# Jamb Physics Questions 2019

## INTRODUCTION

Welcome to poschools in this series I will be revealing likely jamb physics questions to expect in your exam. This jamb physics question will be calculation-based.

I have released jamb physics theory question to expect in 2019 click here

### Jamb Physics Questions 1

1. A 500kg car which was initially at rest travelled with an acceleration of $5m/s^{2}$ , it kinetic energy after 4 seconds was?

(a)$10^{5}J$ (b)$2.5×10^{3}J$ (c)$2×10^{3}J (d)5×10^{3}J$

###### solution
• The question asked for kinetic energy. The first step in solving this jamb physics question is for you to know the formula for finding kinetic energy:

K.E = $\frac{1}{2}mv^{2}$

• The next step is to write down the parameters given in the queston

mass = 500kg

acceleration = $5m/s^{2}$

time  = 4s

u = 0 (From the question, the car was initially at rest thats why u is 0)

we need v (final velocity) so that we can find the kinetic energy of the car. But v wasn’t given.

Recall: v = u + at

making v = 0 + 5*4

v  = 20 m/s

Now we can find the kinetic energy:

K.E = $\frac{1}{2} * 500 * 20^{2}$

K.E  = 0.5*500*400

K.E = $10^{5}J$

Ans : A

### Jamb Physics Questions 2

2. Calculate the resistance of the filament of a lamp rated 240V, 40W.

(a)240 ohms (b)360 ohms (c)720 ohms (d)1440 ohms

###### solution

There are 3 formula’s for calculating electrical power. The formula you are to use depends wholly on the parameters you are given.

Power  = IV…..eq(i)

Power = $I^{2}R$…..eq(ii)

Power = $\frac{V^{2}}{R}$……eq(iii)

Lets write down the parameter given in the question as well as its unknown:

Voltage, V = 240v

Power, P = 40w

Resistance, R = ?

Now from the parameters given, eq….(iii) can be used to solve for R directly, since the equation contains all the parameters as well as the unknown.

Using eq(iii)

P = $\frac{V^{2}}{R}$

40 = $\frac{240^{2}}{R}$

Making R the subject of the formula;

R = $\frac{240^{2}}{40}$

R = 1440 ohms

Ans: D

note: The other equations can also be used. But you will have to find the current first using P = IV

### Jamb Physics Questions 3

An aircraft lands on the run-way with a velocity of 100m/s and decelerates 20$m/s^{2}$ to a velocity of 40 m/s. Calculate the distance travelled on the run-way as it comes to rest.

(a)210 m (b)120 m (c)240 m (d) 260 m

###### solution

First let’s write out the parameters given and the unknown:

Initial velocity, u = 100 m/s

deceleration = 20 $m/s^{2}$

final velocity, v  =40 m/s

distance, s = ?

From the parameters given, it can be seen that the equation of motion:

$v^{2}=u^{2}+2as$ can be used to solve for s, since it contains all the parameters.

Before we proceed, i will like you to notice that deceleration was given not acceleration.

Deceleration is the negative of acceleration.

Therefore, for this question,

a = -20$m/s^{2}$

Putting the parameters into the equation:

$40^{2}=100^{2}-2*20*s$

1600 = 10000 – 40s

1600-10000 = -40s

-8400 = -40s

s = $\frac{-8400}{-40}$

s = 210 m

Ans: A

### Jamb Physics Question 4

4. When taking a penalty kick, a footballer applies a force of 30.0N for a period of 0.05s. If the mass of the ball is 0.075kg. Calculate the speed with which the ball moves off.

(a) 4.5 m/s (b) 11.25 m/s (c) 20.00 m/s (d)45.00 m/s

###### solution

As usual, Lets write the parameter down

Force, F  = 30.0N

time, t = 0.05s

mass = 0.075 kg

initial velocity, u = 0 (This is because the ball was initially at rest before the player took the penalty)

final velocity, v =  0

This problem can be solved from the equation gotten from newtons second law:

$F = \frac{m(v-u)}{t}$

Subtituting the parameters into the equation:

30  = $F = \frac{0.075(v-0)}{0.05}$

cross-multiplication gives:

30*0.05 = 0.075*v

v = $\frac{30*0.05}{0.075}$

v = 20 m/s

Ans: C

### Jamb Physics Questions 5

5. A cell of e.m.f 1.5 V and internal resistance 2.5 ohms is connected in series with an ammeter of resistance 7 ohms. what is the current in the circuit?

(a) 6.5 A (b) 3.5 A (c) 0.16 A (d) 0.3 A

###### solution

Lets write down the parameter as usual:

E  = 1.5v

internal resistance, r = 2.5 ohms

external resistance, R  = 7 ohms

Current, I = ?

The formula to solve this is:

I = $\frac{E}{R+r}$

Substituting

I = $\frac{1.5}{7+2.5}$

I = $\frac{1.5}{9.5}$

I = 0.157 A

Ans: C

### Jamb Physics Question 6

A rectangular tank contains water to a depth of 2.0m. If the base is 4m x 3m, calculate the force on the base. Density of water = $10^{3}kg/m^{3}$, g = $10m/s^{2}$

(a)$2.4×10^{5}N$

(b)$2.4×10^{4}N$

(a)$2.0×10^{4}N$

(a)$1.7×10^{2}N$

###### solution

Writing out the parameters:

depth of water, h = 2 m

dimension of base = 4m x 3m

density of water = $10^{3}kg/m^{3}$

g = $10m/s^{2}$

Force = ?

The question might be quite difficult as there is no direct formula for this problem. We are asked to calculate the force at the base of the tank.

It is important to remember this relation Pressure = Force/Area:

from that relation, Force  = Pressure  x Area.

The pressure at the base can be calculated using :

P = $h\rho g$

P = 2*1000*10

P = $2×10^{4}N/m^{2}$

Force = P x A

A = 4 x 3 = 12 $m^{2}$

Force = $2×10^{4}$ * 12

Force = $2.4×10^{5}N$

Ans: A

### Jamb Physics Question 6

6. The area of the effort and load pistons of a hydraulic press are 0.5$m^{2}$ and 5$m^{2}$. If a force f1 of 100 N is applied on the effort piston, the force f2 on the load is:

(a) 10.0 N (b) 100.0N (c) 500.0 N (d)1000.0 N

###### solution

If you know the principle underlying the principle of the hydraulic press, you should be able to solve this problem by mere looking at it.

Now, the Area of the load piston is 10 times the effort area, therefore, the force at the load should be 10 times the force at the effort.

since the force at the effort is 100 N, therefore, the force at the load should be 10 x 100 = 1000N

Alternatively;

The formula to be used is:

$\frac{F_{E}}{A_{E}} = \frac{F_{L}}{A_{L}}$

Where :

$F_{E}$ is the force applied at the effort piston = 100N

$A_{E}$ is the area of the effort piston = 0.5$m^{2}$

$F_{L}$ is the force applied at the load piston = ?

$A_{L}$ is the area of the load piston = 5$m^{2}$

Substituting the parameters:

$\frac{100}{0.5} = \frac{F_{L}}{5}$

making $F_{L}$ the subject of the formula gives:

$F_{L}$ = 1000N

Ans: D

### Jamb Physics Question 7

7. The speed of light in vacuum is $3 x 10^{8}m/s$. If the refractive index of a transparent liquid is 4/3, then the speed of light in the liquid is?

(a) $0.44 x 10^{8}m/s$ (b) $2.25 x 10^{8}m/s$ (c) $3 x 10^{8}m/s$ (d) $4 x 10^{8}m/s$

###### solution

To solve this problem, you need know the formula for refractive index when considering light:

refractive index, n = $\frac{Sin(i)}{Sin(r)}$….eq(i)

refractive index, n = $\frac{velocity in vacuum}{velocity in medium}$….eq(ii)

writing out the parameters

velocity of light in vacuum, c = $3 x 10^{8}m/s$

velocity of light in medium, v = ?

refractive index ,n = 4/3

using eq…(ii)

$\frac{4}{3} = \frac{3 x 10^{8}}{v}$

solving for v gives:

v = $2.25 x 10^{8}m/s$

Ans: B

### Jamb Physics Question 8

8.A dentist obtains a linear magnification of 4 of a hole in a tooth by placing a concave mirror at a distance of 2.0 cm from the tooth. The radius of curvature of the mirror is

(a) 5.3 cm (b) 3.2 cm (c) 2.7 cm (d) 1.6 cm

###### solution

writing out the parameters:

magnification, m = 4

Object distance = 2.0 cm ( This is the distance between the mirror and the tooth)

we need to get the focal length of the mirror first before finding the radius of curvature.

But the focal length can be gotten from u(image distanc) and v(object distance)

remember: magnification = $\frac{image distance or image height}{object distance or object height}$

therefore, 4 = $\frac{v}{2}$

solving for v gives: v = 4*2 = 8 cm

Now, we can now calculate for focal length using:

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

substituting gives

$\frac{1}{f}=\frac{1}{2}+\frac{1}{8}$

$\frac{1}{f}=\frac{5}{8}$

therefore, f = $\frac{8}{5}$

remember, radius of curvature, R = 2F

R = 2 x $\frac{8}{5}$

R = $\frac{16}{5}$

R = 3.2 cm

Ans: B

### Jamb Physics Question 9

9. Two cells, each of e.m.f 1.5v and an internal resistance of 2 ohms are connected in parallel. Calculate the current flowing when the cells are connected to a 1 ohm resistors

(a) 0.75 A (b) 1.5 A (c) 0.5 A (d) 1.0 A

###### solution

Writing out the parameters:

E.m.f of each cell, E = 1.5 v

internal resistance, r = 2 ohms

External resistance, R = 1 ohm

Current, I = ?

Note, the two cells are connected in parallel, so it is important for us to find their equivalent one-cell value

Equivalent one-cell value simple means what is their resultant voltage and their resultant internal resistance.

Note, when cells are connected in parallel, the equivalent emf is the emf of just a single cell. In this case the equivalent e.m.f is 1.5 v

the equivalent resistance can be gotten by considering resistance in parallel

$\frac{1}{r_{equ}} = \frac{1}{r_{1}} + \frac{1}{r_{2}}$

subtituting the value of internal resistance into the equation gives:

$\frac{1}{r_{equ}} = \frac{1}{2} + \frac{1}{2}$

$r_{equ} = 1$ohm

Therefore, we can now use this formula $I = \frac{E}{R+r}$ to find the current.

Using E as the equivalent e.m.f which is 1.5 v

r as the equivalent internal resistance which is 1 ohm

R as the external resistance which is 1 ohm

substituting gives:

$I = \frac{1.5}{1+1}$

$I = \frac{1.5}{2}$

I = 0.75A

Ans: A

### Jamb Physics Question 10

10. What is the cost of running five 50W lamps and four 100W lamps for 10 hours if electrical energy cost 2 kobo per kWh?

(a)#0.65 (b)#0.13 (c)#3.90 (d)#39.00

###### solution

Lets first calculate the total power consume:

five 50W bulbs gives 5 x 50W = 250W

Four 100W bulbs gives 4 x 100W = 400W

Total power consumed = 250 + 400 = 650W

In kilowatt, total power = 0.65kW

Energy consumed in kwHr becomes Power x time

Energy consumed = 0.65 x 10 = 6.5kwhr

Since 1 kwhr cost 2 kobo, therefore, 6.5kwhr will cost:

6.5 * 2 = 13 kobo

converting to naira gives 13/100 = #0.13

Ans: B

### Jamb Physics Question 11

11. Two forces, each 5N acting at a point at an angle of 60º with each other. Their resultant force is

(a)50N (b)8.66N (c)25N (d)10N

###### solution

Total force can be calculated using this:

$F_{R}^{2} = F_{1}^{2} + F_{2}^{2} + 2F_{1}F_{2}cosθ$

writing this formula:

$F_{1}$ = 5N

$F_{2}$ = 5N

θ = 60°

Substituting in the formula :

$F_{R}^{2} = 5^{2} + 5^{2} + 2*5*5cos60$

$F_{R}^{2} = 25 + 25 + 25$

$F_{R}^{2} = 75$

$F_{R} = \sqrt{75}$

$F_{R} = 7.66N$

### Jamb Physics Question 12

12. A stone thrown vertically upward with an initial speed of 20 m/s. The maximum height reached is (g = 10$m/s^{2}$.

(a)40m (b)60m (c)20m (d)100m

###### solution

Writing out the parameters

Initial velocity, u = 20m/s

At maximum height, final velocity, v = 0

maximum height, h = ?

acceleration due to gravity,g = -10$m/s^{2}$ (this is because the stone was thrown upward i.e against gravity)

Using this formula;

$v^{2} = u^{2}+2gh$

Substituting into the equation;

$0^{2} = 20^{2}-2*10*h$

$-400 = -20h$

h = 20 m

Ans: C

### Jamb Physics Question 13

13. An object of height 5 cm is placed 30 cm from a convex length of focal length 10 cm. the height of the image is.

(a) 2.5 cm (b) 5.0 cm (c) 10.0 cm (d) 3.0 cm (e) 6.0 cm

###### solution

Writing out the parmeters:

object height = 5 cm

object distance, u = 30 cm

focal length = 10 cm

Image height = ?

We need to calculate the image distance, v using the mirror formula

$\frac{1}{f} = \frac{1}{u}+\frac{1}{v}$

substituting the parameters into the formula

$\frac{1}{10} = \frac{1}{30}+\frac{1}{v}$

collecting like terms:

$\frac{1}{10} -\frac{1}{30}=\frac{1}{v}$

simplifying further

$\frac{1}{v} = \frac{2}{30}$

simplifying further gives:

v = 30/2 = 15 cm

v = 15 cm

remember, magnification = $\frac{image height}{object height}=\frac{image distance}{object distance}$

substituting the parameters:

$\frac{image height}{5}=\frac{15}{30}$

solving for image height gives:

image height = 2.5 cm

Ans: A

### Jamb Physics Question 14

14. A pin is positioned at 21 cm from concave mirror of radius of curvature 28 cm. The magnification of the image formed is

(a) 2.0 (b) 1.5 (c) 4.0 (d) 3.0

###### solution

writing the out the parameters:

object distance, u = 21 cm

radius of curvature, R = 28 cm

we can get the focal length from the radius of curvature using:

R = 2f

f = R/2

f = 28/2

f = 14 cm

we can now find the image distance, v, using the mirror formula

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Substituting the parameters

$\frac{1}{14}=\frac{1}{21}+\frac{1}{v}$

$\frac{1}{14}-\frac{1}{21}=\frac{1}{v}$

Using L.C.M of 42 at the left hand side of the equation

evaluating further gives:

$\frac{1}{42}=\frac{1}{v}$

therefore:

v = 42 cm

Magnification = v/u

M =42/21 = 2.0

Ans: A

### Jamb Physics Question 15

15. The echo from a sound wave sent out from a ship towards the seabed is received on the ship 1.6s later. Taking the velocity of sound in water to be 1.5km/s, the depth of the sea is

(a) 1.6 km (b) 2.0 km (c) 1.2 km (d) 2.4 km

###### solution

Remember for echo the formula is:

V = $\frac{2d}{t}$

where v is velocity = 1.5 km/s

t is time = 1.6s

d is distance = ?

Substituting into the formula:

1.5 = $\frac{2d}{1.6}$

2d = 1.5*1.6

d = 1.5*0.8

d = 1.2km

Ans: C

### Jamb Physics Question 16

16. A wave can be represented by the equation Y = 0.20sin[0.4π(x-60t)] where x and y are in cm and  t is in seconds. The velocity of the wave is?

(a) 180 cm/s (b) 150 m/s (c) 120 cm/s (d) 240 cm/s

###### solution

Solving wave problem of this nature requires you to compare it with the standard wave equation which is:

Y = Asin(kx-ωt)

where k is the wave number given by, k = $\frac{2π}{λ}$

Where λ is the wavelength

where ω is the angular velocity, given by, ω = $\frac{2π}{T}$

Where T is the period

Lets rewrite the wave equation to look like the standard wave equation

Y = 0.20sin(0.4πx – 24πt)…eq(i)

comparing this with:

Y = Asin(kx-ωt)…eq(ii)

Careful observation of the two equations shows that:

A = 0.20 cm

kx = 0.4πx……eq(iii)

ωt = 24πt …….eq(iv)

Let’s consider eq(iii)

kx = 0.4πx

x will cancel out so that we will have:

k = 0.4π

Remember that k = $\frac{2π}{λ}$

Therefore;

0.4π = $\frac{2π}{λ}$

Making λ = 4/0.2 = 20 cm

Now let’s consider eq(iv):

ωt = 24πt

the t will cancel out first

making ω = 24π

let’s substitute ω as $\frac{2π}{T}$

Therefore;

$\frac{2π}{T}$ = 24π

solving for T gives

T = 2/24 = 0.08333333s

Remember wave speed, v = $\frac{λ}{T}$

since we have λ as 20 cm

and T as 0.08333

v = 20/0.08333

v = 240 cm/s

Ans: D

### Jamb Physics Question 17

17. A cell emf 1.5v amd internal resistance of 2.5Ω is connected in series with an ammeter of resistance 0.5Ω and a resistance 7.0Ω; calculate the current in the circuit

(a) 0.15 A (b) 0.20 A (c) 0.60 A (d) 3.0 A

###### solution

Writing out the parameter

emf, E = 1.5v

internal resistance, r = 2.5Ω

Ammeter resistance = 0.5Ω

external resistance = 7.0Ω

since the system is connected in series, the total external resistance becomes the sum of the ammeter resistance and the external resistance

Therefore:

R = 0.5 +  7 = 7.5Ω

using I = $\frac{E}{R+r}$

I =

$\frac{1.5}{7.5+2.5}$

I = 0.15 A

Ans: A

### Jamb Physics Question 18

18. Two radioactive elements A and B initially have the same mass. If their half-life is 10 and 5 years respectively, the ratio of their mass (A:B) after 20 years is:

(a) 2:1 (b) 4:1 (c) 1:2 (d)1:4

###### solution

Let both element starts with a mass of m.

Therefore, for element A with a half-life of 10 years;

m kg——after 10 years——–m/2 kg

m/2 kg—–after another 10 years———-m/4 kg

Simply means foe element A, after 20 years, m/4 kg was lefr

For element B:

m kg——–after 5 years——m/skg

m/2 kg——after another 5 years——–m/4 kg

m/4 kg———after another 5 years———m/8

m/8 kg——after another 5 yrs—– m/16kg

ration of their masses (A:B) = [latexx]\frac{m/4}{m/16}[/latex]

evaluating that gives 1:4

Ans: D

### Jamb Physics Question 19

19. A transparent rectangular block 5.0 cm thick is placed on a black dot. The dot when viewed from above is seen 3.0 cm from the top of the block. Calculate the refractive index of the material

(a) 2/5 (b) 3/5 (c) 5/2 (d) 5/3

###### solution

Refractive index, n =$\frac{real depth}{apparent depth}$

Real depth = 5.0 cm

Apparent depth = 3.0 cm

n = $\frac{5}{3}$

Ans: d

### Jamb Physics Question 20

20. A car moving with a speed of 90 km/h was brought uniformly to rest by the application of braked in 10s. How far did the ar travel after the brakes where applied?

(a) 125m (b) 150 m (c)250 m (d) 15 km

###### solution

Writing out the parameters:

Initial velocity = 90 km/h

final velocity = 0  ( since the car was brought to rest)

time = 10 s

distance, s = ?

The equation of motion formula connecting the parameters is given by:

$s= \frac{v+u}{2}$*t[/latex]

Careful observation shows that the initial velocity is given in km/h it is necessary for it to be converted to m/s

initial velocity, u = $\frac{90 * 1000}{3600}$

evaluating that gives: 25 m/

substituting into the formula

$s=\frac{25+0}{2}*10$

s = 125 m

Ans: A

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