bVQhrpracticequestions1_25_mcqs

8/7/2019 bVQhrpracticequestions1_25_mcqs

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Practice Questions and Solutions

Q1. Which of the following pairs of quantities have same dimensions?

A. acceleration and angular acceleration

B. stress and strain

C. latent heat and specific heat

D. momentum and impulse

E. mass and weight

Ans. (D) is correct choice. Momentum= mass x velocity =[MLT-1]

Impulse=Force x time = [MLT-2] x [T]

= [MLT-1

]

Choices A, B, C and E are incorrect.

Q2. Dimensional formula of Gravitational constant is

A. [M-1L2T]

B. [M-1L3T-2]

C. [M0L3T-2]

D. [M-1L-1T-2]

E. [M 1 L1 T 1]


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Ans. (B) is the correct choice,

Gravitational constant G=F x r2/m1 x m

Therefore dimension of G= [MLT-2][L2]/[M2]

= [M-1L3T-2]

Q3. Plancks constant has dimension of

A. Energy

B. Angular momentum

C. Power

D. momentum

E. Impulse.

Ans. (B) is the correct choice.

Plancks constant h=E/= Energy X Time

Dimension of h = [ML 2 T -2][T]

= [ML 2 T -1]

And angular momentum=Linear momentum X distance

Therefore dimension of angular momentum= [M L T -1][L]

= [M L 2 T 1]

Q4. In the fig. shown above two objects are approaching some point Owith equal velocities v. What is the magnitude relative velocity of

object 1 with respect to object 2?


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(A). 2 v

(B). 2 v

(C). v/2

(D). v/2

(E). v

Ans. (B) is the correct choice. Magnitude of relative velocity of object 1with respect to object 2 is v12= (v 2 + v 2)1/2 =2 v

Q5.The velocity of a particle is v= 5+(x + yt), where a and b are

constants and t is the time. The acceleration of the particle is

(A). x

(B). y

(C). 0

(D). xy


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(E) x/y


v = 5 + x + yt

Acceleration dv/dt=d/dt (5 + x + yt) =y

Q6. The following forces are acting simultaneously on a particle

(1). (2 + 3 2k) N

(2). (3 + 3k) N

(3). (-5 -2 -3k) N

So the particle will move in

(A). X-Y Plane

(B). X Plane

(C). Y-Z Plane

(D). Y Plane

(E). X-Z Plane

Ans. (C). is the correct choice.

The total force acting on the particle is the sum of three forces i.e.

F= (2 +3 -2 k) + (3 + -3 k) + (-5 -2 +k)

F = 2-4 k

Therefore the particle will move in Y-Z Plane.

Options (A), (B), (D), (E) are incorrect, because F= 2 -4 k.

is a unit vector in Y direction and k is a unit vector in Z direction. So,particle will move in Y-Z Plane.


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Q7 A projectile is fired from point O at an angle with the

horizontal. What will be the angle between instantaneous velocity and

acceleration at the highest point h?

(A). 45 0

(B). 90 0

(C). 0 0

(D). 180 0

(E). 30 0

Ans. (B) is the correct choice. At point h, acceleration is directed in thedownward direction and velocity is along the tangent at that point. So, the

angle between the two is 90 0.


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Hence the Options (A), (C), (D), (E) are incorrect.

Q8. If y=a +bt +ct2, where y is in meter and t is in second. The unit of

b is

(A). m

(B). s

(C). ms -1

(D) ms -2

(E). m -1s -1

Ans. (C) is the correct choice. According to dimensional analysis

[y]= [bt]

[L]=b [T]

b= [L]/ [T] or unit of b=ms -1


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Q9. The above fig. shows displacement time graph of a body in a

straight like. We can conclude from the graph that

(A). the velocity increases uniformly

(B). the velocity is constant.

(C). the velocity decreases uniformly.

(D). the body is subjected to acceleration from O to A.

(E). the velocity of the body at A is zero

Ans. (E) is the correct choice.

Velocity at any point of the graph is found by drawing tangent to that point.At point A, tangent to the curve is parallel to time axis. It shows that velocity

at A is zero.

option (A),(B),( C),(D) are incorrect since the displacement is first

increasing with time and then starts decreasing, so the velocity of the bodycan not increase or decrease uniformly and nor can it be constant.


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Q10. A ball is dropped from the top of a tower of height h; meters. It

takes T seconds to reach the ground. Where is the ball at the time T/2

second?

(A). at h/2 meter from the ground.

(B). at h/4 meters from the ground.

(C). at 3h/4meter from the ground.

(D).at h/8 meter from the ground.

(E). at h/3 meter from the ground.

Ans. (C) is the correct choice.

h=ut+1/2gT2

The ball is dropped from rest, so u=0

So, h=1/2gT 2

After T/2 seconds, h1=1/2g (T/2) 2

h1=1/2g T2 /4

h1=1/4(1/2gT 2)

h1= h/4

So after T/2 sec. height above the ground =h -h/4=3h/4

Q11.Two resistances when connected in series become 6 ohms and 1

ohm when connected in parallel. The two resistances are

(A). 2, 4(B). 1, 5

(C). 1, 4

(D). 4.7 ,1.3


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(E). 2.5, 3.5

Ans. (D) is the correct choice.

R1+R2=6

R1 R2/R1 +R2=1

Therefore R1 R2/6=1

R1 R2=6

(R1 - R2)2 = ( R1 + R2 )

2 - 4R1 R2=36-24

R1 - R2=2 3

Therefore R1= 4.7R2= 1.3

Q12. Three voltmeters A, B, C are having resistances 2R, 3R, 6R

respectively, are connected as shown. When some potential difference is

applied between X&Y, the voltmeter readings are V A, V B and V Crespectively. Then,


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(A). V A = V B= V C

(B). V A V B = V C

(C).V A =V B V C

(D). V A V B V C

(E). V A V B =V C

Ans. (A) Is the correct choice.

From ohms law V=IR

V A=I.2R=2IR

The division of current in the two parallel branches will be according to the

resistance of each branch.

Current I B = 2/3 I

And I C =1/3 I

Therefore V B= I B RB =2/3 I x 3R=2 IR

AND V C = I C RC =1/3 I x 6R=2IR

Therefore V A= V B= V C

Q13. If the light is incident on a glass slab at polarizing angle then

A). Reflected light is completely polarized.

B). Refracted light is completely polarized.

C). both are completely polarized.

D). both are partially polarized.


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E) Both are un polarized.

Ans. (A) is the correct choice.

When light is incident at polarizing angle, it is partially reflected and

partially refracted. The reflected light is completely polarized.

Q14. The minimum resistance that can be obtained with ten 0.1

resistors is

(A). 1

(B). 0.01

(C). 0.1

(D). 0.001

(E).10


Ten resistors connected in parallel with given, minimum resistance=0.01

Q15. A battery of e.m.f E and internal resistance r is being charged with

current I. The terminal potential will be

(A). E +I r

B). E I r

C).E

D).I r

E).none of the above

Ans. (A) is the correct choice

When the cell is charged, the terminal potential is more than the e.m.fand is equal to E +I r


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Option (B) is incorrect because E I r is the potential when the cell isbeing discharged i.e. when the current is being drawn from the cell.

Options (C) and (E) are also incorrect.

Q16. If potential difference between two plates separated by a

distance d is V, then the electric field is equal to

A).1/Vd

B). V d

C). V/d

D). d/V

E). V 2 d


The electric field E between two plates at a potential difference Vseparated by a distance d is V= E/d

Q17. In the above fig. 9, what is the e.m.f of the given battery if

the e.m.f of each cell is 2 V.?


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A).12V

B). 10V

C). 8V

D). 6V

E). 4V

Ans. (E) is the correct choice.

The e.m.f of four cells is in the same direction and e.m.f of two cells is inopposite direction.

Therefore total e.m.f = (2 x4) 2 x 2=8 -2 =4V.

Q18. Intensity of light depends on its

A). Velocity

B).Wavelength

C).Amplitude

D).Frequency

E).Density


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According to wave theory intensity of light I A 2 (square of amplitude)

Q19. Properties which affect the speed of light in space are

A). only mechanical

B). only electrical

C). only magnetic

D). electrical and magnetic

E). electrical and mechanical


Light waves are electromagnetic. They are produced due to variation ofelectric and magnetic fields. So the properties which affect their speed are

electrical and magnetic.

Options (A), (B), (C),(E) are incorrect.

Q20. If the source of light in Youngs double slit experiment is changed

from red to green then fringes will become

A). Darker

B). Brighter

C). Thinner

D). Broader

E). Remain same



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Fringe width, = D/d. With change in color there is a change inwavelength of light.

Fringe width Wavelength.

So as the wavelength decreases the fringe width also decreases. Since thefringe width decreases the fringes become thinner.

Options (A), (B), (D), (E) are incorrect.

Q21.For obtaining diffraction pattern the aperture of the slit should be

of the order of

A). B) ./2

C). /4

D). 2

E). 2

Ans. (A) is the correct choice.

Condition for diffraction to take place is that the aperture of the slit should

be of the order of the wavelength of light.

Q22. Ordinary light is

A). Plane polarized

B). circularly polarized

C). partially polarized

D). unpolarized

E). elliptically polarized.



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Ordinary light is unpolarized. E.g. sunlight or light coming out of a bulb.

Q23. In the interference experiment spacing between successive maxima

or minima is

A). Dd/

B). D/d

C).d D

D). /dD

E). d/D


Fringe width due to interference is = D/d

Q24. Two waves have intensities in the ratio 1:9. After interference ratio

of maximum to minimum intensity will be

A). 1:3

B). 3:1

C). 4:1

D). 1:4

E). 9:1


Intensity (amplitude) 2

Given I 1 /I 2 =1/9

A 12/ A 2

2 =1/9

A 1 / A 2 =1/3


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Now, I max. / I min. = (a 1 + a 2 )2 / (a 1- a 2)

2

= (3+1) 2/ (3-1) 2

=16/4

So, I max. / I min = 4:1

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