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**Question 1**
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In the figure above if the atmospheric pressure is 760mm, the pressure of the chamber G is

**Question 2**
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When a biro pen rubbed on a dry silk cloth is moved very close to a piece of paper on a dry table, the pen is found to pick up the paper. This is because

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They are all charged either positively or negatively

**Question 3**
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The equation of transverse wave traveling along a string is given by Y = 0.3 sin(0.5x - 50t) where Y and X are in cm and t is in seconds.Find the maximum displacement of the particles from the equilibrium position

**Answer Details**

The equation of the transverse wave traveling along a string is given by Y = 0.3 sin(0.5x - 50t), where Y and X are in cm and t is in seconds. The general form of a sine wave equation is Y = A sin(kx - ωt + φ), where A is the amplitude, k is the wave number, ω is the angular frequency, t is time, x is distance and φ is the phase constant. Comparing the given equation with the general equation, we can see that the amplitude is 0.3. The maximum displacement of particles from the equilibrium position is equal to the amplitude. Hence, the maximum displacement of particles from the equilibrium position is 0.3 cm. Therefore, the correct option is 0.3cm.

**Question 4**
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A cone in an unstable equilibrium has its potential energy

**Question 5**
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The figure above shows a uniform wood of weight 200N and length 50m. It is pivoted at one end and suspended by a cord at the other end at an angle of 30 degrees to the wood. Calculate the tension in the cord if the wood is horizontal.

**Question 6**
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The sound from a source travelled to the bottom of the sea and the echo was heard 4s later. If the speed of sound in sea water is 1500ms-1 the depth of the sea is

**Answer Details**

t = 4s ; v = 150ms-1

depth = X ; but 2X = vt

X = vt2 $\frac{vt}{2}$ = 1500×42 $\frac{1500\times 4}{2}$

X = 3000m

**Question 7**
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A certain radioisotope of 23592U ${}_{92}^{235}U$ emits four alpha particles and three beta particles. The mass number and the atomic number of the resulting elements respectively are

**Answer Details**

23592U ${}_{92}^{235}U$ → $\to $ 4(42α) ${(}_{2}^{4}\alpha )$ + 3(0−1e) ${(}_{-1}^{0}e)$ + bay ${}_{a}^{b}y$

a + (4 + 2) + (3x - 1) = 92 ⇒ $\Rightarrow $ a = 87

b + (4 x 4) + (3 x 0) = 235 ⇒ $\Rightarrow $ b = 219

baY ${}_{a}^{b}Y$ = 21987Y

**Question 8**
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If two parallel conductors carry currents flowing in the same direction, the conducts will

**Answer Details**

Two parallel conductors carrying currents flowing in the same direction will attract each other. This is due to the magnetic fields produced by the currents, which interact with each other to create a force. The direction of the force is determined by the direction of the current and the orientation of the conductors. When the currents are flowing in the same direction, the magnetic fields around the conductors will interact in a way that produces an attractive force between them. This phenomenon is known as the Ampere's law, which states that two parallel conductors carrying currents in the same direction will experience a magnetic attraction force between them.

**Question 9**
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When 100g of liquids L1 at 78oC was mixed with Xg of liquid L2 at 50oC, the final temperature was 60oC. Given that the specific heat capacity of L2is half that of L1, find x

**Answer Details**

Heat lost by L1 = heat gained by L2

cL1 = 2cL2

100 x 2cL2 x (78 - 60)

= xg x cL2 x (60 - 50)

xg = 150g

**Question 10**
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Which of the following is TRUE of a particle moving in a horizontal circle with constant angular velocity?

**Question 11**
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In the figure above, XY is of length 1m, The value of R at balance point Z is

**Question 12**
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Heat is supplied to a test tube containing 100g of ice at its melting point. The ice melts completely in 1 min.What is the power rating of the source of heat? [Latent heat of fusion of ice = 336Jg-1]

**Answer Details**

When heat is supplied to the ice, the ice will absorb heat and its temperature will rise until it reaches its melting point. At this point, the ice will start melting, and the temperature will remain constant until all the ice has melted. The energy required to melt 1 gram of ice at its melting point is called the latent heat of fusion of ice and is given as 336J/g. In this question, we are given that 100g of ice melts completely in 1 min. Therefore, the total energy required to melt the ice can be calculated as follows: Total energy = mass x latent heat of fusion of ice Total energy = 100g x 336J/g Total energy = 33600J Since this energy is supplied in 1 minute, we can calculate the power rating of the source of heat as follows: Power = Total energy supplied / time taken Power = 33600J / 60s Power = 560W Therefore, the power rating of the source of heat is 560W. Option (C) is the correct answer.

**Question 13**
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Which of these statements are TRUE of pressure in liquids? i. pressure acts equally in all direction. ii. pressure decreases with depth. iii. pressure at the same level of a liquid is the same. iv pressure is dependent on the cross-sectional area of the barometer tube

**Question 14**
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A boat travels due east at a speed 40ms-1s across a river flow due south at 30ms-1s. What is the resultant speed of the boat?

**Answer Details**

To find the resultant speed of the boat, we need to use the Pythagorean theorem which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In this case, the boat's velocity and the river's velocity are perpendicular to each other, so we can form a right-angled triangle. The speed of the boat is given as 40 ms^-1 due east, and the speed of the river's flow is 30 ms^-1 due south. Using Pythagoras theorem, we have: Resultant speed = √(40^2 + 30^2) = √(1600 + 900) = √2500 Resultant speed = 50 ms^-1 Therefore, the correct option is 50.0ms^-1.

**Question 15**
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If the Nigeria flag (green, white, green) is viewed in pure yellow light, which of the following set of colours would be observed on the flag?

**Question 16**
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The note produced by a stretched string has a fundamental frequency of 400Hz. If the length of the string is doubled while the tension in the string is increased by a factor of 4, the frequency is

**Answer Details**

F = 12L $\frac{1}{2L}$ √Tm $\sqrt{\frac{T}{m}}$

T2 = 4T1 ; L2 = 2L1

2F1L12F2L2 $\frac{2{F}_{1}{L}_{1}}{2{F}_{2}{L}_{2}}$ = √T1T2 $\sqrt{\frac{{T}_{1}}{{T}_{2}}}$

400×L1F×L1 $\frac{400\times {L}_{1}}{F\times {L}_{1}}$ = √T14T1 $\sqrt{\frac{{T}_{1}}{4{T}_{1}}}$

F = 400Hz

**Question 17**
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Light of frequency 6.0 x 1014 Hz travelling in air is transmitted through glass or refractive index 1.5. Calculate the frequency of the light in the glass

**Answer Details**

1.5 = 6×1014F $\frac{6\times {10}^{14}}{F}$

F = 4 x 1014Hz

**Question 18**
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The melting point of naphthalene is 78oC. What is this temperature in kelvin

**Answer Details**

To convert a temperature from Celsius to Kelvin, we add 273.15 to the Celsius temperature. In this question, the melting point of naphthalene is given in Celsius which is 78°C. To convert it to Kelvin, we add 273.15 to 78°C: 78°C + 273.15 = 351.15 K Therefore, the temperature of the melting point of naphthalene in Kelvin is 351 K. The correct option is: 351k.

**Question 19**
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A pipe of length 45cm is closed at the end. Calculate the fundamental frequency of the sound wave generated in the pipe if the velocity of sound in air is 360ms-1 [Neglect end correction]

**Answer Details**

The formula for calculating the fundamental frequency of a closed-end air column is given by: f = v/4L Where f is the frequency, v is the velocity of sound in air, and L is the length of the air column. Substituting the given values: f = 360/4 x 0.45 f = 200.0 Hz Therefore, the fundamental frequency of the sound wave generated in the pipe is 200.0 Hz.

**Question 20**
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An object is placed directly below a glass block of thickness 3.0cm. Calculate the lateral displacement of the object if the refractive index of the glass is 1.5.

**Answer Details**

1.5 = 3apparent displacement $\frac{3}{\text{apparent displacement}}$

Apparent depth = 2cm

displacement = 3 - 2

= 1cm

**Question 21**
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An object is placed in front of a converging lens of a focal length 20cm. The image is virtual and has a magnification of 2. What is the distance of the object from the lens?

**Answer Details**

V = -2u

F = uvu+v $\frac{uv}{u+v}$

20 = uv−2uu−2u $\frac{uv-2u}{u-2u}$

u = 10cm

**Question 22**
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Which of the following has the lowest internal resistance when new?

**Question 23**
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In a projection lantern of focal length f, the object distance u, is such that

**Question 24**
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Calculate the velocity ratio of a screw jack of pitch 0.3cm if the length of the tommy bar is 21cm

**Answer Details**

V.R = 2πLp $\frac{2\pi L}{p}$

= 2×π×20.3 $\frac{2\times \pi \times 2}{0.3}$

V.R = 140π

**Question 25**
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A force of 15N stretches a spring to a total length of 30cm. An additional force of 10N stretches the spring 5cm further. Find the natural length of the spring

**Answer Details**

We can use Hooke's Law, which states that the force exerted by a spring is proportional to its extension, to solve this problem. Let the natural length of the spring be x. When a force of 15N is applied, the spring is stretched by 30cm - x. Therefore, we have: 15N = k(30cm - x) where k is the spring constant. When an additional force of 10N is applied, the spring is stretched by an additional 5cm. Therefore, we have: 25N = k(35cm - x) Dividing the second equation by the first, we get: 25N/15N = (35cm - x)/(30cm - x) Simplifying this equation, we get: 5/3 = (35cm - x)/(30cm - x) Multiplying both sides by 30cm - x, we get: 5(30cm - x) = 3(35cm - x) 150cm - 5x = 105cm - 3x 2x = 45cm x = 22.5cm Therefore, the natural length of the spring is 22.5cm. Answer: 22.5cm.

**Question 26**
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What is the reading of the vernier calipers shown above?

**Answer Details**

The main scale reading on the vernier caliper is 1.3 cm, which is the closest value to the left of the zero mark on the vernier scale. The first mark on the vernier scale that aligns with a mark on the main scale is the 9th mark, which corresponds to 0.09 cm. The total reading is the sum of the main scale and the vernier scale readings, so: Total reading = 1.3 cm + 0.09 cm = 1.39 cm Therefore, the reading of the vernier calipers shown above is 1.39 cm.

**Question 27**
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A motor vehicle is brought to rest from a speed of 15ms-1 in 20 seconds. Calculate the retardation.

**Answer Details**

Retardation is the rate at which an object slows down. It is also known as deceleration or negative acceleration. To calculate the retardation of the motor vehicle, we can use the following formula: Retardation = (Final Velocity - Initial Velocity) / Time Taken Here, the final velocity is 0 m/s (since the vehicle is brought to rest), the initial velocity is 15 m/s (given in the question), and the time taken is 20 seconds (also given in the question). So, plugging in the values in the formula, we get: Retardation = (0 - 15) / 20 Retardation = -0.75 m/s^2 Therefore, the retardation of the motor vehicle is 0.75 m/s^2 (option A).

**Question 28**
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If it takes 5.0 hrs to drain a container of 540.0cm3 of water, what is the flow rate of water from the container in kg-1? [Density of water = 1000kgm-3]

**Answer Details**

The flow rate of water is the volume of water drained per unit time. To calculate it, we first need to find the volume of water drained, which is given as 540.0 cm^3. To convert this to liters, we divide by 1000: 540.0 cm^3 = 0.54 L We are given the time taken to drain this volume of water as 5.0 hours. To convert this to seconds, we multiply by 3600: 5.0 hours = 18000 s The flow rate of water is therefore: Flow rate = Volume / Time = 0.54 L / 18000 s Now, we are given the density of water as 1000 kg/m^3. To convert the volume of water to mass, we multiply by the density and convert the units from liters to kilograms: 0.54 L × 1000 kg/m^3 × 10^-3 L/kg = 0.54 kg Putting it all together, the flow rate of water is: Flow rate = Volume / Time = 0.54 kg / 18000 s = 0.03 × 10^-3 kg/s = 30.0 kg^-1 s^-1 Therefore, the answer is option D, 30.0 kg^-1 s^-1.

**Question 29**
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If a room is saturated with water vapour, the temperature of the room must be

**Answer Details**

When air holds the maximum amount of water vapor that it can at a given temperature and pressure, it is said to be saturated. The maximum amount of water vapor that air can hold increases with temperature. Thus, if a room is saturated with water vapor, the temperature of the room must be at or below the dew point temperature, which is the temperature at which water vapor in the air begins to condense into liquid. This means that the answer is "below or at the dew point."

**Question 30**
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An object of mass 50kg is released from a height of 2m.Find the kinetic energy just before it strikes the ground

**Answer Details**

When an object is lifted to a height h, it gains potential energy given by: P.E = mgh, where m is the mass of the object, g is the acceleration due to gravity and h is the height. When the object is released, it starts to fall and its potential energy is converted into kinetic energy, which is given by: K.E = (1/2)mv^2, where v is the velocity of the object just before it strikes the ground. At the highest point, the potential energy is maximum and the kinetic energy is minimum. At the instant just before the object strikes the ground, the potential energy is zero and the kinetic energy is maximum. Therefore, the potential energy at the highest point is: P.E = mgh = 50 x 9.81 x 2 = 981 J Since the potential energy is converted to kinetic energy just before the object strikes the ground, the kinetic energy at that point is equal to the potential energy at the highest point: K.E = 981 J. Therefore, the correct option is: 1000J.

**Question 31**
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The fundamental frequency of vibration of a sonometer wire may be halved by

**Answer Details**

The fundamental frequency of a sonometer wire is related to its length, tension, and mass. To halve the fundamental frequency, we need to identify which factor(s) we can change to reduce the frequency by a factor of 2. - Doubling the length of the wire will reduce the frequency by a factor of 2, as the frequency is inversely proportional to the length of the wire. - Doubling the mass of the wire will not affect the frequency, as the frequency is independent of the mass of the wire. - Reducing the tension by half will reduce the frequency by a factor of sqrt(2), not by a factor of 2. - Reducing the absolute temperature by half will not affect the frequency of the wire. Therefore, the correct answer is: "doubling the length of the wire".

**Question 32**
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An object is projected with a velocity of 80ms-1 at an angle of 30o to the horizontal.The maximum height reached is

**Answer Details**

The problem involves finding the maximum height reached by an object that is projected at a given velocity and angle to the horizontal. To solve this, we can use the equations of motion for projectile motion. Firstly, we need to break down the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion and is given by: vx = v * cos(theta) where v is the initial velocity and theta is the angle of projection. The vertical component changes due to the force of gravity and is given by: vy = v * sin(theta) - gt where g is the acceleration due to gravity and t is the time elapsed. At the maximum height, the vertical component of the velocity becomes zero. We can use this fact to determine the time of flight (the time it takes for the object to reach its maximum height and then return to the same height) as follows: 0 = v * sin(theta) - gt_max where t_max is the time taken to reach the maximum height. Rearranging this equation gives: t_max = v * sin(theta) / g Using this value of t_max, we can determine the maximum height reached by the object using the equation: h_max = v * sin(theta) * t_max - 0.5 * g * t_max^2 Substituting the given values in these equations, we get: vx = 80 * cos(30) = 69.28 m/s vy = 80 * sin(30) - 9.81 * t 0 = 80 * sin(30) - 9.81 * t_max t_max = (80 * sin(30)) / 9.81 = 4.08 s h_max = 80 * sin(30) * t_max - 0.5 * 9.81 * t_max^2 = 80 * sin(30) * 4.08 - 0.5 * 9.81 * 4.08^2 = 80 m Therefore, the maximum height reached by the object is 80 meters.

**Question 33**
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Calculate the magnitude of the force required to just move a 20kg object along a horizontal surface if the coefficient of friction is 0.2. [g = 10ms-2]

**Answer Details**

The force required to move an object along a surface is equal to the product of the coefficient of friction and the normal force acting on the object. The normal force is equal to the weight of the object, which is given by the product of the mass and the acceleration due to gravity. Therefore, the force required to move the 20kg object along a horizontal surface is: Force = coefficient of friction * normal force Normal force = mass * acceleration due to gravity = 20kg * 10ms^-2 = 200N Force = 0.2 * 200N = 40N Therefore, the magnitude of the force required to just move the 20kg object along a horizontal surface is 40N. The correct option is: 40.0N

**Question 34**
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If a solid X floats in liquids P of relative density 2.0 and in liquid Q of relative density 1.5, it can be inferred that the

**Answer Details**

R.d α $\alpha $ weight of liquid displaced

R.d of PR.d of Q $\frac{\text{R.d of P}}{\text{R.d of Q}}$ = weight of p displacedweight of Q displaced $\frac{\text{weight of p displaced}}{\text{weight of Q displaced}}$

21.5 $\frac{2}{1.5}$ = weight of p displacedweight of Q displaced

**Question 35**
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The mass of a specific gravity bottle is 15.2g when it is empty. it is 24.8g when filled with kerosene and 27.2g when filled with distilled water. Calculate the relative density of kerosene

**Answer Details**

To calculate the relative density of kerosene, we need to first find the mass of kerosene that fills the specific gravity bottle. We can do this by subtracting the mass of the empty bottle from the mass of the bottle filled with kerosene. Mass of kerosene = Mass of filled bottle - Mass of empty bottle Mass of kerosene = 24.8g - 15.2g Mass of kerosene = 9.6g Next, we need to find the mass of an equal volume of water that fills the specific gravity bottle. We can do this by subtracting the mass of the empty bottle from the mass of the bottle filled with water. Mass of water = Mass of filled bottle with water - Mass of empty bottle Mass of water = 27.2g - 15.2g Mass of water = 12g The relative density of kerosene is then calculated by dividing the density of kerosene by the density of water. Relative density of kerosene = (mass of kerosene / volume of kerosene) / (mass of water / volume of water) We don't have the volumes of the kerosene and water, but since they were both measured using the same specific gravity bottle, their volumes are equal. Therefore, we can simplify the equation to: Relative density of kerosene = (mass of kerosene) / (mass of water) Relative density of kerosene = 9.6g / 12g Relative density of kerosene = 0.8 Therefore, the relative density of kerosene is 0.8, which corresponds to the option: 0.80.

**Question 36**
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In the curve above, PQRST represents an alternating voltage of frequency 50Hz. The time interval between points P and R on the graph is

**Question 37**
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Two capacitors R1 = 4Ω $\Omega $ and R2 = 5Ω $\Omega $ are connected in parallel across a potential difference. If P1 and P2 represent power dissipated in R 1 and R2 respectively, then the ratio P1 : P2 is

**Answer Details**

P1 = V21R1 $\frac{{V}_{1}^{2}}{{R}_{1}}$ and P2 = V22R2 $\frac{{V}_{2}^{2}}{{R}_{2}}$

P1R1 = P2R2

4P1 = 5P2

p1p2 $\frac{{p}_{1}}{{p}_{2}}$ = 54 $\frac{5}{4}$

= 5 : 4

**Question 38**
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Two similar kettles containing equal masses of boiling water are placed on a table. If the surface of one is highly polished and the surface of the other is covered with soot, Which of the following observations is correct

**Answer Details**

The correct observation is: the kettle covered with soot cools down more quickly because it is a good radiator. The process by which a body loses heat to its surroundings is known as radiation, convection or conduction. The surface of the kettle covered with soot is rough and black, and it has a high emissivity that allows it to radiate more heat to its surroundings compared to the highly polished surface of the other kettle. This causes the kettle covered with soot to cool down faster than the polished one.

**Question 39**
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To produce a parallel beam of light from a concave mirror, the distance at a which the lamp should be placed from the mirror is equal to

**Answer Details**

To produce a parallel beam of light from a concave mirror, the distance at which the lamp should be placed from the mirror is equal to the focal length. This is because a concave mirror has a focal point, which is the point where all incoming parallel rays of light converge after reflection. By placing the lamp at a distance equal to the focal length, the rays of light will reflect off the mirror and converge at the focal point, resulting in a parallel beam of light. Placing the lamp at any other distance would result in the rays of light either converging or diverging after reflection, which would not produce a parallel beam. Therefore, option A, the focal length, is the correct answer.

**Question 40**
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Which of the following are found in the receiver but not in the microphone of a telephone handset? i. Diaphragm. ii. soft iron pole pieces. iii. permanent magnet. iv.carbon blocks

**Answer Details**

The receiver and microphone are two essential parts of a telephone handset. The microphone converts sound waves into electrical signals, while the receiver converts electrical signals into sound waves. In the microphone, sound waves cause a diaphragm to vibrate, which then changes the pressure on carbon blocks, varying the electrical resistance of the blocks. This variation in electrical resistance produces electrical signals that correspond to the sound waves. In the receiver, electrical signals produced by the microphone are transmitted to a coil of wire that surrounds a magnet. The electrical signals cause the coil to create a magnetic field that interacts with the permanent magnet. This interaction causes a diaphragm to vibrate, which produces sound waves that correspond to the original sound. From this explanation, we can see that the diaphragm is found in both the receiver and the microphone. The permanent magnet is found in the receiver, but not in the microphone. The soft iron pole pieces are also found in the receiver, but not in the microphone. The only component that is found in the microphone but not in the receiver is the carbon blocks. Therefore, the answer is option (ii) i and iv.

**Question 41**
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A capacitor 8μ $\mu $F is charged to a potential difference of 100V. The energy stored by the capacitor is

**Answer Details**

Energy = 12 $\frac{1}{2}$CV2

= 12 $\frac{1}{2}$ x 8 x 10-6 x (100)2

= 4 x 10-2J

**Question 42**
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A spring of length 25cm is extended to 30cm by a load of 150N attached to one of its end, what is the energy stored in the spring?

**Question 43**
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If light with photon energy 2eV is incident suitably on the surface of a metal with work function 3eV, then

**Answer Details**

An electron can only be emitted if photon energy is equal or greater than work function of the plate

**Question 44**
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A Motor Tyre is inflated to pressure of 2.0 x 105Nm-2 when the temperature of air is 27oC. What will be the pressure in it at 87oC assuming that the volume of the Tyre does not change?

**Answer Details**

The pressure inside a motor tire is directly proportional to the temperature of the air inside it. According to the Gay-Lussac's Law of pressure-temperature, when the temperature of gas increases, its pressure also increases if the volume is constant. Here, the volume of the tire is assumed to be constant. Using the formula P1/T1 = P2/T2, where P1 is the initial pressure, T1 is the initial temperature, P2 is the final pressure, and T2 is the final temperature, we can find the final pressure. Substituting the given values, we get: P1 = 2.0 x 10^5 N/m^2 T1 = 27°C + 273 = 300K T2 = 87°C + 273 = 360K So, P2 = (P1 x T2) / T1 = (2.0 x 10^5 N/m^2 x 360K) / 300K = 2.4 x 10^5 N/m^2. Therefore, the pressure inside the motor tire at 87°C is 2.4 x 10^5 N/m^2. So, the answer is 2.4 x 10^5 N/m^2.

**Question 45**
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In the circuit diagram above, calculate the current in the 12 Ω resistor if the cell has an emf of 12V and an internal resistance of 1Ω

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