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.

81.

A step-down transformer is used on a 2.2 kV line to deliver 110 V. How many turns are on the primary windings if the secondary has 25 turns?

A.

20

B.

250

C.

200

D.

500

Correct answer is D

Ep = 2.2kV = 2200V; Es=110V; Ns=25; Np=?

=\(\frac{E_p}{E_s}\) = \(\frac{N_p}{N_s}\)
 

= \(\frac{2200}{110}\) = \(\frac{N_p}{25}\)

=110 × Np = 2200 × 25

=\(N_p\) = \(\frac{55,000}{110}\)

∴\(N_p\) = 500 turns

82.

The pinhole camera works on

A.

the principle of the reversibility of light

B.

the principle of the rectilinear propagation of light

C.

Huygen’s principle

D.

the principle of total internal reflection of light

Correct answer is B

The pinhole camera works on the principle of the rectilinear propagation of light. This principle states that light travels in straight lines. When light passes through the tiny hole in a pinhole camera, it forms an inverted image on the opposite side of the camera. The size of the image depends on the distance between the object and the pinhole

83.

Which of the following is/are not true about the heat capacity of a substance?
(i) It is an intensive property
(ii) Its S.I unit is \(jK-1\)
(iii) It is an extensive property
(iv) Its S.I unit is \(jkg-1\)

A.

(i) and (iii) only

B.

(i), (ii) and (iv) only

C.

(i) and (iv) only

D.

(ii) and (iii) only

Correct answer is D

The S.I unit of heat capacity is JK-1 and it is an extensive property as it depends on the amount of substance present.

Therefore, (i) and (iv) only are not true

84.

The near point of a patient's eye is 50.0 cm. What power (in diopters) must a corrective lens have to enable the eye to see clearly an object 25.0 cm away?

A.

2 diopters

B.

2.5 diopters

C.

0.5 diopters

D.

3 diopters

Correct answer is A

The patient cannot see clearly an object closer than 50 cm

Therefore, the patient needs a lens that would enable him see clearly, objects placed 25 cm from the lens

So, we take the object to a distance of 25 cm from the lens so that the image forms at 50 cm in front of the lens

u=25cm ;v=-50cm (virtual image); p=

\(\frac{1}{f}\)=\(\frac{1}{u}\) + \(\frac{1}{v}\)

⇒\(\frac{1}{f}\) =\(\frac{1}{50}\)

⇒\(\frac{1}{f}\) = \(\frac{1}{25}\)-\(\frac{1}{50}\)

⇒\(\frac{1}{f}\) = f = 50cm =0.5m

⇒\(\frac{1}{f}\) = \(\frac{2-1}{50}\)

   p = \(\frac{1}{f}\)

   p =  \(\frac{1}{0.5}\)

   p =   2 diopters 
 
  ; The patient needs a converging lens with a power of 2 diopters

85.

A travelling wave of amplitude 0.80 m has a frequency of 16 Hz and a wave speed of 20 \(ms-1\)

Calculate the wave number of the wave.

A.

3

B.

42

C.

5

D.

2

Correct answer is C

A=0.8m; f=16Hz; v=20 \(ms^-1\); k=?

v = fλ

λ =1.25m


= k= \(\frac{2π}{λ}\) =\(\frac{2 × 3.142}{1.25}\)
 
 k=  5m ( to 1 s.f)

 

86.

Three forces with magnitudes 16 N, 12 N and 21 N are shown in the diagram below. Determine the magnitude of their resultant force and angle with the x-axis

A.

7.63N,61°

B.

8.71N,61°

C.

7.63N,29°

D.

8.71N,29°

Correct answer is B

Fx= 12cos32°-16cos26°    =    -4.20N    (right is taken +ve and left is taken -ve)

Fy=12sin32°+16sin26°-21 = -7.63N  (up is taken +ve and down is taken -ve)

\(R^2\)= 4.202+7.632

⇒\(R^2\) = 75.86

⇒ R=√75.86

∴R = 8.71 N

tan θ=\(\frac{opp}{adj}\)


⇒tan θ=\(\frac{7.63}{4.20}\)

⇒tan θ =1.8095

⇒ θ = \(tan^-1\) (1.8095)

   θ = 61°

∴The angle of the resultant force with the x-axis is 61°

 

 

 


 


 

 

 

87.

A simple pendulum, has a period of 5.77 seconds. When the pendulum is shortened by 3 m, the period is 4.60 seconds. Calculate the new length of the pendulum

A.

5.23 m

B.

6.42 m

C.

4.87 m

D.

7.26 m

Correct answer is A

     Let the original length L=xm

      ;New length =( x -3 ) m

       \(T_1\) = 5.77s; \(T_2\) = 4.60s,  

       \(T^2\) α  L

       ⇒\(T_2) = kL  where K is constant

       ⇒K = \(\frac{T^2_1}{L_1}\) = \(\frac{T^2_2}{L_2}\)

      ⇒\(\frac{5.77^2}{x}\)  = \(\frac{4.60^2}{x-3}\)

      ⇒ \(\frac{33.29}{x}\)  = \(\frac{4.60^2}{x-3}\)

      ⇒ 33.29(x-3)  = 21.16x
        
      ⇒ 33.29x - 99.87 =21.16x

      ⇒12.13x = 99.87
     
      ;x =\(\frac{99.87}{12.13}\)  = 8.23m
    
      New length of the pendulum 
     
      =x-3 = 8.23-3
   
      =5.23m

     

 

88.

Light of wavelength 589 nm in vacuum passes through a piece of fused quartz of index of refraction n = 1.458. What is the frequency of the light in fused quartz?

[Speed of light c = \(3.00×10^8 ms^{-1}\)]

A.

\(5.09×10^{15}\)Hz

B.

\(5.09×10^{14}\) Hz

C.

\(1.77×10^{15}\) Hz

D.

\(1.77×10^{14}\) Hz

Correct answer is B

n=1.458, c=\(3.00 ×10^8 ms^-1\) ,λo = 589nm;  f=?

Speed of light in a medium (v)=\(\frac{c}{n}\)  where n is the refractive index of the medium

⇒λn=\(\frac{589}{1.458}\) = 404nm

  v=fλ
  
⇒f=\(\frac{v}{λ}\) 

=\(\frac{2.06×10^8}{404×10^-9}\) \(1nano=10^{-9}\)

∴f=\(5.09×10^{14}\) Hz

89.

A 400 N box is being pushed across a level floor at a constant speed by a force P of 100 N at an angle of 30.0° to the horizontal, as shown in the the diagram below. What is the coefficient of kinetic friction between the box and the floor?
 

A.

0.19

B.

0.24

C.

0.40

D.

0.22

Correct answer is A

W = 400 N; P = 100 N; θ = 30o; μ = ?

Frictional force (Fr) = μR (where R is the normal reaction)

The forces acting along the horizontal direction are Fr and Px

∴ Pcos 30° - Fr = ma (Pcos 30° is acting in the +ve x-axis while Fr in the -ve x-axis)

⇒ 100cos 30° - μR = ma

Since the box is moving at constant speed, its acceleration is zero

⇒ 100cos 30° - μR = 0

⇒ 100cos 30o = μR ----- (i)

The forces acting in the vertical direction are W, Py and R

∴ R - Psin 30° - W = 0 (R is acting upward (+ve) while Py and W are acting downward (-ve) and they are at equilibrium)

⇒ R - 100sin 30° - 400 = 0

⇒ R = 100sin 30° + 400

⇒ R = 50 + 400 = 450 N

From equation (i)

⇒ 100cos 30° = 450μ

⇒μ=100cos30°

  N = \(\frac{100cos30°}{450}\)

        = μ = 0.19

90.

Which of the following thermometers measures temperature from the thermal radiation emitted by objects?

A.

Pyrometer thermometer

B.

Platinum resistance thermometer

C.

Thermocouple thermometer

D.

Constant pressure gas thermometer

Correct answer is A

A pyrometer is a type of remote-sensing thermometer used to measure the temperature of a surface. It does this by measuring the thermal radiation or infrared energy being emitted from the object. Therefore, a pyrometer thermometer measures temperature from the thermal radiation emitted by objects.