Physics questions and answers

Physics Questions and Answers

If you want to learn more about the nature and properties of matter and energy or you're simply preparing for a Physics exam, these Physics past questions and answers are ideal for you.

106.

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

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

107.

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

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

108.

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

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%

109.

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

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


 


110.

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

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.