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JAMB Physics Past Questions & Answers - Page 148

736.

The radioisotope 23592U decays by emitting two alpha particles, three beta particles and a gamma ray. What is the mass and atomic number of the resulting element?

A.

215 and 88

B.

91 and 227

C.

227 and 91

D.

92 and 238

Correct answer is C

Alpha radiation consists of helium- He42

2 Alpha decay --> 2  X  He42 = Element84 

 

- Beta particles are high energy electrons-  e01

3 Beta particles --> 3  X  e01 = Element03

- Gamma rays are waves of electromagnetic energy, or photons.

the mass and atomic number of the resulting element = U23592 - Elements( 84 +  03 )

--> Element22791 

 

737.

An object is weighed at different locations on the Earth. What will be the right observation?

A.

Both the mass and weight vary

B.

The weight is constant while the mass varies

C.

The mass is constant while the weight varies

D.

Both the mass and weight are constant​

Correct answer is C

No explanation has been provided for this answer.

738.

Which of the diagrams above shows the position at which a small ball-bearing is in neutral equilibrium?

A.

A

B.

B

C.

C

D.

D

Correct answer is D

Neutral Equilibrium: If the body neither returns to its original position nor increases its displacement further, it will simply adopt its new position.

Examples of neutral equilibrium

  • Ball moving on the ground.
  • Motion of sphere.
  • A roller.
  • A pencil lying horizontally.
  • An egg lying horizontally on a flat surface.
739.

A progressive wave is represented by y=10Sin(1000πtπx34) . two layers of the wave separated by 153cm have a phase difference of?

A.

2700

B.

450

C.

900

D.

1800

E.

No Correct Option

Correct answer is C

Comparing y=10Sin(1000πtπx34) with y = A sin (ωt - ϕ)

where ϕ = \frac{2\pi x}{λ}

comparing:  \frac{2\pi x}{λ} = \frac{\pi x}{34}

λ = 34 \times 2 = 68

: ϕ =  \frac{2\pi x}{λ}\frac{2\pi \times 153}{68}

ϕ = 4.5\pi rad = 810° or 90°

since sin 810° or 90° = 1

740.

A radioactive substance has a half life of 20 days. What fraction of the original radioactive nuclei will remain after 80 days?

A.

\frac{1}{32}

B.

\frac{1}{16}

C.

\frac{1}{8}

D.

\frac{1}{4}

Correct answer is B

80 days = 4 half lives (i.e 4 x 20dqs = 80dq)

But the fraction always left by any radioactive material after 4 half lives is

\left ( \frac{1}{2} \right)^4 = \frac{1}{16}