JAMB Physics Past Questions & Answers - Page 8

The work done against the gravitational force is calculated using the formula W = mgh, where m is the mass of the object, g is the acceleration due to gravity (approximately $9.8 ms^2$ on Earth), and h is the height. Substituting the given values, we get W = 3.0 kg * $9.8 m/s^2$ * 240 m = 7.2 kJ. Therefore, the correct answer is '7.2 kJ'.

Let mb=mass of empty bottle,

$m_w$=mass of water only and

$m_a$= mass of alcohol only
 

given; $m_b$=19g



$m_b$ + $m_w$ = 66g

$m_b$ + $m_a$ = ?

R.d=0.8

R.d=mass of alcohol

⇒$\frac{mass of alcohol}{mass of equal volume of water}$

⇒ mass of equal volume of water = $m_w$=66-19=47g

⇒0.8 = $\frac{m_a}{47}$

⇒$m_a$=0.8×47 =37.6g

; $m_b$ + $m_a$ = 19+37.6=56.6g

f= -15cm (diverging mirror);  u=35cm; v=?

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

⇒ $\frac{1}{-15}$ = $\frac{1}{35}$ + $\frac{1}{v}$

⇒ $\frac{1}{15}$ - $\frac{1}{35}$ = $\frac{1}{v}$

⇒ $\frac{-2}{21}$ = $\frac{1}{v}$

 = $\frac{-21}{2}$ = -10.5cm

∴The image is 10.5cm behind the mirror

T=800N; I=50cm=0.5m,

m=10g=0.01kg

fundamental freq: $f_o$ =?

$f_o$ = $\frac{1}{21}√{T}{μ}$

μ =$\frac{m}{1}$=$\frac{0.01}{0.5}$

⇒ $f_o$ =$\frac{1}{2×0.5}$√$\frac{800}{0.02}$

                  $f_o$  ⇒√ 40,000

⇒1st overtone = 2$f_o$ =2×200 = 400Hz

⇒2nd overtone =3$f_o$ =3×200=600Hz

∴3rd over tone= 4$f_o$  =4×200=800Hz

 

For the combination in series;

⇒R1 = 35kΩ + 40kΩ = 75kΩ

R is combined with 75kΩ in parallel to give 25kΩ

=  $\frac{1}{R_eq}$ = $\frac{1}{R}$ + $\frac{1}{R}$ 

=   $\frac{1}{25}$     = $\frac{1}{R}$ + $\frac{1}{75}$ 

=  $\frac{1}{25}$     - $\frac{1}{75}$ + $\frac{1}{R}$ 

=   $\frac{3-1}{75}$ = $\frac{1}{R}$ 

=   $\frac{2}{75}$   = $\frac{1}{R}$ 

=   $\frac{75}{2}$  = R

;      R = 37.5k Ω