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##### Physics (2017) QUIZ STUDY

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Which is the incorrect formula for a body accelerating uniformly?

a = v^{2} - u\(\frac{2}{2}\)

v^{2} = u^{2} + 2as

s = \(\frac{1}{2}\)ut + at^{2}

v^{2} - u^{2}= 2as

s = \(\frac{1}{2}\)ut + at^{2} is a wrong equation, it should be s = \(\frac{1}{2}\)ut + at^{2}

= \(\frac{\text{potential difference}}{\text{distance}}\)

= \(\frac{6.5v}{35cm}\)

= \(\frac{6.5}{35 \times 10^{- 2}}\)

= 18.57NC\(^{-1}\)

Calculate the upthrust on an object of volume 50cm^{3} which is immersed in liquid of density 10^{3}kgm^{-3} [g = 10ms^{-2}]

0.8N

2.5N

0.5N

1.0N

Upthrust = change in weight

density = \(\frac{\text{mass}}{\text{volume}}\)

Mass of liquid displaced = Density of Liquid \(\times\) Volume;

= 10^{3} \(\times\) 50 \(\times\) 10^{-6}m^{3}

= 0.05kg

(Note that the volume of 50cm3 was converted to m^{3} by multiplying by 106) i.e 50cm^{3} = 50 \(\times\) 10?6m^{3})

Since mass displaced = 0.05kg

Upthrust = mg = 0.05 \(\times\) 10

= 0.5N

workdone = force \(\times\) distance

Calculate the specific latent heat of vaporization of steam of 1.13 x 10^{6}J, if heat energy is required to convert 15kg of it to water.

7.53 x 10^{5} Jkg\(^{-1}\)

7.53 x 10^{-2}Jkg\(^{-1}\)

7.53 x 10^{4}Jkg\(^{-1}\)

7.53 x 10^{-3}Jkg\(^{-1}\)

Latent Energy = mass(m) \(\times\) specific latent heat (L)

E = mL

L = \(\frac{E}{m}\)

= \(\frac{1.13 \times 10 ^{6}}{15}\)

= 7.53 \(\times\) 10^{4}Jkg\(^{-1}\)

A cell of internal resistance 2 Ω supplies current to a 6 Ω resistor. The efficiency of the cell is

12.00%

25.00%

33.30%

75.00%

Internal resistance determine the maximum current that can be supplied

Efficiency = \(\frac{r}{R}\) \(\times\) 100

= \(\frac{2}{6}\) \(\times\) 100

= 33.3%

Brownian motion is an evidence of particle nature of matter

matter is made up of molecules

The molecules of matter are in constant motion

V.R = number of pulleys = 6

ma = \(\frac{\text{Load}}{\text{Effort}}\) = \(\frac{200}{50}\) = 4

Effiecieny = \(\frac{4}{6}\) \(\times\) 100 = 66.67%

Three 3Ω resistance are connected in parallel what is the equivalent resistance?

1Ω

9Ω

3Ω

0.33Ω

Resistor connected in a parallel have an equivalence given by

\(\frac{1}{\text{Reff}}\) = \(\frac{1}{3}\) + \(\frac{1}{3}\) + \(\frac{1}{3}\)

= \(\frac{1 + 1 + 1}{3}\) = \(\frac{3}{3}\) = 1

Reff = 1Ω

Refrigerator is an appliance to keep things cold

Air blower is used to release air

Air conditioner is used to remove heat and maintain a stable temperature in an occupied space

Moment = force \(\times\) distance

= N \(\times\) m

= Nm

Calculate the light of the image formed by a pinhole camera of length 12cm used to photograph an object 60cm away from the hole and 70cm high

10 cm

16 cm

5 cm

14 cm

Magnification = \(\frac{\text{Length of camera}}{\text{distance of object from pin hole}}\)

m = \(\frac{12}{60}\)

= 0.2

Also magnification m = \(\frac{\text{Height of image}}{\text{Height of object}}\)

ΔHeight of image = m \(\times\) height of object

= 0.2 \(\times\) 70

= 14cm

When two objects P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity pf P to Q is

1:04

4:01

1:01

2:01

H = mcΔ

H_{P} = H_{Q}, Δ_{P} = 2Δq,

m_{P} = \(\frac{1}{2}\)m_{Q}

Δ H_{P} = m_{Q}C_{P}Δ_{P}, also HQ = m_{Q}C_{Q}Δ_{Q}

Since H_{P} = H_{Q}

m_{P}c_{P}Δ_{P} = m_{Q}c_{P}Δ_{q}

Putting in the conditions

\(\frac{1}{2}\)m_{Q} \(\times\) c_{P} \(\times\) 2Δ_{Q} = m_{Q} \(\times\) Δ_{Q} \(\times\) c_{Q}

c_{P} = c_{Q}

\(\frac{C_p}{C_Q}\)= \(\frac{1}{1}\)

= 1:1

When the r.m.s value of a source of electricity supply is given as 240v, it means that the peak value of the supply is

240v

340v

480v

57600v

The relationship between the peak value and r.m.s value of electricity supply is given as

V\(_{r.m.s}\) = \(\frac{V_o}{\sqrt{2}}\)

where V_{o} = peak voltage

V_{o} = V\(_{r.m.s}\) x \(\sqrt{2}\)

= 240 x \(\sqrt{2}\)

= 339.5v

= 340v

The net capacitance in the circuit above is

80µF

6.0µF

4.0µF

2.0µF

their equivalence is 2µF and 2µF = 4µF

The 4µF generated is now in series with the remaining 4µF.

The net capacitance for series connection is

\(\frac{1}{C}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) = 1 + \(\frac{1}{4}\) = \(\frac{2}{4}\)

C = \(\frac{4}{2}\)

= 2µF

Thermometer is used to measure temperature

Hygrometer is used to measure relative density of a liquid

Manometer is used to measure the pressure of a gas

A man of mass 50kg ascends a flight of stains 5m high in 5 seconds. If acceleration due to gravity is 10ms\(^{-2}\) the power expended is

100w

200w

400w

500w

Power is the time rate of doing work

i.e power = \(\frac{\text{energy}}{\text{time}}\)

Since energy = mgH

power = \(\frac{mgH}{t}\)

= \(\frac{50 \times 10 \times 5}{t}\)

= 500W

A radio station broadcast at a frequency of 600KHZ. If the speed of light in air is 3 \(\times\) 10^{8}ms\(^{-1}\). Calculate the wavelength of the radio wave

2.20 \(\times\) 10\(^3\)m

5.0 \(\times\) 10\(^2\)m

5.0 \(\times\) 10\(^5\)m

11.8 \(\times\) 10\(^{11}\)m

V = fλ

λ = \(\frac{v}{f}\)

= \(\frac{3 \times 10^8}{600 \times 10^3}\) = 500m

= 5 \(\times\) 10^{2}m

N.B 600KHz Frequency = 600 \(\times\) 10^{3}Hz

Tension, weight, impulse are vector quantities because they have direction. Mass is the only scalar quantity there

f = focal length

f = \(\frac{1}{p}\)

= \(\frac{1}{5}\)

= 0.2m

A string of length 5cm is extended by 0.04m when a load of 0.8kg is suspended at the end. How far will it extend if a force of 16N is applied? [g = 10ms\(^{-2}\)]

0.04m

0.12m

0.01m

0.08m

From Hoke's Law, F = ke, K = Constant of Force

For he first case, Force = mg,

= 0.8 \(\times\) 10 = 8N

8 = k \(\times\) 0.04

k = \(\frac{8}{0.04}\)

= \(\frac{200N}{m}\)

Since K is constant

For the second case, F = ke

F = 200 \(\times\) e

e = \(\frac{F}{200}\)

= \(\frac{16}{200}\)

= 0.08m

Usage of insulated soft iron wire is to reduce hysteresis loss

Usage of low resistance wire (thick wire) is to reduce I2R loss

Usage of thick wire is to reduce leakage heat loss due to leakage of magnetic flux

Thus the correct answer is to use high resistance wire or thin wire

From the diagram above, calculate the energy stored in the capacitor

4.0 x 10\(^{-2}\)J

4.0 x 10\(^{-4}\)J

8.0 x 10\(^{-4}\)J

8.0 x 10\(^{-2}\)J

c = 8µF, v = 10v

Energy stored = \(\frac{1}{2}\) cv^{2}

= \(\frac{1}{2}\) \(\times\) (8 \(\times\) 10^{-6}) \(\times\) 10^{2}

= 4 \(\times\) 10\(^{-4}\)

A constant force of 5N acts for 5 seconds on a mass of 5kg initially at rest. Calculate the final momentum

125kgms^{-1}

25kgms^{-1}

15kgms^{-1}

0kgms^{-1}

Impulse = momentum

Impulse = Force (f) \(\times\) time (t)

Momentum = mass (m) \(\times\) velocity (v)

Ft = Final momentum - initial momentum

FT = mv - mu

Since it is initially at rest u = 0

5 x 5 = mv ? m(0)

25 = mv

Final momentum = 25kgms\(^{-1}\)

Primary cells are used up gradually often the cells are use and it can't be recharged

Secondary cells or accumulated can be recharged and used for a long period of time and their main advantage is that they have low internal resistance

It is known that a neutron exists in a light atomic nucleus

Which of the following also exists in the nucleus?

An electron

α - particle

β - particle

proton

Nucleus of atom is made up of neutron and proton

Hydrometer is used to measure relative density of liquid.

Spring balance is used to measure the weight of an object

Beaker is used to measure or take volume of a liquid

What is the speed of a body vibrating at 50 cyclic per second

100 rads\(^{-1}\)

200 rads\(^{-1}\)

50 rads\(^{-1}\)

400 rads\(^{-1}\)

1 cycle = 3600 = 2 rad

50 cycle = 2 rad x 50 = 100 rad

angular speed = 50 cycles per seconds = 100 rads\(^{-1}\)

When two mirrors are placed at an angle of 90º to each other, how many images will be formed when an object is placed in front of the mirrors

5

4

3

2

When two mirror are inclined at an angle \(\theta\), the number of images formed, n = \(\frac{360}{\theta}\) - 1

since \(\theta\) = 90^{o}

\(\frac{360}{\theta}\) - 1

= 4 - 1

= 3

Phosphorous is a pentavalent element, so it will produce a o -type semi-conductor

^{o}prism of a refractive index [sin

^{-1}0.75 = 49

^{o}]

^{o}

^{o}

^{o}

^{o}

D = angle of minimum deviation

A = refractive angle of the prism

1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin\frac{1}{2} \times 60}\)

1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin 30}\)

Sin \(\frac{1}{2}\) (60 + D) = 1.5 * Sin 30

Sin \(\frac{1}{2}\) (60 + D) = 0.75

\(\frac{1}{2}\) (60 + D) = Sin

^{-1}(0.75)

but Sin

^{-1}(0.75) = 49

^{o}

\(\frac{1}{2}\) (60 + D) = 49

60 + D = 2 * 49 = 98

^{o}

D = 98

^{o}- 60

^{o}

D = 38

^{o}

Five 80W lamps = 5 \(\times\) 80 = 400W

Total power = 280W + 400W

= 680W

= \(\frac{680KW}{1000}\) = 0.68KW

The kilowatt hour = 0.68 KW \(\times\) 12 hrs

= 8.16 KWh

Cost is N 7 per KWh

So the cost of running them = 8.16KWh \(\times\) N 7/KWh

= 57.12

From the diagram above, calculate the total current in the circuit

5.0A

3.7A

4.5A

4.0A

I = \(\frac{V}{R}\)

Since the resistance are in parallel

\(\frac{1}{Reff}\) = \(\frac{1}{8}\) + \(\frac{1}{12}\) + \(\frac{1}{6}\)

= \(\frac{2 + 3 + 6}{36}\) = \(\frac{11}{36}\)

Reff = \(\frac{36}{11}\) = 3.27

I = I

_{1}+ I

_{2}+ I

_{3}

= \(\frac{V}{Reff}\)= \(\frac{12}{3.27}\)

= 3.669A

I = 3.7A

Calculate the time taken for a mango fruit that fall from the tree 20 m to the ground [g = 10 ms^{-2}]

10s

5s

4s

2s

From equation of motion under gravity

H = ut ± \(\frac{1}{2}\)gt

^{2}, It is falling so g is +ve

H = 0 + \(\frac{1}{2}\)gt

^{2}

H = \(\frac{1}{2}\)gt

^{2}

2H = gt

^{2}

\(\frac{2H}{g}\) = t

^{2}

t = \(\frac{\sqrt{2H}}{g}\)

= \(\frac{\sqrt {2 \times 20}}{10}\)

= \(\sqrt{4}\)

= 2s

V.R = \(\frac{\text{distance moved by effort}}{\text{distance moved by load}}\) = \(\frac{OB}{AB}\)

But Sin? = \(\frac{opp}{hyp}\) = \(\frac{AB}{OB}\)

therefore; = \(\frac{OB}{AB}\) = \(\frac{1}{\sin \theta}\)

^{3}when its pressure is 400cmHg. Determine the volume of the gas when its pressure is 200cmHg.

^{3}

^{3}

^{3}

^{3}

P

_{1}V

_{1}= P

_{2}V

_{2}

V

_{2}= \(\frac{P_1 V_1}{200}\) = \(\frac{400 \times 10}{200}\)

= 20cm

^{3}

\(^{Z}_{A}X\) where Z = mass number

A = atomic number

A = number of proton = number of electrons in a free state = 28

Z = number of protons + number of neutrons = 28 + 30 = 58

\(^{58}_{28}X\)is the representation