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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.

101.

The number of holes in an intrinsic semiconductor

A.

is not equal to the number of free electrons

B.

is greater than the number of free electrons

C.

is equal to the number of free electrons

D.

is less than the number of free electrons

Correct answer is C

An intrinsic semiconductor, also called an undoped semiconductor, is a pure semiconductor. In intrinsic semiconductors the number of excited electrons and the number of holes are equal: n = p

102.

The half life of a radioactive material is 12 days. Calculate the decay constant.

A.

0.8663 day1

B.

0.04331 day1

C.

0.17325 day1

D.

0.05775 day1

Correct answer is D

T1/2 = 12 days ;V=?

T1/2 = \frac{0.693}{γ}

⇒γ ⇒\frac{0.693}{t_{^1/_2}}

⇒γ =\frac{0.693}{12}

⇒γ= 0.05775 day^{-1}

103.

A lorry accelerates uniformly in a straight line with acceleration of  4ms^-2 and covers a distance of 250 m in a time interval of 10 s. How far will it travel in the next 10 s?

A.

650

B.

900

C.

800

D.

250

Correct answer is A

We first find the lorry's initial velocity
 

a = 4ms^-2, s= 250m,  t= 10,  u=?

⇒ ut +  \frac{1}{2}×4×(10^2) ⇒250= 10u + 200 ⇒ 250-200 = 10u ⇒ 50 = 10u ⇒U =\frac{50}{10} = 5ms^-1

For the next 10s, we set t=20s

= s = (5 × 20) + \frac{1}{2}×4×{20^2}

= s  = 100 + 800

⇒ 900 m

;Distance covered:900-250 =650m
 


 

104.

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

105.

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