PHYSICS QUESTIONS AND ANSWERS WITH EXPLANATION

This blog post contain multiple choice questions on
Velocity, acceleration ,power, efficiency of machine, collision of two bodies etc

1.startinh from rest, a Mercedes-Benz car accelerate uniformly at 45m/s² for 70seconds. What distance does it cover in thirty-fifth second of the motion.

(A) 1.5x10²km (B) 1.6 x 10 ⁴km (C) 1.6 X 10³km (D) 1.5 x10³km
Solution
Acceleration ,a =45m/s²
Hint : when ever a body accelerate uniformly from rest initial velocity, u=0
The time taken to cover the distance in thirty-fifth second of the motion is the time interval between
34th – 35th seconds
Therefore time t = t₃₅ - t ₃₄
S₃₅=ut +1/2at²
S₃₄= ut + ½ at²
S = S ₃₅- S₃₄
S=(ut ₃₅+1/2at²₃₅ ) - (ut ₃₄+ 1/2at²₃₄ )
u=0
S = 1/2 a(t²₃₅ – t²₃₄ )
S=1/2 x 45(35² – 34²)
S= 1/2 x45 (1225 – 1156)
S= 1/2 x 45 x69
S = 1552.5 m
S = 1.5525 x10³
S=1.6 x10³ km

2. A machine whose efficiency is 65% is used to move an object of mass 85kg through a vertical height of 8m in 25 seconds . Calculate the power out put.

(A) 320W (B) 302W (C) 203W (D) 230W

Solution

Efficiency ,E = 65% = 65/100 = 0.65

Mass, m= 85kg

Height ,h = 8m

Time ,t = 25s

E= power out put, P₀ x100%

Power input Pᵢ

Power output P₀ ₌ mg x s /t

Power Input Pᵢ = mg x s X 1/ E

Pᵢ = 65x 10 X 8 X 1/0.65

2 5

Pᵢ= 320W

3. An Increase In distance between two charges q₁ and q₂ will result in ----------- attractive force between the two charges

(A) Increase (B) no effect (C) Decrease(D) alternate

SOLUTION

Applying coulomb’s law

F= kq₁ q₂

r²

from the equation above an Increase In distance r will cause the attractive force between the two point charge to reduce by factor of 2

example

let f= k = 24 q₁ =4 q₂ =3 r =5

When distance r is absent

(i)f=24x4x3 = 288N

When r is present

(ii)f = 24 x4x3/5² = 288/25 = 11.52N

BY mare looking at both solution you can see that force f was larger when distance r wasn’t Involved and force f was reduced when r was Introduce

Conclusion f reduce as distance increase

Answer C

4. An ambulance siren has the frequency of 600Hz. If the velocity of sound in air is 240m/s. determine the frequency heard by taxi driver moving at a velocity of 20m/s

Toward the ambulance

(A) 450Hz (B) 551Hz (C) 550Hz (D) 451Hz

Solution

F¹= 600Hz velocity of sound in air Vₐ = 240m/s

The velocity of the taxi car Vₜ= 20m/s

F =?

F¹ (Vₐ-V ₜ)/ Vₐ X f

F¹= (240-20)/240X 600

F¹= 550Hz

N/B An apparent alteration in the frequency of a sound waves is observed whenever there is a relative velocity between the source of sound and the observer or listener .

5. --------- is a feature of light that does not change when light passes through a rectangular block prism

(A) Wavelength (B) Velocity (C) frequency (D) intensity

Answer C

6. A magnetic field 1.4 X 10⁻ ³ Tesla exist at the center of a coil of the radius of 20cm having 15 closely wounded turns. Find the current in the coil.

(A) 30A (B) 40A (C) 50A (D) 60A

Solution

Magnetic flux B = 1.4 X 10⁻³ T

Number of turns N = 15

Radius R = 20cm = 0.2m

μ₀ = 4π X 10⁻⁷Hm⁻¹

current I = ?

B = μ₀N I /2R

I = 2BR/μ₀N

I = 2x 0.2 X 1.4 X 10⁻³/ 4π X 10 ⁻⁷ X 15

I = 30A

ANSWER ⃗ A

7. In photo electric effect, light should have ------- (A) low frequency ( B) high frequency (C) no frequency (D) A and B

Answer

8.An electric fan draws 2.0 A of current from a 220V source .Find the power rating of the fan .

(A) 0.44kw (B) 0.43kw (C) 0.54kw (D) 0.53kw

Solution

Current I = 2.0A

Voltage V = 220V

Power p = ?

P = IV = (2.0A) ( 220V)

P = 440W = 0.44kw

Answer A

9.l heavy duty vehicle of mass 1000kg , moving with acceleration of 12m/s² is acted upon by a constant restricting force of 1200N . calculate the force exerted by the engine to maintain the acceleration.

(A) 14.2 X 10³N (B) 14.1 X 10² N

Solution

M=1000kg

Force exerted by engine F = ?

Restricting force F¹ = 1200N

Acceleration ,a = 12m/s²

Net force acting on the heavy duty vehicle = F – 1200N

F-F¹= ma

F-1200 = 1000 X 12

F=12000 + 1200

F = 13200 = 13.2 X 10³N

Answer D

10. All except on is not a forms of energy

(A) mechanical energy (B) Heat energy (C) Nuclear energy (D) Renewable energy

Answer D

Renewable energy is one of two major group of energy resources.

Renewable energy is define as category of energy which can replenish it’s self within a short period of time

This kinds of energy can be use over and over without finishing.

Example

Wind ,ocean wave, solar energy mechanical energy is form of energy comprising two sub-energy

Potential and kinetic energy.

Potential energy :it is energy possess by a body by virtue of it’s position. Example a stone insert hanged on top of tree will do work when release from the tree. It does work in the sense that when it hits an object it can damage it .glass for instance, if the stone hit glass it will damage it when falling from the tree.

Kinetic energy: it is energy possess by moving body e.g moving Mercedes-Benz possess a kinetic energy.

Heat energy: it is a form of energy release by a radiating body e.g sun

Nuclear energy: it is form of energy possess by nucleus of an atom.

11.An a.c source is connected in series to a capacitor of capacity reactance of 10³ √3 and a resistor of resistance 10³ohm .the Impedance of the circuit

Is ------

(A) 2000√3ohm (B) 2000ohm

Solution

Capacitive reactance x=10³√3

Resistance R =10³ohm

Impedance Z = √R² +X²

Z= √ (10³)² +(10³√3)²

Z=√ 10⁶ +(10³)²(√3 )²

= √10⁶ + 10⁶ x3)

= √4x10⁶ )

= 2000ohm

12. A radioisotope of ²³⁹₉₂U emits six alpha particles and four beta particles . -------and ---

Are the atomic number and mass number.

(A) 211 and 88 (B) 270 and 88

(B) 211 and 87 (D) 201 and 85

Solution

One alpha particle Is equivalent to one helium atom

α—particle = ⁴₂H

Therefore

6α=6⁴₂H

β- particle = ₋₁⁰e

therefore

4β= 4₋₁⁰e

LET Z and A be the atomic number and mass number respectively

²³⁵₉₂U 6⁴He + 4⁰₋₁e +zᴬY

From the equation

92=(4x2)+(4x-1) +z

92=8-4+z

92=4+z

92-4=z

88=z

Atomic number = 88

From the equation

235= (6x4) +(4x0) +A

235= 24+0+A

235-24=A

A = 211

Mass number =211

Answer  211 and 88

13. In 60 days a radioactive isotope reduces in mass from 384g to 24g .what is the decay constant of the radioactive material?

(A) 0.0341day ⁻¹(B) 0.0285day⁻¹

Solution

Decay time, t = 60days

Initial mass,N₁ = 384g

Final mass N ₂= 24g

R= N₁/N₂

R= 384g/24g = 16

Decay constant⃗ λ = 2.303logR/t

=2.303xlog16 )/60days

= 2.303 x1.2041/60days

= 0.0462day⁻¹

14.A planet is two times the size of it’s Moon and the acceleration due to gravity on the planet is 40times that on it’s moon . Calculate the ratio of the mass of the moon to that of the planet.

Solution

Acceleration due to gravity on the moon g ₘ= 1m/s²

Acceleration due to gravity on the planet P ₚ= 40m/s²

Radius rₘ =1

Radius rₚ =2

Acceleration due to gravity ,

g= Gm/r²

Acceleration due to gravity for the planet gₚ=Gₚmₚ/r²ₚ

Making m subject of formula gives

Mₚ= gₚr²ₚ/Gₚ

And that of moon

g ₘ= gₘr²ₘ/G

Mₘ =gₘr²ₘ/G

Mₘ/Pₘ = gₘr²/G ÷ gₚr²ₚ/G

= gₘr²ₘ/G X G/gₚr²ₚ

Mₘ/Mₚ = gₘr²ₘ/gₚr²ₚ

Mₘ/Mₚ = 1x1²/40x2² = 1/160

Therefore M ₘ : Mₚ = 1:160

15. An object is placed in front of a converging lens of focal length 30cm .The image magnification is 4 and it is a virtual image . what is distance of the object from the lens ?

(A) 22cm (B) 23.4cm

Solution

Focal length = 30cm

Image distance = -v (negative because it’s virtual image )

Object distance,u= ?

M= -v/u = 4

V = -4u

f= uv/u+v

f(u+v)= uv

u+v = uv/f

u= uv/f - v

u = u x -4u/30 – (-4u)

u= - 4u²/30 + 4u

30u= -4u²+120u

30u-120u= -4u²

Divide both side by -4u

-90u/-4u = -4u²/-4u

22.5cm = u

Answer u = 22.5cm

16. A spare bullet of mass 8.0gm moving at 80m/s strikes a soldier at a war front and Lodge in his leg. Give that the bullet undergoes uniform acceleration and stops after travelling a distance of 4.0cm .Find the impulse in the leg shoulder and the average force experienced by the soldier.

(A) 0.7 Ns and 1354N (B) 0.90Ns and 1480.51N

(D) 0.86Ns and 1371.40N

Solution

N/B Impulse is a charge of the momentum. Impulse is also a product of force and time it takes to act .

Distance= 4cm =0.04m

M=8gm = 0.008kg

Final velocity ,V=0

Initial velocity U=-120m/s

Ft=?

F=m(v-u)/t

Ft=m(v-u)

Ft=0.008kg(0m/s –(-120m/s)

=0.96Ns

N/B the initial velocity,u has negative magnitude because it is coming from opposite direction .

To find the average force experienced by the soldier ,first find the time the bullet took to travel the distance of 4cm on the soldier's body.

V²=u² + 2as

0²=(-120m/s)²+2 X a X 0.04m

0.08a= 14400m/s²

a = -14400/0.08 m/s²

a= -180000m/s²

v= u-at

0= 120m/s + (-180000m/s²)t

t = 120/180000

t=0.0007seconds

Impulse = force X time it takes act

0.96Ns = F(0.0007s)

F=1371.43N

17 . A body of mass 80kg moving with a velocity of 30m/s from east toward west hit another body from West moving toward east whose mass is 70kg . If the two bodies bounce back from the directions where they were coming from after collision with a final velocity 10m/s and 8 m/s respectively . Calculate the initial velocity of the body from West .

(A) 30.86m/s (B) 30m/s (C) 40m/s (D) 40.86m/s

Let the body from east direction mass be

M1 and it’s velocities

Initial velocity u1

Final velocity v1

And that of the body from West direction

Mass be M 2 and it’s velocities

Initial velocity u2

Final velocity v2

M1= 80kg

u1 = 30m/s

v1 = 10m/s

M2 = 70kg

u2= ?

v2=8m/s

By conservative momentum

M1U1 +M2U2 = M1V1 + M2V2

80x30 - 70xU2 = 80x10– 70x8

2400-70U2 = 800-560

-70u2 = 240-2400

-70u2 /-70= - 2160/-70

u2 = 30.86m/s

18. A bicycle rider of mass 85kg,started from rest and accelerate uniformly pass a circular road of radius 8m with a uniform acceleration 2m/s² in 10seconds .Find the coefficient of friction between the road and the tire .

(A) 5 (B)6 (C) 10 (D) 3

Solution

Mass m = 85kg

Radius, r= 8m

Acceleration = 2m/s²

Time ,t= 10seconds

Velocity ,v = u+at

U=0

V= at = 2x10 = 20m/s

Centripetal force F is given by the formula

F= MV²/r

F= 85 x 20²/8

F=4250N

REACTION R=85x 10 =850N

Coefficient of friction μ =F/R

μ = 4250N/850N = 5

19. A body of mass 15kg incline at an angle 30° to the horizontal . Another body of mass 20kg hung downward with help of a string connected from inclined 15kg mass making the two bodies to be at equilibrium .find the acceleration of the system if the string cut and tension on the string .

(A) 7.86ms⁻² and 192.9N (B)7.68m/s⁻² (C)8.86ms⁻² and 192.9N (D) 6.87m/s⁻² and 189.2N

Solution

The equation for the mass on the inclined plane takes into consideration the angle of inclination .The parallel component of the mass is mgsinθ and that of horizontal component Is mgcosθ

T-mgsinθ=ma

T-15x10sin30=15a

T-75==15a ------- (1)

For 20kg mass mg -T =ma

20x10-T = 20a

200-T=20a----------( 2)

ADD EQUATION 1 AND 2

T-75+ 200-T = 15a +20a

275/35=35a/35

7.86ms⁻²

TENSION ON THE STRING

Substitute the value of acceleration a

In equation (1)

T-75 =15a

T-75=15x7.86ms⁻²

T=117.9+75=192.9N

19 .A body of mass 35kg is moving with initial velocity 10 m/s.If a force of 20 N act on it in 2seconds . calculate the final velocity of the body

(A)11.14m/s (B) 11.15m/s (C) 12.58m/s

(D) 10.14m/s

Solution

Mass,M= 35kg

Initial velocity u= 10m/s

Force ,F= 20N

Time ,t= 2seconds

Final velocity v=?

F=ma

a= v-u/a

F= m(v-u/t)

20= 35(v-10/2)

20/35 x 2 = v-10

1.14=v-10

11.14m/s =v

20. Calculate the instantaneous velocity at 15seconds of an object projected to the space undergoing uniformly accelerated motion if the position is given by the equation

S = 10t² + 25t

Solution

Differentiating s=10t² + 25t

We obtain instantaneous velocity

ds/dt =2( 10t²⁻¹ )+ 25t¹⁻¹

= 20t¹ +25t⁰

For t =15secons

ds/dt =20x15 + 25 =325m/s

21.on top of a spiral spring of force constant 300Nm⁻¹ Is placed a mass of 0.02kg .if the spring Is compressed downwards by a length of 0.02m .IF the spring Is compressed downward by a length of 0.1m and then release calculate the height to which the mass Is projected .

(g=10m/s²)

(A) 7.2M (B) 6.7 M (C) 5.6M (D) 7.5 M

Solution

The work done In compressing he string Is equal to potential energy stored In the string .

1/2 ke²=mgh

Force constant k = 300Nm⁻¹

Extension e =0.1m

Mass m = 0.02kg

1/2 x300x0.1²= 0.02x10xh

1.5 /0.2= 0.2h/0.2

H=7.5 m

22. A wheel and axle is used to raise a load of 600N by the application of an effort of 450N .If the radii of the wheel and the axle are 0.4cm and 0.2cm respectively ,the efficiency of the machine is ?

(A) 66.67% (B) 66.60% (C) 70.50%

(D) 76.53%

Solution

Load ,L = 600N

Effort ,E = 450N

Radius of the wheel ,R= 0.4cm

Radius of axle ,r= 0.2cm

Efficiency ε= L/E X r/R X100%

ε=600/450X 0.2/0.4 X 100%

=66.67%

23.Which of this statement define elastic collision

(A) Kinetic energy equals potential energy before and after collision (B) Total Kinetic and potential energies of the bodies are conserved (C) Total kinetic energy and momentum are conserved.

(D)Only potential energy is conserved

Answer C

Hint :Collision is said to be elastic if the total kinetic energy before collision is equal to total kinetic energy after collision . Therefore kinetic energy of the bodies in collision and that of their momentum are conserved.

24.The relationship between the velocities of two colliding bodies after collision v1 and v2 and their velocities u1 and u2 before collision is given by …………..

(A) Newton’s first law motion

(B) Newton’s second law of motion

(D) Newton’s law of restitution

Answer D

Hint: Newton’s law of restitution is a law that relate velocities of two bodies after collision and before collision.

Whenever two bodies collide, relative velocity after collision =-e (relative before collision)

Therefore

V1-V2 = -e(u1-u2)

e----- denote coefficient of restitution.

25.choose the correct Kepler’s period law. where T and R are period respectively the period of revolution of the planet about the sun and distance, S of the planet from the sun?

(A) T² ~ R²

(B)T²~1/R²

(D) T² ~ 1/R³

Where symbol (~ ) represent proportionality sign.

Answer C

Hint

Kepler’s period law states that the square of the period of revolution of any planet about the sun is proportional to the cube of the planets' mean distance from the sun .

26 . A satellite is said to be in a packing orbit if ?

(A) It’s period is twice the period of the Earth as it turns it’s own orbit

(B) It’s period is exactly equal to the period of the Earth as it turns it’s own orbit .

(D) It’s period is a factor of two that of the period of the Earth.

Answer B

Hint For a satellite to be at it’s packing orbit it’s period must be same with that of Earth.

27.The reading of faulty Mercury in glass thermometer ice and steam point are 0.5° and 99.8° respectively. When used to measure the temperature of a medium,it recorded 20.5°C . Determine the correct temperature of the medium .

(A) 25.22°C

(B) 25.20°C

(D) 35.22°C

Solution

The fixed. Point and recorded reading of the faculty calibrated thermometer is corresponded with a correct calibrated Celsius thermometer as follows.

Let θ be the correct temperature when the faulty thermometer reads 20.5°C

Comparing the faculty thermometer with accurate thermometer readings gives

20.5°C-0.5°C/99.8°C-20.5°C=θ-0 °C/100°C-0°C

20/79.3 = θ/100

Cross multiply

20x100=79.3xθ

2000/79.3 =θ

θ= 25.22°C

Error in the reading of faulty Mercury thermometer is :

θ-20.5 = 25.22-20.5=4.72°C

28.A resistance thermometer has a resistance of 25ohm at 0⁰ and 70ohm at 100°C .If it records 60ohm in a certain medium, calculate the temperature of the medium.

(A) 70.80°C

(B) 77.78°C

(D) 77.78°C

Solution

60ohm-25ohm/70ohm-25ohm

= θ°C-0°C/100°C-0°C

35ohm/45ohms =θ/100

3500/45 = 45θ /45

77.78°C =θ

You can also apply this formula to solve the problem and still get the same answer.

θ=R-R₀/R₁₀₀ - R₀ x 100°C

R= 60ohm

R₀= 25ohm

R₁₀₀= 70ohm

θ=60-25/70-25 x100

θ= 77.78°C

Application of formula is easiest way to solve it . Isn’t it?

29.Which of the following formulas shows a correct formula for the velocity of a geosynchronous satellite of mass, m moving in a circular orbit around the Earth at a steady or constant velocity v and height h above the Earth’s surface.

(A) v= (GMₑ/Rₑ + h)½

(B) v= (Rₑ+h/GMₑ)½

(D) v= (GMₑ/Rₑ+h)

Answer A

30.An object weighs 15N in air and 8 N when completely submerged in liquid of density 650 kgm⁻³.what Is the volume of the solid (g=9.8ms⁻²)

(A)859m³

(B)910m³

(D)810m³

Solution

Weight of the object in are =15N

Weight of the object when completely immersed in water

= 8N

Density of the object = 650kgm⁻³

Acceleration due to gravity= 9.8ms⁻²

First step

Calculate the upthrust on the object.

Upthrust =loss In weight

UPTHRUST=(15-8)N =7N

The volume you are to find In this context Is the volume o f the solid object.

Upthrust = Volume x density x acceleration due to gravity

7N= V x 650kgm⁻³ x 9.8ms⁻²

Making the volume v the subject of formula gives

V= 650kgm⁻³ x 9.8ms⁻²/7N

V= 910m³

32.All except one is incorrect

(A) pressure at any point at the level or depth within the same liquid is the same .

(B) Pressure at any point in a liquid acts equally in all directions.

(D) pressure in different liquids at the same depth varies directly with density.

Answer C

33.A cell of internal resistance supplies current to a 8.0ohm resistor and it’s efficiency 60% . Determine the value of internal resistance ,r.

(A)5.33ohm

(B)8.34ohm

(D)25.07ohm

Solution

Efficiency E=60%

External resistance,R =8.0ohm

E= R/(R+r) x 100%

Making the internal resistance r

r = R(100-E)/E

r =8(100-60)/60

r = 5.33ohm

Alternative method

Imputing your values directly in the equation.

E= R/(R+r) X 100%

60= 8/(8+r) x 100

60 /100 = 8/(8+r)

0.6=8/(8+r)

0.6(8+r) =8

4.8+0.6r=8

0.6r = 8 – 4.8

0.6r/0.6 = 3.2/0.6

r= 5.33ohm

You can use any method you wish to use and you will arrive at the same answer.

34.An object is projected horizontally from the top of a story building with a velocity of 25ms⁻¹

And lands on the ground level at point 125m away from the base of the story building. What Is the height of the story building . (g=9.8ms⁻²)

(A)122.5m

(B)202.8m

(D)136.5m

Solution

Velocity Vₓ =25ms⁻²

Distance Sₓ =125m

Where X denote horizontal

Time taken by the object to hit the ground Is given as

t=125/35 =5 seconds

using the equation o f the motion to find the height

h=ut +1/2gt²

hint Initial velocity =0 not 7m/s

h=0x5 +1/2x10x5²

h=122.5m

35.During Olympic game, a boy projected a javelin horizontally with velocity 20m/s .The javelin Maximum height reached before returning to the level of projection was 650m.Calculate the time of flight.

Solution

U= 20m/s

H=65m

G=10m/s²

V=0

Hint

(I)when ever body is going up final velocity is zero and vise versa

(I) time of flight can be obtain multiple time taken to reach maximum height by 2

t= 2u/g

t=2x20/10 = 4seconds.

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