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JAMB Physics Past Questions & Answers - Page 14

66.

The terminals of a battery of emf 24.0 V and internal resistance of 1.0 Ω is connected to an external resistor 5.0 Ω. Find the terminal p.d.

A.

18.0V

B.

12.0V

C.

16.0V

D.

20.0V

View answer & explanation

Correct answer is D

ε = 24.0 V, r=1Ω, R=5.0Ω

terminal p·d = ?

ε = IR+Ir = I(R+r)

⇒ IR+Ir = I(R+r)

⇒ I = $\frac{24}{5+1}$

⇒ I = $\frac{24}{6}$

⇒ I = 4 A

terminal p·d = 4 × 5

terminal p·d = 20.0 V

67.

Calculate the absolute pressure at the bottom of a lake at a depth of 32.8 m. Assume the density of the water is 1 x 10^3 kgm-3 and the air above is at a pressure of 101.3 kPa.

[Take g = 9.8 ms-2]

A.

422.7

B.

220.14

C.

464.53

D.

321.74

View answer & explanation

Correct answer is A

h=32·8m, p= $1×10^3kgm^-3$

g=9·8 $ms^-2$ , $P_a$ =101·3KPa

$P_T$ = ?

P=hpg

⇒ $32·8×1×10^3 × 9·8$ =321440 $P_a$

⇒321·44KPa

⇒ $P_T$ =321.44+101.3

; $P_T$ =422·7KPa

68.

On a particular hot day, the temperature is 40°C and the partial pressure of water vapor in the air is 38.8 mmHg. What is the relative humidity?

A.

70

B.

62

C.

80

D.

42

View answer & explanation

Correct answer is A

Relative Humidity= $\frac {partial pressure of H2O}{saturated vapor pressure of H2O}$ ×100%

S.v.p. of water at 40°c = 55.3

⇒ Relative Humidity= $\frac{38.8}{55.3}$ ×100%

Relative Humidity= 70%

69.

A missile is launched with a speed of 75 ms-1 at an angle of 22° above the surface of a warship. Find the horizontal range achieved by the missile. Ignore the effects of air resistance.

[Take g = 10 ms-2]

A.

195 m

B.

271 m

C.

391 m

D.

136 m

View answer & explanation

Correct answer is C

U= 75 $ms^-1$ =22º, g= 10 $ms^-2$ , R=?

R= $\frac{U^2sin2\theta}{g}$

⇒R= $\frac{75^2×sin2(22)}{10}$

⇒R= $\frac{5625×Sin44}{10}$

⇒R= $\frac{5625×0.6947}{10}$

⇒R= $\frac{3907.69}{10}$

⇒ 391m

70.

Find the tension in the two cords shown in the figure above. Neglect the mass of the cords, and assume that the angle is 38° and the mass m is 220 kg

[Take g = 9.8 ms-2]

A.

$T_1$ = 2864 N, $T_2$ = 3612 N

B.

$T_1$ = 3612 N, $T_2$ = 2864 N

C.

$T_1$ = 3502 N, $T_2$ = 2760 N

D.

$T_1$ = 2760 N, $T_2$ = 3502 N

View answer & explanation

Correct answer is C

W = mg = 220 x 9.8 = 2156 N

⇒Sin 38º = $\frac{2156}{T_1}$

⇒ $T_1$ = $\frac{2156}{Sin 38}$

⇒ $T_1$ = 3502 N

Cos 38º = $\frac{T_2}{T_1}$

⇒ $T_2$ = 3502 x Cos 38º

⇒ $T_2$ = 2760 N

; $T_1$ = 3502 N, $T_2$ = 2760 N.

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