Sabtu, 10 Mei 2014

Soal Fisika Tentang Alat Optik Beserta Pembahasannya

Diposting oleh Sofi di 02.26
Contoh Soal Fisika Tentang Bab Alat Optik Beserta Pembahasannya



1.Someone is only able to see objects clearly at the furthest distance of 1,5 meters. The lens strength of eye glasses which must be used is ... (medium)
a.      ½ dioptri
b.      2/3 dioptri
c.      -1/2 dioptri
d.      -2/3 dioptri
e.      -1 dioptri
Solution :
PR = 1,5 m / 150 cm
S’   = -PR
                 F = -150 cm
              (D)

2.      A Man observes his KTP  using loop with strength of 10 dioptries. If he has punctum proximum 30 cm and want to obtain the maximum angular magnification. The distance of KTP must be placed to the loop is... (medium)
a.      3,5 cm
b.      4,5 cm
c.      5,5 cm
d.      6,5 cm
e.      7,5 cm
Solution :
Pp= 30cm, f =  =
 = 4
 (E)

3.      A microscope is used to observe an object at distance of 1,5 cm. If the focal length of the objective and ocular lens each is 1 cm and 2,5 cm  and the eye is not in accomodation. The lenght of microscope is ...... (medium)
a.      3,5 cm
b.      4,5 cm
c.      5,5 cm
d.      6,5 cm
e.      7,5 cm
Solution :
Sob      = 1,5 cm
Fob      = 1 cm
Fok      = 2,5 cm
 3 cm
  
4.      Temperature of the Celsius and Fahrenheit scale show the same value in ... (medium)
a.      -40°
b.      -50°
c.      -60°
d.      -70°
e.      -80°
Solution :
          C =
            C=F jadi C =
                          
                        4C = -160 → C = -40 °C
                                   
5.      A hollow ball is made of bronze ( α x 10-6/°C). At temperature 0°C its radius is 1 m. If the ball is warmwd up to 80°C. The area incremen of the ball’s surface is ... (difficult)
a.      1.12 x 10-2
b.      1.15 x 10-2
c.      1.2 x 10-2
d.      1.5 x 10-2
e.      1.25 x 10-2
            Solution :
         
                  = (4
                  = (4
                  = 11520 x 10-6
                  = 1.15 x 10-2  (B)
6.      The water 100 grams in temperature 20°C mixed with the water 50 grams in the temperate 80°C. The mixture temperature of both water is .....( c water = 1 cal/g) (medium)
a.      40°C
b.      50°C
c.      60°C
d.      70°C
e.      80°C
Solution :
M1 = 100g, t1= 20°C
M2 = 100g, t2= 80°C

Qterima = m1. C (tc-t1)
          = 100.1 (tc-20)
          = 100 tc -2000
Qlepas = m1. C (t2-tc)
         = 50.1. (80-tc)
         = 4000 - 50tc
          Qterima = Qlepas
100 tc -2000 = 4000 - 50tc
          150 t= 6000
                 tc  = 40°C (A)

7.      A wall has the relative constant temperature of 25°C in outdoor air temperature of 18°C. The dissipation head in 3 hours because of head convection if the wall area 15m2 and h= 3.5 J/s m2K is ..... (medium)
a.      36,67 x 105 J
b.      37,69 x 105 J
c.      39,69 x 105 J
d.      36,69 x 105 J
e.      39,67 x 105 J

Solution :
h = 3.5 J/s m2K
A = 15m2
°C / 7 K
 3 jam = 10800 s
 
    = 3.5 J/s m2K . 15m2. 7 K . 10800 s
    = 39,69 x 105 J (D)

8.      A voltmeter possesses a measure limit of 100 mV with internal resistance of 1000 ohms. The front resistance that be assemblies in order for the voltmeter to measure voltage up to 10 V is..... (medium)
a.      77.000
b.      88.000
c.      99.000
d.      100.000
e.      111.000
Solution :
V = 100mV = 0,1V
V = 1 V
Rv = 1000 ohm
n =
Rd = (n-1) Rv
     = (100-1)1000 ohm
     = 99.000 ohm (C)

9.      An electrical resistance of 10 is inserted into a water of 100 gr at 20°C. If the resisteance is given a voltage differense of 50 V for 1 minute. The final temperature if the specificheat of the water is cal/g °C is ..... (difficult)
a.      55 °C
b.      56 °C
c.      46 °C
d.      66 °C
e.      45 °C
Solution :
          R = 10                         C = 1 cal/g °C
            M = 100gr                   t = 1 minute
            V = 50 V                      T1 = 20 °C
      = 15 x 103 J =15 x 103 x 0,24 kalori = 3.6 x 103 kalori

 = T1 +  = 20 +36
 (B)
10.  The temperature of a n object is 127and its surface width is 12 m2 for 1 minute. The energy radiated by the object if its considered to be absolutely black is... (medium)
a.      174.104
b.      175.104
c.      176.104
d.      177.104
e.      173.104
Solution :
T = 127 + 273 = 400K
W = e.
     = 5,7 x 10-8 x (400)4
     = 1.459,2 watts/m2
E = W x A x t
    = 1.459,2 x 12 x 60
    = 175.104 J (B)






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