JAMB Physics Past Questions & Answers - Page 379

1,891.

Two liquids L₁ and L₂ are contained in a U - tube. The height and the density of L₁ are 8 cm and 10³ kgm^-3 respectively. If the density of L₂ is 800 kgm^-3, its height measured from the same level is

8 cm

10 cm

16 cm

12 cm

View answer & explanation

Correct answer is B

Generally, from Hares Experiment, density is inversely proportional to height
i.e h ∝1/d; → h₁d₁ = h₂d₂
∴ 8 x 10³ = h₂ x 800

∴ h₂ =	8 x 10³
	_800

= 10cm

1,892.

\( ^{235} _{92}U + ^1 _0n → ^{144} _{56}Ba + ^{90} _{36}Kr +2X \)
In the reaction above, X is

electron

proton

neutrino

neutron

View answer & explanation

Correct answer is D

235U + 1n → 144Ba + 90Kr + 2x9205636
In the reaction above, balancing the left with the right, show that X is neutron

1,893.

In the figure above, the work done by the force of 100N inclined at an angle of 60^o to the object dragged horizontally to a distance of 8m is

800J

600J

100J

400J

View answer & explanation

Correct answer is D

Since the force is inclined at an angle 60^o to the horizontal, the effective component of te force along the horizontal is given by F x cos 60^o
= 100 x 0.5 = 50N.
∴ work done = F x cos 60 x displacement
= 50 x 8.0m
= 400.0J

1,894.

If the system above is in equilibrium, the tension in the string Q is

11N

View answer & explanation

Correct answer is B

If the string is in equilibrium, the the tension in the string Q = 4N + 5N = 9N

1,895.

In the figure above, the coefficient of static friction is 0.5. If the tension in the string is T, the acceleration of the system in motion is

5 ms^-2

15 ms^-2

3 ms^-2

9 ms^-2

View answer & explanation

Correct answer is A

Normal reaction R = 3 kg x 10 = 30N
coefficient of friction , μ = 0.5
∴ acting frictional force = μR = 30 x 0.5
Pulling/sliding force F = 6kg x 10 = 60N.
∴ Net downward force = ma = 60 - 15.0
=> (3 + 6)a = 45
∴ 9a = 45
a = 5m/s2