of a tall cliff claps his hands and hears an echo 0.5s later. If the speed of sound is 340ms−1 ${}^{-1}$, how far is he from the cliff?

**Answer Details**

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

**Question 2**
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A 2H inductor has negligible resistance and is connected to a 50? $\frac{50}{?}$ Hz A.C supply. The reactance of the inductor is?

**Answer Details**

Given: Inductance (𝐿*L*) = 2 H Frequency (𝑓*f*) = 50π Hz

The reactance of the inductor (𝑋𝐿*XL*) can be calculated using the formula:

𝑋𝐿=2𝜋𝑓𝐿*XL*=2*πfL*

Substituting the given values:

𝑋𝐿=2𝜋×50𝜋×2*XL*=2*π*×50*π*×2

𝑋𝐿=200𝜋2*XL*=200*π*2

So, the reactance of the inductor is 200𝜋2200*π*2 ohms.

**Question 3**
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A man stands 4m in front of a plane mirror. If the mirror is moved 1m towards the man, the distance between him and his new image is?

**Answer Details**

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

**Question 4**
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If the stress on a wire is 107
${}^{7}$NM−2
${}^{-2}$ and the wire is stretched from its original length of 10.00m to 10.05m. The young's modulus of the wire is?

**Answer Details**

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

**Question 5**
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A solid weighs 10.00N in air, 6N when forcefully immersed in water, and 7.0N when fully immersed in a liquid, X. Calculate the relative density of the liquid X.

**Answer Details**

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

**Question 6**
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Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss (specific heat capacity of copper = 400Jkg1 ${}^{1}$k−1 ${}^{-1}$, latent heat of fusion of ice = 3.3 x 105 ${}^{5}$Jkg−1 ${}^{-1}$)

**Answer Details**

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

**Question 7**
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The slope of the straight line displacement-time graph indicates?

**Answer Details**

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

**Question 8**
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A source of sound produces waves in air of wavelength 1.65m. If the speed of sound in air is 330ms−1 ${}^{-1}$, the period of vibration in air is?

**Answer Details**

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.

**Question 9**
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A ball of mass 0.5kg moving at 10ms−1 ${}^{-1}$ collides with another ball of equal mass at rest. If the two balls move off together after the impact, calculate their common velocity.

**Answer Details**

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

**Question 10**
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A given mass of gas has a pressure of 80 Nm−2 ${}^{-2}$ at a temperature of 47 C. If the temperature is reduced to 27 C with volume remaining constant, the new pressure is?

**Question 11**
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The motion of a body is simple harmonic if the?

**Answer Details**

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

**Question 12**
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The velocities of light in air and glass are 3.0 x 108 ${}^{8}$ms−1 ${}^{-1}$ and 2.0 x 108 ${}^{8}$ms−1 ${}^{-1}$ respectively. If the angle of refraction is 30º, the sine of the angle of incidence is?

**Answer Details**

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.

**Question 13**
**Report**

A lead bullet of mass 0.05kg is fired with a velocity of 200ms−1 ${}^{-1}$ into a block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is?

**Answer Details**

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

**Question 14**
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Which of the following is not correct about the molecules of a substance in a gaseous state. They?

**Answer Details**

The collision between the gases is perfectly elastic

**Question 15**
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A piece of substance of specific heat capacity 450Jkg?1 ${}^{?1}$k1 ${}^{1}$ falls through a vertical distance of 20m from rest. Calculate the rise in temperature of the substance on hitting the ground when all its energies are converted into heat. [g = 10ms?2 ${}^{?2}$]

**Answer Details**

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

**Question 16**
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Two bodies have masses in the ratio3:1. They experience forces which impart to them, acceleration in the ratio 2:9 respectively. Find the ratio of forces the masses experienced

**Answer Details**

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.

**Question 17**
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The inner diameter of a test tube can be measured accurately using a?

**Answer Details**

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

**Question 18**
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I. A liquid boils when its saturated vapor pressure is equal to the external pressure

II. Dissolved substances in pure water lead to an increase in the boiling point.

III. When the external pressure is increased, the boiling point increases.

IV. Dissolved substances in pure water decreases the boiling point

Which of the above combinations are peculiarities of the boiling point of a liquid?

- A. I, II and III only
- B. I, II, III, and IV
- C. I, II, and IV
- D. II, III, and IV

**Answer Details**

A. I, II, and III only.

**Question 19**
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When the temperature of a liquid increases, its surface tension

**Answer Details**

Surface tension of a liquid decreases when its temperature increases.

**Question 20**
**Report**

0.5kg of water at 10ºC is completely converted to ice at 0ºC by extracting (88000) of heat from it. If the specific heat capacity of water is 4200jkg1 ${}^{1}$ Cº. Calculate the specific latent heat of fusion of ice.

**Answer Details**

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

**Question 21**
**Report**

If the frequency of an emitted x-ray is 1.6 x 1016 ${}^{16}$Hz, the accelerating potential is?

[e = 1.6 x 10−19 ${}^{-19}$J, h = 6.63 x 10-34 ${}^{34}$Js]

**Answer Details**

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

**Question 22**
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A car of mass 800kg attains a speed of 25m/s in 20secs. The power developed in the engine is?

**Answer Details**

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

2.5 x 104 ${}^{4}$W

**Question 23**
**Report**

The inside portion of part of a hollow metal sphere of diameter 20cm is polished. The portion will therefore form a?

**Answer Details**

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

**Question 24**
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If a sound wave goes from a cold air region to a hot air region, its wavelength will?

**Question 25**
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The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20º and 80ºC respectively is?

**Answer Details**

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.

**Question 26**
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A man will exert the greatest pressure when he?

**Answer Details**

Pressure varies inversely as A

i.e P ∝ $\propto $ 1A

**Question 27**
**Report**

A ball of mass 0.1kg is thrown vertically upwards with a speed of 10ms−1 ${}^{-1}$ from the top of a tower 10m high. Neglecting air resistance, its total energy just before hitting the ground is? (Take g = 10ms−2 ${}^{-2}$)

**Answer Details**

Givn

m = 0.1kg

u = 10ms−1
${}^{-1}$

h = 10m

g = 10ms−2
${}^{-2}$

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

Using

v̸
$\mathrm{v\u0338}$ = u̸
$\mathrm{u\u0338}$ = g̸t
$\mathrm{g\u0338}t$

/0
$\text{\u29f8}0$ = w̸
$\mathrm{w\u0338}$ - /10t
$\text{\u29f8}10t$

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

**Question 28**
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The lowest note emitted by a stretched string has a frequency of 40Hz. How many overtones are there between 40Hz and 180Hz?

**Answer Details**

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.

**Question 29**
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An astronomical telescope is said to be in normal adjustment when the?

**Answer Details**

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

**Question 30**
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A certain radioactive source emits radiation that was found to be deflected by both magnetic and electric fields. The radiation is?

**Answer Details**

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

**Question 31**
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In a series R-L-C circuit at resonance, the voltages across the resistor and the inductors are 30V and 40V respectively. What is the voltage across the capacitor?

**Answer Details**

AT resonance; XL ${}_{L}$ = Xc ${}_{c}$

∴ $\therefore $ = VL ${}_{L}$ = Vc ${}_{c}$ = 40V